INSTITUTE of AERONAUTICAL ENGINEERING Dundigal, Hyderabad - 500043

Courtesy IARE INSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad - 500043

UNIT-I 1. Crystallography: Ionic Bond, Covalent Bond, Metallic Bond, Hydrogen Bond, Vander-Waal’s Bond, Calculation of Cohesive Energy of diatomic molecule- Space Lattice, Unit Cell, Lattice Parameters, Crystal Systems, Bravais Lattices, Atomic Radius, Co-ordination Number and Packing Factor of SC, BCC, FCC, Miller Indices, Crystal Planes and Directions, Inter Planar Spacing of Orthogonal Crystal Systems, Structure of Diamond and NaCl. 2.X-ray Diffraction & Defects in Crystals: Bragg’s Law, X-Ray diffraction methods: Laue Method, Powder Method: Point Defects: Vacancies, Substitutional, Interstitial, Frenkel and Schottky Defects, line defects (Qualitative) & Burger’s Vector.

UNIT-II 3. Principles of Quantum Mechanics: Waves and Particles, de Broglie Hypothesis , Matter Waves, Davisson and Germer’s Experiment, Heisenberg’s Uncertainty Principle, Schrödinger’s Time Independent Wave Equation - Physical Significance of the Wave Function – Infinite square well potential extension to three dimensions 4. Elements of Statistical Mechanics& Electron theory of Solids: Phase space, Ensembles, Micro Canonical , Canonical and Grand Canonical Ensembles - Maxwell- Boltzmann, Bose-Einstein and Fermi-Dirac Statistics (Qualitative Treatment), Concept of Electron Gas, , Density of States, Fermi Energy- Electron in a periodic Potential, Bloch Theorem, Kronig-Penny Model (Qualitative Treatment), E-K curve, Origin of Energy Band Formation in Solids, Concept of Effective Mass of an Electron, Classification of Materials into Conductors, Semi Conductors & Insulators.

UNIT-III 5. Dielectric Properties: Electric Dipole, Dipole Moment, Dielectric Constant, Polarizability, Electric Susceptibility, Displacement Vector, Electronic, Ionic and Orientation Polarizations and Calculation of Polarizabilities: Ionic and Electronic - Internal Fields in Solids, Clausius - Mossotti Equation, Piezo -electricity and Ferro- electricity. 6. Magnetic Properties & Superconducting Properties: Permeability, Field Intensity, Magnetic Field Induction, Magnetization, Magnetic Susceptibility, Origin of Magnetic Moment, Bohr Magneton, Classification of Dia, Para and Ferro Magnetic Materials on the basis of Magnetic Moment, Domain Theory of Ferro Magnetism on the basis of Hysteresis Curve, Soft and Hard Magnetic Materials, Properties of Anti-Ferro and Ferri Magnetic Materials and their Applications, Superconductivity, Meissner Effect, Effect of Magnetic field, Type-I & Type-II Superconductors, Applications of Superconductors

UNIT-IV 7.Optics: Interference-Interference in thin films(Reflected light), Newton rings experiment- Fraunhofer diffraction due to single slit, N-slits, Diffraction grating experiment , Double refraction-construction and working, Nicol’s Prism 8. Lasers & Fiber Optics: Characteristics of Lasers, Spontaneous and Stimulated Emission of Radiation, Einstein’s Coefficients and Relation between them, Population Inversion, Lasing Action, Ruby Laser, Helium-Neon Laser, Semiconductor Diode Laser, Applications of Lasers- Principle of Optical Fiber, Construction of fiber, Acceptance Angle and Acceptance Cone, Numerical Aperture, Types of Optical Fibers: Step Index

and Graded Index Fibers, Attenuation in Optical Fibers, Application of Optical Fiber in communication systems.

UNIT-V: 9. Semiconductor Physics: Fermi Level in Intrinsic and Extrinsic Semiconductors, Calculation of carrier concentration in Intrinsic &, Extrinsic Semiconductors, Direct and Indirect Band gap semiconductors, Hall Effect-Formation of PN Junction, Open Circuit PN Junction, Energy Diagram of PN Diode, Diode Equation, I-V Characteristics of PN Junction diode, Solar cell, LED & Photo Diodes. Acoustics of Buildings & Acoustic Quieting:, Reverberation and Time of Reverberation, Sabine’s Formula for Reverberation Time, Measurement of Absorption Coefficient of a Material, Factors Affecting The Architectural Acoustics and their Remedies 10. Nanotechnology: Origin of Nanotechnology, Nano Scale, Surface to Volume Ratio, Quantum Confinement, Bottom-up Fabrication: Sol-gel, Top-down Fabrication: Chemical Vapour Deposition, Characterization by TEM.

TEXT BOOKS: 1. Engineering Physics,K. Malik, A. K. Singh, Tata Mc Graw Hill Book Publishers 2. Engineering Physics, V. Rajendran, Tata Mc Graw Hill Book Publishers

REFERENCES: 1. Fundamentals of Physics, David Halliday, Robert Resnick, Jearl Walker by John Wiley & Sons 2. Sears and Zemansky's University Physics (10th Edition) by Hugh D. Young Roger A. Freedman, T. R. Sandin, A. Lewis FordAddison-Wesley Publishers; 3. Applied Physics for Engineers – P. Madhusudana Rao (Academic Publishing company, 2013) 4. Solid State Physics – M. Armugam (Anuradha Publications). 5. Modern Physics – R. Murugeshan & K. Siva Prasath – S. Chand & Co. (for Statistical Mechanics). 6. A Text Book of Engg Physics – M. N. Avadhanulu & P. G. Khsirsagar– S. Chand & Co. (for acoustics). 7. Modern Physics by K. Vijaya Kumar, S. Chandralingam: S. Chand & Co.Ltd 8. Nanotechnology – M.Ratner & D. Ratner (Pearson Ed.). 9. Introduction to Solid State Physics – C. Kittel (Wiley Eastern). 10. Solid State Physics – A.J. Dekker (Macmillan). 11. Applied Physics – Mani Naidu Pearson Education

INDEX Unit Contents Page 1.Crystallography 4 I 2.X-ray Diffraction & Defects in Crystals 31 3.Principles of Quantum Mechanics 43 II 4.Elements of Statistical Mechanics& Electron 59 theory of Solids 5.Dielectric Properties 94 III 6.Magnetic Properties & Superconducting 116 Properties 7.Optics 151 IV 8.Lasers & Fiber Optics 185 9.Semiconductor Physics 229 V 10.Nanotechnology 297

UNIT-I Chapter-1 CRYSTALLOGRAPHY

1.1.1. Introduction to Bonding in Solids Atom consists of a positively charge nucleus surrounded by a negatively charged electron cloud. When the two atoms are brought closer, there will be both attractive and repulsive forces acting on them. Therefore, the two atoms take relative positions when there is a balance between these two forces. This is known as equilibrium position.

The bonding forces which bind the atoms in a molecule and in the molecules in a solid are mainly four types of forces i.e. gravitational, electrical, and nuclear and weak forces. Chemical bond is defined as inter-atomic or inter-molecular or inter-ionic force of attraction which holds the atoms together.

When the two atoms attain the equilibrium position, then the value of energy needed to move an atom completely away from its equilibrium position is termed as binding energy or cohesive energy. On the basis of the nature of forces which bind solid lattice together, the solids can be divided two mainly groups namely.

 Primary bonds  Secondary bonds

Primary bonds: Primary bonds are the strongest bonds which hold atoms together. The three types of primary bonds are,  Ionic bond  Covalent bond  Metallic bond

Secondary bonds: Secondary bonds are much weaker than primary bonds. Two types of Secondary bonds are,  Hydrogen bond  Vander Waals bond

1.2. Primary bonds 1. Ionic bond: This bond is the simplest type of chemical bond called ‘Hetero polar Bond’. Ionic bonds are mostly insulating in character. These are not pure elements, but they are compounds.

Eg:- NaCl, KCl, KBr etc.

The electronic structure of atoms is relatively stable when the outer shells contain eight electrons (two in the case of 1st shell). Sometimes it is possible by transferring the electrons from one atom to another, electron shells are filled the donor atom will take a positive charge and the acceptance will have a negative charge.

When Na and Cl atoms are placed together, these are transfer of electrons from Na to Cl atoms resulting in a strong electrostatic attraction between the positive Na ion and negative Cl ions. Let us examine Na and Cl electronic configuration how they tends to form an ionic bond.

Atomic number of Na is 11 Electronic configuration of Na is: 1s22s2 2p 63s1

It has one electron in the outer most orbits, this electron is called as valence electron Na atom try to lose this valence electron in order to attain insert gas configuration. The minimum energy required to detach the electron from Na atom is called as ionization energy.

Na + ionization energy → Na+ + e-

Ionization energy required = 5.1 eV Atomic number of Cl is 17 Electronic configuration is 1s22s2 2p 63s2 3s5

It has 7 electrons in its outermost orbit. It needs one electron to insert gas configuration. It readily accepts an electron and release some energy called electron affinity. Cl + e- → Cl- + Electron Affinity

Electron Affinity = 3.6ev In order to produce two ions, NaCl needs energy equal to5.1ev-3.6eV = 1.5ev

Na + Cl + 1.5 eV → Na+ + Cl-

Na+ and Cl- attract each other and form a bond. This bond is called “Ionic Bond”.

Characteristics of ionic bonded Materials  Ionic bonds are strong.  They have high melting points and boiling points.  They are transparent to visible light.  They have closed packed structure.

 They are highly soluble in polar solvents like H20, liquid NH3 and insoluble in non-polar solvent.  Non directional, because the charge distribution is special in nature.

2. Covalent Bond: covalent bonds are called homo polar bounds. In this neighboring atoms share their valence electrons in the formation of a strong covalent bound. Eg: Cl2, O2, etc. Consider the Cl atom, which has seven electrons in the valence shell. Spins of 6 electrons are paired, where as the spin of 7 the electron is unpaired. The unpaired electron is always looking for another unpaired electron. Which comes closer to it. In such case another Cl atom comes near the first Cl atom. The two unpaired electrons of the two atoms get paired.

Characteristics of covalent bonded materials  These bounds are usually hard and brittle.  Binding energy is high.  M.P and B.P are high but low compared to ionic crystal.

 Covalent bonds are insoluble in H2o.

 These materials are soluble in non-polar solvents like C6H6.  They are transparent to longer wavelengths, but opaque to shorter wavelength.

3. Metallic Bond: Metallic bonds are similar to covalent bonds. In this outer electrons have high mobility. In metallic bonding each atom of metal give one or two value electrons to the crystal. These hold the atoms together and are not bound to individual atoms, but move freely throughout the whole metal. In metal, the ionization energies are low, so metallic atoms gives their valence electrons to crystal. The valence electrons will form electron cloud or electron gas, which is occupied throughout the metal space. The positive ions are held together by the electrostatic forces due to the free electrons. The electrostatic interaction between the positive ion and the surrounding electron could hold the metal together. So it is called the metallic bond. Eg; - Al, Ag, Na, Cu etc.

Fig: Atomic arrangement in metallic crystals

Characteristics of Metallic bonded materials  They have high thermal conductivity due to the presence of free electrons.  They have high electrical conductivity.  Metallic bonds are non directional.  The metallic bound is comparatively weaker than the ionic and covalent bounds. 1.3. Secondary bonds 4. Hydrogen bond: covalently bonded atoms often produce an electronic dipole configuration. The hydrogen atom has the positive end of dipole. If the bonds arise as the result of electrostatic attraction between the atoms. It is known as hydrogen bound.

The hydrogen form bounding with electro negative. Ni, O2, Fl2 usually of another molecule.

Let us consider the example of H2O molecule. The greater electro-negative of O2. The electrons tend to stay closer to O2 atom than the hydrogen atoms. The O2 atom acts as negative end of the dipole. While H+ atom acts as the positive end. The positive end

attracts the negative end of another. H2O molecule and thus bonding the molecules together.

Characteristics of Hydrogen bonded materials  The bonding is relating strong as compared to other dipole-dipole interactions.  The hydrogen bounds are directional.  They have low melting points.  They are transparent to light.  They are soluble in both polar and non-polar solvents.  They are good insulators of electricity.  They have low densities.

5. Van der Waal’s Bond Weak and temporary (fluctuating) dipole bounds between hydrogen are known as vandarwaals bounding. They are non-directional. If the symmetrical distribution of electrons around the nucleus is distributed, the center of positive and negative charges may not coincide at that movement giving rise to weak fluctuating dipole.

Fig: Van der Waal’s bond

Characteristics of van der waal’s bonded materials  The molecular crystals have small binding energy.  Molecule structure can be both crystalline and non-crystalline.  They are usually transparent to light.  They have low melting point.

 They are soluble in both polar and non-polar solvents.  They are good insulators.

1.4. Variation of Interatomic Forces with Interatomic Spacing Electrical forces are responsible to binding the atoms gives different solid structures. Magnetic and gravitational forces are negligible in the formation of solids. NaCl crystal is more stable than the Na and Cl atoms. Similarly silicon is more stable than silicon atoms. This shows that silicon atoms attract each other, when they come closer. This is the force which is responsible in the forming are crystal. It means that the energy of crystal is lower than the energy of these energies is called as cohesive energy of the crystal. Sometimes it also called as building energy. Consider two atoms at a separate ion of ‘r’ where both the attractive and repulsive forces are present. The net force between the atoms is given as, F(r) = - → (1)

Where A, B, M, N are constants which are the characteristics of the atom.

Fig: Variaration of interatomic force with interatomic spacing

The first term in the above equation represents attractive force. Where M approximately equal to 2. i.e. M 2 (according to inverse quire law). The second term represents repulsive force which is very strong at small distances and the value is 9. N 9 N>M

The variation of the attractive and repulsive forces with distance of separation r0

Where the separation between atoms = The resultant force = 0

This separation is called equilibrium distance, at equilibrium position, F(r= =0

1.5. Cohesive Energy The potential energy is due to the interaction between the two atoms and it depends on the interaction spacing. The potential energy is, W(r) = = → (1) F(r) = - → (2)

Substitute eq (2) in eq(1)

W(r) = dr

W(r) = A dr – B dr

W(r) = A - + C

W(r) = + +

W(r) = - + + C

= a; = b; M-1 = m; N-1 = n

W(r) = + + C

Applying boundary conditions r , W 0, we get C=0 W(r) = + → (3)

In eq(3), the quality represents attractive potential energy and represents repulsive

potential energy. At equilibrium distance ( ) the potential energy should be minimum mathematically it is represented as,

-a + b = 0

-a (-m) + b(-n) ) =0

= → (4)

Second derivation:-

am - bn > 0

am (-m-1) - bn(-n-1) >0

(n+1) > (m+1)

(n+1)> (m+1) [i.e from eqn (1)]

n+1 > m+1 n>m

Calculation of Cohesive Energy Potential energy for two atoms is given as, W(r) = + → (1)

When the distance of separation is ’ the energy is minimum.( )

= + → (2)

For minimum the first derivation is zero. =0

-a + b = 0

= → (3)

Substitute eq (3) in eq(2)

= = ×

1.6. Calculation of Cohesive Energy for Ionic Solids Electrons are transferred from electro positive atoms like Na,Ca,K,Mg to

electronegative atoms like O, Fe, ,Cl. Let us consider, ions of charges z1e, z2e separated by a distance ‘r’.

Attractive Force =

For complete crystal, coulomb’s potential energy

Where A modeling’s constant. Due to overlap of last orbital’s the negative forces between electrons could be established in a noticeable magnitude.

Repulsive potential energy WR=

The total potential energy of a crystal can be expressed as,

W(r) = - + → (1)

For univalent crystal z1= z2=1

W(r) = +

At equilibrium, =0

Minimum potential energy, = = +

In kilometers

Here Avogadro Number

1.7. Introduction of Crystallography

Matter exists in three different states namely solids, liquids and gaseous states. In liquid and gaseous states the atoms, molecules or ions move from one place to another and there is no fixed position of atoms in the substance. In solids the position of atom or molecules are fixed and they are may or may not be present at regular intervals of distances in three dimensions. If the atoms or molecules in a solid are periodically at regular intervals of distances in 3 dimensions then such a solids are known as crystalline solids. If the atoms or molecules does not have such periodicity in the substances is known as Amorphous solid.

Fig: (a) crystalline solid (b) Amorphous solid

If the periodicity of atoms or molecules are extended throughout the solid such a solid known as single crystalline solid.

Fig: Single crystalline solid

If the periodicity of atoms or molecules are up to small regions called grains. And these grains are large in number of different sizes in the solid. Such a materials are known as polycrystalline solids.

Fig: polycrystalline solid.

Crystalline Solids Amorphous solids

The crystalline solids posses’ regular These solids posses complete random arrangement of ions or atoms or arrangement of ions or atoms or molecules. molecules.

These solids have some physical The solids have different physical properties in all the directions. properties (thermal, electrical conductivity, refractive index) in different directions. These are isotropic in nature. These are anisotropic in nature. These solids posses plasticity except These solids posses elasticity. rubber.

These posses less density. These posses more density.

Eg: calcite, quartz, gold, silver, Al etc... Eg: glass, plastic etc.

1.8. Space lattice (or) Crystal lattice The atomic arrangement in a crystal is called crystal structure. In crystalline materials, the atoms or molecules are arranged periodically in regular intervals of distances. To explain crystal symmetries easily, it is convenient to imagine a points in space about which these atoms are located such points in space are called lattice points and the totality of all such lattice points form space lattice or crystal lattice. (or) The geometric arrangement of lattice points, which describes the three dimensions arrangements of atoms or molecules or ions in a crystal, is called space lattice or crystal lattice. If the points are arranged in a single or single column, then the lattice is solid to be linear lattice. If it is arranged in a two dimension then the lattice is called planar lattice.

1.9.1. Basis: The crystal structure is formed by associating with every lattice point a set of assembly of atoms or molecules or ions identical in composition, arrangement,

orientation is called as Basis. (Or) A group of atoms or molecules identical in composition is called Basis.

Lattice + basis → Crystal structures

Fig: Space lattice + basis = crystal structure

In crystalline solids like aluminium, sodium, copper the basis is single atom. In

case of NaCl, NaBr, KCl, KBr the basis is diatomic and in CaF2 the basis is triatomic.

1.9.2. Unit Cell The smallest portion of space which can generate the complete crystal structure by repeating its own dimensions in various directions is called a “Unit Cell”. (Or) In every crystal some fundamental grouping of particles is repeated. Such fundamental grouping of particles is called a “Unit Cell”.

Primitive Unit Cell It is defined as Unit cell which contains only one lattice point per unit cell.

Multiple Unit Cell (or) Non-Primitive Cell The unit cell which contains more than one lattice point per unit then it is known as Multiple Unit Cell (or) Non-Primitive Unit Cell.

Fig: Unit cell

1.10. Lattice parameters of a Unit Cell The basic lattice parameters are interfacial angles and primitives.

Primitives: The intercepts on x,y,z-axis in a unit cell are called primitives . They are denoted by a,b,c. The primitives will give the knowledge of the size of the unit cell. The interfacial angles give the shape of the unit cell. α β γ ------> interfacial angles a, b, c ------> primitives X,Y,Z ------> crystallographic axis 16

Fig: Unit cell and its lattice parameters 1.11. Crystal Systems The crystal systems are divided into seven typed based on the lattice parameters i.e. primitives and interfacial angles, The different crystal systems are 1). Cubic system 2). Tetragonal system 3). Orthorhombic system 4). Monoclinic system 5). Triclinic system 6). Trigonal system 7). Hexagonal system

1) Cubic system:- In this crystal system all the three edges of the unit cell are equal and right angles to each other a=b=c; α= β= γ=900

Eg :- NaCl , CaF2 etc.

Fig: Cubic Crystal Systems

2) Tetragonal system: -- In this system all the two edges of the unit cell are equal while the third is different. The three axes are mutually perpendicular. a=b≠c; α= β= γ=900

Eg: - TiO2, NiSO4, Sno4 etc.

3).Orthorhombic system:--In this crystal system the three edges are different but three axis are perpendicular each other a ≠ b≠ c ; α = β = γ =900

Eg: BaSO4, KNO3

4).Monoclinic system:-- In this system the edges of the unit cell are different , two axis at right angles and third axis is obliquely inclined. a ≠ b ≠ c; α = β = γ = 900 ≠ γ

Eg: Na2B4O710H2O (Borax), CaSO42H2O (Gypsum )

5).Triclinic system: In this system the edges of unit cell are different and the 3 axis are obliquely inclined to each other a ≠ b ≠ c; α ≠ β ≠ γ ≠ 900

Eg : K2Cr2O7 ,CuSO45H 2O

6).Trigonal system: In this crystal system all the 3 edges of the unit cell are equal and are equally inclined to each other at an angle other than 900.

a = b = c; α = β = γ ≠ 900 Eg: Bismuth, calcite.

7). Hexagonal system: In this system the two axes of the unit cell are equal in length lies in one plate at 1200 with each other and the third axis is perpendicular to this plane . a=b≠c; α=β= 900; γ=1200

Eg: SiO2, Mg, Zn

1.12. Bravais Lattices According to crystal system Bravais showed that there are 14 ways of arranging points in space lattice such that all the lattice points have exactly the same environment. these 14 different lattice types collectively called as Bravais lattices. The different space lattice formed by Primitive (P), Body centered (I), Face centered (F), Base centered(C). S.No Crystal system Bravais No. of Axial Interfacial examples lattice lattice length angles system

1 Cubic P,I,F 3 a=b=c α= β= γ=900 NaCl,

0 2 Tetragonal P,I 2 a=b=c α= β= γ=90 TiO2

0 3 Orthorhombic P,I,F,C 4 a=b=c α= β= γ=90 KNo3

0 4 Monoclinic P,C 2 a=b=c α= β= γ=90 CaSo42H2o

0 5 Triclinic P 1 a=b=c α= β= γ=90 K2Cr2O7

0 6 Trigonal P 1 a=b=c α= β= γ=90 calcite

0 7 Hexagonal P 1 a=b=c α= β= γ=90 SiO2

1.13 The unit cell contains 8 atoms located at 8 corners is called simple cubic structure. In this structure the all atoms touch each other at their edges. Effective number of atoms in unit cell is 1. i.e. 8 corners have 8 atoms; each corner atom is shared by 8 unit cell. So the share of the each atom towards unit cell is 1/8. Therefore n = 8x1/8 N=1. If we consider an atom at one corner as centre it is surrounded by 6 equidistant neighboring atoms. Hence the coordination number of simple cubic is 6.

Nearest neighbor distance: In this structure atoms touch each other at their edges. Hence a’=2r, Where ‘a’ is Lattice constant.

Packing fraction: It is defined as the ratio of volume of all atoms in the unit cell to the volume of unit cell. Volume of the all atoms in the unit cell Therefore packing fraction = ------Volume of the unit cell.

= 0.52 = 52% Eg: polonium.

1.14 Body Centered Cubic (BCC) Structure The structure which 8 atoms are located at 8 corners and one atom is located at centre of body i.e. unit cell such a structure is called body centered cubic structure (BCC). In this structure the corner atoms may not touch each other. But they are in contract with the atom which is located at the centre of the unit cell. Effective number of atoms in BCC is 2 . i.e. 8 atoms are located at 8 corners; one atom is located at centre of the body. So, share of each corner atom towards 8 unit cell is 1/8.

Therefore n = (8x +1

Coordination Number: - If we consider one atom at the centre it is surrounded by 8 equidistant neighboring atoms. Hence the coordination number is 8.

Nearest neibhor Distance (a): From the ∆ ABC, (AC) 2 = (AB) 2 + (BC) 2 (AC) 2 = a2+ a2 (AC) 2 =2 a2 From the ∆ ACD, (AD) 2 = (AC) 2 + (CD) 2 (4r) 2 = 2 a2+ a2 16r2 = 3 a2 4r = √3 a r = √3 a/4 a = 4r/√3.

Packing fraction = nx

= 0.68

= 68%. Eg: Na, Fe etc.

1.15 Face Centered Cubic (FCC) structure The structure in which 8 atoms are located at 8 corners, 6 atoms are located at the centre of the solids. Such a structure is called FCC. In this structure the corner atoms may not touch each other but they are in contract with the atoms located on its faces. Effective Number in F.C.C is 4. i.e. 8 atoms are located at 8 corners, 6 atoms on sides shared by two sides. So they, n = (8× ) + (6× ½) = 4.

Coordination number is 12. If we consider one atom at a face there are 12 atoms of same distance. Therefore, Coordination Number in FCC is 12. Nearest Distance (a): From the ∆ ABC, (AC) 2 = (AB) 2 + (BC) 2 16r2 = a2+ a2 16r2 = 2 a2 8r2 = a2 2√2r = a r = a/2√2.

Packing fraction = nx

= 0.74 = 74%. Eg: Cu, Al, Ag etc.

1.16.1. Miller Indices It is possible for defining a system of parallel and equidistant planes which can be imagined to pass through the atoms in a space lattice, such that they include all the atoms in the crystal. Such a system of planes is called crystal planes. Many different systems of planes could be identified for a given space lattice. The position of a crystal plane can be specified in terms of three integers called Miller indices.If these are enclosed in “( )” as (h, k, l) then it represent a

plane. If they are enclosed in “[ ]” as [h, k, l] then it represent the direction of crystal.Steps to determine Miller Indices for a given plane as shown in the figure. Take a lattice point as origin ‘o’ of crystallographic axis x, y, z in a space lattice. Let A, B, C be the crystal plane intercepts there axis of 4a, 4b, 3c.

s Step (1): Determine the coordinate of intercepts made by the plan along the crystallographic axis x y z 4a 4b 3c In general x y z pa qb rc (where p=4, q=4, r=3). Step (2):-- Divide the intercepts with lattice point translational distances along the axis. 4a/a 4b/b 3c/c 4 4 3 In general Pa/a qb/b rc/c P q r Step (3):-- Determine the reciprocals of these numbers i.e. 1/4 1/4 1/3 In general 1/p 1/q 1/r. Step (4):-- Reduce the reciprocals to smallest integers and enclosed them in “( ) “[by multiplying with L.C.M] 1/4× 12 1/4× 12 1/3× 12 3 3 4 (3 3 4) Miller Indices may be defined as the reciprocals of the intercepts may by the plane on crystallographic axis when reduced to smallest numbers Important features of Miller indices are,

1. When a plane is parallel to any axis, the intercept of the plane on that axis is infinity. Hence its Miller index for that axis is zero. 2. When the intercept of a plane on any axis is negative, a bar is put on the corresponding Miller index. 3. All equally spaced parallel planes have the same index number (h k l) Ex: The planes ( 1 1 2) and (2 2 4) are parallel to each other

1.16.2. Distance of Separation between Successive planes Let us consider a rectangular coordinate system with origin ‘o’ at one of the lattice point. Let (h k l ) be the miller indices of a plane A,B,C which makes intercepts

at OA, OB, OC along x, y, z axis respectively. Let OM be the normal passing through the origin and meet the plane A, B, C at N. such that ON = d, the normal makes an angles α, β, γ along x, y, z axis respectively.

Fig: Inter planar Spacing The intercepts OA, OB, OC are such that OA = a/h, OB = b/k, OC = c/l →(1) We have, Cos α = ON/OA = d1/(a/h) Cos β = ON/OB = d1/ (b/k) ` → (2)

Cos γ = ON/OC = d1/(c/l) We know that, Cos2 α + Cos2 β + Cos 2γ = 1 → (3) Submit eq (2) in eq (3),

d 2 1 = 1

2 d1 = 1

2 d1 =

Let ( h/2 k/2 l/2 ) be the miller indices of next plane A’ B’ C’ making intercepts OA’ OB’ OC’ along x, y, z axis respectively. The normal meets the 2nd plane A’, B’, C’ at N’

such that ON’ = d2. The extension of d1 to d2the normal makes the same along α, β, γ with x, y, z axis respectively. The intercepts of OA’, OB’, OC’ are such that

OA’ = OB’ = , OC’ =

Cos α = ON’/OA’ = hd2/2 Cos β = ON’/OB‘= kd2/2b → (4) Cos γ = ON’/OC’ = ld2/2c

We know that, Cos2 α + Cos2 β + Cos 2γ = 1 → (5)

2 d2 + +

Let the separation between the planes ABC & ABC be‘d ‘.

d = d2 d1

= 2d1 d1 = d1

For cubic system a = b = c.

For Tetragonal system a = b ≠ c.

1.17. Structure of Sodium Chloride (NaCl) Sodium chloride crystal is an ionic one. In the NaCl lattice, positively charged sodium ion and negatively charged chlorine ion are situated side by side. It consists of two FCC sub lattices. One of the chlorine ion has its origin at (0, 0, 0) point while sodium ion has its origin midway along the cube edge at (a/2, 0, 0) point as shown in figure. Due to the electro static force between sodium and chlorine ions, the two ions are attracted towards each other. When the two ions come closer their outer electron shells comes closer giving rise to strong forces of repulsion when attraction and repulsion balance, equilibrium is obtained.

Fig : Structure of Sodium Chloride

Each unit cell contains eight corner ions, six face centered ions, twelve ions at halfway of the edges and one at the center of the body. Number of Cl- ions per unit cell is (8X1/8) + (6X1/2) = 1+3 = 4 ions Number of Na+- ions per unit cell is (12X1/4) + 1 = 3+1 = 4 ions Each unit cell has four NaCl molecules. Each Na+ ion has six Cl- and each Cl- has six Na+ ions. The coordination number of NaCl for opposite kind is six and same kind is twelve. The nearest neighbor’s distance is a/2 for opposite kind and a√2 for the same kind. KCl, KBr, MgO, AgBr etc., have the similar structure as NaCl.

1.18. Diamond Structure Diamond possesses a structure which is a combination of two interpenetrating FCC sub lattices along the body diagonal at a length of quarter of cube edge. One sub lattice, say x, has its origin at the point (0, 0, 0) and other sub lattice has its origin quarter of the way along the body diagonal i.e., at the point (a/4, a/4, a/4). The fractions such as ¼, ½, ¾ denote the height above the base since the coordination number is 4, the structure is loosely packed. In the unit cell, in addition to the eight corner atoms, there are six face centered atoms and four atoms are located inside the unit cell. Each corner atom is shared by eight adjacent unit cells and each face centered atom is shared by two unit cells.

Effective number of atoms per unit cell In the unit cell in additional to the 8 corner atoms there are 6 face centered atoms and four more atoms located inside the unit cell. Each corner atom is shared by 8 adjacent unit cells and each face centered atom is shared by 2 cells hence the total number of atoms unit cell is (8 × ) + (6 × ) + 4 = 8

Fig: Diamond Structure

Coordination number: In this lattice each atoms has 4 nearest neighbors with it forms covalent bonds. Thus the coordination number of diamond crystal is 4.

Atomic Radius: From the diagram

(XY) 2 = +

(XY) 2 = =

(XZ) 2 = (XY) 2 + (YZ) 2

(XZ) 2 = +

Nearest neighbor distance: r =

Atomic packing fraction:

Packing fraction =

= = 0.34 = 34%

Problems

1) In sodium crystal the equilibrium distance between ion is r0=2.81 A° and A=1.748 Taking n=9. Calculate the potential energy per ion pair (June 2008)

Solution: We know that,

W = -1.27× Joule

= -7.97 eV 2) Calculate the binding energy of NaCl of which the nearest neighbor distance is 0.324nm. express the energy in eV and also in KJ/K mol. Modeling constant for NaCl=1.748 and n=9.5 (June 2010) Solution: We know that,

W0 = Joule

= - (9× )

= -6.9eV The binding energy per mol

Wk mol = NAW0= 6.9 eV = × (6.9) × ( Joule = 667× 103 KJ/k mol

3) Calculate the interplanar spacing for (3 2 1) plane in a simple cubic lattice whose lattice contact a = 4.2× m.

Solution: (h k l) = (3 2 1) a = 4.2× m.

= 1.13× m.

4) The atomic radius of copper is 1.278 A0. It has atomic weight 63.54. Find the density of copper.

Solution: we know that, a = = = 3.16 A0

ρ = =

= 8.98 gm/cm3 = 8980 kg/ m3

Objective Questions

1) At equilibrium condition the atoms posses [ a ] a) Minimum potential energy b) Maximum potential energy c) Maximum kinetic energy d) Maximum total energy

2) Ionic bonds are mainly formed in [ b ] a) Organic compounds b) Inorganic compounds d) Metals d) None of these

3) The nature of binding for a crystal with alternate and evenly spaced positive and negative ion is [ a ] a) Ionic b) Covalent c) Metallic d) Dipole

4) Ionic solids have [ b ] a) Low melting point b) Moderate melting point c) High melting point d) none of these

5) What is the nature of binding in OH4? [ b ] a) Ionic b) Covalent c) Van der waals d) Metallic

6) Among the following the strongest bond is [ a ] a)Ionic bond b)Covalent bond c)Metallic bond d)Hydrogen bond

7) A crystal is [ c ]

(a) A three dimensional representation of a solid (b) A three dimensional regular arrangement of U.C (c) A basis attached to the lattice point (d) All the above 8) The number of lattice points in a primitive cell is [ a ] (a) 1 (b) ½ (c) 2 (d) 3/2 9).The atomic radius of BCC lattice is [ a ] (a) (b)

(c) a/2 (d)

10). Ratio of number of atoms per unit cell for Sc, Bcc, Fcc crystal is [ a ] (a) 1:2:4 (b) 2:1:4 (c) 1:2:2 (d) 4:2:1

11). In a simple cubic lattice, : is [ d ] (a) 6:3:2 (b) 2:3:6 (c) : : (d) : : 12). The Basis structure of Nacl is [ c ] (a) Simple cubic (b) BCC (c) FCC (d) HCP

Chapeter-2 X-RAY DIFFRACTION & DEFECTS IN CRYSTALS

2.1. Introduction to X – ray Diffraction X-Rays are electromagnetic waves like ordinary light, therefore, they should exhibit interference and diffraction .Diffraction occurs when waves pass across an object whose dimensions are of the order of their own wavelengths. The wavelength of X-rays is of the order of 0.1nm or 10 -8 cm so that ordinary devices such as ruled diffraction gratings do not produce observable effects with X-rays. Laue suggested that a crystal which consisted of a 3-dimensional array of regularly spaced atoms could serve the purpose of a grating. The crystal differs from ordinary grating in the sense that the diffracting centers in the crystal are not in one plane. Hence the crystal acts as a space grating rather than a plane grating. There are 3 main X-ray diffraction methods by which the crystal structure can be analyzed. Laue Method – for single crystal Powder Method- for finely divided crystalline or polycrystalline powder Rotating crystal Method - for single crystal 2.2. Bragg′s Law The monochromatic x-ray beam of wavelength λ is incident at an angle θ to Bragg’s planes. Let the inter planer spacing of crystal planes is‘d’. The dots in the planes represent positions of atoms in the crystal. Every atom in the crystal is a source of scattering for the X-ray incident on it.

Fig: X-Ray Scattering by Crystal

Apart of the incident x-ray beam AB incident an atom at B in plane 1 is scattered along the direction BC. Similarly apart of incident x-ray DE fall on atom at E in plane 2 and is scattered in the direction EF and it is parallel to BC. Let the beam AB and DE make an angle θ with the Bragg’s planes. This angle θ is called the angle of diffraction or glancing angle. If the path difference between the rays ABC & DEF is equal to nλ, than

the rays reflected from constructive plane or phase. So constructive interference takes place among the reflected ray BC and EF, then the path difference between the rays is, PE+QE= nλ From the Δle PBE, = =

PE = d From the Δle QBE, = =

EQ = d For constructive interference, PE + EQ = nλ d + d = nλ 2d = nλ The above equation is called Bragg’s Law.

2.3. Laue method

S1 & S2 are two lead screens in which two pin holes act as slits. X-ray beam from an X –ray tube is allowed to pass through these two slits S1 & S2 . The beam transmitted through S2 will be a narrow pencil of X – rays . The beam proceeds further to fall on a single crystal such that Zinc blended (ZnS) which is mounted suitably on a support . The single crystal acts as a 3 – dimensional diffraction grating to the incident beam. Thus, the beam undergoes diffraction in the crystal and then falls on the photographic film. The diffracted waves undergo constructive interference in certain directions, and fall on the photographic film with reinforced intensity. In all other directions, the interference will be destructive and the photographic film remains unaffected. The resultant interference pattern due to diffraction through the crystal as a whole will be recorded on the photographic film (which requires many hours of exposure to the incident beam). When the film is developed, it reveals a pattern of fine spots, known as Laue spots.

The distribution spots follow a particular way of arrangement that is the characteristic of the specimen used in the form of crystal to diffract the beam. The Laue

spot photograph obtained by diffracting the beam at several orientations of the crystal to the incident beam are used for determining the symmetry and orientations of the internal arrangement of atoms, molecules in the crystal lattice . it is also used to study the imperfections in the crystal .

2.4. Powder Method (or) Debye Scherrer method X-ray powder method is usually carried out for polycrystalline materials. The given polycrystalline material is grounded to fine powder and this powder is taken in capillary tube. This tube is made up of non diffracting material, and fixed at the centre of cylindrical Debye Scherrer cylindrical camera as shown in figure. The principle under this technique is that millions of tiny crystals in powder have completely random orientation all the possible diffraction phases are available for Bragg reflection to takes place. All the orientations are equally reflected ray will form a cone. whole axis likes along the direction of incident beam and whose semi vertical angle is twice the glancing angle for that particular planes.

Fig: Powder method - apparatus

The different cones intercept the film in a series of concentric circular from the radial of these arcs. The angle can be calculated and hence the spacing between the atoms can be evaluated as shown in figure. The photographic film is in cylindrical shape, whose axis is perpendicular to the beam. Let s be the distance of particular arc from the centre. Let R be the radius of camera. Then 4θ =

If S1, S2, S3, are the distances between symmetrical lines on the stretched film, then

θ2 = ……………………….

Using these values of θn in Bragg’s equation nλ =2d Sinθn.

Where, n=1, 2, 3, … is the Order of diffraction d= Interplanar spacing th θn = Angle of diffraction for n order. The inter-planar spacing d can be calculated. 2.5. Introduction to Defects in crystals The structural imperfections of crystal can be classified on the basis of their geometric structures as point defect, line defect, surface defect and volume defect.

2.6. Point defects As the name indicates, these defects are at same point in the crystal .so these are called as point defects point defects are also called as zero dimensional defects. The common types of point defect in a solid material are grouped into 3 categories. 1. Lattice site defect a) Vacancies (Schottky defect ) b) Interstitial vacancies (Frenkel defect ) 2. Compositional defect a) Substitutional impurity defect b) Interstitial impurity defect 3. Electronic defect

1. Lattice site defects In this type of defect, some atoms may not present in their regular atomic sites (or) positions. They are,

a) Vacancies As shown in the diagram at a point one or two or three atoms are missed. This is referred as single or double or treble vacancies. The vacancies are formed due to imperfect pattern during crystallization or due to thermal vibration at high temperature.

Fig: Vacancy

Schottky defect In an ionic crystal if a action vacancy exist, then in the very nearby place. On anion vacancy will also exist that is moved to surface of the crystal. So that change neutrality is maintained in vacancy region as shown in diagram. This is known as Schottky defect. Eg: Crystals such as NaCl, KCl, KBr etc,

Fig: Schottky Defect

b) Interstitial vacancies If an atom is moved to an interstitial space in the crystal then the defect is known as interstitial defect.

Frenkel defect In an ionic crystal, if a cation (positive ion) moves to an interstitial space so that a vacancy is formed in its atomic position Here change neutrality is maintained in the defect region as shown in the diagram. This type of defect is known as Frenkel defect. Eg; crystal such as CaF2, AgCl, AgI, AgBr etc.

2. Compositional defect The presence of impurity atoms in the crystal leads to compositional defects. Impurity atoms are present at the sites of regular parent atoms are in the sites of regular parent atoms are in the interstitial spaces. These defects are described as follows.

a) Substitutional Impurity defect:

As shown in diagram during crystallization free foreign atoms occupies the regular parental positions or atomic sites. Eg: In extrinsic semiconductors either trivalent or pentavalent atoms he sites of silicon or germanium atoms.

Fig: Substitutional impurity

b) Interstitial impurity defect The spaces between the parents the parental atoms in the crystal are known as interstitial spaces small sized (lower atomic number) atoms, such as hydrogen etc. may fit in these interstitial spaces. These atoms are known as interstitial atoms and the defect formed due to presence of interstitial atoms is known as interstitial impurity defect.

Fig: Interstitial impurity defect 3. Electronic defect At absolute zero in a purely covalent crystal (silicon) the electrons are tightly bound to the core and all are set to be in the value band. Absolute zero sum of the electrons likely to occupy. So in the crystal of pure silicon some temperature .so in the crystal of pure of silicon some of the electrons from the covalent bonds get thermally released and become free to move. In this way the deficiency of an electron creates a hole. Then the electrons & holes give rise to electronic imperfection.

2.7. Line defects

If a crystal plane ends somewhere in the crystal, that along the edge of the incomplete. Plane produces defect in the crystal called line defect the line defect is of two types, a) Edge dislocation b) Screw dislocation

a) Edge dislocation In a perfect crystal , the atoms arranged in both vertical and horizontal plane parallel to the side faces as shown in figure if one of these vertical planes does not extend to full length but ends in between within the crystal. Then it is called edge dislocation therefore in this defect, a line of atoms is not in proper removed.

Figure: Edge dislocation

The following points are observed. i) In the perfect crystal, the atoms are in equilibrium position. ii) In the imperfect crystal, just above the discontinuity the atoms are prell hard and are in a stat of completion. iii) Just below the discontinuity, the atoms are pulled apart and are in a state of tension. iv) This distorted configuration extends all along the edge inside the crystal, perpendicular to the cross section. v) Edge dislocations are represented by perpendicular or T depending on whether the incomplete plane stale from the top or from the bottom of the crystal. These two configurations are referred as positive or negative edge dislocation respectively.

Burger’s Vector To understand the concept of burgers vector let us consider two crystal one perfect and the other edge dislocation as shown figure

Figure: Burger’s Vector

The burgers vectors of a dislocation are determined by tracing a burgers circuit. Let us consider the burgers circuit in case of perfect crystal. The burgers circuit is realized by drawing a rectangular from point p. starting from point p, we go up by 6 steps to a point Q then move towards right by 6 steps to a point R, further move down by 6 steps to a point S. Finally move towards left by 6 steps to reach the starting point p. This gives a closed circuit PQRS known as Burger circuit.

b) Screw dislocation Figure shows when a part of the crystal is displaced relative to the rest of the crystal. It is important to mention here that the displacement terminates with the crystal. Screw dislocation is defined as displacement of the atoms in one part of a crystal relative to the rest of the crystal. The screw dislocation forms a ramp around the dislocation line. The Buggers vector is also in the figure. Buggers vector in parallel dislocation line. Normally the real dislocations in the crystal are mixture of edge quell screw dislocation.

Figure: Screw dislocation

Problems

1) Calculate the ratio d100:d110:d111 for a single cubic structure.

Sol: d =

d100 =

d110 =

: : = 1: :

= √6:√3:√2 2) The Bragg’s angle in the 1st order for [2, 2, 0] reflection from Ni (BCC) is 38.2o. When x-rays of wavelength λ=1.54 Ao are employed in a diffraction experiment. Determine the lattice parameter of Ni. Sol: (h, k, l) = (2, 2, 0) Θ=38.2o λ=1.54 Ao a=? 2d sinθ = nλ 20×d× =1×1.54×10-10 d=1.245Ao d=

1.245×10-12 =

a= × 1.245 × 10-12 a=3.52Ao 3) Monochromatic x-rays of λ=1.5Ao is incident on a crystal phase having an inter planar spacing of 1.6Ao. Find the highest order for which Bragg’s reflection maximum can be seen. Sol: d=1.6 Ao, θ=90o, λ=1.5 Ao, n=? We know that, nλ = 2d sinθ o For the highest order of diffraction n = nmax and θ = 90 o nmax ×1.5 = 2(1.6) sin90

nmax = 2.13

Highest order of reflection is nmax=2 4) Calculate the glancing angle at (110) plane of a cubic crystal having axial length a=0.2nm.corresponding to the 2nd order diffraction maximum for the x-rays of wavelength 0.065nm Solution: (h, k, l) = (1, 1, 0 ) a = 0.26m λ=0.065nm We know that 2d sinθ = nλ d =

d=0.18 2(0.18) sinθ=2(0.065) 0.368sinθ=0.13 Sinθ =

Θ=20041’13’’ 5) The Bragg’s angle for reflection from the (111) plane in a BCC crystal is 19.20. For an X-ray of wavelength 1.54 A0. Compute the cube edge of the unit cell. [June 2010] Solution: θ=19.2, order = 1 2d = nλ 2d×sin (19.2) = 1(1.54) 2d × 0.32 = 1.54 d = 2.341 A0 We know that, d =

a = 4.05 A0 6) What is the angle at which the 3rd order reflection of x-rays of 0.79A0 wavelength can occur in a crystal of 3.04X10-8cm? Sol: 2d = nλ λ=3.04×10-8cm

θ = 22. 940 7) A beam of x-ray is incident on an ionic crystal with lattice spacing 0.313nm. Calculate the wavelength of x-rays, if the first order Bragg’s reflection takes place at a glancing angle of 7048’. [June 2011] Sol: 2d Sinθ = nλ Given data, d=0.313nm n=1 θ= 7048’

λ = λ = 0.0836nm.

Objective Type Questions

1) X-ray are used for crystal diffraction studies because [ c ] a) They have higher penetrating power b) Crystals are transparent to x-rays c) The inter atomic spacing is of the order of x-ray wavelength d) They have high resolving power 2) Inter atomic spacing is of the order of [ a ] a) 2 to 3 A0 b) 2 to 3µm c) 2 to 3nm d) 2 to 3mm 3) For simple cubic the ratio : is given by [ d ]

c) 1:√2:

d) √1:√2:√3 4) For body centered cubic the ratio : is given by [ a ]

c) 1:√2:

d) √1:√2:√3 5) For face centered cubic ratio : is given by [ c ]

c) 1:√2:

d) √1:√2:√3

6) Substitutional impurity is [ a ]

(a) Point defect (b) line defect

(c) Surface defect (d) volume defect

7) Interstitial impurity is [ d ]

(a) Surface defect (b) volume defect

(c) Line defect (d) point defect

8) These are point defects [ d ] (a) Vacancies (b) interstitial atoms (c) substitutional (d) all of these 9) Point defects influence [ d ] (a) Mechanical properties (b) optical properties (c) Electrical properties (d) all of these 10) Frenkel and Schottky defect are exhibited [ c ] (a) by metals (b) ceramics (c) ionic crystals (d) b and c 11) Change in density is caused by [ b ] (a) Frenkel defects (b) Schottky defect (c) Edge dislocation (d) screw dislocation 12) One dimensional imperfections are called [ b ] (a) Point defects (b) line defect (c) Surface defect (d) volume defect 13) Screw dislocation belongs to [ b ] (a) Point defects (b) line defect (c) Surface defect (d) volume defect 14) Burger’s vector define [ c ] (a) The magnitude of point defect (b) The orientation of different crystal planes vertically (c) The managitude and direction of dislocation (d) None of these 15) A grain boundary is ------[ c ] (a) A point defect (b) The combination of edge dislocation and screw dislocation (c) The region where the crystal orientation changes sharply (d) The region where the crystal orientation changes gradually 16) The number of vacancies in a crystal [ d ] (a) First increases and then decreases (b) Does not change with temperature (c) Decrease as the temperature increases (d) Increase as the temperature increases

UNIT-II Chapter3 PRINCIPLES OF QUANTUM MECHANICS

3.1.1 Wave and particle duality A particle has mass, it is located at some definite point, it can move from one place to another, it gives energy when slowed down or stopped. Thus, the particle is specified by (1) Mass m (2) Velocity v (3) Momentum p and (4) Energy E.

A wave is spread out over a relatively large region of space , it cannot be said to be located just here and there, it is hard to think of mass being associated with a wave. Actually a wave is nothing but a rather spread out disturbance. A wave is specified by its (1) Frequency (2) Wavelength, (3) Phase of wave velocity, (4) Amplitude and (5) Intensity

Considering the above facts, it appears difficult to accept the conflicting ideas that radiation has a dual nature, i.e., radiation is a wave which is spread out over space and also a particle which is localized at a point in space. However, this acceptance is essential because radiation sometimes behaves as a wave and at other times as a particle as explained below: (1) Radiations including visible light, infra-red, ultraviolet, X-rays, etc. behave as waves in experiments based on interference, diffraction, etc. This is due to the fact that these phenomena require the presence of two waves at the same position at the same time. Obviously, it is difficult for the two particles to occupy the same position at the same time. Thus, we conclude that radiations behave like wave.. (2) Planck’s quantum theory was successful in explaining black body radiation, the photo electric effect, the Compton Effect, etc. and had clearly established that the radiant energy, in its interaction with matter, behaves as though it consists of corpuscles. Here radiation interacts with matter in the form of photon or quanta. Thus, we conclude that radiations behave like particle.

3.1.2 de–Broglie Hypothesis In the Newton time, matter and radiation both were assumed to consist of particles. With the discovery of phenomena like interference, diffraction and polarization it was established that light is a kind of a wave motion. In the beginning of 20th century some new phenomena (photoelectric effect, Compton effect, etc.) were discovered which could not be explained on the basis of wave theory. These phenomena were explained on the basis of quantum theory in which light quanta or photons are endowed corpuscular properties – mass (hν/c2), velocity v and 43

momentum hν/c. But, when the photon theory was applied to phenomenon such as interference, diffraction, etc. (which have been fully explained on wave theory) it proved helpless to explain them. Thus, light has a dual nature, i.e., it possesses both particle and wave properties. In some phenomena corpuscular nature predominates while in others wave nature predominates. The manifest of properties depends upon the conditions under which the particular phenomena occur. But wave and particle never expected to appear together. Louis de- Broglie in 1924 extended the wave particle parallelism of light radiations to all the fundamental entities of Physics such as electrons, protons, neutrons, atoms and molecules etc. He put a bold suggestion that the correspondence between wave and particle should not confine only to electromagnetic radiation, but it should also be valid for material practices, i.e. like radiation, matter also has a dual (i.e., particle like and wave like ) character. In his doctoral thesis de-Broglie wrote that there is an intimate connection between waves and corpuscles not only in the case of radiation but also in the case of matter. A moving particle is always associated with the wave and the particle is controlled by waves. This suggestion was based on the fact that nature loves symmetry, if radiation like light can act like wave some times and like a particle at other times, then the material particles (e.g., electron, neutron, etc.) should act as waves at some other times. These waves associated with particles are named de- Broglie waves or matter waves.

3.1.3 Expression for de- Broglie wavelength The expression of the wavelength associated with a material particle can be derived on the analogy of radiation as follows: Considering the plank’s theory of radiation, the energy of photon (quantum) is E = h = → (1)

Where c is the velocity of light in vacuum and is its wave length. According to Einstein energy – mass relation E = mc2 → (2) = → (3)

Where mc = p is momentum associated with photon. If we consider the case of material particle of mass m and moving with a velocity v , i.e momentum mv, then the wave length associated with this particle ( in analogy to wave length associated with photon ) is given by = → (4)

Different expressions for de-Broglie wavelength (a) If E is the kinetic energy of the material particle then

E = mv2 = =

p2 = 2mE or p = Therefore de- Broglie wave length → (5)

(b) When a charged particle carrying a charge ‘q’ is accelerated by potential difference v, then its kinetic energy K .E is given by E = qV Hence the de-Broglie wavelength associated with this particle is → (6)

For an electron q = 1.602 10-19 Mass m = 9.1 X 10-31 kg

= = A0 → (7)

3.1.4 Properties of Matter Waves

Following are the properties of matter waves: (a) Lighter is the particle, greater is the wavelength associated with it. (b) Smaller is the velocity of the particle, greater is the wavelength associated with it. (c) When v = 0, then = , i.e. wave becomes indeterminate and if v = then = 0. This shows that matter waves are generated only when material particles are in motion. (d) Matter waves are produced whether the particles are charged particles or not ( = is independent of charge). i.e., matter waves are not electromagnetic

waves but they are a new kind of waves . (e) It can be shown that the matter waves can travel faster than light i.e. the velocity of matter waves can be greater than the velocity of light. (f) No single phenomenon exhibits both particle nature and wave nature simultaneously.

3.1.5 Distinction between matter waves and electromagnetic waves S.No Matter Waves Electromagnetic Waves 1 Matter waves are associated with Electromagnetic waves are produced moving particles (charged or only by accelerated charged particles. uncharged)

2 Wavelength depends on the mass of Wavelength depends on the energy of the particle and its velocity, photon

Matter waves can travel with a 3 Travel with velocity of light velocity greater than the velocity of c= 3×108 m/s 45

light.

4. Matter wave is not electromagnetic Electric field and magnetic field oscillate wave. perpendicular to each other.

5. Matter wave require medium for Electromagnetic waves do not require propagation, i.e, they cannot travel any medium for propagation, i.e., they through vacuum. can pass through vacuum.

3.2.1 Davisson and Germer’s Experiment The first experimental evidence of matter waves was given by two American physicists, Davisson and Germer in 1927. The experimental arrangement is shown in figure 3.1(a). The apparatus consists of an electron gun G where the electrons are produced. When the filament of electron gun is heated to dull red electrons are emitted due to thermionic emissions. Now, the electrons are accelerated in the electric field of known potential difference. These electrons are collimated by suitable slits to obtain a fine beam which is then directed to fall on a large single crystal of nickel, known as target T which is rotated about an angle along the direction of the beam is detected by an electron detector (Faraday cylinder) which is connected to a galvanometer. The Faraday cylinder ‘c’ can move on a circular graduated scale s between 290c to 900 to receive the scattered electrons.

Fig 3.1(a) Davisson and Germer’s experimental arrangement for verification of matter waves

First of all, the accelerating potential V is given a low value and the crystal is set at any orbital azimuth (θ). Now the Faraday cylinder is moved to various positions on the scale’s’ and galvanometer current is measured for each position. A graph is plotted between galvanometer current against angle θ between incident beam and beam entering the cylinder [Figure3.1(b)]. The observations are repeated for different acceleration potentials.

Fig. 3.1(b) Variation of Galvanometer current with variation of angle θ between incident beam and beam entering the cylinder

It is observed that a ‘bump’ begins to appear in the curve for 44 volts. Following points are observed. (a) With increasing potential, the bump moves upwards. (b) The bump becomes most prominent in the curve for 54 volts at θ = 500. (c) At higher potentials, the bumps gradually disappear.

The bump in its most prominent state verifies the existence of electron waves. According to de- Broglie, the wavelength associated with electron accelerated through a potential V is given by A0.

Hence, the wavelength associated with an electron accelerated through 54 volt is = 1.67 A0

From X-ray analysis, it is known that a nickel crystal acts as a plane diffraction grating with space d = 0.91 A0 [see Figure 3.1(c)]. According to experiment, we have diffracted electron beam at θ = 500. The corresponding angle of incidence relative to the family of Bragg plane θ1 = = 650

Using Bragg’s equation (taking n=1), we have 2dsinθ = 2(0.91A0) 0 This is in good agreement with the wavelength computed from de-Broglie hypothesis.

Fig. 3.1(c) Bragg planes in Nickel crystal

As the two values are in good agreement, hence, confirms the de-Broglie concept of matter waves.

3.2.2 G .P. Thomson′s Experiment The experimental arrangement is as shown in Figure 3.2. High energy electron beam produced by the cathode ‘C’ is accelerated with a potential up to 50kV. A fine pencil of accelerated beam is obtained by allowing it to pass through a narrow slit S and is made to fall on a very thin metallic film F of gold (or silver or aluminum). The electron beam is scattered in different directions by the metallic film and incident on photographic plate P. The entire apparatus is exhausted to a high vacuum so that the electrons may not lose their energy in collision with the molecules of air.

Fig 3.2: (a) G.P.Thomson’s Apparatus, (b) pattern recorded on the photographic plate

Since ordinary metals like gold are micro crystalline in structure, the diffraction pattern produced by them are similar in appearance to the X- ray diffraction powered pattern and consist of a series of well-defined concentric rings about a central spot as shown in figure 3.2(b). To make sure that this pattern is due to electrons and not due to any possible X - rays generated, the cathode rays in the discharge tube are deflected by magnetic field. It was observed that the diffraction pattern observed on the fluorescent screen placed instead of photographic plate also shifted. This confirmed that the pattern is due to the electrons. Thomson calculated the wavelength of the de-Broglie waves associated with the cathode rays using the equation A0

and determined the spacing between the planes in the foils using Bragg’s equation. The results obtained are in good agreement with those from X-ray studies.

3.3 Heisenberg′s Uncertainty Principle

As a direct consequence of the dual nature of matter, in 1927, Heisenberg proposed a very interesting principle known as uncertainty principle. If a particle is moving, based on classical mechanics, at any instant we can find its momentum at any position. In wave mechanics, a moving particle can be regarded as a

wave group. The particle that corresponds to this wave group may be located anywhere within the group at any given time. In the middle of the group, the probability of finding the particle is more but the probability of finding the particle at any other point inside the wave group is not zero. Narrower the wave group higher will be the accuracy of locating the particle. At the same time, one cannot define the wavelength of the wave accurately when the wave group is narrower. Since ( ), measurement of particles momentum (mv = ) also

becomes less accurate.

Fig 3.3 (a) A narrow de – Broglie wave group. Since the position can be precisely determined, Δ x is small. As the measurement of is less accurate, Δ p is large. (b) A wide de –Broglie wave group. Measurement of λ is accurate and hence Δ p is small whereas, since wave group is wide, Δ x is large

On the other hand, when we consider a wide wave group, wavelength can be well defined hence measurement of momentum becomes more accurate. At the same time, since the width of the wave group is large, locating the position of the particle becomes less accurate. Thus the uncertainty principle can be stated as “it is impossible to know both exact position and exact momentum of an object at the same time”. If ∆x and ∆p are the uncertainties in the position and momentum respectively when they are simultaneously measured, then ∆x ∆p

Another form of the uncertainty concerns energy and time. In an atomic process let energy ‘E’ be emitted during the time interval ∆t, If the energy is emitted in the form of electromagnetic waves, we cannot measure frequency of the waves accurately in the limited time available . Let the minimum uncertainty in the number of waves that we count in a wave group be one wave.

Since frequency =

Hence the corresponding uncertainty in energy ∆E = h ∆ ∆E or ∆E ∆t

A more precise calculation based on the nature of wave of wave groups modifies this result to ∆E ∆t

This gives uncertainty in the measurement of energy and time of a process. Thus the more generalized statement of Heisenberg’s uncertainty principle is “It is impossible to specify precisely and simultaneously the values of both members of particular pair of physical variables that describe the behavior of an atomic system”.

3.4.1 Schrodinger's time independent wave equation

Schrodinger developed a differential equation whose solutions yield the possible wave functions that can be associated with a particle in a given situation. This equation is popularly known as Schrodinger equation. The equation tells us how the wave function changes as a result of forces acting on the particle. One of its forms can be derived by simply incorporating the de-Broglie wavelength expression into the classical wave equation. If a particle of mass ‘m’ moving with velocity v is associated with a group of waves, let be the wave function of the particle. Also let us consider a simple form of progressing wave represented by the equation

0 → (1) Where (x, t)

0 is amplitude Differentiating eq (1) partially with respect to ‘x’, we get

= -K 0

Again differentiating equation (1) with respect to ‘x’

2 = K 0 sin ( t-kx)

+ k2 = 0 → (2)

Since k = , + = 0 → (3)

Eq (2) or Eq (3) is the differential form of the classical wave equation. Now, incorporating de- Broglie wavelength expression in to eq (3), we get

+ = 0 → (4)

The total energy E of the particle is the sum of its kinetic energy k and potential energy V i.e., E = K + V But K = mv2

E = mv2 + V

mv2 = E – V

m2v2 = 2m (E - V) → (5) Substituting eq (5) in eq (4), we get

+ = 0 → (6)

In quantum mechanics, the value occurs most frequently. Hence we denote

ђ = using this notation, we have

+ = 0 → (7)

For simplicity, we have considered only one dimensional wave extending eq(7) for a 3 – dimensional wave

+ + + = 0 → (8)

Where (x, y, z); here, we have considered only stationary states of after separating the time dependence of The Laplacian operator is defined as

2 = + + → (9)

Hence eq (10) can be written as 2 + = 0 → (10)

This is Schrodinger wave equation. Since time factor doesn’t appear, eq(8) or eq(10) is called ‘time independent Schrodinger wave equation’ in three dimensions.

3.4.2 Physical significance of wave function (1) The wave function has no direct physical meaning. It is a complex quantity representing the variation of matter wave. (2) It connects the practical nature and its associated wave nature statically. (3) | |2 (or * if function is complex) at a point is proportional to the probability of finding the particle at that point at any given time. The probability density at any point is represented by | |2. (4) If the particle is present in a volume dxdydz, then | |2 dxdydz =1

If a particle is present somewhere in space

The wave function satisfying the above condition is said to be normalized. 51

3.5.1 Particle in Infinite square potential well

A free electron trapped in a metal or charge carriers trapped by barriers trapped by the potential barriers of a double hetero junction can be approximated by an electron in an infinitely deep one- dimensional potential well. Consider one – dimensional potential well of width L as shown in fig. Let the potential V = 0 inside well and V = outside the well.

Fig.3.4 Square potential well infinite height

The time independent Schrödinger wave equation in one dimensional case

+ = 0 → (1)

For a particle present inside the well where V=0 and

+ = 0 → (2)

Let the general solution of eq (2) be = A + B → (3) Where A and B are constants which can be determined from boundary conditions = 0 at x = 0 And = 0 at x = L → (4) Since = 0 at x = 0 0= A + B B = 0 → (5) Since = 0 at x = L 0 = A Which means A =0 or = 0 since both A and B cannot be zero, A≠ 0. If A = 0, then everywhere. This means that the particle is not in the well. The only meaningful way to satisfy the condition is = 0, or kL = nπ ; n = 1,2,3,…

k = → (6)

Thus, eq (3) simplifies to = A → (7)

Differentiating in eq (7)

Again Differentiating, we get

+ = 0 → (8)

Comparing eq (2) and eq (8), we get

= = k2

n is called the quantum number. Thus we obtain an important result. The particle cannot possess any value of energy as assumed in classical case, but it possesses only discrete set of energy values. The energy of the nth quantum level,

En = = (since = ) → (9)

The wave functions and the corresponding energy levels of the particles are as suggested in Figure 3.5

Fig.3.5 Ground state and first two excited states of an electron in a potential well: a) the electron wave functions and b) the corresponding probability density functions. The energies of these three states are shown on the right.

We are still left with an arbitrary constant ‘A’ in eq (7). It can be obtained by applying normalization condition i.e.; the probability of finding the particle inside the box is unity.

L [x ] 0 = 1

= 1 or A = → 10)

The normalized wave function is

n = sin → (11)

3.5.2 Particle in Three dimensional potential box Let us consider the case of a single particle, i.e., a gas molecule of mass m, confined within a rectangular box with edges parallel to X, Y and Z axes as shown in figure.

Fig 3.6 Three dimensional potential box

Let the sides of the rectangular box be a, b and c respectively. The particle can move freely within the region 0 < x< a, 0 < y < b and 0 < z < c, i.e., inside the box where potential V is zero, i.e., V (x, y, z) = 0, for 0 < x< a 54

V (x, y, z) = 0, for 0 < y < b and V (x, y, z) = 0, for 0 < z < c

The potential rises suddenly to have a large value at the boundaries, i.e., the potential outside the box is infinite. The Schrodinger wave equation inside the box is given by

+ + + E = 0 → (1)

This is partial differential equation in three independent variables and may be solved by the method of separation of variables. The solution of equation (1) is of the form

(x, y, z) = X (x) Y(y) Z (z) → (2) Where X (x) is a function of x alone, Y(y) is a function of y alone and Z (z) is a function of z alone. Differentiating in eq (2) partially with respect to ‘x’, we get = Y(y) Z (z) → (3)

Again differentiating eq (3) partially with respect to ‘x’, we get

= Y(y) Z (z)

= → (4)

Where X = X(x) and = (x, y, z)

Similarly, = → (5)

And = → (6)

Substituting eqs.(4),(5) and (6) in eq (1), we get

+ E = 0

+ E = 0 → (7)

This equation can be written as

= → (8)

The left hand side of eq (8) is a function of x alone, while the right hand side is a function of y and z and is independent of x. both sides are equal to each other. Both sides are equal to each other. This is possible only when they are separately equal to a constant quantity, i.e.

= → (9)

Or = → (10)

In eq (10), the left hand side is a function of y alone while right hand side is a function of z and is independent of y. If the above equation is to be satisfied, both sides must be

equal to a constant say , i.e.

= → (11)

Or = → (12)

Again, we have = → (13)

Or = → (14)

For convenience, introduce

= , = and =

Now, the differential equations in x, y and z coordinates may be written as

+ X = 0 → (15a)

+ Y = 0 → (15b)

And + Z = 0 → (15c)

The general solution of eq [(15a)] will be a sine function of the arbitrary amplitude, frequency and phase, i.e., X(x) = A sin (Bx + C) → (16) Where A, B and C are constants whose values are determined by boundary conditions. 2 represents the probability of finding the particle at any point within the box. Therefore, 2 which is a function of x coordinates only represents the probability of Finding the particle at any point along the X-axis. As the potential is very high at the walls of the box, the probability of finding the particle at the walls will be zero, i.e.,

2 = 0 when x = 0 and x= a = 0 when x = 0 and x= a Using the above boundary conditions in eq (16), we have 0 = A sin (0 + C); A 0 Sin C = 0 and 0 = A (Ba +C) \ Or 0 = . + . 0 = Sin Ba = 0 [since = 0, is not zero]

Ba = π 56

Or B = → (17)

X(x) = A Sin x → (18)

Applying the normalization condition between x=0 to x=a, we have 2 dx = 1

2 dx = 1

dx = 1

[a 0] = 1

= or A =

Therefore, X(x) = → (19)

Similarly, we can solve equations [15(b)] and [15(c)] to obtain

Y(y) = → (20)

Z (z) = → (21)

The complete wave function , has the form

= → (22)

The wave function of a particle in a finite box and its probability density are shown in Figure 3.7.

Fig 3.7 a) wave functions and b) probability densities of a particle in three dimensional box 57

Differentiating eq (19) twice with respect to ‘x’, we get

= = X(x) → (23)

Substituting eq (23) in eq [15(a)], we get X(x) + X (x) = 0

= → (24)

Similarly, = → (25)

And = → (26)

The allowed values of total energy are given by

E = + +

= [ + + ] → (27)

Where , and denote any set of three positive numbers. When the box is a cube, i.e., a = b= c, the energy expression is given by

E = [ ] → (28)

With , , = 1,2,3……..

Chapter – 4 ELEMENTS OF STATISTICAL MECHANICS & ELECTRON THEORY OF SOLIDS

4.0 Introduction to Elements of Statistical mechanics Up to the end of seventeenth century, the physical phenomena were explained with the help of ordinary laws of classical mechanics. Consider the following situations in which the system consists of a large number of particles. i) Consider the case of a gas consisting of 1023 molecules. In order to obtain the complete information about the motion of this system, we have to solve 1023 equations or more. Moreover, it is not possible to have a complete knowledge regarding the positions and velocities of all the molecules. Therefore, it is impossible to apply ordinary laws of mechanics to such a physical system. ii) Consider the radioactive decay of an atom. In this case, one cannot say which atom of the radioactive material will decay first and when. iii) Consider the case of an atom with two electrons present in it. To consider its behavior, no one has solved it completely by the use of ordinary laws of mechanics.

Therefore, it is impossible to solve the problem of a system consisting of a large number of particles with the help of ordinary laws of mechanics. Such problems have been successfully solved by the method of statistical mechanics. Statistical mechanics is a branch of science which deals with the relationship between the overall behavior of a system of many particles and the properties of the particles. It is not concerned with the dynamics or interactions of individual particles, but it is concerned with the happening to the entire system. Since many phenomena in the physical world involve systems of very large number of particles, statistical approach is being made. Statistical mechanics can be applied to classical systems such as molecules in a gas as well as photons in a cavity and free electrons in a metal.

4.1. Phase Space Let us consider a system consisting of N particles distributed in a given volume V. when we consider a static system, all the particles are fixed at various points in space. With three coordinates X, Y and Z mutually perpendicular to each other, we can specify the location of any particles in three dimensional space. Since there are N particles, 3N coordinates give complete information about such a system. The three dimensional space in which the location of a particle is completely specified by the three coordinates is known as position space. A small volume dV in position space is given by

dV = dxdydz If the system is dynamic, the particles move with various velocities and possess moments. Hence to specify such a system, in addition to three position coordinates x, y

and z, we need three components of momentum px, py and pz which are given by px = m.vx, py = m.vy and pz = m.vy

Where vx, vy and vz are the three velocity components. The three dimensional space in which the momentum of a particle is completely

specified by the three momentum coordinates px, py and pz is known as momentum space. A small volume element d is given by

d = dpx.dpy.dpz

A combination of the position space and momentum space is known as phase space. A point phase space is completely specified by six coordinates x, y, z, px, py and pz. Since there are N particles, 6N coordinates provide complete information regarding the position and momentum of all the N particles in the phase space in a dynamic system. A small volume dτ is given by

dτ = dxdydz.dpxdpydpz

i.e., dτ = dV d Total phase space volume of the system is obtained by integrating the above equation i.e., τ = = τ = V Where V is the physical volume occupied by all the particles of the system and Π is the momentum space volume occupied by all the particles.

4.2.0 Ensembles A system is defined as a collection of a number of particles. An ensemble is defined as a collection of a large number of macroscopically identical, but essentially independent systems. By the term microscopically identical we mean that each of the systems constituting an ensemble satisfies the same macroscopic conditions, e.g., volume, energy, pressure, total number of particles etc. By the term independent systems we mean that the systems constituting an ensemble are mutually non-interacting. In an ensemble the systems play the same role as the non-interacting molecules in a gas.

4.2.1 The Canonical ensemble In the canonical ensemble, the assemblies are having the same volume V, number of particles N and are in thermal contact with each other so that they are in thermal equilibrium and have the same temperature T.

Fig4.1.The canonical ensemble

4.2.2. The Grand Canonical Ensemble The grand canonical ensemble is a collection of independent assemblies having the same temperature T, volume V and the chemical potential μ.

Fig4.2. the Grand Canonical ensemble

4.2.3. The Micro Canonical Ensemble The ensemble in which a system has the same fixed energy, fixed volume and also the same number of particles is called micro canonical ensemble.

Fig4.3.The Micro Canonical ensemble

4.3.0 Statistical Distribution Statistical mechanics determines the most probable way of distribution of total energy E among the N particles of a system in thermal equilibrium at absolute

temperature. It establishes how many particles are likely to have the energy Є1, how many particles are likely to have the energy Є2, and so on. In statistical mechanics, one finds the number of ways W in which the particles can be arranged among the available states. If g(Є) is the number of states of energy Є and f(Є) is the distribution function i.e., the probability of occupancy of each state of energy Є, then the number of particles of energy Є is given by n(Є) = g(Є) f(Є) When the energy distribution is continuous g (Є) is replaced by g (Є) dЄ. Three kinds of distributions are possible corresponding to three different kinds of particles. Maxwell- Boltzmann Statistics Bose – Einstein Statistics Fermi – Dirac Statistics

4.3.1 Maxwell – Boltzmann Statistics Maxwell – Boltzmann Statistics is also known as classical statistics. The main assumptions of Maxwell – Boltzmann statistics are: (i) The particles are identical and distinguishable (ii) The volume of each phase space cell chosen is extremely small and hence chosen volume has very large number of cells. (iii) Since cells are extremely small, each cell can have either one particle or no particle. (iv) The system is isolated i.e., the total number of particles of the system and their energy remain constant. (v) The particles are distributed among the energy levels that are spread from zero to infinity on the energy scale i.e., energy levels are continuous. (vi) Molecules of gases follow Maxwell – Boltzmann statistics. (vii) The number of particles with energies between Є and Є + dЄ in a sample of a gas that contains N molecules is given by

n(Є) dЄ = dЄ

4.3.2 Bose – Einstein Statistics Bose – Einstein Statistics is also known as Quantum statistics. In Bose – Einstein Statistics, the following assumptions are made. 62

(i) The particles are identical, indistinguishable and have integral spin. These particles are called Bosons. (ii) Bosons obey uncertainty principle. (iii) Any number of Bosons can occupy a single cell in phase space. (iv) Bosons do not obey Pauli’s exclusion principle. (v) Wave functions representing Bosons are symmetric. i.e., Ψ(2,1) = Ψ(1,2) (vi) The wave functions of Bosons do overlap slightly. i.e., Weak interaction exists. (vii) Energy states are discrete. (viii) The probability that a Boson occupies a state of energy Є is given by

f B-E(Є) =

(ix) This is called Bose – Einstein distribution function. (x) The examples of particles obeying Bose- Einstein statistics are Photons, phonons, pions, gluons, He4 etc.

4.3.3 Fermi – Dirac Statistics Fermi – Dirac Statistics is also known as Quantum statistics. The main assumptions are: (i) The particles obeying Fermi –Dirac statistics are identical, indistinguishable and have half integral spin. These particles are known as Fermions. (ii) Fermions obey Pauli’s exclusion principle. (i.e. There cannot be more than one particle in a single cell in phase space). (iii) Fermions obey uncertainty principle. (iv) Energy states are discrete. (v) Wave functions representing Fermions are antisymmetric. i.e., Ψ(2,1) = -Ψ(1,2) (vi) Weak interaction exists between the particles. (vii) Electrons, protons, neutrons and He3 are the examples of Fermions. (viii) The probability that a Fermion occupies a state of energy Є is given by

fF-D (Є) =

4.4.0 Black body and its radiation A perfectly black body is one which completely absorbs all the radiations of all wavelengths incident on it. Since it neither reflects nor transmits any radiation, it appears black whatever the color of incident radiation is. According to Kirchhoff’s law, a body which is capable of absorbing radiation must also be capable of emitting radiation of all possible wavelengths. So, a perfectly black body is a good absorber as well as a good radiator. When it is heated to a suitable temperature, it emits radiations of all wavelengths (continuous spectrum). The wavelength of emitted radiation by a black body depends only on its temperature and is independent of the material of the body.

The distribution of energy in black body radiation for different wavelengths and at various temperatures was determined experimentally by Lummer and Pringsheim 1899. The intensity of radiation corresponding to different wavelengths is measured at different temperatures and is plotted as shown in figure.

Fig4.4 Energy distribution in black body radiation From this study, it is observed that: 1) The emission from a black body at any temperature is composed of radiation from all wavelengths. 2) At a given temperature, the energy is not uniformly distributed. As the temperature of the black body increases, the intensity of radiation for each wavelength increases. This shows that the total amount of energy radiated per unit area per unit time increases with increase of temperature. 3) The total energy of radiation at any temperature is given by the area between the curve corresponding to that temperature and the horizontal axis. 4) The amount of radiant energy emitted is small at very short and very long

wavelengths. At a particular temperature, the spectral radiancy Eλ is maximum at a particular temperature. 5) The wavelength corresponding to the maximum energy represented by the peak of the curve shifts towards shorter wavelengths as the temperature increases. This is

called Wien’s displacement law. According to this law, λmT = constant.

4.4.1 Laws of black body radiation Wien’s law Wien showed that the maximum energy point in the black body energy distribution shifts towards shorter wavelengths when the temperature of the body is raised. He showed that

λmT = constant Where λm is the wavelength corresponding to maximum energy emission from a black body at absolute temperature T. He also showed that the maximum energy emitted by a black body is proportional to the fifth power of its absolute temperature. 5 5 (Eλ) max ∞ T or (Eλ) max/ T = constant. Wien by applying Maxwell’s law for distribution of velocities and the

principle of equipartition of kinetic energy gave the expression for Eλ as

-5 Eλ = C1λ

Where C1 and C2 are constants. This law agrees with experimental values at shorter wavelengths.

Rayleigh-Jean’s Law According to Rayleigh-Jean’s law, the energy distribution in the thermal spectrum is given by

This law agrees with experimental values at longer wavelengths only.

Planck’s radiation law In 1901, Max Plank derived a theoretical expression for the energy distribution in the black body on the basis of quantum theory of radiation. He made the following assumptions:

1) A black body radiator contains simple harmonic oscillators of possible frequencies. 2) The oscillators cannot emit or absorb energy continuously. Emission or Absorption of energy takes place in discrete amounts, i.e., energy of oscillator is quantized. 3) The energy of an atomic oscillator of frequency ν can have only certain values like 0, hν, 2hν, 3hν… This is an integral multiple of a small unit of energy hν called the quantum or photon. Let the total energy E be distributed among N number of particles. Let there be

N0, N1, N2, N3, Nr…etc oscillators having energy Є, 2 Є, 3Є, + …. rЄ, ….. Etc. respectively. Now we have

N = N0 + N1+ N2+ ….. + Nr + …. . → (1)

And E =0xN0 + Є N1+ 2 Є N2+ ….. + rЄ Nr + …. . → (2) Where Є = hν According to Boltzmann’s distribution formula, the number of oscillators having energy rЄ is given by

Nr = N0 → (3) Where k is Boltzmann’s constant.

Substituting the values of N1, N2, N3 … from eq (3) in eq (1), we get

N = N0 + N0 + N0 + ….. + N0 + ….. → (4)

= N0 [1 + + + + ….. + …..….

Let x = 2 3 r N = N0 [1 + x + x + x + …… x + ……… = → (5) [since 1 + x + x2 + x3 + …… xr + ……… = ]

Substituting the values of N1, N2, N3 … from eq (3) in eq (2), we get 2 3 r E = N0 (0) + Є N0x + 2Є N0x + 3Є N0x + …….. rЄx +…… 2 r-1 = Є N0x [1+ 2x + 3x + …….. + rx + …….. = [since 1+ 2x + 3x2 + …….. + rxr-1 + …….. = ]

= [From eq (5)]

= → (6)

Now the average energy of an oscillator is given by ⋶ = =

= → (7)

The number of photons present in the wavelength range λ and λ+ dλ per unit volume (i.e. photon density) = 8π λ-4kT → (8) Hence, total energy of photons within the wavelength range λ and λ+ dλ is given by -4 E λ d λ = 8π λ kT

E λ d λ = = → (9)

This is called Planck’s radiation law.

Deduction of Wien’s law and Rayleigh Jean’s law from Planck’s radiation law

Case (i): For shorter wavelengths, i.e., when λ is very small, >> 1, i.e., ‘1’ in the

denominator on the right hand side of eq(9) can be neglected in comparison to .

E λ d λ =

-5 E λ d λ = C1λ d λ → (10)

Where C1 = 8πhc and C2 = are constants.

This is Wien’s law which agrees with experiment at short wavelengths.

Case (ii):

For longer wavelengths, i.e., when λ is very large, is small and can be expanded as 1 = 1 + + + …………. 1

= 1 + [neglecting higher order terms]

Eλd λ = = d λ → (11)

This is Raleigh-Jean’s formula which agrees with experiment at longer wavelengths. Therefore, Planck’s radiation law is valid for all wavelengths.

4.5. Density of energy states The density of states (DOS) of a system describes the number of energy states at each energy level that are available to be occupied. A high density of states at a specific energy level s means that there are many states available for occupation. If the density of states is zero, no states can be occupied at that energy level. The number of energy states with a particular energy value depends on how many combinations of the quantum number s result in the same value n. 2 2 2 2 Let us consider a sphere of radius n (where n =nx +ny +nz ) in three dimensional space as shown in figure4.5. n z dn

n x E dE E

Fig.4.5 Density of energy states

Every point (nx, ny, nz) within this space represents an energy state. Also as every integer represents one energy state, unit volume of this space contains exactly one state. Hence

the number of states in any volume is equal to the volume expressed in units of cubes of lattice parameters.

The number of energy states within a sphere of radius n, = π → (1)

Since nx, ny, nz can have only positive integer values, we have to consider only one octant of the sphere. Hence available energy states = [ π ] → (2)

If we consider another sphere of radius (n+dn) within a small energy interval E+dE, the number of energy states having energy values between E and E+dE (i.e., in the energy interval dE) is given by Z′ (E) dE = [ π ] - [ π ]

= x [ - ]

[ ] [ is very small, higher order terms are

neglected] = [ ]

Z' (E) dE = [ ] → (3)

The expression for the energy of electron is

E = → (4)

= E → (5)

Or n = → (6)

Differentiating eq (5) taking n and E as variables

2n dn = E

n dn= E → (7)

Substituting eq (6) and eq (7) in eq (3)

Z' (E)dE = [ ] dE

= dE → (8)

According to Pauli’s exclusion principle, each energy level contains two electrons. This means each energy level will have two sub energy levels. Therefore, the above equation should be multiplied by 2.

= 2 x dE

= dE → (9)

The number of energy states per unit volume, i.e., density of states is given by

Z (E) dE =

= (2 dE

Z (E) dE = (2 dE → (10)

4.6.1 Fermi energy 1. We have seen that quantization lead to the discrete energy levels. Pauli's exclusion principle explains the distribution of electrons amongst the allowed energy levels. It allows a maximum of two electrons (with spins in opposite directions) in any energy level. 2. Thus, a pair of electrons, one with spin up and the other with spin down occupy the lowest energy level. The next pair occupies the next level. This process goes on until all electrons in the metal occupy their positions. 3. But, there will be many more allowed energy levels available for occupation.

“The highest energy level that can be occupied at 0K" is called Fermi level.

Fig 4.6: Distribution of electrons at different energy levels at a temperature a) 0K

b) T1K c) T2K (T1 > T2 K) At 0K, when the metal is not under the influence of an external field, all the levels above the Fermi level are empty, those lying below Fermi level are completely filled. According to Fermi – Dirac distribution, the probability F (E) of an electron occupying an energy level of energy E is given by

F (E) =

At any temperature other than 0k, if E= EF, F (E) =

Hence Fermi level is that state at which the probability of occupancy at any temperature above 0K is and also, it is the highest level of the filled energy states at

zero Kelvin. Fermi energy is “the energy of the state at which the probability of occupation of electron is at any temperature above 0K”. It is also “the maximum energy of filled

states at 0K”.

4.6.2 Fermi energy of electron at zero Kelvin

In the case of a metal at absolute zero, the upper occupied level is EF and for all

the levels below EF, F (E) =1. Let EF = at 0K. Hence, the concentration of electrons is

given by dE

= [2m ]

[2m ] = =

This is the expression for the Fermi-Energy of electrons in solids at absolute zero.

4.7.0 Introduction to Electron Theory of Solids

The electrons in the outermost orbital of the atom which constitute the solid, determine its electrical properties. On the basis of electrical properties, solid can be broadly classified into insulators, semi conductors and conductors. The electron theory of solids (which is applicable to all solids both metals and non metals) aims to explain the structures and properties of solids through their electronic structure. It explains the electrical, thermal and magnetic properties of solids. The theory has been developed in three main stages. i) The classical free electron theory (CFET) ii) The quantum free electron theory (QFET) iii) The Zone theory or Band theory of solids.

4.7.1 The classical free electron theory

This theory was proposed to account for the electrical conduction in metals by Drude in 1900. It was extended by H.A. Lorentz in 1909. This theory was also known as Free electron theory of metals. Metal atoms have one or more loosely bound valence electrons. Such electrons get detached easily even with small thermal energy. The detached electrons are neither shared nor acquired by any of the atoms of the metal. Hence they are free and just form a common fool. The electrostatic interaction of detached electrons with the positive ion cores and with other electrons is assumed to be negligible. A consequence of this assumption is that the detached electrons can move freely everywhere within the confinements of the metal piece. Hence these electrons are called as free electrons or conduction electrons. When an electric field is applied, free electrons will experience slow drift motion in the positive direction of the field and produce a current in the metal.

Successes of Classical Free Electron Theory 1) It proved the validity of ohms law 2) It could give a satisfactory explanation to the mechanism of electrical current in conductors and thermal conductivity of metals. 3) It explains optical properties of metals. It derives Widemann-Franz law i.e the relation between electrical conductivity and thermal conductivity.

Failures: 1) Temperature dependence of resistivity of metals could not be established correctly. 2) Heat capacity and paramagnetic susceptibility of conduction electrons could not be explained successfully. 3) It could not predict correct values to the mean free path of electrons.

4.7.2 The quantum free electron theory Somerfeld developed this theory during 1928. He considerd Drude’s assumption on the free electrons as it is. In addition to that, the applied Pauli’s exclusion principle to the electron gas and applied Fermi-Dirac statistics in place of Maxwell-Boltzmann statistics. The results of these modifications indicated that no all free electrons would contribute to the processes like electrical and thermal conductivities of metals. Rather only a small fraction of the free electron gas of the metal would participate in such properties.

Successes of quantum free electron theory 1) It could explain the specific heat and paramagnetic properties. 2) It could predict the temperature dependence of the resistivity correctly. 3) It could explain the thermionic emission of metals.

Failures of quantum free electron theory 1) This model could not explain why certain solids behave as conductors and why certain others behave as insulators. 2) The properties of the semiconductors could not be explained on the basis of this theory.

4.7.3 Band Theory of solids The failure of the free electron models (both classical and quantum) is due to the over simplified assumption that the interaction between a free or detached electrons and positive ion cores is negligible. This assumption implies that the potential due to a positive ion core is zero or rather constant throughout the crystal, as had been taken in the potential well. Which means the detached electrons can move freely within the metal. But, in reality, the free electron-ion core interaction cannot be neglected. As such the potential due to the positive ion cores cannot be taken as constant throughout the crystal. Therefore detached electrons cannot be completely free. The actual situation of potential in the crystal is roughly as follows: the potential field of a single positive ion core seen by an electron is as shown in figure 4.7(a).

Fig.4.7 Potential fields of (a) single atom, (b) a diatomic molecule and (c) crystal

The corresponding potential due to that of a diatomic molecule is as in figure 4.7(b). Extending the structure, the potential field of a crystal defined as regular periodic

arrangement of atoms would be as shown in fig 4.7(c). That is the potential field now becomes a periodic function with a period ‘a’. Thus, while the actual structure of the potential field is something like this or even more complicated; free electron model takes just a constant potential ignoring the periodic variations. As compared to this, the potential field assumed in free electron theory is very crude approximation. Kronig and penny had tried to rectify this drawback by taking the periodic variation of the potential field into account. But, they did not take the potential field of figure 4.7 as it would be too complex for mathematical analysis. Rather they have taken a rectangular periodic potential of Figure4.2. Though, this is not an exact representation of the potential given in fig 4.8, this is much closer to the actual situation than the previous model. In fact this model lead to marvelous conclusions and gave clue to several of the properties of the solids that are not explained by the previous model.

\ Fig4.8. Potential energy model used by Kronig and Penny

4.8. Electron in a periodic potential While the Somerfeld theory accounts satisfactorily for electrical conductivity in most metals, it failed to explain why other substances that also contain free electrons have virtually no conductivity and are considered to be excellent insulators. A solution to this problem is given by the zone theory or Band theory of solids. The effect of the periodic lattice field on the motion of the electrons leads to the zone or band theory of solids, which is of the greatest importance for the understanding of the structure and properties of metals, alloys and non metallic solids. According to zone theory, the electrons move in a periodic potential provided by the lattice. ` The potential of the solid varies periodically with the periodicity of space lattice and the potential energy of the electron is zero near the nuclei which it is halfway between the adjacent nuclei which are separated by a interatomic spacing ‘a’. This model was first postulated by Kronig and Penny. So taking this model and solving the Schrödinger equation for this case, we can find the existence of energy gap between the allowed values of energy of electron. So, if we use classical theory, we can get a parabola when we plat the curve between the electron’s energy and its momentum. Since the curve is a parabola, we can infer that the energy varies continuously. But by Kronig-Penny model, we can get a parabola with some discontinuities in it as shown in figure 4.9 73

Fig 4.9 Energy versus momentum curve

4.9 Bloch′s theorem Crystalline solid consists of a lattice which is composed of a large number of ionic cores at regular intervals [Figure 4.10(a)] and the conduction electrons move throughout the lattice. Let us consider the picture of the lattice in only one dimension, i.e., only an array of ionic cores along x-axis. If we plot the potential energy of a conduction electron as a position in the lattice, the variation of potential energy is as shown in figure. The potential is minimum at the positive ion sites and maximum between the two ions.

Fig 4.10 a) periodic positive ion cores inside metallic crystals. b) One dimensional periodic potential in crystal.

The one dimension Schrodinger equation corresponding to this can be written as

+ [E-V(x)] ψ=0 → (1)

The periodic potential V(x) may be defined by means of the lattice constant ‘a’ as

V(x)= V(x+a) → (2) Bloch considered the solution as

ψK (x)= (ikx)Uk(x) → (3)

Eqn (2) is known as Bloch function. Uk (x) is periodic with the periodicity of the crystal lattice. The free electron wave is modulated by periodic function Uĸ(x) is periodic with the periodicity of the crystal lattice. The free electron wave is modulated by periodic function Uĸ(x). For a linear chain of atoms of length ‘L’ in one dimensional case with ‘N’ (=even) number of atoms in the chain,

Uĸ(x) = Uĸ (x+Na) → (4) From eqn (3) and eqn (4)

Ψĸ(x +na) = Uĸ(x+Na)

= Uĸ(x)

= ψĸ(x) → (5) This is referred to as Bloch condition. Now, ikNa). (-ikNa) ψĸ(x+Na)ψĸ*(x+Na) = ψĸ(x) ψĸ*(x) e (0) = ψĸ(x) ψĸ*(x) e Ψĸ(x+Na) Ψĸ*(x+na) = Ψĸ(x) Ψĸ*(x) → (6) This means that the electron is not located around any particular atom and the probability of finding the electron is same throughout the crystal.

4.10 The Kronig-Penny Model The periodic potential assumed by Kronig and Penny is shown in Fiure4.11. i.e., a series of rectangular wells of width ‘a’ and are placed at a separation of b. in the regions where 0

Fig.4.11 One dimensional periodic potential assumed by Kronig and Penny

The main features of the model and its predictions can be explained qualitatively.

4.10.1. Main features of the model A. Schrodinger equation: The dynamical behavior of electrons in the Kronig-Penny model is represented by the following Schrodinger equation, + ] Eψ=0 for 0

And + [ → (1)

Let us assume that total energy ‘E’ of the electron under consideration is less than V₀. Further, let us substitute α²= and β² = (V₀-E) → (2)

Where α and β are real quantities. Now Eq(1) becomes + α²ψ=0, for 0

And – β²ψ=0, for –b

These equations can be solved with the help of block theorem. The final solution of eq (3) is given in the form of the following condition. P + cosαa =coska → (4)

Where P= V₀a is scattering power of the potential barrier and V₀ is barrier strength.

That means, eq (3) will have a solution only when the condition (4) is satisfied.

B. Graph of αa versus + cosαa

For the best understanding of the meaning of eq(4), let us consider the plot of the condition(4) i.e. L.H.S versus αa. Since the values of coska on R.H.s of eq (4) lie between +1 and -1, αa (which is a measure of energy) can take only those values for which the total left hand side (L.H.S) value lies between -1 and +1. Other values are not allowed. This means that energy E is restricted to lie within certain ranges which form the allowed energy bands or zones.

Fig 4.12. Plot of the left hand side of eq(4) as a function of αa for p = . The solid and broken lines on the

abscissa (αa- axis) correspond to allowed and forbidden energy regions of the energy spectrum respectively that are plotted in fig 4.13.

Conclusions of the graph 1. The energy spectrum consists of alternative regions of allowed and vacant bands. Forbidden band implies that the energy levels that lie in this region are not occupied by the electrons. 2. The allowed (shaded) bands are narrowest for low values of energy and become broader as energy increases, the unallowed (forbidden) bands becoming narrower. 3. a) For P=0 (i.e. on the extreme left), the whole energy spectrum is quasi- continuous. That is all allowed bands are joined together forming an almost continuum. b) However, the width of a particular allowed band decreases with increase in the value of P. As P , the allowed energy bands compress into simple energy levels and thus result in a line spectrum.

Fig4.13 Plot of P vs E revealing band structure of energy spectrum. The whole energy spectrum gets divided into allowed and forbidden bands.

C. P versus energy relation i. Tight binding limit For P , eq (4) will have solution only if sin αa = 0. That implies αa= n (or) α= → (5)

Where ‘n’ is a positive integer Substituting eq (5) in eq (2)

Or E = = → (6)

The energy spectrum given by eq(6) is same as that of an electron confined to a potential well of width ‘a’. Hence P is also known as the tight binding limit.

ii. Free electron limit (P In the limit of P , eq (4) becomes cos a = cos Ka, Or α= K

E= → (7)

The energy spectrum given by eq (7) is same as that of free electron. That means, all values of energy are allowed for the electron like in the case of a classical particle. Hence this limit is called free electron limit.

D. Energy versus wave vector (E-K) relationship: Brillouin zones As can be seen from the figure, E is not continuous with respect to K, but it shows discontinuity at K= . Thus these discontinuous divide the whole E-K curve into

different regions called as Brillouin zones.

Fig4.14 The relation between energy and wave number for a one-dimensional lattice

The limits of the first Brillouin zones are as follows. 1st zone: K = to

2nd zone K= to and to

3rd zone K=- to - and to

4.11.1. Origin of Energy band formation in solids In an isolated atom, the electrons are tightly bound and have discrete, sharp energy levels [Figure]. When two identical atoms are brought closer, the outermost orbits of these atoms overlap and interact. When the wave functions of the electrons of the different atoms begin to overlap considerably, the energy levels split into two [Figure 5(b)]

Fig4.15. Splitting of energy levels due to interatomic interaction

If more atoms are bought together, more levels are formed and for a solid of N atoms, each of the energy levels of an atom splits into N levels of energy [Figure]. The levels are so close together that they form an almost continuous band. The width of this band depends on the degree of overlap of the electrons of adjacent atoms and is largest for the outermost atomic electrons. In a solid, many atoms are brought together that the split energy levels form a set of energy bands of very closely spaced levels with forbidden energy gaps between them.

Overlapping of these atoms occurs for smaller equilibrium spacing ro.

Fig4.16. With decrease of interatomic spacing overlapping of energy bands take place

The band corresponding to outermost orbit is called conduction band and the next band is called valence band. The gap between these two allowed bands is called forbidden energy gap or band gap. According to the width of the gap between the bands and band occupation by electrons all solids can be classified broadly into three groups namely, conductors, semiconductors and insulators

4.11.2 Classification of materials into conductors, semiconductors and insulators On the basis of band theory, solids can be broadly classified into three categories, viz, insulators, semiconductors and conductors. Their band structures can be as shown in figure4.17

Insulators 1. In case of insulators, the forbidden gap is very wide. Due to this fact electrons cannot jump from valence band to conduction band. 2. They have completely filled valence band and completely empty conduction band. 3. The resistivity of insulators is very high. 4. Insulators are bad conductors of electricity.

Fig 4.17.Valence and conduction bands of insulator separated by large band gap

Semiconductors 1. In semiconductors, the band gap is very small (0.7 eV for germanium and 1.1 eV for silicon). 2. At 0k, these are no electrons in the conduction band and the valence band is completely filled. As the temperature increases, electrons from the valence band jump into conduction band. 3. The resistivity varies from 10 -14 to 107Ω meter.

4. They have electrical properties between those of insulators and conductors.

Fig4.18. Valence and conduction bands of semiconductor separated by small band gap

Conductors 1. In case of conductors, there is no forbidden gap and the valence band conduction band overlaps each other. 2. Plenty of free electrons are available for electrical conduction. 3. They posses very low resistivity and very high conductivity values. 4. Metals c like copper, iron etc. are best examples of conductors.

Fig4.19. Metals having (a) partially filled valence band and (b) overlap of completely filled valence band 81

4.12. Concept of effective mass of electron When a crystal is subjected to the influence of an electric field, a free electron in the crystal moves under the combined influence of the applied electric field and that of a potential due to lattice ions. Because of such effect, the electron responds as if it possesses a mass called ‘effective mass’ which is different from its true mass with which it is accounted if it were to be under the influence of external filed alone.

Definition: when an electron in a periodic potential of lattice is accelerated by an electric field or magnetic field, the mass of the electron is called effective mass. It is denoted by m*. When a free electron of mass ‘m’ moves under the influence of an external force F, the equation of motion is F= m = ma → (1)

Where ‘a’ is acceleration If the force is due to electric field of intensity E, then ma = eE

a= → (2) Or But the acceleration a = is not a constant in the periodic lattice of the crystal through

the applied electric field is constant. If can be considered that this variation in acceleration is caused by the variation of electron’s mass when it moves in the crystal lattice. Acceleration a= → (3)

The electrical force on the electron F=m*a → (4) In the crystal, the electron is bounded between very small spacing since the lattice constant ‘a’ is approximately 1 A0. Hence the wave nature like a group of waves and is governed by wave mechanics. Its velocity ‘V’ is the group velocity of the electron waves. From the concept of de-Broglie waves, we are familiar that

p =ЋK → (5) And E= ω

= → (6)

= → (7)

According to wave theory, gives the group velocity.

Hence, v = = = → (8)

Differentiating eq (8), we get a= =

a= . → (9)

Substituting the value of from eq (6)

a = → (10)

The effective mass m* varies with k as shown in figure 4.20. From the curve, it is clear

that m*is positive between –k0 and +k0 and negative at other values of k up to . It

means that effective mass is negative at the top of the band. It is in similar form of Newton’s law of force F = ma or a= . Therefore by comparision,

we can write

m* = → (11)

This is the expression for effective mass of the electron. The effective mass m* of the electron is a fictitious mass. It differs from its actual mass m of the electron. Depending

upon the value of , m* can be less or more than ‘m’. Interestingly it may take on

negative values also.

Variation of effective mass of electron with wave number The variation of effective mass as a function of wave number k is shown in figure (4.20). From the graph, we observe the following points: (i) Near k = 0, the effective mass approaches m. (ii) As the value of k increases, the effective mass m* increases, and reaches its maximum value at the point of inflection on E-k curve. (iii) Above point of inflection, m* is negative and as k→π/a, it decreases to a small negative value. Therefore, near the bottom of the band, the effective mass m* has constant positive value. Beyond the point of inflection, m* becomes negative. So,

effective mass m* may be greater, or smaller or even negative than the mass m of the electron.

Fig. 4.20.Variation of effective mass of electron with wave number

Objective Questions I choose the correct answer 1. Davisson and germens experiment relates to: [ ] (a) Interference (b) Polarization (c) Phosphorescence (d) electron diffraction

2. Uncertainty principle was discovered by ______[ ] (a) Bohr (b) de-Broglie (c) Heisenberg (d) Schrodinger

3. The wavelength if de-Broglie wave associated with an electron accelerated through 50 volts is [ ] (a)1 A0 (b) 1.5 A0 (c) 1.27 A0 (d) 1.28 A0

4. Among the following practices moving with velocity, the particle having shortest wave associated with it is: [ ] a) Proton (b) Neutron (c) -particle (d) Particle 5. The wavelength associated with a particle moving with velocity v is: [ ] (a) h/m (b) h2/m

(c) m (d)

6.the quantized energy do a particular of mass ‘m’ confined in one dimensional box of length L is ______[ ]

(a) (b)

(c) (d)

7. The statistics which does not permit more than one particle per state is ______[ ] (a)Maxwell-Boltzmann (b) Bose-Einstein (c) Fermi-Dirac (d) None of the above 8. Fermi energy is the of the state at which the probability of electron is______at any temperature above ok [ ] (a) 1 (b) 0 (c) 0.5 (d) Any value between 0 and 1

9. Statics which deal with energy levels which are discrete are ______[ ] (a) Maxwell-Boltzmann and Bose-Einstein (b) Bose-Einstein and Fermi-Dirac (c) Fermi-Dirac and Maxwell-Boltzmann (d) None of the above 10. If E is the kinetic energy of the material particle of mass ‘ m’ then the de-Broglie wavelength is given by ______[ ] (a) (b)

(c) (d)

11. If E is the kinetic energy of the material particle of mass ‘m’ then de-broglie wavelength is given by [ ] (a) Lesser than velocity of light (b) Equal to velocity of light (c) Greater than velocity of light (d) none of these 12. The statistics which deals with identical particles which are distinguishable is: [ ] (a)Fermi-Dirac (b) Bose-Einstein (c) Maxwell-Boltzmann (d) Fermi-Boltzmann 13. Statistics which deals with particles with half integral spin is [ ] (a) Maxwell-Boltzmann (b) Bose-Einstein (c) Fermi-Dirac (d) Maxwell-Dirac 14. Existence of matter waves was experimentally first demonstrated by [ ] (a) Newton (b) Planck (c) de-Broglie (d) Davisson and Germer 15. Wave nature and particle nature called dual nature is exhibited by [ ] (a)Waves only (b) Particles only (c)Photons (d) By both particles and waves 16. Statistics dealing with particles with integral spin is: [ ] (a) Maxwell-Boltzmann (b) Bose-Einstein (c)Fermi-Dirac (d) Maxwell-Dirac 17. The following come under quantum statistics [ ] (a) Maxwell-Boltzmann and Bose-Einstein (b) Bose-Einstein and Fermi-Dirac (c) Fermi-Dirac and Maxwell-Boltzmann (d) None of these 18. The lowest energy state of a particle of mass M confined in a linear box of size L is: [ ]

(a) (b) (c) (d)

19. The wavelength associated with an electron which has been accelerated from rest application of potential of 900V is [ ] (a) 0. 613A0 (b) 0. 365A0 (c) 0. 4086A0 (d) None of these

20. The normalized wave function of a particle n a one dimensional potential well is: [ ]

(a) (b)

(c) (d)

21. The theory which proposes the free electrons inside periodic lattice field is: [ ] (a) Classical free electron theory (b) Huygens wave theory (c) Quantum free electron theory (d) Zone theory

22. When the potential barrier strength is zero, then E-k curve is: [ ] (a) Continuous parabola (b) Discrete energy levels (c) Parabola with discontinuities (d) Exponentially decaying function of k

23. When the potential barrier strength is not zero, then E-k curve is: [ ] (a) Continuous parabola (b) Discrete energy levels (c) Parabola with discontinuities (d) Exponentially decaying function of k

24. The expression of effective mass m* is: [ ]

(a) m*=

(b) m*=

(c) m*=

(d) m*=

25. The discontinuities in E-k curve, occurs at: [ ] (a) K =

(b) K =

(c) K =

(d) k =

26. An increase in the potential barrier strength leads to: [ ] (a) Allowed energy bands separated by forbidden energy values (b) Narrowing of the allowed energy bands (c) Discontinuities in E-k curve at k=

(d) None of these

27. Existence of energy bands in a solid is a consequence of ______[ ] (a) Random motion of electrons (b) Uncertainty principle (c) Interaction of atoms (d) Interference of electron waves

28. Effective mass of an electron can have value: [ ] (a) Positive (b) Negative (c) Infinity (d) All the above

29. First Brillouin zone corresponds to k value extending form: [ ] (a) 0 to

(b) to

(c) to

(d) to

30. Zone theory of solids was developed by: [ ] (a) Drude and Lorentz (b) Sommerfeld (c) Bloch (d) Newton

II. Fill in the blanks: 1. Particles which obey Bose- Einstein statistics are known as ______2. Uncertainty principle was proposed by ______3. Lighter the particle, ______the de-Broglie wavelength associated with it. 4. Statistics dealing with molecules of a gas is ______5. At shorter wavelength range, Planck’s law reduces to ______6. According to uncertainty principle, it is not possible to measure both energy and ______of a process very accurately and simultaneously. 7. According to Planck’s law, the expression for energy of photons per unit volume in a black body is ______87

8. Particles which obey Fermi-Dirac statistics are known as ______9. ______can travel faster than velocity of light. 10. According to Planck’s quantum theory, energy is emitted in the form of packets or quanta called ______11. One dimensional time independent Schrodinger wave equation is ______12. The statistics dealing with phonons in a solid is ______13. The statistics dealing with electrons in a conductor is ______14. The wavelength associated with an electron moving under a potential of 1600V is ______15. Fermi energy is the energy of the state at which the probability of electron occupation is ______at any temperature above 0K. 16. At 0K, all the energy levels above Fermi level are ______17. Lighter is the particle, ______is the wavelength associated with it. 18. Three dimensional time independent Schrodinger’s wave equation is ______19. Matter wave is associated with ______particle. 20. Wave functions representing ______are ant symmetric. 21. He4 atoms obey ______statistics. 22. ______is the velocity of the particle greater is the wavelength associated with it 23. Wave functions representing Bosons are ______24. The subject which deals with the relationship between the overall behavior of the system and the properties of the particles is called ______25. Single crystal target of nickel is used in ______experiment to prove wave nature of electrons. 26. Protons obey ______statistics.

27. If is the Fermi-energy of an electron at 0K, then its Fermi – energy at any temperature above 0K is ______28. Bosons ______Pauli’s exclusion principle. 29. All the energy levels below Fermi-level are completely filled at a temperature ______30. Thin metallic film of gold or silver or aluminium is used as target in ______experiment to prove wave nature of electrons. 31. When an electron moves in a periodic potential of lattice, its mass varies and this mass is called ______of the electron. 32. ______Model proposed a simpler potential in the form of an array of square wells. 33. The representation of permissible values of K of the electrons in one, two or three dimensions is known as ______34. The periodicity of the potential of zone theory of solids is given by ______theorem. 35. ______have relatively wide forbidden gaps. 88

36. ______have relatively narrow band gaps. 37. According to zone theory of solids, free electrons move inside ______potential field 38. When the band gap is in the order of 1ev in a solid, it behaves as a ______39. ______have either partially filled valence band or overlap of completely filled valence band with partially filled conduction band. 40. When the band gap is in the order of 7ev in a solid, it behaves as a ______

Key I. Choose the correct answer 1)d 2)c 3)a 4)c 5)a 6)a 7)c 8)c 9)b 10)a 11)c 12)c 13)c 14) d 15)d 16)c 17)c 18)a 19)c 20)d 21) d 22) a 23) c 24)b 25) a 26) b 27) d 28) d 29) d 30)c

II. Fill in the blanks 1. Bosons 2. Heisenberg 3. Greater 4. Maxwell-Boltzmann 5. Rayleigh-Jean’s 6. Time

7. = or =

8. Fermions 9. Matter waves 10. Photons

11. + =0 (or) + =0

12. Bose-Einstein 13. Fermi-Dirac 0 0 14. 0.3065A [solution: = = 0.3065A ]

15. (Or) 0.5 (or) half

16. Completely empty 17. Greater

18. + =0 (Or) ) + =0

(Or) + =0 (Or) + =0

19. Moving

20. Fermions 21. Bose-Einstein 22. Lighter 23. Symmetric 24. Statistical Mechanics 25. Davisson-Germer 26. Fermi-Dirac

28. Do not Thomson 29. Zero Kelvin (0 K) 30. G.P Thomson 31. Effective mass 32. Kronig and penny 33. Brillouin Zones 34. Bloch 35. Insulators 36. Semiconductors 37. Periodic 38. Semi conductors 39. Metals 40. Insulator

Problems

1. A body at 1500K emits maximum energy at a wavelength 20,000 A0.If the sun emits maximum temperature of wavelength 5500 A0, what would be the temperature of the Sun.

Solution: According to Wien’s displacement law,

=constant

= = =5454K

2. At what temperature we can expect a 10% probability that electrons in silver have an energy which is 1% above the Fermi energy? (The Fermi energy of silver is 5.5eV.)

Solution: probability function

F (E) =

Given data: F (E) =10%=0.1

EF = 5.5eV

E = EF+ EF

(5.5+0.055) = 5.555eV -19 E- EF = 0.055eV=0.055×1.6×10 J Substituting in the formula

i.e. 0.1 =

+1 = 10

T= = 290.2K

3. Evaluate the Fermi function for an energy KT above the Fermi energy.

Solution: Fermi function F (E) =

Given that E-EF =KT F (E) = =

= 0.269 4. Calculate the wavelength of an electron raised to a potential 1600V.

Solution: de-Broglie wavelength A0

0 = 0.3065A

5. If the kinetic energy of the neutron is 0.025eV calculate its de-Broglie wavelength (mass of neutron =1.674X10-27 Kg)

Solution: Kinetic energy of neutron E = = 0.025eV

= 0.025 × 1.6x10-19 J

0.2186x104m/s e- Broglie wavelength

=0.181nm 6. Calculate the energies that can be possessed by a particle of mass 8.50 x10- 31kg which is placed in an infinite potential box of width 10-9cm.[June 2012]

Solution: The possible energies of a particle in an infinite potential box of width L is

given by En =

M=8.50 x 10-31Kg L=1 x 10-11m h=6.626× 10-34J-s For ground state n=1

=6.456 × 10-16 joule -16 For first excited state E2 = 4 x 6.4456 x 10 = 258.268 x10-16Joule

7. Calculate the wavelength of matter wave associated with a neutron whose kinetic energy is 1.5times the rest mass of electron.[Given that mass of neutron=1.676×10-27kg,mass of electron 9.1×10-31Kg,Mass of electron 9.1×10- 31J-Sec,velocity of light is 3 108m/s](June 2010,set1).

Solution: For neutron m =1.5×9.1 10-31Joules

= 16.288x10-4

= 4.046 10-2m/s The de-Broglie wavelength expression is

= 9.76X10-6m

Questions 1) Distinguish between Maxwell-Boltzmann statistics and Fermi-dirac statistics [June-12] 2) Write notes on black body radiation [June11, set1] 3) Drive Schrodinger’s wave equation for the motion of an electron [June-11, Sep3; Jan12, Set4] 4) describe the experimental verification of matter waves using Davisson-Germen experiment [June 11, set 4, June 10, Set 1] 5) Derive an expression for the de-Broglie wavelength of an electron [Dec 2011, Set 1] 6) What are the important conclusions of G.P Thomson’s experiment [Dec 11,set 2, Dec- 10, Set 3] (Or) Describe the experimental verification of matter waves using G.P Thomson experiment 7) Write the physical significance of wave function [Jan-12, Set 4] 8) Derive an expression for density of states of electrons [Dec 10 Set 1, June 10. Set 4] 9) Write short notes on: i)De Broglie wavelength and ii)Heisenberg’s uncertainty principle[Dec 10,Set 1,June 10, Set 4] 10) What are matter waves? Derive an expression for the wavelength of matter waves [Set 4, Dec -10] 11) Describe an experiment to establish the wave nature of electron [Set 4, Dec -10] 12) Explain Fermi-Dirac distribution function .Illustrate the effect of temperature on the distribution [Sep3, june10] 13) Derive an expression for Fermi energy [Set 4, June 10] 14) Show that the energies of a particle in a potential box are quantized [June 09, Set, Set3, June09] 15) Using Kronig-Penny model show that the energy spectrum of an electron contains a number of allowed energy bands separated by forbidden bands. 16) On the basis of band theory how the crystalline solids are classified into conductors, semiconductors and insulators. 17) What is Bloch’s theorem? Explain. 18) What is effective mass of an electron? Derive an expression for it.

UNIT-III

Chapter – 5 DIELECTRIC PROPERTIES

5.1 Introduction  Dielectrics are insulating materials. In dielectrics, all electrons are bound to their parent molecules and there are no free charges.  Even with normal voltage or thermal energy electrons are not released.  Dielectrics are non metallic materials of high specific resistance and have negative temperature coefficient of resistance.  Dielectrics are electrical insulators. They possess high resistivity values within the range 106 m to 1016 m. Under high voltage bias, they allow very little current. They with stand for very high voltages. The conduction is mostly associated with ionic motion through defects or hopping of charges. They have no free charges. The electrical properties of a dielectric are associated with inherent property of possessing electric dipoles.  Dielectrics are the materials having electric dipole moment permanently or temporarily by applying electric field. These are mainly used to store electrical energy and as electrical insulators. All dielectrics are electrical insulators. But all electrical insulators need not be dielectrics. For example the vacuum is a perfect insulator. But it is not a dielectric. The study of dielectrics is essentially study of insulators.

5.2 Basic Definitions

Electric dipole

Fig. 5.1 Electric dipole

Two equal and opposite charges separated by a distance ‘r’ constitute a dipole.

Electric dipole moment (μ) The product of charge and distance between two charges is called electric dipole moment. μ = q × r Units: coulomb - meter or Debye.

1 Debye = 3.33 × 10-30 coulomb - meter

Non-polar dielectrics Mono atomic materials are made up of atoms. The centre of gravity of negative charge and the centre of gravity of positive charge of an atom coincide. That means even though there are two equal and opposite charges are not separated. Their dipole moment is zero. μ = q × r = q × 0 = 0 Such dielectrics are called Non- polar dielectrics.

Non-Polar Molecule  Consider an atom. The positive charge of nucleus may be concentrated at a single point called as centre of gravity of the positive charge.  The negative charge of electrons may be supposed to be concentrated at a single point called as Center of gravity of the positive charge.  When the two centre of gravity coincide, the molecule is known as Non-polar molecule. The Non-polar molecules have symmetrical structure and zero electric dipole moment.

 Examples: H2, N2, O2, CO2, Benzene.

Fig. 5.2 Non polar molecules with zero dipole moment (μ=0)

 The electric dipole moment has a direction from positive charge to negative charge.

Polar dielectrics  In polyatomic molecules, the center of gravity of negative charge distribution may not coincide with the center of positive charge distributions.  There is an effective separation between centers of negative and positive charge distributions.

 The molecule has a net dipole moment. Such dielectrics are called Polar dielectrics.

Polar Molecules They have unsymmetrical structure and have a permanent electric dipole moment. The Center of gravity of positive and negative charges do not coincide, the molecule is called as polar molecule.

e.g.:- H2O, HCl, CO, N2, NH3 etc.

Fig.5.3 Polar molecules with net dipole moment (μ 0)

Dielectric constant (or) Relative permittivity of the medium  It is the ratio between the permittivity of the medium and the permittivity of free space.

Єr = . It has no units.

 Єr is also called as relative permittivity of the medium. It is a measure of polarization in the dielectric material.

Єr = Єr where Є = absolute permittivity of the medium Where Єo = permittivity of free space=8.854 x 10-12 F/m

Electric Polarization When a dielectric substance is placed in an electric field, then positive and negative charges are displaced in opposite direction.  The displacement of charges produces local dipoles.  This process of producing dipoles by the influence of electric field is called electric polarization. dielectric Polarization = P =

P= surface charge density ‘σ’

Polarizability (α) The average dipole moment μ is directly proportional to the electric field (E) applied. μ α E μ = E = Polarizability = Farad/m2

Polarization Vector (P) It is defined as the average dipole moment per unit volume of a dielectric. If ‘N’ molecules are present per unit volume, Then polarization vector P =

P = N μ coulomb/m2

Electric flux density (or) Electric displacement (D) The number of electric lines of forces received by unit area is called Electric flux density. D α E D = ЄE

But Є = Є0 Єr D = Є0 Єr E

Relation between Polarization (P), Electric field (E) & Dielectric

constant Єr

We know that electric flux density ‘D’ is written as D = Є E → (1) Where Є= absolute permittivity of medium E = electric field

Where E = Є0 Єr → (2) Hence substitute (2) in (1)

D = Є0 Єr → (3)

-12 Where = permittivity of free space = 8.854 x10 F/m Єr = Dielectric constant or relative permittivity of the medium.

Fig. 5.4 Electric lines of forces in a polar dielectric

If ‘P’ is the polarization of the dielectric material due to the applied electric field (E), then the flux density ‘D’ is equal to the flux density in vacuum plus polarization of the material

D = Є0 E + P ` → (4)

Єr E = E + P

P = E [Єr-1] P = Є0 Єr E E → (5)

Put [Єr-1] = χ → (6) Where χ = electric susceptibility From equation (5) & (6)

P = E χ χ = → (7)

Conclusion:

= E [Єr-1]

χ = Єr-1 χ =

Electric Susceptibility (χ) The electric susceptibility ‘χ’ is defined as the ratio of polarization vector to the applied electric field ‘E’. χ = P/ E χ has no units.

P = χ E

Where χ = Єr-1

Dielectric strength It is defined as the minimum voltage required producing dielectric break down. Dielectric strength decreases with rising of temperature, humidity and age of the material.

5.3 Non-Polar Dielectric in an electric field When a dielectric is placed in an electric field, say between the plates of a charged Condenser; the positive and negative charges are re oriented i.e. the center of gravity of positive charges is pulled towards the negative plate of the condenser and vice versa. Thus the net effect of the applied field is to separate the positive charges from the negative charges. This is known as Polarization of dielectric. The dielectrics which are polarized only when they are placed in an Electric field are called Non- polar dielectrics.

Fig. 5.5 (a) non polar dielectric in E=0 (b) Non polar dielectric when ‘E’ is applied (c) Totally polarized non polar dielectric with net field

Thus if the dielectric is placed in an electric field, induced surface charges appear which tend to weaken the original field with in the dielectric. That means E1 opposes the

original field Eo.

5.4 Polar dielectric in electric field We know that polar dielectric have permanent dipole moments with their random orientations. In the presence of an electric field, the partial alignment of dipoles takes place.

Fig. 5.6 Polar dielectric orientation without field and with field.

Polar dielectrics already possess some dipole moment inside due to the presence of permanent atomic dipoles. But these are randomly oriented when no field is applied. Their dipole moment and polarization increases since dipoles align along the field direction gives some extra polarization. Hence

P= Polarization already Polarization induced Existing due to permanent + due to electric dipoles field

P = Pp+Pi

5.5. The Local field (or) Internal field Ei (or) Elocal Definition: In dielectric solids, the atoms or molecules experienced not only the external applied electric field but also the electric field produced by the dipoles. Thus the resultant electric field acting on the atoms or molecules of dielectric substance is called the “Local field or an internal field.” Derivation:

Consider a dielectric material placed in an External field ‘E1’, placed between the parallel plates of a capacitor. As a result opposite type of charges are induced on the surface of dielectric.

Fig. 5.7 (a) Polar dielectric in electric field (b) Enlarged view of spherical cavity

Imagine a small spherical cavity of radius ‘r’. In this sphere inside dipoles are present. Consider a dipole at the center of spherical cavity. This dipole experiences

the following fields, in addition to the externally applied field ‘ ’. The total internal field experienced by the dipole

Elocal = Ei = + + + → (1)

Where = External applied field. Here, (a) The field ‘E2’ produced by induced charges on the dielectric sample near the surface.

(b) The field E3 arising from dipoles inside the sphere. E3depends on crystal symmetry. [For isotropic materials E3=0] (c) The field E4 is due to polarization of charges on the surface of spherical cavity. It is called the Lorentz cavity field.

The surface charge density on the surface of the spherical cavity is Pcosθ. If ‘ds’ is the area of the surface element shaded in figure shown.

Then charge on the surface element ( ) is = (normal component of polarization) x (area of the surface element)

= (P ) (ds) (2)

Let a test charge q2 = q placed at the center of cavity. From coulombs’ law, the force experienced between the surface charges.

dF = dF =

= → (3)

The resulting electric field E4 = = → (4)

The electric field is resolved into two components: One component is along the direction of ‘P’ & other perpendicular to it. The Perpendicular components cancel themselves out leaving only the horizontal components. Hence the sum of all such horizontal components of electric field for the whole Surface is:

E4= cosθ = → (5)

The surface area of the ring ds = 2 sinθdθ → (6) Substitute (6) in (5)

Limits are θ = 0 to

E4 = → (7)

Let cosθ = z -sindθ = dz & Limits are z = 1 to z = -1

Equation (7) becomes E4 =

E4 = dz

E4 = → (8)

Substitute the value of E4 in equation (1) Total internal field (or) local field

Ei = Eloc = E1 + E2 + E3+ E4 Here E3 = 0

Ei = E1 + E2 + E4 Let E1 + E2 = E

Ei = E + E4

Ei =E +

5.6 Clausius – Mosotti relation This makes the relation between microscopic & macroscopic quantities of Polarization. From Polarization Vector

P = N μ → (1) Where N = No. of molecules per unit volume, μ = average dipole moment

μ = α E = α Ei → (2)

Where Ei = local (or) internal field = → (3)

Substituting (2) & (3) in (1)

P = N α Ei P = N α

P = N α E +

= NαE

P = NαE

P = ———(4)

We know that P = E → (5) Equating (4) & (5)

From Clausius Mosotti relation

5.7 Types of Polarization Polarization is the process of inducing dipole moment in a molecule. There are four types of polarization. They are: (1) Electronic Polarization (2) Ionic Polarization (3) Orientation (or) Dipolar Polarization (4) Space charge polarization

5.7.1 Electronic Polarization:

Definition: When an electric field is applied on a dielectric material then all the positive nuclei of atoms move in the field direction and all the negative electron cloud of atoms move in opposite directions, hence dipoles are formed to produce dipole moment.  The electron cloud readily shifts towards the positive end of the field. The extent of shift by electrons is proportional to field strength.  Hence dipole moment is the product of charge and shift distance.

Expression for Electronic Polarizability

Fig. 5.8 (a) Un polarized atom in the absence of field (b)Electronic polarization due to distortion of electron cloud by the field E

 Consider an atom of dielectric material such that its atomic radius is ‘R’ & atomic number ‘Z’. When no field is applied the charge centers of electron cloud and positive nucleus are at the same point & hence there is no dipole moment.  Suppose an atom is placed in an electric field of strength E, due to Lorentz force the positive nucleus move towards the field direction and electron cloud will move in opposite direction of the field.  When nucleus and electron cloud are displaced from their equilibrium positions, an attractive Coulomb force act between them to maintain original position. But Lorentz force will tend to separate them from them equilibrium position.  When these forces are equal and opposite produces a net dipole moment.

Let the displacement between centers of nucleus and electron could be ‘X’. Let ‘ze’ is charge of nucleus. ‘-ze’ is charge of electron cloud in a sphere of radius ‘R’. The charge density due to negative electron cloud of radius R (E=0) is

ρ = → (1)

Lorentz force which tends to separate positive nucleus from negative electron cloud is = (-ze) E → (2) Coulomb force which tends force of attraction between them is (ze) → (3)

But the charge enclosed in the sphere of radius ‘x’ due to electrons of radius ‘x’

= → (4)

Substitute equation (4) in (3) we get

Coulomb force =

Coulomb force = → (5)

In equilibrium position coulomb & Lorentz forces are equal. Comparing equations (5) & (2) we get

= (-ze) E

Hence electron cloud displacement ‘X’ = E → (6)

The two charges (+ze) of nucleus & (-ze) of electron cloud are displaced through ‘X’.

Hence electric dipole moment is =│charge│×(displacement) =│ze│(X)

= (ze) E

= (4 ) E → (7) We know that electronic polarizability =

Electronic Polarisability Where R= radius of an atom.

5.7.2 Ionic Polarization Ionic polarization takes place in ionic dielectrics due to displacement of positive and negative ions by the influence of external electric field. Expression for Ionic Polarisability When an electron field is applied on an ionic dielectric then positive ions move in the field direction & negative ions move in opposite direction, hence dipoles will be formed. This phenomenon is known as ionic polarization. Na+ Cl- Cl- Na+ M M m

m X x1 x2

Fig. 5.9 (a) In the absence of field Fig. 5.9(b) when field is applied

Let ‘e’ the charge of ions and M and m be the masses of negative and positive ions respectively. When an electric field ‘E’ is applied on an ionic dielectric then positive ions

displaces in the field direction through distance and negative ion displaces in opposite direction through distance. The induced dipole moment μ= │charge│× displacement

μ=│e│ ( ) → (1)

But against to the displacement of positive and negative ions, restoring force acts and opposes the displacements of cation and anion. Under equilibrium conditions, the restoring force

F = [for negative ion] F = [for positive ion]

= → (2)

where are force constants But

= m → (3)

where = angular velocity of the ions Substituting (3) in (2) we get

Hence = & =

But F = e E

= → (4)

Substituting equation (4) in equation (1) Electric dipole moment

μ= → (5)

We know that the Ionic Polarizability = =

Here = natural frequency of the ionic molecule.

5.7.3 Orientational Polarization Definition: When Electric field is applied on a polar dielectric then all the dipoles tend to rotate In the field direction, hence dipole moment increases gradually. This phenomenon is known as dipolar (or) orientational polarization.

Expression for Orientational (or) dipolar Polarisability Orientation Polarisation takes place only in polar dielectrics in which dipoles orient in random manner such that the net dipole moment is zero. When Electric field is applied, all the dipoles try to rotate in the field direction as shown in the figure 5.10.

Fig. 5.10: Orientational polarization

To derive an expression for dipolar Polarisability of a system, we have to know the total dipole moment and total number of dipoles present in the system since Average dipole moment μ= (1)

According to statistical mechanics, the Average component of dipole moment along the field direction is equal to μ‹cosθ›. It is given by

μ‹cosθ›= → (2)

To evaluate the integral Let = a

Let cosθ = x

a cosθ = x -a sinθdθ = dx

Then equation (2) becomes & the limits changes from –a to +a

μ‹ cosθ › =

‹ cosθ › = = = L (a)

‹ cosθ ›=L (a) →(3) (or) μ‹ cosθ ›=μ L(a)

The function L (a) is called the Langevin function A graph L (a) has been plotted as a function of a = [i.e a=1]

Fig. 5.11 Langevin curve

From graph for very large values of ‘a’, i.e for very high field strength, the function approaches the saturation value unity. This saturation corresponds to complete alignment of dipoles in the field direction, so That =1 Hence μ = μ Case i: But if the field strength is not too high and temperature is not too low, then a 1 1

[Also ]

Under these conditions the large in function reaches to: L(a) =

From equation (3) 108

Hence μ = μ L(a) [i.e from equation (3)] μ = μ

= [i.e a= ]

μ= → (4)

But from definition Polarisability i.e orientational (or) dipolar polarisability (or) = (or)

(or) μ = E →(5) Equating equations (4) & (5)

= Orientational (or) dipolar polarisability μ= dipole moment

= Boltzmann constant T= absolute temperature.

The total polarizability, α = + +

α= + +

5.8. Ferro Electricity Ferro Electric Materials & Applications: Definition: Some materials (below a certain temperature) posses spontaneous polarization even when electric field is not applied. These materials have acquired electric dipole moment even no field is applied. These materials are known as Ferro electrics. And this phenomena is known as Ferro electricity.

Ferro Electric Materials There are 3 main types of crystal structure in Ferro electric materials. They are (1) Rochelle salt structure also called as Sodium Potassium Tartrate i.e

(2) The perovskite group consisting of Titanates & Niobates

Eg:- Barium Titanate i.e (3) The dihydrogen Phosphates & Arsenates

Eg:- 109

Some other examples of ferro electric materials are:

Ammonium dihydrogen phosphate (N P ) Lithium Niobate (LiNb ) Potassium Niobate (KNb ) etc

Properties of Ferro electric materials  The Ferro electric materials posses’ spontaneous polarization below a certain temperature.  As temperature increases the spontaneous polarization decreases and at a particular temp, the spontaneous polarization vanishes. This temperature is called Curie temperature. It is defined as the temperature at which Ferro electric material converts into a Para electric material.

Fig. 5.12 (a) Spontaneous polarization with respect to temperature (b) Variation of with respect to temperature

 Below Tc - curie temperature; the dielectric material possesses Ferro electric

property. As temperature increases the dielectric constant reduces after ‘Tc’. It is shown in the figure 5.13

Fig. 5.13 Variation of dielectric constant with temperature

From curie-Weiss law =

= Dielectric constant, =curie constant, = absolute temperature, =curie temperature 110

 The Ferro electrics exhibit the property called Piezo–electricity & Pyroelectricity.  “Piezo electricity” means when mechanical stress is applied to Ferro electric materials then the opposite charges are formed on the surface of crystal generating electricity.[In greek ‘piezo’ means pressure]  “Pyro electricity” means the change of Spontaneous Polarisation of Ferro electric materials by the application of temperature.[In Greek “Pyro means Heat]  Ferro electric materials also exhibit the property called “Di-electric Hysteresis”  Dielectric Hysteresis: It is defined as the lagging of polarization behind the electric field applied. The area of the loop represents the dielectric loss.

Applications of Ferro Electric Materials

 The Ferro electrics have high dielectric constant . They are used in small sized capacitors to produce large capacitance.  As they satisfy Dielectric Hysteresis property; they can be used as memory devices in computers.  Piezo-electric transducers are used to produce & to detect sound waves.  Pyro electric behavior in Ferro electrics is used to detect infrared radiation.  They are used in electro mechanical filters.

5.9 Dielectric Hysteresis Definition: It is a property of lagging of polarization behind the applied electric field. The area of the loop represents the dielectric loss in the dielectric Ferro electric materials.

Fig. 5.14 Dielectric Hysteresis (Variation of polarization with electric field)

Explanation:

 When electric field ‘E’ is applied the polarization rapidly increases and reaches a point called saturation polarization which is constant.  If electric field is made zero [E=0] the polarization will not reach the original point

‘O’. But takes a new path at point ‘ ’ known as Remanent (or) remained polarization.

 To destroy ‘ ’ a negative electric field is applied, hence =0

 The field required to make =0 is known as negative coercive electric field i.e ‘- ’.

 Further if ‘E’ is increased, negative polarization takes place and reaches to ‘- ’ known as negative saturation polarization.

 If negative electric field (-E) is decreased back to zero, the negative saturation (- ) will not go to the point ‘O’ but creates a new path at point ‘ ’. This is known as negative remanent polarization.

 To destroy this ‘ ’ a positive electric field is applied. This field ‘ ’ is positive coercive field.  At last increase of electric field leads to reach again, positive saturation polarization. This is a cyclic process.  In the above mechanism the polarization started at point ‘O’ does not reach back to that point. Hence polarization lags behind the electric field applied which is known as “Dielectric Hysteresis”.

5.10 Structure of BaTi - A Ferro electric material

Fig. 5.15 Structure of BaTi above TC

Fig. 5.16 Variation of dielectric constant with temperature

 We know that BaTi is a ferro electric material. The cubic unit cell of BaTi crystal structure consists of Barium cations at the corners & oxygen anions at the face centers. Titanium ion is in the octahedral void at the body center.

 Above 120°C, BaTi is a cubic crystal. In this state, the centers of positive and negative charges coincide. Hence there is no spontaneous dipole moment & polarization is zero.  If the crystal is cooled below 120°C, the titanium ions shift to one side of the body centre and neighboring oxygen ions are also get shifted. The centers of positive charge due to titanium cat ions and centers of negative charges due to oxygen anions gets shifted (o) displaced. Hence dipoles are created to give dipole moment.

5.11Piezo Electricity (or) Piezo electric effect The development of electrical charges and electric polarization when subjected to stress is called “Piezo-electric effect”. This phenomenon was discovered by “Curie Brothers in 1880”.  Crystals like quartz, Tourmaline, Rochelle salt when subjected to compression or tension then opposite charges develop on their surfaces.  The materials also exhibit “Inverse Piezo Electric Effect”. That is on these substances when electric field is applied, they get strained. That means the di2513mensions gets changed. The crystals either expand (or) contract.  001 also under do vibrations.  All Ferro electrics are Piezo electric. But all Piezo electric materials need not necessarily be Ferro electric. Example:- Barium Titanate exhibits both Ferro electricity and Piezo electricity.  Quartz is only Piezo electric.

5.11 .1 Piezo electricity in quartz crystal

Quartz is silicon dioxide (SiO2). The atoms are arranged in the form of Hexagon. The quartz consists of three axes: X-axis known as Electrical axis, Y-axis known as Mechanical axis & Z- axis known as Optical axis. When external stress is not applied, all the charges get balanced. Hence net polarization is zero. When external stress is applied, the balance is disturbed. Hence crystal is polarized. As a result, electrical charge is developed on the faces. The structure of quartz is shown in the figure below.

Fig. 5.17 Piezo electric effect in quartz crystal

5.11.2 Applications of Piezo electric materials  In case of Inverse Piezo electric effect, If a quartz plate is subjected to an alternating electric field, the quartz crystal expands or contracts. This also undergoes vibrations. If the frequency of vibration of quartz crystal is equal to ultrasonic frequency, then the ultrasonic waves are produced. Hence quartz oscillator is used to produce ultrasonic waves.  Rochelle salt- a dielectric material is used in gramophones, ear phones, hearing aids, microphones etc.  Barium titanate, lead zirconate, lead titanate is used high voltage generation.  A transducer is a device which converts one form of energy into another. Piezo electric effect can be utilized to convert force, pressure and strain into electrical voltages.  Acceleration transducers are used to measure acceleration.  Piezo electric crystals are used to produce radio frequency waves for broad casting.  They are used in quartz watches to maintain accurate time. 5.12 Pyro Electricity Definition: Pyro electric effect is the change in spontaneous polarization when the temperature of the specimen is changed.

Explanation  Many Piezo-electric materials, such as tourmaline, have the tendency to exhibit a change in internal electrical polarization as a response to temperature changes.  An electric field develops in a pyro-electric crystal as a result of temperature change.  Pyro-electricity requires the pressure of permanent electric dipoles (whose magnitude is affected as a result of temperature changes).  It is possible to detect a change of coulomb with a suitable electrometer, temperature changes as small as °C can be measured using the pyro electric effect.

Applications

 The Pyro- electric materials such as BaTiO3, LiNibO3 etc are used to make very good Infra-red detectors which can operate at room temperature.

 Materials such as TGS, NaNO2 and PZT ceramics are used in the construction of pyroelectric image tubes.  Used in burglar alarms, which detect the thermal radiation from Human body.

Chapter- 6 MAGNETIC PROPERTIES & SUPERCONDUCTING PROPERTIES

6.1 Introduction  The magnetic effects in magnetic materials are due to atomic magnetic dipoles in the materials. These dipoles result from effective current loops of electrons in atomic orbits, from effects of electron spin & from the magnetic moments of atomic nuclei.  The electric currents in an atom are caused by orbital and spin motions of electrons and those of its nucleus. Since all these motions of charged particles form closed electric currents, they are equivalent to “magnetic dipoles”. When such dipoles are subjected to an external electric field, they experience a torque which tends to align their magnetic moments in the direction of the externally applied field.

6.2. Definitions Magnetic dipole Each tiny dimension of a magnetic material (or) atoms in magnetic materials is called magnetic dipole. This magnetic dipole produces magnetic moment depending on the alignment with respect to the applied magnetic field.

Magnetic flux (Ф) It is defined as the amount of magnetic lines of forces passing perpendicularly through unit area of a given material. It is denoted by ‘Ф’ Ф = AB Where A= Area of cross section of the material in m2 B = magnetic Induction in Wb/ m2 Units: Weber (Wb)

Intensity of Magnetization (M) When a material is magnetized, it develops a net magnetic moment. The magnetic moment per unit volume is called Intensity of magnetization Magnetization (M) =

Units: Amp/m Magnetic Induction (B) Magnetic induction at a point is defined as the force experienced by a unit North Pole Placed at that point. It is denoted by ‘B’ i.e. B = weber / m2

Magnetizing field strength (H) When a medium is exposed to a magnetic field of intensity ‘H’, it causes an induction ‘B’ in the medium. i.e. B  H B = µH Where µ = absolute permeability of the medium. If the medium is air or vacuum

B= H µ0=permeability of free space i.e. air or vacuum -7 µ0 =4 10 H/m Units for H: Amp /m.

Permeability (µ) It indicates, with which the material allows magnetic lines of force to pass through it. Or It is the ability of the medium to pass magnetic lines of forces through it.

There are three Permeabilities i.e. µ1, µ0, µr

µ = µ0 µr Where µ = Absolute permeability of the medium

µ0 = Permeability of free space i.e. air or vaccum

µr = Relative permeability of the medium

Magnetic moment

Magnetic moment µm = (current) × (area of circulating orbit) 2 µm = (I) × ( r ) Units: Amp-m2

Fig.6.1 Revolving electron in an atom produces magnetic moment

When the magnetic dipoles (atoms consisting of charged particles like protons &

neutrons) undergo orbital motion (or) spin motion produces a magnetic moment. Since motion of charged particles is considered as closed electric current loops which inturn produces a magnetic moment.

Relation between B, H, M We know that B = µH → (1)

But µ = µ0 µr

B = µ0 µr H → (2)

Adding &subtracting with µ0H on right hand side of equation (2)

B = [µ0 µr H] + µ0 H µ0 H

= [µ0 µr H µ0 H] + µ0 H

= µ0 H [µr 1] + µ0H

But M = H [µr-1] → (3) Now eq(3) becomes B = µ0M + µ0H B = µ0 [H+M] → (4) Consider equation (3), M = H [µr-1]

= µr 1 → (5)

But magnetic susceptibility χ =

From equations (5) and (6)

χ = = µr 1

µr =1+ χ Magnetic susceptibility (χ) If H is the applied magnetizing field intensity and M is the amount of magnetization of the material, Then χ =

χ = 0 in vacuum χ = +ve for paramagnetic and Ferro magnetic materials χ = -ve for diamagnetic materials Units:It has no units.

6.3. Origin of magnetic moment (Or) Sources of magnetic moment In atoms, the permanent magnetic moment arises due to a) Orbital motion of electrons and its magnetic moment is called orbit magnetic

moment of electrons ( ) b) The spin of electrons and its magnetic moment is called spin magnetic moment

of electrons ( ) c) The spin of nucleus (due to protons) and its magnetic moment is called spin

magnetic moment of the nucleus. ( ). 118

Explanation

a) Magnetic moment due to orbital motion of the electrons

Let us consider an electron of charge ‘e’ revolving around a nucleus in time period ‘T’ in a circular orbit of radius ‘r’. Then a magnitude of circular current ‘I’ is given by I = = → (1)

But T =

Where = angular velocity of electron I =

But magnetic moment of electron is = I A = current area of circulating orbit =

= → (2)

Fig. 6.2 Orbital magnetic moment of electrons

We know that angular momentum of any particle, L = m r2 Substituting eq.(4) in eq.(3) we get Orbital magnetic moment, = (- ).L → (5)

[-ve sign indicates µl and are in opposite directions] = (- ) L

But from Bohr’s atomic model mvr =

L = Where = orbital quantum number

L = orbital angular momentum The values of = 0, 1, 2------(n-1) Hence = (- ) ( )

= - ( ) → (6)

Where = is a constant called Bohr magneton and its value is 9.27 10-24amp-m2

Hence eq(6) becomes

= → (7) Bohr magneton is the fundamental unit of magnetic moment. It is clear from eq (7) that electron can take only certain specified values of magnetic moment depending on the value of ‘ ’. Bohr suggested that both magnitude and direction of ‘l’ are quantified. It is known as “Spatial quantization”. The spatial quantization introduces a new set of quantum numbers.

(a) Orbital magnetic quantum number ( ) (b) Spin magnetic quantum number (ms)

For example: If electron is in‘p’ shell. Then n = 2, L= 0 to n-1, L = 0, 1, If electron is placed in external magnetic field then eqn (6) can be written as

( ) → (8)

Hence for‘p’ shell electron, = 0 to ± L. The values are = -1, 0, 1 Hence eqn (8) becomes ( ), 0, ( )

Therefore we have “Three” possible orientations for electron in d-shell which is shown in the figure 6.3.

Fig. 6.3 The three possible orientations of electron

‘OC’ represents the orientation of electron if = 0 ‘OB’ represents the orientation of electron if = +1 ‘OD’ represents the orientation of electron if = -1

b) Magnetic moment of electrons due to spin of electrons (µs) According to quantum theory; electrons should have intrinsic angular momemtum due to spin. Spin is also quantized both in magnitude and direction spin can take only one value i.e ½ or - ½. The magnetic moment produced due to spin of electrons is 120

called spin magnetic moment (µs). It is given by

Spin magnetic moment µs = -2( ) S → (9)

Where S=spin angular momentum, e = charge of electron, m = mass of electron S =

where S = spin quantum number h = Planck’s constant.

Fig. 6.4 Spin magnetic moment of electrons

From equation (9), µs = -2( ) S

Since S =

µs= -2( ) ( )

s = ± , µs = ±

µs = , -

µs = + , -

Hence spin magnetic moment of electron is equal to µB. That is one Bohr magneton Hence there are two possible orientations of electron. Conclusion: Para magnetism, Ferro magnetism is due to spin magnetic moment. Diamagnetism is due to orbital magnetic moment.

(c) Magnetic moment due to Nuclear spin or spin of all protons (µn)

The magnetic moment of the nucleus is given by µn = → (10)

Where mp= mass of proton The constant is called nuclear magneton.

The value of nuclear magneton =5x10-27A-m2

This is small when compared to Bohr magneton.

6.4. Classification of Magnetic Materials Magnetic materials are classified as follows: a) Diamagnetic b) Paramagnetic c) Ferro magnetic d) Anti Ferro magnetic e) Ferric magnetic or ferrites

Diamagnetic Paramagnetic Ferromagnetic Anti ferro Ferri magnetic materials materials materials magnetic materials materials (or)ferrites

1.Diamagnetism: 1.Paramagnetism: 1.Ferromagnetism 1. Anti Ferro 1.Ferrimagnetism: magnetism: It is the property of It is the property It is the property of It is property of the It is the property the material which of the material the material which material which has of the material has repulsive nature which has weak has strong strong attractive which has weak (or) opposing attractive force. attractive force. force. attractive force. magnetization

2.The property is 2.The property is 2.The property is 2. The property 2.The property is due to orbital due to spin of due to spin of is due to spin of due to spin of motion of electrons electrons electrons electrons. electrons

3.There is no spin 3.Spin is random 3.Spin is parallel 3. Spin is anti 3. Spin is anti parallel. parallel but of different magnitudes.

4. These materials 4.These materials 4.They have 4.They have 4. They have are lack of have permanent permanent permanent permanent magnetic dipoles dipoles magnetic dipoles magnetic dipoles magnetic dipoles.

1. 1. 5. They do not2. 5. They possess2. 5. They possess 5. They do not 5. They possess possess permanent permanent permanent possess permanent dipole magnetic magnetic dipole magnetic dipole permanent magnetic moment (it is zero). moment. moment. Also in magnetic dipole diploement. Also Hence spontaneous But there is no the absence of moment. Since in the absence of magnetization is spontaneous field they have in the absence of field they have zero. magnetization in spontaneous field they have spontaneous the absence of magnetization no spontaneous magnetization. external field. even in the magnetization Since spin is anti Due to random absence of due to anti parallel but of spin. external field due parallel spin different to parallel magnitudes.

6. 6. 6. 6. 6.

7.The relative 7. The relative 7.The relative 7.The relative 7.The relative

permeability µr<1 permeability µr>1. permeability permeability permeability µr>>1 µr>1 µr>>1

8.Susceptibility χ is 8. Susceptibility is 8. Susceptibility is 8.Susceptibility χ 8. Susceptibility is small and negative small but positive large and positive is small but large and positive negative

9. does not 9. depends on 9. depends on 9. depends on 9. depend on depend on temperature temperature temperature temperature temperature. No particular is graph is drawn.

10. does not 10. χ = c/T 10. χ = c/T-θ 10. χ = C/T+θ 10. χ = C/T±TN

depend on (curie law) curie-Weiss law = C/T+TN T = absolute temperature C=curie constant θ = curie TN = Neel temperature T = absolute temperature temperature temperature

11.Examples 11.Examples: 11.Examples: 11.Examples 11.Examples

Cu, Au, Zn, H20, Bi Al, Pt, Mn,CuCl2 Fe, Ni, Co, MnO, FeO,MnO,Cr205 ZnFe204, Ni etc. organic etc. Fe2 03, Zn ferrite, etc. ferrite, Cu ferrite, materials. Alkali & Ni ferrite, Mn Mn ferrite, Ferrous transition metals. ferrite Ferrite

6.5. Domain theory (or) Weiss theory of Ferromagnetism  According to Weiss, Ferromagnetic material consists of a number of regions -6 called “Domains” [~10 mts] which are spontaneously magnetized.

Fig. 6.5: Different possible orientations of domains

 Spontaneous magnetization is due to parallel alignment of all magnetic dipoles (in each domain) even when no external field is applied.  Different domains possess different orientations hence net magnetization is zero.  When an external field is applied there are two possible ways of alignment of domains. They are 1. By motion of domain walls 2. By rotation of domain walls  Domains arise to minimize the energy of the material. The total internal energy is minimum. The alignment of domains, parallel to field is discussed as follows: a) The domains which are parallel to the direction of applied magnetic field will grow in size than other domains. This is called “Motion of domain walls”. Also other domains which are opposite to the field direction are reduced. This is shown in the figure 6.6.

Fig. 6.6 (a) Domain orientation in the absence of the magnetic field (b) Domain enhancement shrinkage due to weak fields (c) Domain ratation due to strong fields (d) Saturation due to very high fields

b) As the magnetic field is strong, the magnetic moments of the domains can rotate in the applied field direction. This is called “rotation of domain walls”.

6.6 Domain theory of ferromagnetism based on the basis of B-H curve (or) Hysteresis curve

Definition: Hysteresis means the lagging of magnetization “B” behind the applied magnetizing field “H”. The energy supplied to the specimen during magnetization is not fully used. The balance of energy left in the material is produced as heat i.e. loss of heat called” Hysteresis Loss”.  This phenomenon of magnetic Hysteresis is an “Irreversible” characteristic of ferromagnetic material. The loop (or) area refers to the hysteresis loop. Hysteresis loss occurs in ferromagnetic materials below Curie temperature.

Fig. 6.7: Magnetic Hysteresis curve (or) B – H curves of a Ferro magnetic material

The complete cycle of operation is discussed as follows:  When the magnetic field is applied on a ferromagnetic material the magnetization

increases slowly and reaches a constant Ms called saturation magnetization  In fig (6.7) from point O to A, the displacement of domain walls takes place. When the field is suddenly off, the domains again go for original position.  From point A to B, as the field is further increased, the magnetization also increases. Here when the field is made off the domain displacement does not return back to original condition.  For higher fields the magnetization reaches maximum ie saturation magnetization

Ms due to rotation of domain walls.  In this case at the region B to P, if the field is suddenly made off, the domains does not return back to original direction. But there is some magnetic field remained inside the specimen.

 The point Ms is called saturation of magnetization. When the field is off, the curve does not go back to ‘O’[as shown in fig] but creates a new path to a point

Mr called “retentivity” (or) residual (or) remanence Magnetization  To reduce the residual magnetism to zero, a negative field ‘Hc’ has to be applied. When the sufficient negative field is applied, the residual; magnetization

becomes zero and this field is known as “negative coercive field” (-Hc) or coercively. Further again if the negative field is applied then magnetization increases but in negative direction. This is known as negative saturation

magnetization (-Ms).  If the negative field is decreased back to zero the negative saturation of magnetization will not reach the initial path at ‘0’ but creates a new path and

reaches a point called negative residual magnetism ‘-Mr’ know as negative receptivity.  To decrease the negative residual magnetism to zero some positive field is applied. The amount of magnetic field required to bring residual magnetization to

zero is known as positive coercive field (Hc).  Further the increase of positive magnetic field the magnetization reaches again to

positive saturation (Ms) and this is a cyclic process.  The final conclusion is that when the magnetization vector is started from origin ‘O’ will not reach back to that point. “The magnetization lags behind H”. This is called magnetic hysteresis loss measured in the area of the loop (or) curve. If the loop area of a ferromagnetic material is large, more energy is wasted. This is also called as “dielectric loss” for one complete cyclic operation.

6.7. Anti Ferro Magnetism and properties  Anti Ferro magnetism arises when the spin magnetic moments of neighboring atoms of the crystal are anti parallel.  In these materials the distance between two atoms is very small, hence the exchange force (or) exchange coupling (or interaction) produce anti parallel alignment of the adjacent spin of electrons.  The important property of anti Ferro magnetic substance is that its susceptibility ‘χ’

shows a maximum as a function of temperature below Neel temperature TN as shown in fig(b).  This features is explained on the basis of “two sub-lattice model”

Fig. 6.8 Anti ferromagnetic two sub-lattice model

Because of opposite spin moments; we consider that the antiferromagnetic material consists of two different types of atoms i.e. A-type atoms, B-type atoms.  The crystal consists of interpenetration of two cubic sub lattices, one with A-atoms and other with ‘B’ atoms. One sub lattice is spontaneously magnetized in one direction and the other sub-lattice is spontaneously magnetised in other direction.  These materials possess BCC structure with ‘A’ atoms occupying at corner points and ‘B’ atoms occupying at the centre of the cell as shown in fig (a).  In the absence of external applied magnetic field the magnetization of the specimen will be zero, because of antiparallel spins of equal magnitudes.

Fig. 6.9 Variation of χ with temperature  As the temperature is increased, the susceptibility increases with temperature up to

Neel temperature TN.  At Neel temperature χ is maximum. For further increase of temperacute ‘χ’ decreases. This shows paramagnetic behavior.  At low temperature the interaction between the adjacent spins is more hence the resulting magnetization is small, χ is small [due to anti-parallels spins] as temperature increases, the interaction between adjacent atoms decreases

[antiparallel spins become parallel spin] and hence χ increases up to Neel (TN) temperature.

 Above Neel temperature all the atoms are free’ hence the material shows paramagnetic behavior.

The susceptibility χ =

C =Curie constant, T=absolute temperature, TN= Neel temperature.

6.8. Ferri magnetic materials (or) Ferrites 6.8.1Ferrimagnetism 1. It is the property of the material which has strong attractive force. 2. The property is due to spin of electrons. 3. Spin is anti parallel but of different magnitudes.

4. They have permanent magnetic dipoles. 5. They possess permanent magnetic dipole moment. Also in the absence of field they have spontaneous magnetization. Since spin is anti parallel but of different magnitudes.

6. Bin >> Bout

7. The relative permeability µr>>1 8. Susceptibility is large and positive 9. Magnetic susceptibility (χ) depends on temperature. 10. Hence χ = . Shown in the figure below.

11. Examples: ZnFe204, Ni ferrite, Cu ferrite, Mn ferrite, Ferrous Ferrite

6.8.2 Applications of ferrites a) Production of ultrasonic waves Ferrites exhibit magnetostriction method. As such of ferrite rod kept in a varying magnetic field (B applying current through it) produces vibrations of the order of ultrasonic range. Hence ultrasonic waves are produced. b) Ferrites as core materials Ferrite rods are used as cores for audio and TV transformers. They are also used in will of radio receivers to improve sensitivity and selectivity of audio radio receivers. Ferrites are used in Inductors, antenna cores, magnetic amplifiers, tuners etc, ex: Mn Zn ferrite, Ni Zn ferrite. c) Ferrites as switches In Ni-Zn ferrite the permeability drops sharply to zero near the Curie temperature. This property is utilized in thermal sensing switches used in refrigerates, air conditioners, electric ovens and fix alarms. d) Ferrite minimizes eddy current losses in core materials. e) Ferrites in digital computers  Ferrites are used for rapid storage and retrieval of digital information.  Mn-Zn ferrites are used in antennas for long wave broadcasting. Ni-zinc ferrites are used in short wave broad casting in telecommunication  They are also used in calculations, thermal sensing switches,  These are used in induction cores, antennas variable inductors etc.

6.9. Differences between soft and hard magnetic materials Based on area of the hysteresis loop, they are classified as soft and hard magnetic Materials.

Soft magnetic materials Hard magnetic materials 1. They have small hysteresis loss due 1. They have large hysteresis loss due to to small area of hysteresis loop large area of hysteresis loop.

2. The domain wall movement is 2. The domain wall movement is difficult easier due to presence of impurities crystal imperfections 3. The coercively (-Hc)and retentively 3. The coercively(-Hc) and retentively (Mr) (Mr) are small are large 4. They can be easily magnetized and 4. They cannot be easily magnetized and demagnetized demagnetized 5. The value of permeability and 5. The values of permeability and susceptibility are large susceptibility are small

6. Magnetization energy is small 6. Magnetostriction energy is large 7. They are used as temporary 7. They are used as permanent magnets magnetic or electro magnetic 8. Examples: ferrites, Fe-si alloy, Fe- Ni alloy 8. Example: Al-Ni-Co, Cu-Ni-Fe, Cu-Ni-Co etc 9. In magnetic inductors, cores, 9. For production of permanent magnets storage of data, switching circuits, which are used in magnetic detectors, audio frequency applications, microphones flux meters, voltage magnetic amplifier etc as electro regulators etc. magnets.

6.10 Superconductivity

1. It is a property of metals, alloys exhibit “zero” resistivity or infinite conductivity when these metals and alloys are cooled to sufficiently low temperatures in the absence of magnetic field.

Fig. 6.10: Resistivity of mercury falls to zero at TC=4.2K

2. This was first observed by H. Kammerlingh ones in 1911 that the electrical resistance of pure mercury vanishes suddenly at 4.2k as shown in fig (a) below. This temperature where the resistivity drops to zero is called superconducting transition

temperature ie Tc. 3. For mercury Tc = 4.2 k 4. Niobium Tc = 9.3 k

5. Above Tc, the material possesses resistance and is said to be in normal state. 6. Below Tc, résistance become zero and the conductivity reaches infinity. This state is called superconducting state. ‘Tc’ is defined as the temperature at which a material changes from normal state to superconducting state. The materials are called superconductors. Explanation The electrical resistivity of all metals and alloys decreases when they are cooled since when the temperature is lowered, the thermal vibrations of the atoms decrease and the conduction electrons are less frequently scattered. Hence the resistance decreases and drops to zero.

6.11 Concept of perfect diamagnetism Super conductors in the Meisser state exhibit perfect diamagnetism or super diamagnetism such that their magnetic susceptibility, χ = -1. Diamagnetism is defined as “the generation of a spontaneous magnetization of a material which directly opposes the direction of an applied field”. In superconductors the diamagnetism arises from the persistent screening currents. Which flow to oppose the applied field. Hence the magnetic field inside the superconductor is zero i.e.

B=0 We know that

B = µ0 [H+M]

B = µ0 H [1+ ]

B = µ0 H [1+χ] { since }

But B = 0 inside a superconductor

0 = µ0 H [1+χ]

= [1+χ ] =0 since {µ0 H 0} Or χ = -1 Hence χ is negative which nothing but perfect diamagnetism is.

6.12 Meissner Effect Definition In 1933, Meissner and Ochsenfield found the exclusion of magnetic flux lines by a

superconductor below Tc. “When a sample had been cooled below their transition temperatures (Tc) in a weak magnetic field, they found that the magnetic flux is pushed out of the superconducting material showing that the material had become perfectly diamagnetic. Inside the superconducting material B = 0. This is called “Meissner effect” Reason When a superconductor is cooled in a weak magnetic field, at the transition temperature Persistent currents arise on the surface and circulate so as to cancel the flux density inside the material. Hence the field inside the superconductor is zero (for

B = [H+M]

B = H [1+ ]

B = H [1+χ] { χ=M/H} As B = 0 inside a superconductor

0 = H [1+χ] If H 0 [1+χ] = 0 χ = -1 i.e. ‘χ’ is negative which is nothing but the property of diamagnetism [opposing the applied magnetic flux].

Fig. 6.11 (a) Normal conductor passes magnetic lines at T >TC (b)Superconductor repels magnetic lines at

This is called Meissner effect.

6.13. Effect of magnetic field (HC) By applying magnetic field of sufficient strength, the superconductivity of a material can be destroyed. The minimum magnetic field strength required to destroy

superconductivity of a substance, below TC is called critical magnetic field (HC) at

that temperature. HC varies with temperature. This variation has been shown for number of superconducting element in the figure. 6.12.

Fig. 6.12 Variation of HC with temperature

From the graph, the critical magnetic field for different elements will be different at different temperatures. Also for a material, the critical magnetic field increases with

decrease of temperature below TC. At TC, no magnetic field is required to change the material from superconducting to normal state. Maximum magnetic field is required to

destroy superconductivity at 0K. The critical magnetic field at 0K is H0. The critical magnetic field, HC at any temperature ‘T’ below TC can be represented as:

Where is the field required to destroy the super conducting property at 0K, TC is the

critical temperature of the superconductor and T is the temperature of the super conductor.

When T = TC, then HC = 0. When T = 0K then HC = H0.

6.14. Types of superconductors Depending on the way of transition from superconducting state to normal state by the application of magnetic field, superconductors are divided into 1) Type – I superconductor 2) Type – II superconductor

 Type – I superconductors In this case of superconductors, as we increase the intensity of applied magnetic field,inside the material opposing magnetization takes place up to some applied magnetic field, up to applied magnetic field, the material is completely diamagnetic and it is

Fig. 6.13 Type – I Superconductor

in superconducting state. As the applied field reaches the critical value HC, all of a sudden the magnetic force lines pass through the material and the material changes sharply from superconducting state to normal state as shown in the figure 6.13. This effect was first observed by Silsbee in 1916. So, it is called Silsbee effect. The critical

magnetic field HC for Type – I superconductors is of the order of 0.1 Tesla or less. These are called soft superconductors. Almost all the elements show Type– I superconductivity.

Consider a cylindrical superconductor to which a magnetic field H is applied along the axis of it. It will not allow the magnetic lines and obeys Meissner effect. i.e H = -M. Inside the superconductor, the magnetization acting opposite to H oppose the

field. When the magnetic field is equal to the critical field HC then the magnetic lines of force immediately penetrate the entire superconductor and convert it into normal conductor. It is shown in the figure. Examples: Sn, Hg, Nb, V etc.

 Type – II superconductors In case of Type II superconductors, as we increase the intensity of applied magnetic field, in the material opposing magnetization takes place up to some applied magnetic

field , called lower critical magnetic field. Up to this magnetic field, the materia completely expels the magnetic force lines. The material is completely diamagnetic and it is

superconducting state. If the applied magnetic field , slowly the magnetic force lines pass through the material and the transition from superconducting state to normal state takes place gradually. The penetration of magnetic force lines through the material

increases gradually from to . At , the magnetic force lines completely pass through the material and the material changes from superconducting to normal state.

Above , the material is in normal state. The material is in mixed state from to . The variation of magnetization with applied magnetic field in Type – II is shown in figure below. Type –II superconductivity was discovered by Shubnikov in 1930. The critical

magnetic field type – II superconductors is of the order of 10 Tesla. is upper critical magnetic field. Type - II with a large amount of magnetic hysteresis are called hard superconductors.

Consider a spherical superconductor in the magnetic field H. As per Meissner effect, it expels the magnetic line and H = -M, i.e. the magnetization inside a superconductor opposes the passage the lines through it in figure ( 6.14 ). The curved magnetic lines are concentrated at the two ends marked as in figure [6.14(b)] such

that intensity H is predominant at those ends. Up to lower critical field it behaves as a

superconductor. When the magnetic field exceeds then the penetration starts at those

ends marked as . This is because the intensity at those ends reaches value at first. As a result –M value decreases and the penetrated portion lose superconductivity and

becomes a normal conductor. This conversion is usually completed at (upper critical

field) where the penetration covers the entire surface. Thus between to it has a

mixed state of superconducting as well as normal conducting state. After , it is converted into normal conductor figure 6.13. The variation of –M with H is as shown in figure 14. In this case the conversion

is slow and is known as type – II superconductors. Examples Nb3, Sn, Nb3 Ge, etc.

Fig. 6.14 Type – II Superconductor

6.15 Applications of super conductors Generation of high magnetic fields as superconducting magnets a) Superconducting materials are used to produce very high magnetic fields of the order of 20T to 50T eg: Nb-Zr, Nb-T are used Superconducting magnets can be formed by using coils of wire made up of superconducting materials. Once the small amount of current is passed, it produces high magnetic field and the current will remain for very long periods. The cost of power is less. The size of superconductor is also less. But the superconducting magnetic are always maintained in very low temperatures. b) Superconducting coils are also used in electric machines, transformers and magnetic resonance imaging (MRI) instruments. MRI is used to scan human body. c) They are used to determine the unknown structure of chemical compounds, molecular structure of unknown compounds.

Low loss transmission lines and transformers Since resistance is zero, the power loss during transmission is negligible. Hence electric cables are designed with superconducting wires. A.C superconductor shows

some resistance but DC current through superconductors shows zero resistance.

Data storage (Memory Cell) in Computers Flux is quantized in superconductors. A closed superconducting ring is used as memory cell. When persistent current is passed through it, then it is said to be in ‘1’ state [ON STATE]. In normal state, current will not pass through it, then it is said to be ‘0’ state [OFF STATE] Electric generators Superconducting generators are very smaller in size and weight. Motors with very high powers as large as 2500kw can be constructed at very low voltage as low as 450 volts. Magnetic bearings: The Meissner effect is used in these bearings, mutual repulsion between two superconducting materials due to opposing. A magnetic field is used in the construction of magnetic bearings. There is no friction in these bearings. Josephson junction devices A very small gap made between two superconductors form a junction called Josephson junction. The devices which use such junctions are called Josephson junction devices. The dc Josephson effect is used in the construction sensitive gallon meters. These devices can measure magnetic fields accurately up to 10-11gauss. The A.C Josephson effect is used to generate electromagnetic radiations [EM waves] i.e. from RF to IR frequencies can be produced. Super conducting quantum Interference devices [SQUIDS] a) SQUIDS are used in construction of superconducting sensitive magnetometers. They are used in magnetic flux storage devices. b) SQUIDS are used to study and decect tiny magnetic signals. It measures the strengths which are less than 1/1000 of a quantum in of magnetic flux. c) SQUIDS are used in medical diagnosis the magnetic fields generated by heart and brain of the order of 10-14 teals in the form of magnetic signals are dected by SQUIDS . SQUIDS can be used to measure the voltage associated with brain, chest and cardiac activity. d) Used in detection of minerals, oil deposits inside the earth. Super conductors in Medicine a) Used in MRI instruments to scan human body instead of x-rays. b) Used to detect minute magnetic fields generated by brain in the form of signal up to 10-14tesla. c) Used to measure voltage associated with brain, chest and cardiac activity. d) Human blood contains iron in certain percentage. Iron supplies oxygen to various parts of the body. If iron content is less or more than oxygen supply than heart

attack occurs. These are detected by SQUIDS. These are actually known as superconducting susceptometer. e) A disease that produces disorder in nerrous system of brain is call “epilepsy”. This will causes fits and mal functions. The damaged region of the brain is called short circulated epileptic centre which produces magnetic signals. f) SQUIDS are used to detect these signals. Again they are sending to computer which gives a 3-D picture of activity of brain. This technique is known as “Magneto encephalography” g) As a cryotron Switch In this device, the principle is that: A superconductor has two states, the superconducting state and normal state. The application of magnetic field greater than

critical field ‘Hc’ can change from superconducting to normal state. And removal of field changes from normal to superconducting state.

Fig. 6.15 A Cryotron switch

This device consists of two superconductors.

 Tantalum: which acts as core of super conductor has Tc=4.5k  Niobium: a super conductor which acts as a coil (or) in the form of wire wound around tantalum.

 The transition temperature ‘Tc’ of niobium Tc=9.5k. As shown in the fig 6.15, on the tantalum superconducting core, Niobium as a superconducting wire is wound on the core of tantalum. This set up is placed in very low temperatures.  The current flowing through the coil is adjusted such that the magnetic field produced is very close to the critical magnetic field of core of tantalum.  Now by slightly increasing the current through the coil of the core, the core can be changed from superconducting to normal state and by reducing the current through the coil the core can be brought back to super conducting state.  The reason is that by increasing the current through the coil, the magnetic field produced coil exceeds the critical magnetic field of the core hence the superconducting state of core is changed into normal state.  By decreasing the current through the coil, the magnetic field can be brought below the critical magnetic field of the core. Hence the core changes to its original superconducting state.

 Thus the resistance of core is made ON (or) OFF by external control I ie coil of wire [superconducting state means ‘ON’. Normal state means ‘OFF’] this acts as ON/OFF switch.

Superconducting Magnetic-levitating Trains Principle  In a superconductor the magnetization is in the direction opposite to that of external magnetic field. This is called diamagnetism. When a super conductor is brought near a permanent magnet, a strong repulsive force between the two causes the lighter one to float over the other. This is known as “magnetic levitation”.  Magnetically levitated trains in Japan were constructed in 1979. Which record speed of 321miles per hour? The train floated without touching the ground as a result of repulsion between its superconducting magnetic materials.  These trains are based on Meissner effect. The magnetic levitating vehicle consists of superconducting magnet at its base. There is a segmental aluminum guide way above which the magnetic levitating vehicle can be made to float by magnetic repulsion.  The magnetic repulsion is in between superconducting magnet attached at the bottom of the vehicle and the magnetic field is obtained by passing current through electric coils arranged is aluminum guide way. The current in the segmented guide way not only tevitate the vehicle but also helps to move.  The vehicle is levitated above the guide way by 10 to 15 cm. the vehicle is provided with wheals are pulled in to the body, while slowly settles on the guide way after running certain distance.  The magnetic levitating trains has been constitutes in Japan, which runs at a speed of 500km/hr.

Fig. 6.16 Superconducting Magnetic-levitating Train

PROBLEMS & SOLUTIONS

(1) Find the electric suspectibility of dielectric gas having dielectric constant of 1.000041

Solution: Given that: Dielectric constant

=1.000041 Electric susceptibility χ =?

Formula: χ = 1 χ = 1.000041 1 χ = 4.1×

(2) The electronic polarizability of dielectric material having permanent dipoles no ions is 1.5 10-40 F/m2. The density of material is 2.5 1028 atoms/m3. Calculate the di electric constant of the material.

Note: [here density of material is considered as ‘n’. since for unit volume of the dielectric material]. Density ρ = = n

Solution: Given data: -40 2 = electronic polarisability = 1.5 0 F/m N= Density of material 2.5 1028 atoms/m3

dielectric constant = ? - =

where = permittivity of free space -12 = 8.854 10 F/m Density =

N=2.5 1028 atoms/m3 =

1.5x10-40 =

1 =1.423

(3) A parallel capacitor has an area of 100cm2, a plate separation of 1 cm & is charged to a potential of 100 Volts. calculate the capacitance of the capacitor & the change on the plates.

Solution: Given data: Area = 100cm2 = 100 10-4m2 d= 1cm = 10-2m V=100volts Hence C =

= 8.854 10-12 F

Capacitance C=8.854 10-12F Also C=Q/V Hence charge Q = CV = 8.854 10-12 100 Q=8.854 10-10 Coulomb

(4) The relative dielectric constant of sulphur is 3.75 when measured at 27°c. assuming the internal field constant υ= , calculate the electronic

polarisability of sulphur if the density at this temperature is 2050kg/m3. The atomic weight of sulphur being 32.

Solution: No. of atoms / m3 =

N=385.66 1026 atoms/m3 -12 =8.854 10 f/m =

-40 2 = 3.181 10 Fm

(5) A parallel plate capacitor having area 6.45 10-4 m2 and plate separation of 2x10-3m which a potential of 12volts having dielectric constant 5.0. compute polarization.

Solution: Area A = 6.45x10-4 m2

Separation between plates d = 2 10-3 m

Potential V =12volts

Dielectric constant =5.0 Polarization P = E [ -1] Electric field, E=

E= volt/m

Hence polarization P=8.854 10-12 C-m.

(6) The relative dielectric constant of sulphur is 3.75. Calculate electronic polarisability of sulphur if its density at this temperature is 2050 kg/m3. The atomic weight of sulphur being 32.

Solution: Dielectric constant of sulphur = 3.75 Density of sulphur, D= 2050 kg/m3 Atomic weight of sulphur, M = 3.2

Electric polarization =? No. of atoms/m3 N= where = Avogadro number

N= 385.66 1026 atoms/m3 -12 = 8.854 0 f/m From Clausius - Mossotti relation =

-40 2 = 3.18 10 F/m

7) The magnetic susceptibility of silicon is -0.4x105. calculate the flux density and magnetic moment permit volume when placed in magnetic field intensity of 5x105a/m

Solution: Given data: χ = -.4 05 Magnetic flux density B=? Magnetization M=magnetic moment/unit volume=? Magnetic field intensity H=5 105A/m

B = µ0 (H+M) -7 µ0 = 4 10 H/M

B = µ0 H [1+M/H] = µ0 H [1+χ]

B = µ0 H [1+χ]

B = 4 10-7 5 105[1 - 0.4 10-5]

B = 0.62Wb/m2

χ = M/H

M = Hχ

M = -5 105 0.4 10-5 = -2 Amp/m

8) Calculate magnetization and magnetic flux density if magnetic field intensity 250amp/m and relative permeability is 15. Given data

Μr =15 & H = 250 a/m M =? B =?

Formula B = μ0 [H+M] But M = H (μr-1) M = 250 (15-1) M = 3500A/m B = 4 10-7 [250 + 3500] B = 4.7 0-3 Wb/m2

9) A circular 100p of copper having a diameter of 10cm carries a current of 500mA. Calculate the magnetic moment. Given data: Diameter of 1oop 2r =10cm r = 5cm r = 5 0-2 mt

Current, I= 500 10-3amp

I = 0.5A Formula: magnetic moment 2 Μm = ( r ) i = ( ) (5 10-2)2 0.5 -3 2 Μm = 3.93 10 a-m

10) Find relative permeability if H=220amp/m and M=3300 amp/m Given H =220amp/m M = 3300amp/m χ = M/H =3300/220 144

Μr = χ +1 = 16

11) A paramagnetic material has 1028atoms/m3.its susceptibility at 350k is 2.8 x10-4. Calculate susptibility at 300k. Given -4 Χ1=2.8 0 T1=350k χ 2=?

T2=300k Formula χ = C/T

χ1 /χ2 =T2/t1 -4 χ 2 = χ1T1/T2 =2.8 10 50/300) -4 χ 2 = 3.26 10

12) A superconducting tin has critical tempature of 3.7k at zero magnetic field and critical field of 0.0306 t at ok. Find critical field at 2k. Find critical current also if r =1 mts

Given Tc = 3.7k, H0 = 0.0306T

Hc =? T = 2k 2 Hc = H0 [1-(T/Tc) ] 2 Hc = 0.0306[1-(2/3.7) ] = 0.02166T

Critical current Ic= 2 rHc

Ic = 2 1 0.02166 tesla.

PREVIOUS UNIVERSITY QUESTIONS 1.(a) Obtain the relevant expressions for (i) Electric polarizability (ii) Ionic polarizability

(b)Describe domain theory of Ferro magnetism on the basis of B-H curve (or) Hysteresis curve (c) Super conductor exhibits perfect diamagnetism. Explain [May/June – 2012]

2. (a) Define the terms magnetic induction (B), magnetization (M) and magnetic field ‘H’. Obtain an expression relating to these quantities. (b) What are ferrites? Prove that ferrites are superior to ferro-magnetic materials. Write the applications of ferrites.

(c) The magnetic susceptibility of aluminium is 2.3x10-15. Find the permeability and relative permeability. [June-2010] 3.(a) What is internal field in a dielectric material? (b) Derive an expression for calculation of internal field for a cubic. (c) Write notes on dielectric theory of ferro-electricity [June 2011] 4. (a) What is meant by polarization mechanism in dielectrics? Discuss the different polarization mechanism in dielectrics. (b) Distinguish between ferro-electricity and piezo electricity. (c) Find the electric susceptibility of a dielectric gas having dielectric constant of 1.000041. [June-2011] 5.(a) Derive an expression between electronic polarization and electric susceptibility of the dielectric medium. (b) Describe Lorentz method to calculate the internal field of a cubic structure. (c) What is pyro electricity. Write the applications of pyro-electric materials. [June-2011] 6. (a) What is electronic polarization. Derive an expression for electronic polarizability interms of the radius of the atom. (b) What is piezo electricity. Write the applications of piezo-electric materials. (c) The electronic polarizability o a dielectric material having no ions & permanent dipoles is 1.510-40 f/m2. The density of material is 2.5 1028 atoms/m3. Calculate the dielectric constant of the material. [June-2011] 7.(a) Describe the sources of permanent dipole moment in magnetic materials. (b) Distinguish between diamagnetic & para magnetic materials. (c) Explain in details the B-H curve of ferro magnetic material. What are Hysteresis losses. Explain [Jan-2012] 8.(a) Explain B,χ,M & μ. (b) Explain domain theory of ferro magnetism. (c) For a para magnetic material the susceptibility at 340K is 1.76 10-4. Calculate the susceptibility at 310k. [jan-2012] 9.(a) The susceptibility of aluminium is 2.3 10-5. Find its permeability (μ) and

relative permeability( ) (b) Explain the origin of spontaneous magnetism in ferro magnets on the basis of spin-spin interaction. Explain B-H curve. [Jan-2012] 10. (a) What are hard & soft magnetic materials. Give their properties and applications. (b) Explain Meissner effect. Explain in detail. [June 2010] 11. (a)Distinguish between superconductor & normal metal. (b) Write notes on magnetic levitation. (c) Explain concept of perfect diamagnetism. [June-2010] 12. (a) Define magnetic moment. What is Bohr magneton. Explain. (b) Explain the characteristics of dia, para, ferro magnetic substances. 146

(c) If a magnetic field strength of 300 amp/mt produces a magnetism of 4200A/m in a ferro magnetic material. Find relative permeability of the material. 13. (a) Explain (i) Polarization (ii) Polarization vector (iii) Electric displacement (iv) Dielectric constant (v) Dielectric susceptibility

(b) Prove that P= E [ -1] [June-2010] (c) Derive clausius-Mossotti relation. 14. (a) Explain the applications of super conductors

(b) Prove the relation B= [H+M] (c) What is dielectric break down. Explain how it can be achieved. [June-2009]

(15) a) Explain the following i) Dielectric constant ii) electric polarization ii) Displacement vector iv) electric susceptibility

b) Find the electric susceptibility of a dielectric gas having dielectric constant of 1.00041.

c) Define the magnetization and show that B=µ0(H+M) June 2013

OBJECTIVE QUESTIONS 1. Dielectrics are insulating materials. 2. The process of orientation of induced dipoles along the field direction is called Polarization. 3. The dipole moment per unit volume is a measure of Polarization. 4. The ratio of the permittivity of the medium to the permittivity of free space is called Dielectric constant (or) relative permittivity 5. If α is the polarizability and E is electric field applied, then the expression for induced dipole moment is µ=αE

6. Electric flux density D=εrε0E.

7. Polarization P= ε0(εr-1)E.

8. Electric susceptibility χe=(εr-1).

9. Displacement of positively charged nucleus and the negatively charged electron cloud of an atom in opposite directions on application of an electric field is called electronic polarization.

10. Electronic polarizability αe= ε0(εr-1)/N 11. In ionic solids, on application of the electric field, displacement of cations and anions in opposite directions is called ionic Polarization. 12. A superconductor exhibits zero resistance. 13. In some materials which have molecules with permanent dipole moment, applied electric field results in orienting these molecular dipoles along the field direction. This phenomenon is called Orientational polarization.

14. Internal field or Lorentz field Ei=E+ P/3ε0

15. Clausius-Mosotti equation (εr-1)/ (εr+2)=Nαe/3ε0 16. The time taken for a polarization process to reach 1/e of its maximum value is called relaxation time. -1 17. The unit for permittivity of free space ε0 is Fm 18. Insertion of a dielectric material between the plates of a capacitor increases the capacitance. 19. Electronic polarization is independent of temperature. 20. Dielectric loss occurs when the dielectric is subjected to a.c. voltage. 21. Orientational polarization is inversely proportional to absolute temperature. 22. Dielectrics are insulators. 23. In the absence of external electric field, the electric dipoles in a dipolar substance are randomly oriented. 24. Orientational polarization is due to rotation of polar molecules in dielectric substances. 25. The property of becoming electrically polarized when mechanical stress is applied is known as piezoelectric. 26. Hysteresis in polarization-electric field relation is exhibited by Ferro electric materials. 27. The material in which spontaneous polarization changes with temperature is known as Pyro electric. 28. Posistors are used to measure and control temperature. 29. Using pyroelectrometer one can measure temperature changes as small as 10-6 0C. 30. If N is the number of dipoles per unit volume and E is the electric field applied Orientational

polarizability αo is related to Orientational polarization Po by αo= Po/NE. 31. The polarizations which are dependent of temperature are orientation and space charge. 32. The change in spontaneous polarization per unit temperature change is called pyroelectric coefficient. 33. When a dielectric material loses its resistivity and permits very large current to flow through it, then the phenomenon is called dielectric breakdown. 34. Materials do not having permanent magnetic dipoles are diamagnetic.

35. When a diamagnetic material is placed inside an external magnetic field, induced magnetic dipoles act opposite to the applied field direction. 36. Diamagnetic property is independent of temperature. 37. Paramagnetic property is dependent of temperature. 38. Ferromagnetic property is dependent of temperature. 39. The ratio of the intensity of magnetization produced in the sample to the magnetic field intensity producing it is called magnetic susceptibility

40. Relative permeability is related to magnetic susceptibility by μr=1+ χ. 41. Materials possessing permanent dipoles which do not interact among themselves are called paramagnetic. 42. Materials possessing permanent dipoles which strongly interact among themselves are called ferromagnetic. 43. The unit of magnetic field intensity is Am-1

44. Magnetic induction B is related to magnetic field intensity H by the relation B=μ0(H+M) 45. One Bohr magneton equals 9.27×10-24A-m2 46. The temperature at which the transition of antiferro to Para magnetism takes place is called Neel temperature. 47. The transition from the ferromagnetic to the paramagnetic state is named after Curie.

48. One Bohr magneton μB is equal to eh/4πm. 2 49. Orbital magnetic dipole moment μ0 of a revolving electron is given by - (er ω0/2) 50. Para magnetic susceptibility varies as 1/T 51. Diamagnetic materials possess no permanent magnetic dipoles. 52. For an ferromagnetic material, in the graph plotted between 1/x versus T, the extrapolated line (on x-axis) makes positive intercept at a temperature above 0 K. 53. For a Para magnetic material, in the graph plotted between 1/x versus T, the extrapolated line makes an intercept which is passing through origin at 0 k 54. Ferrox cube is the commercial name of ferrimagnetic materials. 55. Magnetic ceramics are ferrimagnetic materials. 56. The magnetization retained by the specimen when the magnetizing field is reduced from saturation value to zero is known as retentivity. 57. The dimension of ferromagnetic domains is in the order of 10-6m. 58. One nuclear magneton equals 5.05× 1029A-m2 59. The permeability of free space is 4 ×10-7 Hm-1 60. Below transition temperature, a superconducting material exhibits zero resistance and diamagnetism. 61. The magnetization M of a superconductor is equal to –H 62. Examples of type-I superconductors are Al, Zn and Hg. 63. In a superconducting state both entropy and electronic specific heat change. 64. In type-I superconductors the coherence length is of the order of 10-6m

65. When a material becomes superconductor the properties of lattice structure do not change. 66. In superconducting state the energy gap is very small as compared to semiconductors and insulators. 67. In type-II superconductors the magnetic flux does not suddenly drop to zero but decreases exponentially. 68. Switching times with a Josephson junction are in the order of magnitude 10-2ns. 69. Energy gap of a superconductor is maximum at 0 K.

70. Energy gap of a superconductor is minimum at Tc. 71. Energy gap of a superconductor is found to decrease with increase of temperature. 72. The correlation of wave functions of super electrons on both sides of a thin insulating layer sandwiched between two superconductors is known as Josephson Effect. 73. The magnetic susceptibility of a superconductor is -1. 74. The region between low critical field and upper critical field is called Vortex state. 75. BCS theory explains superconductivity

Multiple choice questions 1. Polarization vector P= [ a ] a) Nµ b) N/µ c) µ/N d)

2. Orientational polarizability αo= ------[ c ] a) 3µ2/kT b) µ2/kT c) µ2/3kT d) µ/3kT e) none of these 3. The unit for relative permittivity is ------[ d ] a) F/m b) N/m c) H/m d) dimensionless 4. Orientational polarization with increase of temperature. [ b ] a) increases b) decreases c) increases exponentially d) decreases exponentially e) remains constant 5. At normal temperatures, the polarizations which are independent of temperature are ------[ c ] a) ionic and orientation b) electronic and orientational c) electronic and ionic d) space charge and ionic e) none of these 6. With increase of temperature dielectric constant ------[ b ] a) increases b) decreases c) decreases exponentially d)increases exponentially e) none of these 7. Diamagnetic susceptibility is ------[ b ] a) small, positive b) small, negative c) large, positive d) none of these

UNIT-IV Chapter 7 OPTICS

7.1 Interference-Introduction Wave Theory of light attempts to understand the various optical phenomena exhibited by light waves. Interference constituted the first proof of the wave nature of light. Thomas Young first experimentally demonstrated interference in light waves. The superposition principle forms the conceptual basis for the explanation of interference. To produce interference, the light waves should be coherent, i.e., the light waves should have constant phase difference and same frequencies.

7.2 Interference When two or more light waves superimpose in the medium then according to superposition principle, the resultant displacement at any point is equal to the algebraic sum of the displacements due to individual waves. The variation of resultant displacement influences amplitude variation, which causes intensity variations. This modification in the distribution of intensity in the region of superposition is known as interference. When the resultant amplitude is the sum of the amplitudes due to two light waves, the interference is constructive interference. If the resultant amplitude is equal to the difference of two amplitudes, the interference becomes destructive interference. The intensity variations are studied as interference fringes or patterns.

7.2.1 Conditions for Interference 1. The two light sources emitting light waves should be coherent. 2. The two sources must emit continuous light waves of same wavelengths and frequency. 3. The separation between the two sources should be small. 4. The distance between the two sources and the screen should be large. 5. To view interference fringes, the background should be dark. 6. The amplitudes of the light waves should be equal or nearly equal. 7. The sources should be narrow, i.e., they must be small. 8. The sources should be monochromatic.

7.2.2 Interference in Thin Film by Reflection When light is incident on a plane parallel thin film, some portion gets reflected from the upper surface and the remaining portion is transmitted into the film. Again, some portion of the transmitted light is reflected back into the film by the lower surface and emerges through the upper surface. These reflected light beams superimpose with each other, producing interference and forming interference patterns.

Fig7.1 Interference in thin film

Consider a transparent plane parallel thin film of thickness‘t ‘with refractive index . Let a monochromatic light ray AB be incident at an angle of incidence of ‘i’ on the upper surface of the film. BE and BC are the reflected and transmitted light rays. Let the angle of refraction is ‘r’. The ray BC will be reflected into the film and emerge through the film in the form of light ray DF. These two light rays superimpose and depending upon the path difference between them, they produce interference patterns. To know the path difference, draw the normal DG to BE. From the points D and G onwards, the light rays travel equal distances. By the time the light ray travels from B to G, the transmitted light ray has to travel from B to C and C to D. The path difference between light rays (1) and (2) is,

Path difference= (BC+CD) in film – BG in air (1) Consider the BCH, cos r =

BC = =

Similarly, from DCH, CD=

BC = CD = (2)

To calculate BG, first BD which is equal to (BH+HD) has to be obtained. From BHC tan r = =

BH = t tan r Similarly, HD = t tan r BD = BH+ HD = 2t tan r ( BH = HD) (3) From BGD, sin i =

BG=BD sin i = 2t tan r sini From Snell’s law, sin i = sin r BG= 2 t tan r sin r (4) Substituting the above values in Eq.(1) Path difference= - 2 t tanr sinr

= (1- )= r

= 2 t cos r (5)

At the point B, reflection occurs from the upper surface of the thin film (denser medium). Light ray (1) undergoes an additional phase change of or an additional path difference of

Total path difference = 2 t cos r +

When the path difference is equal to integral multiple of then the rays (1) and (2) meet in phase and undergo constructive interference. The condition for bright fringe is 2 t cos r + = n

2 t cos r = (2n-1) where n= 0,1,2,3… (6)

When the path difference is equal to half integral multiple of then the rays (1) and (2) meet in out of phase and undergo destructive interference. The condition for dark fringe is 2 t cos r + = (2n-1)

2 t cos r = n where n=0, 1, 2, 3… (7) Depending on the above conditions, the interference pattern consists of bright and dark fringes.

7.3 Newton’s Rings Newton’s rings are one of the best examples for the interference in a nonuniform thin film. When a Plano-convex lens with its convex surface is placed on a plane glass plate, an air film of increasing thickness is formed between the two. The thickness of the film at the point of contact is zero. If monochromatic light is allowed to fall normally and the film is viewed in the reflected light, alternate dark and bright rings concentric around the point of contact between the lens and glass plate are seen. These circular rings were discovered by Newton and are called Newton’s rings.

7.3.1 Experimental Arrangement

Fig7.2 Newton’s rings (Experimental set up)

The experimental arrangement is as shown in fig. 7.2. The Plano-convex lens (L) of large radius of curvature is placed with its convex surface on a plane glass plate P). The lens makes the contact with the plate at ‘O’. The monochromatic light falls on a glass plate G held at an angle of 45° with the vertical. The glass plate G reflects normally a part of the incident light towards the air film enclosed by the lens L and the glass plate P. A part of the light is reflected by the curved surface of the lens L and a part is transmitted which is reflected back from the plane surface of the plate. These reflected rays interfere and give rise to an interference pattern in the form of circular rings. These rings are seen near the upper surface of the air film through the microscope.

7.3.2 Explanation of Newton’s Rings

Fig 7.3 Formation of Newton’s rings

Newton’s rings are formed due to interference between the light rays reflected from the top and bottom surfaces of air film between the plate and the lens. The formation of Newton’s rings can be explained with the help of Fig 7.3. A part of the incident monochromatic light AB is reflected at B (glass-air boundary) in the form of the ray (1) with any additional phase (or path) change. The other part of light is refracted along BC. Then at C (air-glass boundary), it is again reflected in the form of the ray (2) with additional phase change of or path change of .

As the rings are observed in the reflected light, the path difference between them is 2 t cos r + . For air film =1 and for normal incidence r=0, path difference is 2t + .

At the point of contact t=0, path difference is , i.e., the reflected light at the point of

contact suffers phase change of . Then the incident and reflected lights are out of phase and interfere destructively. Hence the central spot is dark. The condition for bright ring is 2t + = n

2t= (2n-1) where n=1, 2, 3…

The condition for dark ring is 2t + = (2n+1)

2t = n where n= 0,1,2,3…

For monochromatic light, the bright and dark rings depend on thickness of the air film. For a Newton’s rings system, the focus of points having same thickness lie on a circle 154

having its centre at the point of contact. Thus, we get bright and dark circular rings with the point of contact as the centre.

7.3.3 Theory of Newton’s Rings

Fig 7.4 Theory of Newton’s rings

To find the diameters of dark and bright rings, let ‘L’ be a lens placed on a glass plate P. The convex surface of the lens is the part of spherical surface (Fig 7.4) with center at ‘C’. Let R be the radius of curvature and r be the radius of Newton’s ring corresponding to the film thickness ‘t’. From the property of a circle, NA x NB = NO x ND Substituting the values, r x r = t x (2R-t) = 2Rt- As‘t’ is small, will be negligible = 2Rt

For bright ring, the condition is 2t= (2n-1)

2 = (2n-1)

Replacing r by , the diameter of nth bright ring will be

D = D α D

Thus, the diameters of the bright rings are proportional to the square root of odd natural numbers.

Fig 7.5: Newton’s ring pattern For dark ring, the condition is

2t = n

= n R = 4 n R D= D

Thus, the diameters of dark rings are proportional to the square root of natural numbers. With the increase in the order (n), the rings get closer and the fringe width decreases and are shown in Fig 7.5.

7.3.4 Determination of Wavelength of a Light Source

Let R be the radius of curvature of a Plano-convex be the wavelength of th th light used. Let and are the diameters of m and n dark rings respectively. Then

And =4(n) R

=4(m-n) R

Fig 7.6 Plot of D2 with respect to number of rings

Newton’s rings are formed with suitable experimental setup. With the help of travelling microscope, the readings for different orders of dark rings were noted from one edge of the rings to the other edge. The diameters of different orders of the rings can be known. A plot between and the number of rings gives a straight line as shown in the fig 7.6. From the graph,

The radius R of the Plano-convex lens can be obtained with the help of a Spherometer.. Substituting these values in the formula, can be calculated.

7.3.5 Determination of Refractive Index of a Liquid

The experiment is performed when there is an air film between glass plate and the Plano-convex lens. The diameters of mth and nth dark rings are determined with the help of travelling microscope. We have

= 4(m-n) R (8)

The system is placed into the container which consists of the liquid whose refractive index ( ) is to be determined. Now, the air film is replaced by the liquid film. Again, the diameters of the same mth and nth dark rings are to be obtained. Then we have = (9)

(10) Using the above formula, ‘μ’ can be calculated.

7.4 Introduction to Diffraction The wave nature of light is further confirmed by the optical phenomenon of diffraction. The word ‘diffraction’ is derived from the Latin word diffractus which means to break to pieces. It is common experience that waves bend around obstacles placed in their path. When light waves encounter an obstacle, they bend round the edges of the obstacle. This bending is predominant when the size of the obstacle is comparable to the wavelength of light. The bending of light waves around the edge of an obstacle is diffraction. It was first observed by Gremaldy.

7.5 DIFFRACTION

Fig7.7 Diffraction

As shown in fig.7.7, when light falls on an obstacle, the corresponding geometrical shadow on the screen should be completely dark. In practice, the geometrical shadow consists of bright and dark fringes. These fringes are due to the superimposition of bended light waves around the corners of an obstacle. The amount of bending depends upon the size of the obstacle and wavelength of light. When the light falls on an obstacle whose size is comparable with the wavelength of light, then the light bends around the edges or corners of the obstacle and enters into the geometrical shadow. This bending of light is known as diffraction. The bright and dark fringes in the shadow form a diffraction pattern.

The diffraction phenomena are broadly classified into two types:

Fresnel diffraction

To study diffraction, there should be a light source, obstacle and screen. In this class of diffraction, the source and screen are placed at finite distances from the obstacle. To study this diffraction, lenses are not necessary as the source and screen are at a finite distance. This diffraction can be studied in the direction of propagation of light. The incident wave fronts are either spherical or cylindrical.

Fraunhofer diffraction

In this class of diffraction, the source and screen are placed at infinite distances from the obstacle. Due to the above fact, lenses are needed to study the diffraction. This diffraction can be studied in any direction. In this case, the incident wave front is plane.

7.6 Fraunhofer Diffraction at Single Slit

Consider a slit AB of width ‘e’. Let a plane wave front WW’ of monochromatic light of wavelength propagating normally towards the slit is incident on it. The diffracted light through the slit is focused by means of a convex lens on a screen placed in the focal plane of the lens. According to Huygens-Fresnel, every point on the wave front in the plane of the slit is a source of secondary wavelets, which spread out to the right in all directions. Those wavelets travelling normal to the slit, i.e., along the direction

OP0 are brought to focus at P0 by the lens. Thus, P0 is a bright central image. The secondary wavelets travelling at an angle Ѳ with the normal are focused at a point P1 on the screen. Depending on path difference, the point P1 may have maximum or minimum intensities. To find the intensity at P1, draw the normal AC from A to the light ray at B.

Fig7.8 Fraunhofer diffraction - Single slit

The path difference between the wavelets from A and B in the direction Ѳ is given by Path diff =BC=AB sinѲ = e sin Ѳ Corresponding phase diff = x path difference

= e sinѲ

Let the width of the slit be divided into n equal parts and the amplitude of the wave from each part is ‘a’. The phase difference between any two successive waves from these parts would be [Total phase] = Ѳ = d (say)

Using the method of vector addition of amplitudes, the resultant amplitude R is given by R=

= Ѳ Ѳ = a where = Ѳ

= a ( is very small)

= n ( na =A)

Intensity = I=R2=A2 (1)

7.6.1 Principal Maximum

The resultant amplitude R can be written in ascending powers of as

I will be maximum, when value of R is maximum. For maximum value of R, the negative terms must vanish, i.e., =0 159

sinѲ =0 Ѳ=0 (2) Then R=A 2 2 Imax =R =A (3) The condition Ѳ=0 means that the maximum intensity is formed at P0 and is known as principal maximum.

7.6.2 Minimum Intensity Positions

‘I’ will be maximum, when sin =0 = π, 2π, 3π = mπ Ѳ= mπ

esinѲ= mλ where m = 1,2,3… (4)

Thus, the points of minimum intensity are obtained on either side of the principal maximum. For m=0, sinѲ=0, which corresponds to principal maximum.

7.6.3 Secondary Maxima

In between these minima, we get secondary maxima. The positions can be obtained by differentiating the expression of I w.r.t and equating to zero. We get

. . =0

Either sin =0 or =0 Sin =0 gives positions of minima. Hence the positions of secondary maxima are given by =0 =tan (5)

The values of α satisfying the above equation are obtained graphically by plotting the curves y= and y=tan on the same graph. The points of intersection of the two curves give the values of which satisfy the above equation. The plots of y= and y= tan are shown in fig 2.3.

Fig7.9 Plots of y=α and y= tanα

The points of intersections are =0, , …

Substituting the above values in equations (1), the intensities in various maxima are obtained.

= 0, I0= (principal maximum)

st = , I1= (1 secondary maximum)

nd = , I2= (2 secondary maximum)

and so on.

From the above expressions, it is evident that most of the incident light is concentrated in the principal maximum.

7.6.4 Intensity Distribution

Fig7.10 Intensity distribution

The variation of I with respect to is shown in fig 7.10. The diffraction pattern consists of a central principal maximum for =0. There are secondary maxima of

decreasing intensity on either sides of it at positions = , . Between secondary

maxima, there are minima at positions = π, 2π, 3π,…..

7.7 Fraunhofer Diffraction at Double Slit

Let S1 and S2 be double slits of equal widths e and separated by a distance d. The distance between the middle points of the two slits is (e +d). A monochromatic light of wave length λ is incident normally on the two slits. The light diffracted from these slits is focused by a lens on the screen placed in the focal plane of the lens. The diffraction at two slits is the combination of diffraction as well as interference, i.e., the pattern on the screen is the diffraction pattern due to a single slit on which a system of interference fringes is superimposed. When a plane wave front is incident normally on two slits, the secondary wavelets from the slits travel uniformly in all directions. The wavelets travelling in the direction of incident light come to a focus at while the wavelets travelling in a direction making an angle , come to focus at .

Fig7.11 Fraunhofer diffraction-double slit

From the study of diffraction due to single slit, the resultant amplitude =

Where, = .

Since a double slit is used, from each slit a wavelet of amplitude in a

direction These two wavelets interfere and meet at a point on the screen. To calculate the path difference between the wavelets, draw a normal to the wavelet through .

Path difference = = sin Phase difference = sin (6)

Fig7.12 vector addition method

To find the resultant amplitude at vector addition method is used. (Fig. 7.12) in which the two sides of a triangle are represented by the amplitudes through and . The third side gives the resultant amplitude. From the figure,

= + + 2

= [2+2 ]

The resultant intensity I = = 4 (7)

From the above expression, it is clear that the resultant intensity is the product of two factors, i.e,

1. which represents the diffraction pattern due to a single slit

2. which gives the interference pattern due to wavelets from double slits. The resultant intensity is due to both diffraction and interference effects.

7.7.1 Diffraction Effect

The diffraction term gives the principal maximum as the centre

of the screen with alternate minima and secondary maxima of decreasing intensity. We get principal maximum for = 0. We get minima for = 0

= , where m = 1, 2, 3…

The positions of secondary maxima occurs for

, , …..

7.7.2 Interference Effect

The interference term cos2β gives the equidistant bright and dark fringes. The maxima will occur for cos2β = 1 β = , where n = 0, 1, 2, 3… β = 0, π, 2π, 3π…………

The minima will occur for cos2β =0

β = 0, 1, 2, 3…

(e+d) sin

7.7.3 Intensity Distribution

Figure 7.13 (a) and (b) represents the intensity variations due to diffractions as well as interference effect individually. When both effects are combined then the resultant intensity variation is obtained. (Fig.7.13(c)). From the figure it is clear that the resultant minima are not equal to zero, still they have some minimum intensity due to interference effect.

Fig7.13 (a) Diffraction effect (b) Interference effect (c) Resultant intensity

7.8 Diffraction Grating-Normal incidence–(Diffraction at N parallel slits) 7.8.1 Construction

An arrangement consisting of large number of parallel slits of the same width and separated by equal opaque spaces is known as diffraction grating. Fraunhofer used the first grating consisting of a large number of parallel wires placed very closely side by side at regular intervals. The diameter of the wires was of the order of 0.05mm and their spacing varied from0.0533 mm to 0.687 mm. Now gratings are constructed by ruling equidistant parallel lines on a transparent material such as glass with a fine diamond point. The ruled lines are opaque to light while the space between any two lines is transparent to light and acts as a slit. This is known as Plane transmission grating. On the other hand, if the lines are drawn on a silvered surface (plane or concave) then the light is reflected from the positions of mirrors in between any two lines and it forms a plane or concave reflection grating. When the spacing between the lines is of the order of the wavelength of light, then an appreciable deviation of light is produced.

7.8.2 Theory

Fig.7.14 represents the section of a plane transmission grating placed perpendicular to the plane of the paper. Let ‘e’ be the width of each slit and‘d’ be the width of each opaque part. Then (e+d) is known as grating element. XY is the screen placed perpendicular to the plane of a paper. Suppose a parallel beam of monochromatic light of wavelength be incident normally on the grating. By Huygen’s principle, each of the slit sends secondary wavelets in all directions. The secondary wavelets travelling 165

in the same direction of incident light will come to a focus at a point P0 of the screen as the screen is placed at the focal plane of the convex lens. The point P0 will be the central maximum. Now, consider the secondary waves travelling in a direction inclined at an angle with the direction of the incident light. These waves reach the point P1 on passing through the convex lens in different phases. As a result, dark and bright bands on both sides of the central maximum are obtained.

Fig7.14 Section of a Plane transmission grating

The intensity at point P1 may be considered by applying the theory of Fraunhofer diffraction at a single slit. The wavelets proceeding from all points in a slit along the direction ϴ are equivalent to a single wave of amplitude (A sin /α) starting from the middle point of the slit, where α = (π e sinϴ/λ).

If there are N slits, then there will be N diffracted waves, one each from the middle points of the slits. The path difference between two consecutive slits is (e+d) sinϴ. Therefore, there is a corresponding phase difference of (2π/λ). (e+d) sinϴ between the two consecutive waves. The phase difference is constant and it is 2β.

Hence, the problem of determining the intensity in the direction ϴ reduces to finding the resultant amplitude of N vibrations each of amplitude (A sin / ) and having a common phase difference (e+d) sinϴ = 2β (1)

Now, by the method of vector addition of amplitudes, the direction of ϴ will be R*= .

2 2 2 And I = R = ( ( ) = Io ( ) ( ) (2)

The factor ( )2 gives the distribution of intensity due to single slit while the factor

(sin2 Nβ/ sin2β) gives the distribution of intensity as a combined effect of all the slits.

7.8.3 Intensity distribution in N-Slits

Principle maxima

The intensity would be maximum when sinβ = 0. or β= n=0, 1, 2, 3,… but at the same time sin Nβ = 0, so that the factor (sinNβ/sinβ) becomes indeterminate. It may be evaluated by applying the usual method of differentiating the numerator and the denominator, i.e., by applying the Hospital’s rule. Thus,

Hence, = 2

The resultant intensity is ( ) N2. The maxima are most intense and are called as

principal maxima. The maxima are obtained for = (e+d) sin ϴ =

Where, n= 0, 1, 2, 3….

corresponds to zero order maximum. For n=1, 2, 3...etc., the first, second, third, etc., principal maxima are obtained respectively. The sign shows that there are two principal maxima of the same order lying on the either side of zero order maximum.

Minima

A series of minima occur, when Sin Nβ = 0 but sinβ For minima, Sin Nβ = 0 Nβ = N (e+d) sinϴ =

N (e+d) sinϴ =

Where m has all integral values except 0, N, 2N …nN, because for these values sinβ becomes zero and the principal maxima is obtained. Thus, m = 1, 2, 3… (N-1). Hence, they are adjacent principal maxima.

Secondary maxima As there are (N-1) minima between two adjacent principal maxima, there must be (N-2) other maxima between two principal maxima. To find out the position of these secondary maxima, equation (2) must be differentiated with respect to β and then equate it to zero. Thus, = 2. 2( )

[ ] = 0 or

N cos N β sinβ – sinNβ cosβ = 0 N tan = tan Nβ The roots of this equation other than those for which β = (which correspond to principal maxima) give the positions of secondary maxima. To find out the value of (sin2 Nβ/sin2β) from equation N tanβ = tan Nβ, a triangle shown below, in the figure 7.15 is used.

Fig 7.15

Sin Nβ =

Therefore, =

As N increases, the intensity of secondary maxima relative to principal maxima decreases and becomes negligible when N becomes large. Figures 7.16 (a) and (b) show the graphs of variation of intensity due to the factors and respectively. The resultant is shown in figure 7.16 (c).

Fig7.16 (a) & (b) Graphs showing the variation of intensity (c) The Resultant

7.9 Diffraction Grating

An arrangement which consists of a large number of parallel slits of the same width and separated by equal opaque spaces is known as diffraction grating. Fraunhofer used the first grating consisting of a large number of parallel wires placed side by side very closely at regular intervals. Now gratings are constructed by ruling equidistant parallel lines on a transparent material such as glass with a fine diamond point. The ruled lines are opaque to light while the space between any two lines is transparent to light and acts as a slit (Fig.7.17). Commercial gratings are produced by taking the cast of an actual grating on a transparent film like that of cellulose acetate. Solution of cellulose acetate is poured on the ruled surface and allowed to dry to form a thin film, detachable from the surface. These impressions of a grating are preserved by mounting the film between two glass sheets.

Fig7.17 Diffraction Grating

Let ‘e’ be the width of the line and‘d’ be the width of the slit. Then (e+d) is known as grating element. If ‘N’ is the number of lines per inch on the grating, then

N (e+d) =1”=2.54 cm e+d= cm

There will be nearly 30,000 lines per inch of a grating. Due to the above fact, the width of the slit is very narrow and is comparable to the wavelength of light. When light falls on the grating, the light gets diffracted through each slit. As a result, both diffraction and interference of diffracted light gets enhanced and forms a diffraction pattern. This pattern is known as diffraction spectrum.

7.9.1 Grating Spectrum

The condition to form the principal maxima in a grating is given by (e+d) sin Ѳ = nλ Where (e+d) is the grating element and the above equation is known as grating equation. From the grating equation, the following is clear. 1. For a particular wavelength λ, the angle of diffraction Ѳ is different for principal maxima of different orders. 2. As the number of lines in the grating is large, maxima appear as sharp, bright parallel lines and are termed as spectral lines. 3. For white light and for a particular order of n, the light of different wavelengths will be diffracted in different directions. 4. At the center, Ѳ=0 gives the maxima of all wavelengths which coincides to form the central image of the same colour as that of the light source. This forms zero order.(Fig.2.9) 5. The principal maxima of all wavelengths forms the first, second,… order spectra for n=1,2,… 6. The longer the wavelength, greater is the angle of diffraction. Thus, the spectrum consists of violet being in the innermost position and red being in the outermost positions.

Fig7.18 Grating Spectrum

7. Most of the intensity goes to zero order and the rest is distributed among other orders. 8. Spectra of different orders are situated symmetrically on both sides of zero order.

9. The maximum number of orders available with the grating is nmax =

7.10.0 Introduction to Polarisation

Interference and diffraction phenomenon prove that the propagation of light is in the form of wave motion. But these phenomenon do not reveal the character of these wave motion, i.e., whether it is longitudinal or transverse. In longitudinal waves, the particles of the medium move to and fro in the direction of propagation of the wave as in case of sound waves. In a transverse wave, the particles of the medium vibrate up and down perpendicular to the direction of propagation of the wave as ripples on a water surface. When the phenomenon of polarization was discovered, it was established that light waves are transverse in nature.

7.10 Fundamentals of Polarized light

7.10.1 Polarization of a wave

Consider a string AB passing through two slits S1 and S2. The two slits are parallel to each other. Let one end of the string B be fixed. If one end of the string is vibrated, up and down in all possible directions, to produce transverse waves in all directions, then the wave having vibrations parallel to S1 will pass through S1 even as it is rotated and others are stopped. The waves from S1 will pass through S2 provided S2 is parallel to S1 (Fig 7.19 (a)). The slit S2 does not allow the wave to pass through it when it is at right angles to the slit S1 (Fig 7.19(b)). The transverse wave at A is not affected by rotations of S1 since the wave has vibrations in all directions and is known as unpolarised wave. The transverse wave between S1 and S2 has acquired vibrations in one direction parallel to S1 S2, it is stopped by S2. The wave with vibrations in one direction is called a polarized wave. Similar observations can be made when a ray of light is passed through a tourmaline crystal.

Fig7.19 (a) vibration of a string between two slits parallel to each other

Fig 7.19 (b) vibration of a string between two slits perpendicular to each other

7.9.2 Polarization of light wave

Consider an ordinary beam of light is passed through a tourmaline crystal as shown in Fig 7.20. If the Tourmaline crystal is rotated by taking the incident beam as an axis, there is no variation in the intensity of the emergent beam. The tourmaline crystal has the property of allowing the components of light that are vibrating parallel to its axis to pass through it, and it does not allow all the other components of light that are not parallel to the axis of the crystal. The emergent beam from the tourmaline crystal vibrates in only one direction, i.e., it vibrates parallel to the axis of the crystal. The light rays that are vibrating in only one direction are known as polarized light. The first tourmaline crystal that produces the polarized light is called the polarizer. Consider that the emergent beam from the first tourmaline crystal is passed through another tourmaline crystal kept perpendicular to the first crystal. No light is emerging from the second tourmaline crystal.

Fig7.20 Tourmaline crystal--polarized light

Consider the second tourmaline crystal is rotated by taking the incident beam as an axis. The intensity of the emergent beam from the second tourmaline crystal varies. The intensity of the emergent beam from the second tourmaline crystal has maximum intensity two times and minimum intensity two times within one complete rotation. When the second tourmaline crystal is parallel to the first one, the emergent beam from the second crystal is maximum. The intensity of the emergent beam from the second crystal is minimum when these two crystals are perpendicular of each other. The second tourmaline crystal is used to analyze the polarized light. Therefore, the second crystal is said to be an analyzer. This property of light ray confirms the transverse nature of light waves. There are three different types of polarized lights, as given below.

(i) Plane polarized light (ii) Circularly polarized light (iii) Elliptically polarized light

7.9.3 Pictorial representation of Light The plane polarized light, vibrating parallel to the plane of this paper, is represented as shown in Fig.7.21 (a). When the plane polarized light is vibrating perpendicular to the plane of this paper, it is represented as shown in Fig.7.21 (b). The unpolarized or ordinary light has vibration both parallel and perpendicular to the plane of the paper or in all directions, and it is represented as shown in Fig. 7.21(c).

(a) Plane polarized light—parallel to plane of this paper

(b) Plane polarized light---perpendicular to the plane of paper

(c) Unpolarized light

Fig 7.21 Representation of plane polarized light

7.9.3 Plane of vibration and plane of polarization

The plane in which the vibrations of polarized light takes place is known as the plane of vibration, and the plane perpendicular to the plane of vibration is known as the plane of polarization and is shown in Fig.7.22. The emergent beam vibrates parallel to the axis of the polarizer. The plane ABCD is parallel to the axis of the polarizer and containing vibrations of polarized light and hence, it is known as plane of vibration. The plane perpendicular to the plane of vibration is known as plane of polarization. The plane EFGH is perpendicular to the axis of the polarizer. It should be noted that no vibration takes place in this plane EFGH.

Fig 7.22 Plane of vibration and polarization

7.9.4 Geometry of Calcite crystal

Calcite is a transparent colorless crystal. It is chemically known as hydrated calcium carbonate (CaCO3). It can be reduced into a rhombohedron by cleavage or breakage. The Rhombohedron-shaped calcite crystal is shown in Fig. 7.23. It consists of six faces. These six faces are parallelograms having angles of and . Strictly speaking, the angles of the parallelograms are and the corners C and F consist of three obtuse angles and hence, these corners are said to be blunt corners. The other corners of the crystal consist of two acute and one obtuse angle.

Fig 7.23 Calcite crystal—Rhombohedron

7.9.6 Optic Axis A line joining any two blunt corners is known as optic axis of the crystal. An optic axis is not an axis, but it is a direction in the crystal. Since optic axis is a direction, any parallel direction (line) to the optic axis, is also known as optic axis. Certain crystals like calcite, quartz, tourmaline, etc., have only one optic axis. Hence, they are called uniaxial crystals. The crystals like borax, mica and selenite possess two optic axes and hence, they are called biaxial crystals.

7.9.7 Principal section and Principal planes

A plane containing the optic axis and perpendicular to two opposite faces of the crystal is known as the principal section of the crystal. The principal section of the calcite crystal is a parallelogram having angles of and The principal plane of the ordinary ray is the plane containing the optic axis and the ordinary ray. The plane consisting of the extraordinary ray and the optic axis the crystal is known as the principal plane of the extraordinary ray.

7.9.8 Double Refraction

When unpolarized light passes through a calcite, it is split into two plane polarized refracted lights. The one which obeys laws of refraction and has vibrations perpendicular to the principal section is known as ordinary ray. The other which does not obey the laws of refraction and has vibrations in the principal section is called extraordinary ray. This phenomenon is known as double refraction and was discovered

by Bartholinus. From Fig.7.24, the refractive indices of O and E rays for calcite crystal are given by = and

Since r1 r2,

Fig 7.24 Calcite crystal

Thus the velocity of light for O-ray inside the crystal will be less than the E-ray. It is clear that µo is same for all the angles of incidence while µe varies with angles of incidence. Hence, the O-ray travels with the same velocity and it is represented by a spherical wave front, whereas the E-ray travels with different velocity in different directions and it is given by ellipsoidal wave front. Both the rays travel with same velocities along optic axes.

7.10 Nicol Prism It is an optical device made from calcite and is used to produce and analyse plane polarized light. It was invented by William Nicol, a Scotch Physicist, in 1828, and is known Nicol Prism. It is based on the principle of double refraction exhibited by calcite. The Nicol prism is designed in such a way that it can transmit the extraordinary rays and cut off the ordinary rays. The emergent beam from a Nicol prism is a plane polarized light.

7.10.1 Construction The Nicol prism is constructed using a calcite crystal having a length three times as that of its breadth. Consider a calcite crystal AB’CDEFGH’ in which B’ and H’ are blunt corners. It is Rhombohedron shaped and is shown in Fig. 7.25. The plane B’DH’F is known as the principal section of the crystal since it is perpendicular to the planes ABCD and EFGH’, and it consists of the optic axis. The plane B’DH’F is a parallelogram and its angles are 71o and 108o. The angles B’DH’ and B’FH’ are 71o. The upper and lower faces of the crystal, AB’CD and EFGH’, are grounded in such a way that the angles B’DH’ and B’FH’ are reduced from 71o to 68o. Let ABCD and EFGH be the new upper and lower faces of the crystal. The new principal section is BDHF. The crystal is cut into two along the plane JHKB. The plane JH’KB’ is perpendicular to the principal section B’DH’F of the crystal. The two cut surfaces are highly polished into optical flatness and then cemented together by means of Canada balsam.

Fig 7.25 Calcite rhombohedron

The refractive indices of Nicol prism for sodium light of wavelength λ=5893 Ao are 1.55 (µCB), 1.65837(µo) and 1.48641 (µe), respectively, for Canada balsam layer, ordinary rays and extraordinary rays. The two end faces of the crystals are kept open, while the sides are blackened so as to absorb the totally internally reflected rays.

7.10.2 Action of a Nicol Prism

Consider a ray of light SM is made to b incident on the front face DB’ of the Nicol prism. The incident beam is split into two beams, namely, ordinary and extraordinary rays. The Canada balsam layer is a rarer medium for the ordinary ray. Therefore, the ordinary ray is traveling from a denser (calcite) to a rarer (Canada balsam) medium. Hence, it undergoes total internal reflection, if the angle of incidence made by the ordinary ray at the Canada balsam layer is greater than the critical angle. The critical angle for the ordinary ray for calcite to Canada balsam layer is

ϴc=sin

=69.2o

Fig 7.26 Nicol prism

In order to make the angle of incidence made by the ordinary ray at the calcite and Canada balsam interface more than the critical angle, the angle of the upper and bottom surfaces of the crystal is grounded from 71o to 68o and the length of the crystal is taken as

three times as that of its breadth. For the extraordinary rays, the Canada balsam layer is a denser medium. Therefore it travels from a rarer (calcite) to a denser (Canada balsam) medium and hence it travels in a straight line and emerges our from the Nicol prism. The emergent beam is a plane polarized light.

7.10.3 Plane Polarised light by Nicol prism

A Nicol is used to produce and analyse plane polarized light. Consider an unpolarized light is passed through a Nicol prism, P. It is split into ordinary and extraordinary rays. The ordinary ray is cut off when it is passing from calcite to Canada balsam layer. The extraordinary ray is passed through it. If it is passed through another Nicol prism, A which is kept parallel to the first one, the extraordinary ray is passed through it.

Fig 7.27 Nicol prism (a) production of plane polarised light (b) analysis of plane polarised light

If the second Nicol prism is rotated by taking the incident beam as an axis, the emergent beam from the second Nicol prism is extinguished two times and it has maximum intensity two times within one complete rotation. The first Nicol prism is known as polarizer and the second Nicol prism is known as analyser. Thus, a Nicol prism can be used as polarizer as well as analyser.

7.10.4 Limitations of Nicol Prism

The Nicol prism acts as a polarizer only for the light rays having slightly convergent and slightly divergent beam. If the incident beam is more divergent or more convergent, the Nicol prism does not act as a polariser. Let the incident beam be divergent and making a larger angle with the front face DB’. In this case, the extraordinary ray will become parallel to the optic axis of the crystal, and hence it travels along with the ordinary ray. It is also totally internally reflected back by the Canada balsam layer. Therefore, no light emerges from the Nicol prism. Let the incident beam be convergent and making a smaller angle with the face DB’. The ordinary ray makes an angle less than the critical angle at the Canada balsam layer.

Therefore, the ordinary ray is also transmitted with the extraordinary rays. This limits the angle of incidence for the crystal to 14o. The angle between the extreme incident rays is limited to 28o, so as to avoid the transmission of the ordinary ray with the extraordinary rays and to avoid the total internal reflection of the extraordinary rays. Hence, a Nicol prism cannot be used for highly convergent and divergent light beams.

Objective Questions

1. Two sources are said to be coherent if their emitted waves have same amplitude, same wave length and constant phase difference. 2. When a light wave is reflected at the glass-air interface, the change of phase of the reflected wave is equal to zero. 3. When a light wave is reflected at the air-glass interface, the change of phase of the reflected wave is equal to π. 4. In the presence of a plane parallel film, the path difference is equal to 2µtcosr. 5. If a light wave is refracted from air to denser medium, then the path difference is π or .

6. In a plane parallel film due to reflected light, the condition for maxima for the air film to appear bright is 2µtcosr = (2n-1) .

7. In a plane parallel film due to reflected light, the condition for minima for the air film to appear dark is 2µtcosr = nλ. 8. In Newton’s rings experiment, interference is due to light rays reflected from lower surface of lens and upper surface of glass plate. 9. In Newton’s rings experiment, the condition for bright fringes in case of reflected light is D α . 10. In Newton’s rings experiment, the condition for dark fringes in case of reflected light is D α . 11. Diffraction phenomenon is usually divided into two classes. 12. The screen and the source are at finite distance in Fresnel’s diffraction experiment. 13. In Fraunhofer’s diffraction, incident light waves have plane type of wave front. 14. In Fraunhofer diffraction, lenses are required. 15. In single slit experiment, if the red colour is replaced by blue, then the diffraction pattern becomes narrower and crowded together. 16. Diffraction grating has a large number of equidistant slits. 17. In a diffraction grating, the condition for principal maxima is (e+d) sinϴ= nλ. 18. The characteristics of grating spectra are, the spectral lines are almost straight and outer sharp, the spectral lines are more and more dispersed as order increases, it is situated symmetrically on both sides of zero order image. 19. In Fraunhofer diffraction, the wave front undergoing diffraction has to be plane.

20. The penetration of the waves into the regions of geometric shadow is called diffraction. 21. In single slit diffraction, the first diffraction minima are observed at an angle of 30o, when a light having a wave length of 500nm is used. The width of the slit is 10×10-5cm. 22. In a single slit experiment, if the slit width is reduced, the fringes become wider. 23. The diffraction pattern of a single slit consists of wider and brighter band at the center with alternate dark and bright bands of decreasing intensity on either side. 24. In diffraction due to double slit, wider diffraction bands and narrow interference fringes within that are observed. 25. A parallel beam of monochromatic light falls normally on a plane diffraction grating having 5000 lines/cm. a second order spectral line is diffracted through an angle of 30o. The wave length of light is 5×10-4 cm. 26. When white light is incident on a diffraction grating, the light diffracted more is red. 27. Maximum number of orders possible with a grating is directly proportional to the grating element. 28. Plane polarized light can be produced by Nicol’s prism, simple reflection and pile of plates. 29. Along the optic axis, O-ray and E-ray travel with same velocity. 30. The working of Nicol prism is based on the action of double refraction. 31. Polarization is the phenomenon which demonstrates the transverse nature of light. 32. Double refraction phenomenon causes the polarization of light. 33. When a polariser and an analyser are kept in a crossed position then no light comes out from analyser. 34. Polarisation of light conclusively proves that light waves are transverse. 35. On rotating the analyser, the intensity of emergent light varies between a maximum and minimum which is zero, and then the light incident on the analyser is plane polarized. 36. On rotating the analyser, the intensity of emergent light varies between a maximum and minimum which is not zero, and then the light incident on the analyser is partially plane polarized. 37. If on rotating the analyser, the intensity of emergent light does not vary then the light incident on the analyser is either unpolarized or circularly polarised. 38. If on rotating the analyser, the intensity of emergent light varies between a maximum and minimum but does not fall to zero, then the light incident on the analyser is elliptically polarized. 39. To analyse a plane polarized light, an analyser is rotated. In one complete rotation of the analyser, two times brightness and two times darkness occurs.

40. Two plane polarized beams of equal amplitude having a phase difference of π/2, are mutually perpendicular. The superposition of the two results in circularly polarized light. Review questions 1. What is interference? 2. What are the conditions to get interference? 3. Discuss the interference in a thin film by reflection. 4. Explain the formation of Newton’s rings with suitable theory. 5. Explain how the wave length of a light source can be found using Newton’s rings. 6. Explain the determination of refractive index of a liquid using Newton’s rings. 7. Derive the expressions for the diameters of bright and dark rings in Newton’s rings. 8. What is diffraction? 9. Distinguish between Fraunhofer and Fresnel’s diffraction. 10. Describe Fraunhofer diffraction due to single slit. 11. Describe Fraunhofer diffraction due to double slit. 12. Explain the importance of diffraction grating. 13. Explain the formation of grating spectrum. 14. Explain polarisation of light wave. 15. Mention the representation of polarised light. 16. Explain the terms: a. Plane of polarisation b. Plane of vibration c. Optic axis d. Principal section 17. Explain double refraction in calcite crystal. 18. Describe the construction and working of a Nicol’s prism. 19. Explain the theory of circular and elliptically polarized lights.

Solved Problems 1.A parallel beam of light of 6000A is incident on thin glass plate of refractive index 1.5 such that the angle of refraction into the plate is 50 . Find the least thickness of the glass plate which will appear dark by reflection.

Given data: wavelength of the given light =6000A =6000×10-8 cm

Refractive index of the glass plate =1.5 Angle of refraction r=50 2 t =n Solution: For least thickness, n=1 Thickness t=

t= =3.111 cm

=3.11× cm 2. In a Newton’s rings experiment, the diameter of the 5th ring is 0.30 cm and the diameter of the 15th ring is 0.62 cm. Find the diameter of the 25th ring. Given data: Diameter of the 5th ring =0.3cm Diameter of the 15th ring =0.62cm Solution: we want diameter of the 25th ring

We know that - =4 R (m-n)

- =4 R (15-5) (0.62)2-(0.3)2=4 R 0.3844-0.09=4 R

Similarly, and -

- =4 R (25-5)

-0.09=8 R Dividing Eq (ii) by Eq. (i)

=2×0.2944+0.09=0.6788 D25= =0.824cm Diameter of the 25th ring is 0.814 cm.

3. A convex lens on plane glass plate is exposed to a monochromatic light. The diameter of the 10th dark ring is 0.433 cm. Find the wavelength of the light use if the radius of curvature of the lens is 70 cm. Given data: Radius of curvature, R=70cm Diameter of the 10th ring =0.433cm Solution:

= =6.6786×10-5cm

4. A soap with refractive index of 1.33 and thickness of 500 is exposed to white light what wavelengths in the visible region are reflected? Given data: Refractive index of the soap film, =1.33 Thickness of the film, t=5000 A =5000*10-10m Solution: Condition for maximum reflection is 2 t (2n+1) , where n=0, 1, 2, etc.

Assume

Then = =

-10 When n=0, =26600×10 m(infrared region)

When n=1, = =8866.7×10-10m(IR)

n=2, = =3800×10-10m(ultraviolet region)

All wavelength reflected, at wavelength 5320A the visible regions are reflected.

5. What is the thickness of the thinnest film of 1.33 refractive index in which destructive interference of the yellow light (6000A ) of a normally incident beam in air can take place by reflection? Given data: Refractive index of the film, =1.33 Wavelength of yellow light =6000A =6000×10-8cm Angle of incidence i=0 , r=0 (reflection) For least thickness, n=1 Solution: 2 t =n t=

= =2.255×10-5cm

The thickness of the thinnest film =2.255×10-5cm. 6. In Newton’s rings experiment, the diameter of the ring was 0.35 cm and the diameter of the ring was 0.65 cm. If the wavelength of the light used is 6000 then find the radius of curvature of the Plano- convex lens. Given data: Diameter of the ring = 0.65 cm Diameter of the ring = 0.35 cm Wavelength of the light, λ = 6000 Ǻ

Solution: Radius of curvature of the Plano-convex lens is

 R=

= = × = 125cm

Radius of curvature of the Plano-convex lens is 125cm.

7. In a Newton’s rings experiment, the diameter of the ring changes from 1.45 cm to 1.25 cm. when a liquid is introduced between the lens and the glass plate. Find the refractive index of the liquid.

Given data: Diameter of the ring in air medium, = 1.45 cm

Diameter of the ring in a liquide, = 1.25 cm Solution: =

2 (D12) = = 1 for air

2 (D12) =

8. A plane transmission grating having 4250 lines per cm is illuminated with sodium light normally. In the second – order spectrum, the spectral lines are deviated by . What is the wavelength of the spectral line? Given data: Grating lines, N = 4250 lines/cm Deviated angle, = Order of spectrum ‘n’ = 2 Solution: = n (i) Each slit width = = = 2.353× cm

Then substituting the value of e in (i) 2.353× × sin = 2λ 2.353× × 0.5 = 2λ 1.1765 × = 2λ => λ = 0.5882 × cm = 5.882 × cm λ = 5882 . 9. A source of light having a wavelength of 600 nm is incident on a slit with a width of 1 m. Find the angular separation between the first – order minima and maxima of either side. Given data: Wavelength of the light, λ = 600nm = 600 × m Width of the slit, e + d = 1× m Order n = 1 Solution: = nλ => sin = => =

Angular separation is 10. A plane grating having 1052 lines per cm is illuminated with light having a wavelength of 5× cm at normal incidence. How many orders are visible in the grating spectra? Given data: No. of lines in the grating N = 10520 lines/cm Width of each slit, =

Wavelength of the light, λ = 5× cm Angle = Solution: = nλ Where = = = 9.506×

n= = 1.901

n= 2. Therefore, in this grating, two orders can be seen. 11. A grating has 6000lines/cm. find the angular separation between two wavelengths of 500nm and 510nm in the order. Given data: No. of lines in the grating, N = 6000 lines/cm =

6000× lines/m Order n = 3 Solution: We know that sin = nλ Sin = = nNλ

For first wavelength, = 500nm = 500 × m

Sin = 3×6000× ×500× Sin = 0.9

Similarly = 510nm = 510× m

Sin = 3×6000× ×510× = 0.918

= Angular separation between two wavelengths in Third order = - = - = . 12. Find the highest order that can be seen with a grating having 15000 lines/inches. The wavelength of the light used is 600 nm. Given data: Wave length of the given light λ = 600nm =600× m Grating lines N = = 15000 lines/inch

=15000/2.54 lines/cm = 5906 lines/cm = 5906 × lines/m Solution: We know that = =

= 2.822 The highest order that can be seen is 2.

Chapter-8 LASERS

8.1 Introduction LASER stands for light amplification by stimulated emission of radiation. It is different from conventional light (such as tube light or electric bulb), there is no coordination among different atoms emitting radiation. Laser is a device that emits light (electromagnetic radiation) through a process is called stimulated emission.

8.2 Spontaneous and stimulated emission In lasers, the interaction between matter and light is of three different types. They are: absorption, spontaneous emission and stimulates emission .Let E1 and E2 be ground and excited states of an atom. The dot represents an atom. Transition between these states involves absorption and emission of a photon of energy E2-E1=hν12. Where ‘h’ is Planck’s constant. (a) Absorption: As shown in fig8.1(a), if a photon of energy hν12(E2-E1) collides with an atom present in the ground state of energy E1 then the atom completely absorbs the incident photon and makes transition to excited state E2. (b) Spontaneous emission: As shown in fig8. 1. (b), an atom initially present in the excited state makes transition voluntarily on its own. Without any aid of external stimulus or an agency to the ground. State and emits a photon of energy h ν 12(=E2-E1).this is called spontaneous emission. These are incoherent. (c) Stimulated emission: As shown in fig8.1.(c), a photon having energy hν12(E2- E1)impinges on an atom present in the excited state and the atom is stimulated to make transition to the ground state and gives off a photon of energy hν12. The emitted photon is in phase with the incident photon. These are coherent. This type of emission is known as stimulated emission.

Fig.8.1 (a) Absorption ;(b) Spontaneous emission;(c) Stimulated emission

Differences between Spontaneous emission and stimulated emission of radiation

Spontaneous emission Stimulated emission 1. Polychromatic radiation 1. Monochromatic radiation 2. Less intensity 2. High intensity 3. Less directionality, more angular 3. High directionality, so less angular spread during propagation spread during propagation. 4. Spatially and temporally in 4. Specially and temporally coherent coherent radiation radiation. 5. Spontaneous emission takes place 5. Stimulated emission takes place when excited atoms make a when a photon of energy equal to transition to lower energy level h ν12 (=E2-E1) stimulates an excited voluntarily without any external atom to make transition to lower stimulation. energy level.

8.3 Characteristics of Laser Light

(i). Coherence: Coherence is one of the unique properties of laser light. It arises from the stimulated emission process. Since a common stimulus triggers the emission events

which provide the amplified light, the emitted photons are in step and have a definite phase relation to each other. This coherence is described in terms of temporal and spatial coherence. (ii). Monochromaticity: A laser beam is more or less in single wave length. I.e. the line width of laser beams is extremely narrow. The wavelengths spread of conventional light sources is usually 1 in 106, where as in case of laser light it will be 1 in 105.I.e. if the frequency of radiation is 1015 Hz., then the width of line will be 1 Hz. So, laser radiation is said to be highly monochromatic. The degree of non-monochromaticity has been expressed as ξ = (dλ/λ) =dν/ν, where dλ or dν is the variation in wavelength or variation in frequency of radiation. (iii) Directionality: Laser beam is highly directional because laser emits light only in one direction. It can travel very long distances without divergence. The directionality of a laser beam has been expressed in terms of divergence. Suppose r1 and r2 are the radii of laser beam at distances D1 and D2 from a laser, and then we have. Then the divergence, ∆θ= (r1 - r2)/ D2-D1 The divergence for a laser beam is 0.01 mille radian where as incase of search light it is 0.5 radian. (iv) High intensity: In a laser beam lot of energy is concentrated in a small region. This concentration of energy exists both spatially and spectrally, hence there is enormous intensity for laser beam. The power range of laser is about 10-13w for gas laser and is about 109 w for pulsed solid state laser and the diameter of the laser beam is about 1 mm. then the number of photons coming out from a laser per second per unit area is given by 2 -2 N l =P/ hνπr ≈1022to1034photons/m -sec By assuming h ν =10-19 Joule, Power P=10-3to109watt r=0.5×10-3meter Based on Planck’s black body radiation, the number of photons emitted per second per unit area by a body with temperature T is given by 4 ( ) 16 2 Nth= (2hπC/ λ ) (1/e h ν/k T -1) dλ≈10 photons/m .sec By assuming T=1000k, λ=6000A0 This comparison shows that laser is a highly intensive beam.

8.4. Einstein’s Coefficients LetN1 be the number of atoms per unit volume with energy E1 and N2 the number of atoms per unit volume with energy E2. Let ‘n’ be the number of photons per unit volume at frequency ν such that h ν=E2-E1. Then the energy density of interacting photons ρ (v) is given by ρ(ν) = n h ν →(1) When these photons interact with atoms, both upward (absorption) and downward (emission) transitions occur. At equilibrium these transition rates must be equal. Upward Transition Stimulated absorption rate depends on the number of atoms available in the lower energy state for absorption of these photons as well as the energy density of interacting radiation. I.e. stimulated absorption rate α N1 α ρ (ν) = B12N1 ρ (ν) → (2)

Where the constant of proportionality B12 is the Einstein coefficient of stimulated absorption

Downward transition Once the atoms are excited by stimulated absorption, they stay in the excited state for a short duration of time called the lifetime of the excited state. After their life time they move to their lower energy level spontaneous by emitting photons. This spontaneous emission rate depends on the number of atoms in the excited energy state. i.e., spontaneous emission rate α N2

=N2 A21 → (3) Where the constant of proportionality A21 is the Einstein coefficient of spontaneous emission. Stimulated emission Before excited atoms de excites to their lower energy states by spontaneous emission they may interact with photons resulting in stimulated emission of photons. Therefore stimulated emission rate depends on the number of atoms available in the excited state as well as energy density of interacting photons I.e. stimulated emission rate α N2 α ρ(ν) =N2ρ (ν) B21 → (4) Where the constant of proportionality B21 is the Einstein coefficient of stimulated emission. During stimulated emission, the interacting photon called the stimulating photon and the photon due to stimulated emission are in phase with each other. Please note that during stimulated absorption, the photon density decreases where as during stimulated emission it increases. For a system in equilibrium, the upward and down word transition rates must be equal and hence we have N1 ρ (ν) B12= N2ρ(ν)B21 +N 2A21 → (5)

Hence ρ(ν) =

ρ (ν) =

The population of various energy levels in thermal equilibrium is given by Boltzmann distribution law.

Fig.8.2 Boltzmann distribution for several energy levels

Ni=gi No exp (-E i/KT) Where Ni is the population density of the energy level E i, N0 is the population density of th the ground state at temperature T1, g i is the degeneracy of the i level and k is the Boltzmann constant (=1.38x10-23joule/k). The concept of degeneracy occurs since more than one level has the same energy. Hence N1= N o exp (-E 1/k T)

= exp → (7)

Substituting e q (6) in e q (4) ρ (ν) = → (8)

From Planck’s law of blackbody radiation, the radiation density is given by

ρ(ν) = → (9)

Comparing equations (8) and (9), we get

g1B12=g2B21 → (10)

And → (11)

Equation’s (10) and (11) are referred to as the Einstein relations. The ratio of spontaneous emission rate to the stimulated emission rate is given by

From equation (9)

R= -1] → (12)

In practice, the absorption and emission phase occur simultaneously. Let us calculate the rates of spontaneous emission and stimulating emission for a tungsten filament lamp

operating at a temperature of 2000k. Taking the average frequency to be 5x1014Hz, this ratio is

R= 1 → (13) This confirms that under conditions of thermal equilibrium, even for sources operating at higher temperatures and lower frequencies, spontaneous emission predominates. From equation (12), we understand that to make R smaller ρ(ν) the energy density of interacting radiation has to be made larger. Let us consider the relation of stimulated emission rate to stimulated absorption rate Thus at thermal equilibrium stimulated absorption predominates over stimulated emission. Instead if we create a situation that N2>N1. Stimulated emission will predominate over stimulated absorption. If stimulated emission predominates the photon density increases and light amplifies the photon density increases and light amplification by stimulated emission of radiation (LASER) occurs. Therefore, in order to achieve more stimulated emission, population of the excited state (N2) should be made larger than the population of the lower state (N1) and this condition is called population inversion. Hence if we wish to amplify a beam of light by stimulated emission, then we must create population inversion and increase the energy density of interacting radiation.

8.4.1 Population inversion Usually in a system the number of atoms (N1) present in the ground state (E1) is larger than the number of atoms (N2) present in the higher energy state. The process of making N2>N1 called population inversion. Conditions for population inversion are: a) The system should posses at least a pair of energy levels (E2>E1), separated by an energy of equal to the energy of a photon (hν). b) There should be a continuous supply of energy to the system such that the atoms must be raised continuously to the excited state.

Population inversion can be achieved by a number of ways. Some of them are (i) optical pumping (ii) electrical discharge (iii) inelastic collision of atoms (iv) chemical reaction and (v) direct conversion

8.5 Helium-Neon gas laser Helium-Neon gas laser is a continuous four level gas laser. It consists of a long, narrow cylindrical tube made up of fused quartz. The diameter of the tube will vary from 2 to 8 mm and length will vary from 10 to 100 cm. The tube is filled with helium and neon gases in the ratio of 10:1. The partial pressure of helium gas is 1mm of Hg and neon gas is 0.1mm of Hg so that the pressure of the mixture of gases inside the tube is nearly 1 mm of Hg. Laser action is due to the neon atoms. Helium is used for selective pumping of neon atoms to upper energy levels. Two electrodes are fixed near the ends of the tube to pass electric discharge through the gas. Two optically plane mirrors are fixed at the two ends of the tube at Brewster angle normal to its axis. One of the mirrors is fully silvered so that nearly 100% reflection takes place and the other is partially silvered so that 1%of the light incident on it will be transmitted. Optical resources column is formed between these mirrors.

. Fig.8.3 Helium-Neon gas laser Working When a discharge is passed through the gaseous mixture, electrons are accelerated down the tube. These accelerated electrons collide with the helium atoms and excite them to higher energy levels. The different energy levels of Helium atoms and Neon atoms is shown in fig.8.3 the helium atoms are excited to the levels F2 and F3 these levels happen to be meta stable energy states. Energy levels and hence Helium atoms exited levels spend sufficiently large amount of time before getting de excited. As shown in the fig 8.5(a), some of the excited states of neon can correspond approximately to the same energy of excited levels F2 and F3. Thus, when Helium atoms in level F2 and F3 collide with Neon atoms in the ground level E1, an energy exchange takes place. This results in the excitation of Neon atoms to the levels E4 and E6 and de excitation of Helium atoms to the ground level (F1). Because of long life times of the atoms in levels F2 and F3, this process of energy transfer has a high probability. Thus the discharge through the gas mixture continuously populates the neon atoms in the excited energy levels E4 and E6. This helps to create a state of population inversion between the levels E4 (E6) to the lower energy level (E3 and E5). The various transitions E6→E5, E4→E3, E6→E3 leads to the emission of wave lengths 3.39mm, 1.15 um and 6328 A0. Specific frequency selection may be obtained by employing mirrors The excited Neon atoms drop down from the level E3 to the E2 by spontaneously emitting a photon around wavelength 6000A0. The pressures of the two gases in the mixture are so chosen that there is an effective transfer of energy from the Helium to the Neon atoms. Since the level E2 is a meta stable state, there is a finite probability of the excitation of Neon, atoms from E2 to E3 leading to population inversion, when a narrow tube is used, the neon atoms in the level E2 collide with the walls of the tube and get excited to the level E1. The transition from E5 to E3 may be non radioactive. The typical power outputs of He-Ne laser lie between 1 and 50 mw of continuous wave for inputs of 5-10W.

Fig.8.4. Energy level diagram of He-Ne atoms.

8.6 Ruby Laser

Ruby Laser is a solid state pulsed, three level lasers. It consists of a cylindrical shaped ruby crystal rod of length varying from 2 to 20cms and diameter varying 0.1 to 2cms. This end faces of the rod are highly flat and parallel. One of the faces is highly silvered and the other face is partially silvered so that it transmits 10 to 25% of incident light and reflects the rest so as to make the rod-resonant cavity. Basically, ruby crystal is aluminum oxide [Al 2O3] doped with 0.05 to 0.5% of chromium atom. These chromium atoms serve as activators. Due to presence of 0.05% of chromium, the ruby crystal appears in pink color. The ruby crystal is placed along the axis of a helical xenon or krypton flash lamp of high intensity.

Fig.8.5 Ruby laser

Fig.8.6 Energy level diagram of chromium ions in a ruby crystal Construction: +3 Ruby (Al2O3+Cr2O3) is a crystal of Aluminum oxide in which some of Al ions are replaced by Cr +3 ions. When the doping concentration of Cr+3 is about 0.05%, the color of the rod becomes pink. The active medium in ruby rod is Cr+3ions. In ruby laser a rod of 4cm long and 5mm diameter is used and the ends of the rod are highly polished. Both ends are silvered such that one end is fully reflecting and the other end is partially reflecting. The ruby rod is surrounded by helical xenon flash lamp tube which provides the optical pumping to raise the Chromium ions to upper energy level (rather energy band). The xenon flash lamp tube which emits intense pulses lasts only few milliseconds and the tube consumes several thousands of joules of energy. Only a part of this energy is used in pumping Chromium ions while the rest goes as heat to the apparatus which should be cooled with cooling arrangements as shown in fig.8.5. The energy level diagram of ruby laser is shown in fig.8.6

Working: Ruby crystal is made up of aluminum oxide as host lattice with small percentage of Chromium ions replacing aluminum ions in the crystal chromium acts as do pant. A do pant actually produces lasing action while the host material sustains this action. The pumping source for ruby material is xenon flash lamp which will be operated by some external power supply. Chromium ions will respond to this flash light having 0 +3 wavelength of 5600A . When the Cr ions are excited to energy level E3 from E1 the population in E3 increases. Chromium ions stay here for a very short time of the order of 10-8 seconds then they drop to the level E2 which is mat stable state of life time -3 10 s. Here the level E3 is rather a band, which helps the pumping to be more effective. The transitions from E3 to E2 are non-radioactive in nature. During this process heat is given to crystal lattice. Hence cooling the rod is an essential feature in this method. The life time in mete stable state is 10 5times greater than the lifetime in E3. As the life of the state E2 is much longer, the number of ions in this state goes on increasing while ions. In this state goes on increasing while in the ground state (E1) goes on decreasing. By this process population inversion is achieved between the exited Meta stable state E2 and the ground state E1. When an excited ion passes spontaneously from the metastable state E2 to the ground state E1, it emits a photon of wave length 6943A0. This photon travels through the rod and if it is moving parallel

to the axis of the crystal, is reflected back and forth by the silvered ends until it stimulates an excited ion in E2 and causes it to emit fresh photon in phase with the earlier photon. This stimulated transition triggers the laser transition. This process is repeated again and again because the photons repeatedly move along the crystal being reflected from its ends. The photons thus get multiplied. When the photon beam becomes sufficiently intense, such that part of it emerges through the partially silvered end of the crystal.

Drawbacks of ruby laser: 1. The laser requires high pumping power to achieve population inversion. 2. It is a pulsed laser.

. Fig.8.7 the output pulses with time.

8.7 Semi conductor laser The semiconductor laser is also called diode laser. Among the different semiconductors there are direct band gap semiconductors and indirect band gap semiconductors. In the case of direct band gap semiconductors, there is a large possibility for direct recombination of holes and electrons which emit photons. But in indirect band gap semi conductors like silicon and germanium, direct recombination of holes and electrons is less possible and hence there is no effective emission of photons. A well know example of a direct band gap semiconductor is GaAs. Let E g be the energy gap of a material, then it emits a photon of wavelength (λ) is given the relation λ = → (1)

Where C is the velocity of light and h is the Planck’s constant The equation (1) is reduced to λ= µm → (2)

Where Eg is expressed in eV As Eg increases, it emits shorter wavelengths. Diode lasers are always operated in forward bias. If p and n type materials are prepared from the same material then the p-n junction, is called as Homo junction semiconductor laser source. If p and n type materials are prepared from different materials then they are called as Hetero junction semi conductor laser source.

Construction of GaAs Semiconductor diode laser

The basic mechanism responsible for light emission from a semi conductor is the recombination of electrons and holes at p-n junction when a current is passed through the diode. The active medium is a p-n junction diode made from crystalline Gallium Arsenide (fig1). The p-region and n-region in the diode are obtained by heavily doping with suitable do pants. At the junction the sides through which emitted light is coming out are well polished and parallel to each other. Since the refractive index of GaAs is high, the reflectance at the material air interface is sufficiently large so that the external mirrors are not necessary to produce multiple reflections. When a current is passed through a p-n junction p region being positively biased, holes are injected from p-region into n-region and n-region being negatively biased electrons are injected from n-region into p-region and is shown in fig8.9.

Fig.8.8 GaAs semiconductor diode laser

The connections in this p-n junction circuit are called forward bias. The electrons and holes recombine and release of light energy takes place in or near the junction. In the case of GaAs homo junction which has an energy gap of 1.44ev gives a laser beam of wavelength around 8600A0 the electron-hole recombination takes place in the active region of the device. When the junction is forward biased, a large amount of the order of 104 amp/cm2 is passed through the narrow junction. Thus the electrons (holes) are injected from n side to p side (p side to n side) respectively.

Fig8.9 (a) Band diagram of heavily doped p-n junction in equilibrium. (b) Band diagram of heavily doped p-n junction with forward bias

The electrons are minority charge carriers in p-side and holes are the minority charge carriers in n-side. The continuous injection of charge carriers creates the population inversion of minority carriers is n and p side respectively. The excess minority charge carries diffuse away from the region recombining with majority carriers of the n and p

type material, resulting in the release of photons. Further, the emitted photons increase the recombination of injected electrons from the n-region and holes in p- region by inducing more recombinations thus the stimulated emission takes place more effectively. The efficiency of a laser emission increases when the device is cooled.

The drawbacks of homojunction lasers 1. Only output is obtained 2. The threshold current density is very large(400amps/mm2) 3. Electromagnetic confinement is poor. 4. Output has more beam divergence, poor coherence. To overcome these deficiencies, scientists have developed heterojunction laser sources like GaAs/GaAlAs structures. The advantages of heterojunction laser structures are: 1. Low threshold current density(5-10amp/mm2) 2. The output is continuous 3. Carrier confinement is more effective there by less beam divergence. 4. High output power. 5. Narrow beam, high coherence, high mono chromaticity is achieved. 6. Long lifetime of the device 7. Very stable. Hence hetero junction laser diode used extensively in optical fiber communication

8.8. Applications of Lasers Lasers find applications in various fields. They are described below.

a) In Communications : Lasers are used in optical fiber communications. In optical fiber communications, lasers are used as light source to transmit audio, video signals and data to long distances without attention and distortion.

b) The narrow angular spread of laser beam can be used for communication between earth and moon or to satellites. c) As laser radiation is not absorbed by water, so laser beam can be used in under water (inside sea) communication networks.

2. Industrial Applications a) Lasers are used in metal cutting, welding, surface treatment and hole drilling. Using lasers cutting can be obtained to any desired shape and the curved surface is very smooth. b) Welding has been carried by using laser beam. c) Dissimilar metals can be welded and micro welding is done with great case. d) Lasers beam is used in selective heat treatment for tempering the desired parts in automobile industry e) Lasers are widely used in electronic industry in trimming the components of ICs

3. Medical Applications 1. Lasers are used in medicine to improve precision work like surgery. Brain surgery is an example of precision surgery Birthmarks, warts and discoloring of the skin can easily be removed with an unfocussed laser. The operations are quick and heal quickly and, best of all, they are less painful than ordinary surgery performed with a scalpel. 2. Cosmetic surgery (removing tattoos, scars, stretch marks, sun spots, wrinkles, birthmarks and hairs) see lasers hair removal. 3.Laser types used in dermatology include ruby(694nm),alexandrite(755nm),pulsed diode array(810nm), Nd:YAG(1064nm), HO:YAG(2090nm), and Er:YAG(2940nm) 4. Eye surgery and refracting surgery. 5. Soft tissue surgery: Co2 Er :YAG laser. 6. Laser scalpel (general surgery, gynecological, urology, laparoscopic). 7. Dental procedures. 8. Photo bio modulation (i.e. laser therapy) 9. “No-touch” removal of tumors, especially of the brain and spinal cord. 10. In dentistry for caries removal, endodontic/periodontic, procedures, tooth whitening, and oral surgery.

4. Military Applications The various military applications are: a) Death rays: By focusing high energetic laser beam for few seconds to aircraft, missile, etc can be destroyed. So, these rays are called death rays or war weapons. b) Laser gun: The vital part of energy body can be evaporated at short range by focusing highly convergent beam from a laser gun. c) LIDAR (Light detecting and ranging): In place of RADAR, we can use LIDAR to estimate the size and shape of distant objects or war weapons. The differences between RADAR and LIDAR are that, in case of RADAR, Radio waves are used where as incase of LIDAR light is used.

5. In Computers: By using lasers a large amount of information or data can be stored in CD-ROM or their storage capacity can be increased. Lasers are also used in computer printers. 6. In Thermonuclear fusion: To initiate nuclear fusion reaction, very high temperature and pressure is required. This can be created by concentrating large amount of laser energy in a small volume. In the fusion of deuterium and tritium, irradiation with a high energy laser beam pulse of 1 nano second duration develops a temperature of 1017 0c, this temperature is sufficient to initiate nuclear fusion reaction.

7. In Scientific Research: In scientific, lasers are used in many ways including a) A wide variety of interferometrie techniques. b) Raman spectroscopy. c) Laser induced breakdown spectroscopy. d) Atmospheric remote sensing. e) Investigating non linear optics phenomena

f) Holographic techniques employing lasers also contribute to a number of measurement techniques. g) Laser (LADAR) technology has application in geology, seismology, remote sensing and atmospheric physics. h) Lasers have been used abroad spacecraft such as in the cassini-huygens mission. i) In astronomy lasers have been used to create artificial laser guide stars, used as reference objects for adaptive optics telescope.

Multiple Choice Questions 1. Laser action is found in ______semiconductor ( ) (a) Direct band gap (b) indirect band gap (c) Germanium (d) silicon 2.In computer ______laser is used ( ) (a) He-Ne gas (b) ruby (c) Semi conductor (d) CO2 3. Laser radiation is ______( ) (a) Highly monochromatic (b) partially monochromatic (c) White light (d) none 4. Under population inversion, the number of atoms in the higher energy state is _____ Than in the lower energy state ( ) (a) Laser (b) larger (c) both a and b (d) none 5. Laser radiation is ____ ( ) (a) Highly directional (b) monochromatic (c) Coherent and stimulated (d) all 6. In conventional light sources ( ) (a) Different atoms emit radiation at different times (b) There is no phase relation between the emitted photons (c) Different atoms emit photons in different directions (d) all

7. In laser source ( ) (a) Photons emitted by different atoms are in phase or maintain constant phase relationship (b) Different atoms emit photons in the same direction (c) Both a and b (d) none 8. In spontaneous emissions ( ) (a) Atoms are initially in the excited state (b) Transitions are without any aid of an external agency (c) Both a and b (d) none 9. In conventional light sources, the ratio of spontaneous emission rate to stimulated emission rate is nearly ( ) (a)1010 (b)1020 (c)105 (d)103 10. In excited states, the atoms will remain for a time of ( )

(a)108 s (b)10 -8 s (c) 10-3 s (d)10-5 s 11. He-Ne gas laser is a______( ) (a) Continuous laser (b) pulsed laser (c) Both a and b (d) none 12. The ratio of the helium and neon gaseous atoms are ( ) (a) 1:10 (b)10:1 (c)1:1 (d)1:20 13. Ruby laser is a solid state: ( ) (a) Pulsed, three level laser (b) pulsed, four level laser (c) Continuous, three level laser (d) continuous four level lasers 14. if the ruby rod contains 0.005% of chromium atoms then it appears in ___ color ( ) (a) Read (b) yellow (c) pink (d) green 15. At room temperatures, the threshold current density in hetero structure is of the order of _____ A/m2 ( ) (a) 105 (b) 10 3 (c) 102 (d) 104

16. In hetero structure strip geometry semiconductor lasers, the strip width will vary from ( )

(a) 5 to 30 µm (b) 50 to 100 µm (c) 5 to 150 µm (d) 1 to 5 µm

ANSWERS 1.a 2.c 3.a 4.b 5.d 6.d 7.c 8.c 9.a 10.b 11.a 12.b 13.a 14.c 15.b 16.a

SOLVED PROBLEMS 1. A semi conductor laser diode laser has a peak emission wavelength of 1.55 µm. find its energy gap in e V. (2009) Solution. Energy gap of a semiconductor,

Energy of emitted photon, Eg= hν =

Where C= velocity of light=3×108m/s Wavelength, λ=1.55 µm=1.55×10-6m Energy gap, Eg=?

2. Calculate the wavelength of emitted radiation from GaAs which has a band gap of 1.44eV (2010) Solution: Energy gap of a semiconductor, Eg = hν

Where h is Planck’s constant=6.63×10-34J.S

Eg= hν= or λ=

-19 Eg=1.44eV=1.44×1.6×10 J

-10 0 Eg= = =8633×10 m=8633A

3. Calculate the ratio of the stimulation rate to spontaneous emission rate for an in candescent lamp operating at a temperature of 1100k. Assume average operating length λ=0.55 µm. (Model problem) Solution: We know that C= νλ

ν = =

= 5.455×1014Hz. -11 4.5×10

Where h=6.63×10-34J.S, ν=5.455×1014Hz KB=1.38×10-23J/K, T=1100K.

FIBER OPTICS 8.9 Introduction 1. An optical fiber (or fiber) is a glass or plastic fiber that carries light along its length. 2. Fiber optics is the overlap of applied science and engineering concerned with the design and application of optical fibers.

3. Optical fibers are widely used in fiber-optic communications, which permits transmission over long distances and at higher band widths (data rates) than other forms of communications. 4. Specially designed fibers are used for a variety of other applications, including sensors and fiber lasers. Fiber optics, though used extensively in the modern world, is a fairly simple and old technology. 8.10 Principle of Optical Fiber Optical fiber is a cylinder of transparent dielectric medium and designed to guide visible and infrared light over long distances. Optical fibers work on the principle of total internal reflection. Optical fiber is very thin and flexible medium having a cylindrical shape consisting of three sections 1) The core material 2) The cladding material 3) The outer jacket

The structure of an optical is shown in figure.8.10. The fiber has a core surrounded by a cladding material whose reflective index is slightly less than that of the core material to satisfy the condition for total internal reflection. To protect the fiber material and also to give mechanical support there is a protective cover called outer jacket. In order to avoid damages there will be some cushion between cladding protective cover.

Fig.8.10.Structure of an optical fiber

When a ray of light passes from an optically denser medium into an optically rarer medium the refracted ray bends away from the normal. When the angle of incidence is increased angle of refraction also increases and a stage is reached when the refracted ray just grazes the surface of separation of core and cladding. At this position the angle of refraction is 90 degrees. This angle of incidence in the denser medium is called the critical angle (θc) of the denser medium with respect to the rarer medium and is shown in the fig2. If the angle of incidence is further increased then the totally reflected. This is called total internal reflection. Let the reflective indices of core and cladding materials be n1 and n2 respectively.

Fig.8.11 Total internal reflection.

When a light ray, travelling from an optically denser medium into an optically rarer medium is incident at angle greater than the critical angle for the two media. The ray is totally reflected back into the medium by obeying the loss of reflection. This phenomenon is known as totally internal reflection. According to law of refraction, n1 sinθ1= n 2 sinθ2 Here θ1= θc, θ2=90 n 1 sin θc =n2 sin 90

Sinθc =

θc = sinˉ¹( ) →(1)

Equation (1) is the expression for condition for total internal reflection. In case of total internal reflection, there is absolutely no absorption of light energy at the reflecting surface. Since the entire incident light energy is returned along the reflected light it is called total internal reflection. As there is no loss of light energy during reflection, hence optical fibers are designed to guide light wave over very long distances.

8.11 Acceptance Angle and Acceptance Cone Acceptance angle: It is the angle at which we have to launch the beam at its end to enable the entire light to propagate through the core. Fig.8.12 shows longitudinal cross section of the launch of a fiber with a ray entering it. The light is entered from a medium of refractive index n0 (for air n0=1) into the core of refractive index n1. The ray (OA) enters with an angle of incidence to the fiber end face i.e. the incident ray makes angle with the fiber axis which is nothing but the normal to the end face of the core. Let a right ray OA enters the fiber at an angle to the axis of the fiber. The end at which light enter the fiber is called the launching pad.

Fig.8.12 Path of atypical light ray launched into fiber.

Let the refractive index of the core be n1 and the refractive index of cladding be n2. Here n1>n2. The light ray reflects at an angle and strikes the core cladding interface at angle θ. If the angle θ is greater than its critical angle θc, the light ray undergoes total internal reflection at the interface. According to Snell’s law n0sinαi=n1sinαr → (2) From the right angled triangle ABC 0 α r+θ=90

0 αr=90 –θ → (3) Substituting (3) in (2), we get 0 n0sinαi =n1sin (90 –θ) = n1cos θ

sinαi=( ) cos θ →(4)

When θ= θc, αi= αm=maximum α value

sinαm=( ) cos θc →(5)

From equation (1) Sinθc =

cos θc= = = →(6)

Substitute equation (6) in equation (5)

sinαm=( ) = →(7)

If the medium surrounding fiber is air, then n0=1

sinαm= →(8)

This maximum angle is called the acceptance angle or the acceptance cone half angle of the fiber. The acceptance angle may be defined as the maximum angle that a light ray can have with the axis of the fiber and propagate through the fiber. Rotating the acceptance angle about the fiber axis (fig .8.13) describes the acceptance cone of the fiber. Light 203

launched at the fiber end within this acceptance cone alone will be accepted and propagated to the other end of the fiber by total internal reflection. Larger acceptance angles make launching easier. Light gathering capacity of the fiber is expressed in terms of maximum acceptance angle and is termed as “Numerical Aperture”.

Fig.8.13 Acceptance cone 8.12 Numerical Aperture Numerical Aperture of a fiber is measure of its light gathering power. The numerical aperture (NA) is defined as the sign of the maximum acceptance angle.

Numerical aperture (NA) = sinαm= →(9)

→ (10)

Where ∆= called as fractional differences in refractive indices and are the

refractive indices of core and cladding material respectively. As n1≈ n2, we can take n1+ n2= 2n1 2 1/2 1/2 Then numerical aperture= (2n1 ∆) = n1 (2∆) → (11) Numerical aperture is a measure of amount of light that can be accepted by a fiber. From equation (9) it is seen that numerical aperture depends only on the refractive indices of core and cladding materials and it is independent on the fiber dimensions. Its value ranges from 0.1 to 0.5. A large NA means that the fiber will accept large amount of light from the source. Proof: We shall show that the light emitted by a small defuse source situated on the fiber axis near to one end face, only a fractionNA2 can be collected by the fiber and propagated along the fiber. Consider a small defuse light source such as the isotropic radiator in which the power radiated per unit solid angle in a direction θ to the normal to the surface

is given by Fig8.12. Diffuse light source I(θ) =I0Cos θ→ (12)

The total power φ0 emitted such source is obtained by integrating I (θ) over all forward directions

d θ → (13) Where dΩ=small element of solid angle=2πSinθ dθ → (14)

= 2πSinθ d θ

= =πI0 → (15) But the power from such source that can be collected by an adjacent fiber whose core diameter is greater than the diameter of the source is given by φ, where

Φ= 2 Sinθ dθ = 2 2 2 (Φ /φ0)= (sinαm) = (NA) =2n1 ∆ → (16) Where ∆=

Equation (16) represents the light gathering power of the fiber. In order to collect as much light as possible it is necessary to make n1 and ∆ large. In some cases it is customary to use ∆n which can be defined as ∆n = n1- n2 But ∆= =

2 2 (Φ /φ0)= (sinαm) =(NA) =2n1∆n→(17) NA= (2n1∆n) → (18)

8.13. Step index fibers and graded index fiber -transmission of signals in them: Based on the variation of refractive index of core, optical fibers are divided into: (1) step index and (2) graded index fibers. Again based on the mode of propagation, all these fibers are divided into: (1) single mode and (2) multimode fibers. In all optical fibers, the refractive index of cladding material is uniform. Now, we will see the construction, refractive index of core and cladding with radial distance of fiber, ray propagation and applications of above optical fibers. i. Step index fiber: The refractive index is uniform throughout the core of this fiber. As we go radially in this fiber, the refractive index undergoes a step change at the core-cladding interface. Based on the mode of propagation of light rays, step index fibers are of 2 types: a) single mode step index fiber & b) multimode step index fibers. Mode means, the number of paths available for light propagation of fiber. We describes the different types of fiber below a) Single mode step index fiber: The core diameter of this fiber is about 8 to 10µm and outer diameter of cladding is 60 to 70 µm. There is only one path for ray propagation. So, it is called single mode fiber. The cross sectional view, refractive index profile and ray propagation are shown in fig.8.13(i). In this fiber, the transmission of light is by successive total internal reflections i.e. it is a reflective type fiber. Nearly 80% of the fibers

manufactured today in the world are single mode fibers. So, they are extensively used.

Fig8.13 (i).Single mode step index fiber ;( a) Cross sectional view and refractive index profile ;( b) Ray propagation Multimode step index fiber: The construction of multimode step index fiber is similar to single mode step index fiber except that its core and cladding diameters are much larger to have many paths for light propagation. The core diameter of this fiber varies from 50 to 200 µm and the outer diameter of cladding varies from 100 to 250 µm. The cross-sectional view, refractive index profile and ray propagations are shown in fig 2. Light propagation in this fiber is by multiple total internal reflections i.e it is a reflective type fiber. b) Transmission of signal in step index fiber: Generally the signal is transmitted through the fiber in digital form i.e. in the form of 1’s and 0’s. The propagation of pulses through the multimode fiber is shown in fig8.13 (i)(b). The pulse which travels along path 1(straight) will reach first at the other end of fiber. Next the pulse that travels along with path 2(zig-zag) reaches the other end. Hence, the pulsed signal received at the other end is broadened. This is known as intermodal dispersion. This imposes limitation on the separation between pulses and reduces the transmission rate and capacity. To overcome this problem, graded index fibers are used.

2) Graded index fiber: In this fiber, the refractive index decreases continuously from center radially to the surface of the core. The refractive index is maximum at the center and minimum at the surface of core. This fiber can be single mode or multimode fiber. The cross sectional view, refractive index profile and ray propagation of multimode

graded index fiber are shown in fig8.13.(ii)(a). The diameter of core varies from 50 to 200µm and outer diameter of cladding varies from 100 to 250 µm. The refractive index profile is circularly symmetric. As refractive index changes continuously radially in core, light rays suffer continuous refraction in core. The propagation of light ray is not due to total internal reflection but by refraction as shown in fig.8.13 (ii). (b). in graded index fiber, light rays travel at different speed in different paths of the fiber. Near the surface of the core, the refractive index is lower, so rays near the outer surface travel faster than the rays travel at the center. Because of this, all the rays arrive at the receiving end of the fiber approximately at the same time. This fiber is costly. . Transmission of signal graded index fiber: In multimode graded index fiber, large number of paths is available for light ray propagation. To discuss about inter modal dispersion, we consider ray path 1 along the axis of fiber.

Fig.8.13(ii).Multimode step index fibre(a)Cross sectional view and refractive index profle(b)Ray propagation As shown in fig.8.13. (ii)(b) and another ray path 2. Along with the axis of fiber, the refractive index of core is maximum, so the speed of ray along path 1 is less. Path 2 is sinusoidal and it is longer, along this path refractive index varies. The ray mostly travels in low refractive region, so the ray 2 moves slightly faster. Hence, the pulses of signals

that travel along path 1 and path 2 reach other end of fiber simultaneously. Thus, the problem of intermodal dispersion can be reduced to a large extent using graded index fibers

8.13.1. Differences between step index fiber and graded index fibers: Step index fiber Graded index fiber

1.The refractive index of core 1.The refractive index of core is non Is uniform and step or abrupt uniform, the refractive index of core Change in refractive index decreases Para- Takes place at the interface of core Felicity from the axis of the fiber to its and cladding in step index fibers surface.

2. The light rays propagate in zigzag 2. Light rays propagate in the form of skew manner inside the core. The rays travel in rays or helical rays. They will not cross the the fiber as meridional rays they cross the fiber axis fiber axis for every reflection.

8.13.2. Difference between single mode fibers and multi mode fibers: Single mode fiber Multimode fiber 1. In single mode fiber there is only 1. In multimode fiber, large number of one path for ray propagation paths is available for light ray propagation. 2. A single mode step index fiber has 2. Multi mode fibers, large number of less core diameter (< 10 µm) and paths are available for light ray the difference between the propagation. reflective indices of core and cladding is very small.

3. In single mode fibers, there is no 3. There is signal distortion and dispersion. dispersion takes place in multimode fibers.

4. The band width is about 50 MHz 4. The band width of the fiber lies in for multimode step index fiber between 200 MHz km to 600 MHz where as it is more than 1000 MHz km even though theoretically it has km in case of single mode step an infinite bandwidth. index fiber. 5. NA of multimode step index fiber is 5. NA of graded index fibers is less. more where as in single mode step index fibers, it is very less. 6. Launching of light into single mode 6. Launching of light into multimode

fibers is difficult. fibers is easy. 7. Fabrication cost is very high. 7. Fabrication cost is less

8.14. Advantage of optical fibers in communication: Let us see the advantages of optical fiber communication over conventional communication system. 1. Enormous bandwidths:

The information carrying capacity of a transmission system is directly proportional to the carrier frequency of the transmitted signals. The optimal carrier frequency is in the range of10 ¹⁴ Hz while the radio frequency is about 10⁶ Hz and microwave frequency is about 10¹⁰ Hz. Thus the optical fibers have enormous transmission bandwidths and high data rate. 2. Low transmission loss:

Due to the usage of ultra low loss fibers and the erbium doped silica fibers as optical amplifiers, one can achieve almost loss less transmission. Hence for long distance communication fibers of 0.002db/km are used. Thus the repeater spacing is more than 100km. 3. Immunity to cross talk:

Since optical fibers are dielectric wave guides, they are free from any electromagnetic interference (EMI) and radio frequency interference (RFI). Since optical interference among different fibers is not possible, cross talk is negligible even many fibers are cabled together. 4. Electrical isolation: Optical fibers are made from silica which is an electrical insulator. Therefore they do not pick up any electromagnetic wave or any high current lightening. It is also suitable in explosive environment. 5. The size and weight: The size of the fiber ranges from 10µm to 50µm which is very very small. The space occupied by the fiber cable is negligibly small compared to conventional electrical cables. Optical fibers are light in weight. These advantages make them to use in air craft’s and satellites more effectively. 6. Signal security: The transmitted signal through the fiber does not radiate. Unlike in copper cables, a transmitted signal cannot be drawn from a fiber without tampering it. Thus, the optical fiber communication provides 100% signals security. 7. Ruggedness and flexibility:

The fiber cable can be easily bend or twisted without damaging it. Further the fiber cables are superior than the copper cables in terms of handling, installation, storage, transportation, maintenance, strength and durability. 8. Low cost and availability: Since the fibers are made of silica which is available in abundance. Hence, there is no shortage of material and optical fibers offer the potential for low cost communication. 9. Reliability: The optical fibers are made from silicon glass which does not undergo any chemical reaction or corrosion. Its quality is not affected by external radiation. Further due to its negligible attenuation and dispersion, optical fiber communication has high reliability. All the above factors also tend high to reduce the expenditure on its maintenance.

Application of fibers in medical field: Fiber scopes are employed widely in endoscopic applications. Bundles of optical fibers form part of gastro scopes and other medical instruments. These instruments enable physicians to view internal body parts without performing surgery. Fiber optic endoscopes are classified according to the application. Gastro scope to examine the stomach, bronchoscope to view the upper passages of lungs and orthoscope to study the small spaces with in joints. A gastro scope consists of a long, flexible rubber tube and a rigid metal section that has a lens and various controls. Two bundles of long, thin fiber glass threads called optical fibers extend along the inside of gastro scope. A gastro scope can be fitted with various special attachments to photograph tumors, and ulcers to cut out growths called polyps or to remove objects that have been swallowed. A gastro scope can also guide a laser used to destroy tumors. In ophthalmology laser guided by the fibers is used to retain detached retina and to correct defects in vision. In cardiology, laser angioplasty and bypass surgery. Gastro scope is used to examine female felvic organs. Peritoneoscope is used to examine the abdominal cavity, lower parts of liver and gall bladder. Fiber optic sensors: Sensors are devices used to measure or monitor quantities such as displacement pressure, temperature, flow rate liquid level, chemical composition etc.,A smoke detector and pollution detector can be made from fibers. There are two types of fiber optics sensors. 1. Intrinsic or active sensor 2. Extrinsic or passive sensors

Intrinsic sensors: Here guided like in the fiber gauge modulated by the variable to be measured and then de modulated. Thus sensing and simultaneous measurements of taking place in the active sensors. They are i. Intensity modulate sensors:

Which are based on the change in absorption or transmission of light, refractive index, temperature etc. ii. Phase modulated sensors: Which involve the interference between signal and reference in the interferometer leading to a shift in the interference fringes . These will have high sensitivity. iii. Polarization modulated sensors: Which involve the change in polarization state of the guided signal by the variable.

Extrinsic sensors: Here the modulation takes place outside the optical fiber which acts merely as a convenient transmission channel for the light radiation. It has a sensor head and the sensed optical signal is transferred to remove point for signal processing. For magnetic field sensors polarization maintaining fibers are used. Longitudinal magnetic field applied on the length of an optical fibers a change in the state of polarization in the optical beam (faraday effect). Table: 1. Fiber optics sensors: Measured parameter Modulation effect in optical fibers 1. Temperature 1. Thermo luminescence 2. Magnetic field 2. Magneto optic effect 3. Pressure 3. Piezo-optic effect 4. Mechanical force 4. Stress birefringence 5. Electric current 5. Electro luminescence 6. Electric field 6. Electro optic effect 7. Density 7. Triboluminscence 8. Nuclear radiation 8. Radiation induced luminescence

Based on the parameters listed in the table different fiber sensors are prepared. For ex: micro bending displacement sensor, laser Doppler velocity sensor, fiber optic sensor for the measurement of electric field or voltage, faraday reaction sensor based on the measurement of magnetic field. Fiber optic temperature sensors are used to measure temperature differences of the order 0.001°C with much low response time.

OBJECTIVE TYPE QUESTIONS

1. Cladding in the optical fiber is mainly due to ( ) (a)To protect the fiber from mechanical stress (b) To protect the fiber from corrosion (c) To the proper mechanical strength (d) To provide proper electromagnetic guidance 2. Self focusing effect is produced in ( ) (a) Step index fiber (b) step index single mode fiber (c) Single mode graded index fiber (d) multimode graded index fiber

3. Propagation of light through fiber core is due to ( ) (a)diffraction (b) interference (c)Total internal reflection (d) Refraction. 4. In an optical fiber, let n1 and n2 be the refractive indices of core and cladding respectively. then (a) >1 (b) <1 ( )

(c) = 0 (d) = ∞

5. The numerical aperture of a fiber is given by ( )

(c) (d)

6. Step index fiber can be ( ) (a) Monomode fiber only (b) multimode fiber only (c) Monomod as well as multimode (d) Either monomod or multimode

7. Inter modal dispersion is negligible in ( ) (a) Multimode step index fiber (b) multi mode graded index fiber (c) Mono mode step index fiber (d) all the above

8. Let n1 and n2 be the refractive indices of core and cladding respectively. Then ( ) (a) n1=n2-1 (b) n1>n2 (c) n1

9. At the critical angle (θc), the angle of refraction is ( ) (a) 900 (b) 00 (c) 450 (d) 600

10. The trajectory of alighting step index fiber is given as ( ) (a) zigzag path (b) straight line path (c) Sinusoidal path (d) curved path ANSWERS 1. d 2.d 3. c 4.b 5. a 6.d 7.c 8. b 9.a 10.a SOLVED PROBLEMS 1. The refractive indices of core and cladding of a fiber are 1.54 and 1.5 respectively. Calculate its numerical aperture and maximum accepting angle. (2010) Solution: Refractive index of core=1.54 Refractive index of the cladding=1.50 2 2 1/2 Numerical aperture of fiber NA=( n1 - n2 ) =(1.542-1.502)1/2 =0.349 0 But= sinαm=NA=0.349 => αm=20.43

2. Calculate the numerical aperture, accepting angle and critical angle of an optical fiber, having refractive index of the core is 1.5 and refractive index of cladding is 1.45. (2013) 1/2 Solution: Numerical aperture= n1(2∆) Where ∆= = =0.0333

1/2 NA= n1(2∆) =1.5(2×0.0333)=0.387 0 Maximum accepting angle sinαm=0.387 αm=22.8 Let θc be the critical angle

= = =0.966.

3. Calculate the V-number and number of modes propagating through the fiber of diameter 100μm having core and cladding refractive indices of 1.53 and 1.50 respectively. The propagating wave has wavelength of 1 μm. (Model problem)

Solution: The V- number = (NA)

NA= = =0.301

V-number= =94.6

The maximum number of modes propagating through the SI fiber is

ASSIGNMENT QUESTIONS 1.a) Explain the important characteristics of Laser light. Explain i) Stimulated Absorption ii) Spontaneous Emission iii) Stimulated Emission iv) Population inversion. (Sep2004,June2006,Sep2007,June2008,June2009,May2011) Sol: Characteristics of Laser Light (i). coherent: Coherence is one of the unique properties of laser light. It arises from the stimulated emission process . Since a common stimulus triggers the emission events which provide the amplified light, the emitted photons are in step and have a definite phase relation to each other. This coherence is described interms of temporal and spatial coherence. (ii). Monochromacity: A laser beam is more or less in signle wave length. i.e the line width of laser beams are extremely narrow. The wavelengths spread of conventional light sources is usually 1 in 106, where as in case of laser light it will be 1 in 105. i.e if the frequency of radiation is 1015hz., then the width of line will be 1 Hz. So,laser radiation is said to be highly monochromatic. The degree of non-monochromacity has been expressed as ξ =(dλ/λ)=dν/ν , where dλ or dν is the variation in wavelength or variation in frequency of radiation.

(iii) Directionality:Laser beam is highly directional because laser emit light only in one direction. It can travel very long distances without divergence. The directionality of a laser beam has been expressed interms of divergence. Suppose r1 and r2 are the radii of laser beam at distances D1 and D2 from a laser, and then we have. Then the divergence,∆θ=( r1 - r2)/( D2-Dr1) The divergence for a laser beam is 0.01 milliradition where as incase of search light it is 0.5 radian. (iv) High intensity: In a laser beam lot of energy is concentrated in a small region. This concentration of energy exists both spatially and spectrally, hence there is enormous intensity for laser beam. The power range of laser is about 10-13w for gas laser and is about 109 w for pulsed solid state laser and the diameter of the laser beam is about 1 mm. then the number of photons coming out from a laser per second per unit area is given by 2 -2 Nl=P/ hνПr ≈1022to1034photons/m -sec By assuminghν=10-19 Joule,Power P=10-3to109watts, r=0.5×10-3meter Based on plancks’s black body radiation, the number of photons emitted per second per unit area by a body with temperature T is given by 4 ( ) 16 2 Nth= (2hПC/ λ )(1/e hν/kT -1) dλ≈10 photons/m .sec By assuming T=1000k, λ=6000A0 This comparison shows that laser is a highly intensive beam. In lasers, the interaction between matter and light is of three different types. They are: absorption, spontaneous emission and stimulates emission.Let E1 and E2 be ground and excited states of an atom. The dot represents an atom. Transition between these states involves absorption and emission of a photon of energy E2-E1=hν12. Where ‘h’ is Planck’s constant.

(i)Stimulated Absorption: As shown in fig1(a), if a photon of energy hν12(E2-E1) collides with an atom present in the ground state of energy E1 then the atom completely absorbs the incident photon and makes transition to excited state E2.

(ii)Stimulated emission: As shown in fig1(c), a photon having energy hν12(E2- E1)impinges on an atom present in the excited state and the atom is stimulated to make transition to the ground state and gives off a photon of energy hν12. The emitted photon is in phase with the incident photon. These are coherent. This type of emission is known as stimulated emission. (iii)Spontaneous emission: As shown in fig(b), an atom initially present in the excited state makes transition voluntarily on its own. Without any aid of external stimulus or an agency to the ground. State and emits a photon of energy hν12(=E2-E1).this is called spontaneous emission. these are incoherent.

Population inversion: Usually in a system the number of atoms (N1) present in the ground state (E1) is larger than the number of atoms(N2) present in the higher energy state. The process of making N2>N1 called population inversion.

Fig1.(a) absorption (b) Spontaneous emission(c) stimulated emission

b) Derive the relation between the probabilities of spontaneous emission and stimulated emission in terms of Einstein’s coefficients.(June2004,June2007,Sep2008, May2012) Sol: LetN1 be the number of atoms per unit volume with energy E1 and N2 the number of atoms per unit volume with energy E2. let ‘n’ be the number of photons per unit volume at frequency ν such that hν=E2-E1. then the energy density of interacting photons ρ (v) is given by ρ (ν)=nhν------(1) When these photons interact with atoms, both upward (absorption) and downward (emission) transitions occur. At equilibrium these transition rates must be equal. Upward Transition: Stimulated absorption rate depends on the number of atoms available in the lower energy state for absorption of these photons as well as the energy density of interacting radiation. i.e stimulated absorption rate where the constant of proportionality B12 is the Einstein coefficient of stimulated absorption Downward transition:

Once the atoms are excited by stimulated absorption, they stay in the excited state for a short duration of time called the lifetime of the excited state. After their life time they move to their lower energy level spontaneous by emitting photons. This spontaneous emission rate depends on the number of atoms in the excited energy state. i.e spontaneous emission rate α N1 α ρ(ν) =N1ρ(ν)B12 where the constant of propositionally A21 is the Einstein coefficient of spontaneous emission. Stimulated emission: Before exicited atoms de excite to their lower energy states by spontaneous emission they may interact with photons resulting in stimulated emission of photons. Therefore stimulated emission rate depends on the number of atoms available in the excited state as well as energy density of interacting photons i.e stimulated emission rate α N2 α ρ(ν) =N2ρ(ν) B21 Where the constant of proportionality B21 is the Einstein coefficient of stimulated emission. During stimulated emission, the interacting photon called the stimulating photon and the photon due to stimulated emission are in phase with each other. Please note that during stimulated absorption, the photon density decreases where as during stimulated emission it increases. For a system in equilibrium, the upward and down word transition rates must be equal and hence we have N1ρ(ν)B12=N2ρ(ν)B21 +N 2A21 ------(2) Hence Ρ(ν)= N 2A21 / N1B12- N 2A21 ------(3) Ρ(ν)= A21/ B21 B12/ B21N1/N2-1 The population of various energy levels of in thermal equilibrium is given by Boltzmann

distribution law. Fig1. Boltzmann distribution for several energy levels Ni=giNoexp(-Ei/KT)

Where Ni is the population density of the energy level Ei, N0 is the population density of th the ground state at temperature T1, gi is the degeneracy of the i level and k is the Boltzmann constant.(=1.38x10-23joule/k) The concept of degeneracy occurs since more than one level has the same energy.

Hence N1=giNoexp(-Ei/kT)

= exp (E2-E1/kT)

= exp (hγ/kT)------(6)

Substituting eq(6) in eq(4) Ρ(ν)= /[ exp(hν/KT)-1]------(7)

From planck,s law of blackbody radiation, the radiation density is given by Ρ(ν)=8∏hν3/e3[1/exp(hν/kT)-1]------(8) Comparing eqs (7) and (8) =1

G1B12=g2B21------(9) 3 3------(10) And A21/B21=8∏hν /C Eqs( 9 )and (10) are referred to as the Einstein relations. 2. a) what is lasing action? Describe the construction and working of a ruby laser. (June2004,May2005,Sep2006,June2007,June2008) Sol: Ruby Laser: Ruby Laser is a solid state pulsed, three level lasers. It consists of a cylindrical shaped ruby crystal rod of length varying from 2 to 20 cms and diameter varying 0.1 to 2 cms. This end faces of the rod are highly flat and parallel. One of the faces is highly silvered and the other face is partially silvered so that it transmits 10 to 25% of incident light and reflects the rest so as to make the rod-resonant cavity. Basically, ruby crystal is aluminum oxide [ Al 2O3] doped with 0.05 to 0.5% of chromium atom. These chromium atoms serve as activators. Due to presence of 0.05% of chromium, the ruby crystal appears in pink color. The ruby crystal is placed along the axis of a helical xenon or krypton flash lamp of high intensity.

Fig1.Ruby laser

Fig2.Energy level diagram of chromium ions ina ruby crystal Construction: Ruby(Al2O3+Cr2O3) is a crystal of Aluminium oxide in which some of Al+3 ions are replaced by Cr +3ions when the doping concentration of Cr+3 is about 0.05%. the colour of the rod becomes pink. The active medium in ruby rod is Cr+3ion. In ruby laser a rod of 4cm long and 5mm diameter is used and the ends of the rod are highly polished. Both ends are silvered such that one end is fully reflecting and the other end is partially reflecting. The ruby rod is surrounded by helical xenon flash lamp tube which provides the optical pumping to raise the Chromium ions to upper energy level(rather energy band). The xenon flash lamp tube which emits intense pulses lasts only few milliseconds and the tube consumes several thousands of joules of energy. Only a part of this enology is used in pumping Chromium ion while the rest goes as heat to the apparatus which should be cooled with cooling arrangements as shown in fig 1. The energy level diagram of ruby laser is shown in fig2. Working: Ruby crystal is made up of aluminum oxide as host lattice with small percentage of Chromium ions replacing aluminum ions in the crystal chromium acts as dopant. A dopant actually produces lasing action while the host material sustains this action. The pumping source for ruby material is xenon flash lamp which will be operated by some external power supply. Chromium ions will respond to this flash light having wavelength 0 +3 of 5600A . When the cr ions are excited to energy level E3 from E1 the population in E3 increases. Chromium ions stay here for a very short time of the order of 10-8 seconds -3 then they drop to the level E2 which is matastable state of life time 10 s. Here the level E3 is rather a band, which helps the pumping to be more effective. The transitions from E3toE2 are non-radioactive in nature. During this process heat is given to crystal lattice. Hence cooling the rod is an essential feature in this method. The life time in mete stable 5 state is 10 times greater than the lifetime in E3. As the life of the state E2 is much longer, the number of ions in this state goes on increasing while ions. In this state goes on increasing while in the ground state (E1)goes on decreasing. By this process population inversion is achieved between the exited matestable state E2 and the ground state E1. When an excited ion passes spontaneously from the metastable state E2 to the ground 0 state E1, it emits a photon of wave length 6943A . This photon travels through the rod and if it is moving parallel to the axis of the crystal is reflected back and forth by the

silvered ends until it stimulates an excited ion in E2 and causes it to emit fresh photon in phase with the earlier photon. This stimulated transition triggers the laser transition. This process is repeated again and again because the photons repeatedly move along the crystal being reflected from its ends. The photons thus get multiplied. When the photon beam becomes sufficiently intense, such that part of it emerges through the partially silverved end of the crystal.

Fig.The output pulses with time ©.Describe the construction and working of Semi-conductor laser( June2008,June2009,’1 Sol:Semi conductor laser: The semiconductor laser is also called diode laser. Among the different semiconductors there are direct band gap semiconductors and indirect band gap semiconductors. In the case of direct band gap semiconductors, there is a large possibility for direct recombination of holes and electrons which emit photons. But in indirect band gap semi conductors like silicon and germanium, direct recombination of holes and electrons is less possible and hence there is no effective emission of photons. A well know example of a direct band gap semiconductor is GaAs . Let Eg be the energy gap of a material, then it emits a photon of wavelength(λ) is given the relation λ =ch/Eg ------(1) where c is the velocity of light and h is the planck’s constant the equation (1) is reduced to λ=1.24/Eg µm------(2) where Eg is expressed in ev as Eg increases, it emits shorter wavelengths. diode lasers are always operated in forward bias. If p and n type materials are prepared from the same material then the p-n junction, is called as Homo junction semiconductor laser source. If p and n type materials are prepared from different materials then they are called as Hetero junction semi conductor laser source. Construction of GaAs Semiconductor diode laser: The basic mechanism responsible for light emission from a semi conductor is the recombination of electrons and holes at p-n junction when a current is passed through the diode. The active medium is a p-n junction diode made from crystalline Gallium Arsenide(fig1). The p-region and n-region in the diode are obtained by heavily doping with suitable dopants. At the junction the sides through which emitted light is coming out are well polished and parallel to each other. Since the refractive index of GaAs is high, the reflectance at the material air interface is sufficiently large so that 219

the external mirrors are not necessary to produce multiple reflections. When a current is passed through a p-n junction p region being positively biased, holes are injected from p-region into n-region and n-region being negatively biased electrons are injected from n-region into p-region and is shown in fig2.

Fig1.GaAs semiconductor diode laser

The connections in this p-n junction circuit is called forward bias. The electrons and holes recombine and release of light energy takesplace in or near the junction. In the case of GaAs homo junction which has an energy gap of 1.44ev gives a laser beam of wavelength around 8600A0 the electron-hole recombination takes place in the active region of the device. When the junction is forward biased, a large amount of the order of 104 amp/cm2 is passed through the narrow junction. Thus the electrons (holes) are injected from n side to p side (p side to n side) respectively.

The electrons are minority charge carriers in p-side and holes are the minority charge carriers in n-side. The continuous injection of charge carriers creates the population inversion of minority carriers is n and p side respectively. The excess minority charge carries diffuse away from the region recombining with majority carriers of the n and p type material, resulting in the release of photons. Further, the emitted photons

increase the recombination of injected electrons from the n-region and holes in p- region by inducing more recombinations thus the stimulated emission takes place more effectively. The efficiency of a laser emission increases when the device is cooled. 3. a)Describe the construction and working of He-Ne laser (June2006,Sep2007,June2008) Sol: Helium-Neongas[He-Ne]laser: Helium-Neon gas laser is a continuous four level gas laser. It consists of a long, narrow cylindrical tube made up of fused quartz. The diameter of the tube will vary from 2 to 8 mm and length will vary from 10 to 100 cm. the tube is filled with helium and neon gases in the ratio of 10:1. The partial pressure of helium gas is 1mm of Hg and neon gas is 0.1mm of Hg so that the pressure of the mixture of gases inside the tube is nearly 1 mm of Hg. Laser action is due to the neon atoms helium is used for selective pumping of neon atoms to upper energy levels. Two electrodes are fixed near the ends of the tube to pass electric discharge through the gas. Two optically plane mirrors are fixed at the two ends of the tube at Brewster angle normal to its axis. One of the mirrors is fully silvered so that nearly 100%. Reflection takes place and the other is partially silvered. So that 1%of the light incidedent on it will be transmitted. Optical resources column is formed between these mirrors.

Working: When a discharge is passed through the gaseous mixture, electrons are accelerated down the tube. These accelerated electrons collide with the helium atoms and excite them to higher energy levels. The different energy levels of Helium atoms and Neon atoms is shown in fig. the helium atoms are excited to the levels F2 and F3 these levels happen to be meta stable energy states. Energy levels and hence Helium atoms exited levels spend sufficiently large amount of time before getting de excited. As shown in the fig some of the excited states of neon can correspond approximately to the same energy of excited levels F2 and F3. Thus, when Helium atoms in level F2 and F3 collide with Neon atoms in the ground level E1, an energy exchange takes place. This results in the excitation of Neon atoms to the levels E4 and E6 a nd de excitation of Helium atoms to the ground level (F1). Because of long life times of the atoms in levels F2 and F3, this process of energy transfer has a high probability. Thus the discharge through the gas mixture continuously populates the neon atoms in the excited energy levels E4 and E6. This helps to create a state of population inversion between the levels E4(E6) to the lower energy level (E3 and E5). The various transitions E6--->E5,E4--->E3, E6--->E3 leads to

the emission of wave lengths 3.39mm, 1.15 um and 6328 A0. specific frequency selection may be obtained by employing mirrors The excited Neon atoms drop down from the level E3 to the E2 by spontaneously emitting a photon around wavelength 6000A0. the pressures of the two gases in the mixture are so chosen that there is an effective transfer of energy from the Helium to the Neon atoms. Since the level E2 is a meta stable state, there is a finite probability of the excitation of Neon, atoms from E2 to E3 leading to population inversion, when a narrow tube is used, the neon atoms in the level E2 collide with the walls of the tube and get excited to the level E1. The transition from E5 to E3 may be non radioactive. The typical power outputs of He-Ne laser lie between 1 and 50 mw of continuous wave for inputs of 5-10W.

b) Fig.2.Energy level diagram of He-Ne atoms. (b)The applications of lasers in different fields of engineering. Sol: 1.In Communications : (a).lasers are used in optical fiber communications. In optical fiber communications, lasers are used as light source to transmit audio, video signals and data to long distances with out attention and distortion. (b).The narrow angular spread of laser beam can be used for communication between earth and moon or to satellites 2. Industrial Applications: (a).lasers are used in metal cutting,welding,surface treatment and hole drilling. Using lasers cutting can be obtained to any desired shape and the curved surface is very smooth. (b)Welding has been carried by using laser beam. 3. In Computers: By using lasers a large amount of information or data can be stored in CD-ROM or their storage capacity can be increased. Lasers are also used in computer printers. 4. a) Explain the working principle of Optical fiber.

Sol: PRINCIPLE OF OPTICAL FIBER: Optical fiber is a cylinder of transparent dielectric medium and designed to guide visible and infrared light over long distances. Optical fibers work on the principle of total internal reflection. When a ray of light passes from an optically denser medium into an optically rarer medium the refracted ray bends away from the normal. When the angle of incidence is increased angle of refraction also increases and a stage is reached when the refracted ray just grazes the surface of separation of core and cladding. At this position the angle of refraction is 90 degrees. This angle of incidence in the denser medium is called the critical angle (θc) of the denser medium with respect to the rarer medium and is shown in the fig2. If the angle of incidence is further increased then the totally reflected.This is called total internal reflection.Let the reflective indices of core and cladding materials be n1 and n2 respectively.

When a light ray, travelling from an optically denser medium into an optically rarer medium is incident at angle greater than the critical angle for the two media. The ray is totally reflected back into the medium by obeying the loss of reflection. This phenomenon is known as totally internal reflection. According to law of refraction, n 1 sinθ1= n 2 sinθ2 Here θ1=θc, θ2=90⁰ n 1 sinθc=n 2 sin 90⁰ Sinθc = n2/n1 θc = sinˉ¹(n2/n1)————>(1) Equation (1) is the expression for condition for total internal reflection. In case of total internal reflection, there is absolutely no absorption of light energy at the reflecting surface. Since the entire incident light energy is returned along the reflected light it is called total internal reflection. As there is no loss of light energy during reflection, hence optical fibers are designed to guide light wave over very long distances.

b)Give the construction of an Optical fiber( June2008,June2009,May2011)

Sol: Optical fiber is very thin and flexible medium having a cylindrical shape consisting of three sections 4) The core material 5) The cladding material 6) The outer jacket

The structure of an optical is shown in figure1. The fiber has a core surrounded by a cladding material whose reflective index is slightly less than that of the core material to satisfy the condition for total internal reflection. To protect the fiber material and also to give mechanical support there is a protective cover called outer jacket. In order to avoid damages there will be some cushion between cladding protective cover.

Optical fiber is a cylinder of transparent dielectric medium and designed to guide visible and infrared light over long distances. Optical fibers work on the principle of total internal reflection.

c) Define acceptance angle and Numerical aperture. Derive the expressions for them. Sol: ACCEPTANCE ANGLE AND ACCEPTANCE CONE: Acceptance angle is the angle at which we have to launch the beam at its end to enable the entire light to propagate through the core. Fig.1 shows longitudinal cross section of the launch of a fiber with a ray entering it. The light is entered from a medium of refractive index n₀ (for air n₀ =1) into the core of refractive index n₁ . The ray (OA) enters with an angle of incidence to the fiber end face i.e the incident ray makes angle with the fiber axis which is nothing but the normal to the end face of the core. Let a right ray OA enters the fiber at an angle to the axis of the fiber. The end at which light enter the fiber is called the launching pad.

Fig1.Path of a typical light ray launched into fiber

Let the refractive index of the core be n₁ and the refractive index of cladding be n₂ . Here n₁ >n₂ . the light ray reflects at an angle and strikes the core cladding interface at angle θ. If the angleθ is greater than its critical angle θc, the light ray undergoes total internal reflection at the interface. According to Snell’s law n₀ sinαi=n1sinαr------>(2) From theright angled triangle ABC 0 αr+θ=90 0 αr=90 –θ------(3) substituting (3) in (2),we get 0 n₀ sinαi=n1sin(90 –θ)= n1cos θ sinαi=( n1/ n₀ ) cos θ------(4) when θ= θc, αi= αm=maximum α value sinαm=( n1/ n₀ ) cos θc------(5) From eqution(1) Sinθc = n2/n1 1/2 2 2 2 1/2 cos θc=(1- Sin2θc) =(1-( n2/n1) )=(( n1 - n2 ) / n1)------(6) substitute equation(6) in equation(5) 2 2 1/2 2 2 1/2 sinαm=( n1/ n₀ )(( n1 - n2 ) / n1)= )=(( n1 - n2 ) / n0)------(7) If the medium surrounding fibre is air, then n0=1 2 2 1/2 sinαm=( n1 - n2 ) ------(8)

This maximum angle is called the acceptance angle or the acceptance cone half angle of the fiber. The acceptance angle may be defined as the maximum angle that a light ray can have with the axis of the fiber and propagate through the fiber. Rotating the acceptance angle about the fiber axis(fig 2) describes the acceptance cone of the fiber. Light launched at the fiber end within this acceptance cone alone will be accepted and propagated to the other end of the fiber by total internal reflection. Larger acceptance angles make launching easier. Light gathering capacity of the fiber is expressed in terms of maximum acceptance angle and is termed as “Numerical Aperture”.

Numerical Aperture: Numerical Aperture of a fiber is measure of its light gathering power. The numerical aperture (NA) is defined as the sign of the maximum acceptance angle. 2 2 1/2 Numerical aperture(NA)= sinαm=( n1 - n2 ) ------(9) 1/2 =(( n1- n2) ( n1+ n2)) 1/2 =(( n1+ n2) n1∆) ------(10) Where ∆= ( n1- n2)/ n1 called as fractional differences in refractive indices n₁ and n₂ are the refractive indices of core and cladding material respectively. 5.a) Discuss the different types of optical fibers based on i) refractive index profile and ii) mode of propagation. Sol: Based on the variation of refractive index of core, optical fibers are divided into: (1) step index and (2) graded index fibers. Again based on the mode of propagation, all these fibres are divided into: (1) single mode and (2) multimode fibers. In all optical fibers, the refractive index of cladding material is uniform.. (1)Step index fiber: The refractive index is uniform throughout the core of this fiber. As we go radially in this fiber, the refractive index undergoes a step change at the core- cladding interface. Based on the mode of propagation of light rays, step index fibers are of 2 types: a)single mode step index fiber & b)multimode step index fibers. Mode means ,the number of paths available for light propagation of fiber. We describes the different types of fiber below (a)Single mode step index fiber : The core diameter of this fiber is about 8 to 10µm and outer diameter of cladding is 60 to 70 µm. There is only one path for ray propagation. So ,it is called single mode fiber.

Multimode step index fiber:The construction of multimode step index fiber is similar to single mode step index fiber except that its core and cladding diameters are much larger to have many paths for light propagation. The core diameter of this fiber varies from 50 to 200 µm and the outer diameter of cladding varies from 100 to 250 µm. The cross-sectional view, refractive index profile and ray propagations are shown in fig 2. Light propagation in this fiber is by multiple total internal reflections i.e it is a reflective type fiber. 2) Graded index fiber: In this fiber, the refractive index decreases continuously from center radially to the surface of the core. The refractive index is maximum at the center and minimum at the surface of core. This fiber can be single mode or multimode fiber. The diameter of core varies from 50 to 200µm and outer diameter of cladding varies from 100 to 250 µm. The refractive index profile is circularly symmetric. As refractive index changes continuously radially in core, light rays suffer continuous refraction in core.In graded index fiber, light rays travel at different speed in different paths of the fiber. Near the surface of the core, the refractive index is lower, so rays near the outer surface travel faster than the rays travel at the center. Because of this, all the rays arrive at the receiving end of the fiber approximately at the same time. This fiber is costly. . c) What are the different losses in optical fibers?

Sol:The signal attenuation is defined as the ratio of the optical power(Pi) into the fiber to the power of light coming out at the output end(Po).The attenuation coefficient s is given as α= (10/L)log10((Pi/Po)dB/Km , where L is the length of the optical fiber. The attenuation is mainly due to i) scattering and (ii)absorption (iii)Bending losses( iv) micro bending and wave guide losses c) Give the applications of Optical fibers. Discuss fiber optic communication system using the Block diagram. (June2007,June2008) Sol: Application of fibers in medical field: Fiber scopes are employed widely in endo scopic applications. Bundles of optical fibers form part of gastro scopes and other medical instruments. These instruments enable physicians to view internal body parts without performing surgery. Fiber optic endoscopes are classified according to the application. Gastroscope to examine the stomach, bronchoscope to view the upper passages of lungs and orthoscope to study the small spaces with in joints. A gastroscope consists of a long, flexible rubber tube and a rigid metal section that has a lens and various controls. Two bundles of long, thin fiber glass threads called optical fibers extend along the inside of gastro scope. A gastroscope can be fitted with various special attachements to photograph tumors, and ulcers to cut out growths called polyps or to remove objects that have been swallowed. A gastro scope can also guide a laser used to destroy tumoures. In ophthalmology laser guided by the fibers is used to retain detached retina and to correct defects in vision. In cardiology, laser angioplasty and bypass surgery. Couldo scope is used to examine female felvic organs. Peritoneoscope is used to examine the abdominal cavity, lower parts of liver and gall baldder. Fiber optic sensors: Sensors are devices used to measure or monitor quantities such as displacement pressure, temperature, flow rate liquid level, chemical composition etc.,A smoke detector and pollution detector can be made from fibers. There are two types of fiber optics sensors. 3. Intrinsic or active sensor 4. Extrinsic or passive sensors

Intrinsic sensors: Here guided like in the fiber guage modulated by the variable to be measured and then de modulated. Thus sensing and simultaneous measurements of taking place in the active sensors. They are iv. Intensity modulate sensors: Which are based on the change in absorption or transmission of light, refractive index, temperature etc. v. Phase modulated sensors:which involve the interference between signal and reference in the interferometer leading to a shift in the interference fringes. These will have high sensitivity. vi. Polarization modulated sensorsWwhich involve the change in polarization state of the guided signal by the variable.

Extrinsic sensor: Here the modulation takes place out side the optical fiber which acts merely as a convienient transmission channel for the light radiation. It has a sensor head and the sensed optical signal is transferred to removed point for signal processing. For a magnetic field sensors polarization maintaining fibers are used. Longitudinal magnetic field applied on the length of an optical fibers a change in the state of polarization in the optical beam (faraday effect ) Fiber optic communication system : Fiber opptics essentially deals of with the communication(including voice signals,video signals or digital data) by transmission of light through optical fibers. Optical fiber communication system essentially consists of three parts:(a)transmitter (b) optical fiber and (c) receiver.The transmitter includes modulator, encoder, light source, drive circuits and couplers. The receiver includes amplifier and decoder,binary electrical signal and light decoder.

Fig.1.Block diagram represents optical fibre communication system

UNIT – V

Chapter – 9 SEMICONDUCTOR PHYSICS & ACOUSTICS OF BUILDINGS AND ACOUSTIC QUIETING

a. Introduction to Semiconductor Physics

Based on electrical conductivity, materials are divided into conductors, insulators, & semiconductors. Metals are good conductors of electricity. All dielectrics are insulators (non-conducting). Semiconductors electrical conductivity lies between metals & insulators. Good examples are Silicon and Germanium. They belong to IV-A group. At 0K, Semiconductors behave like Insulators. At temperature greater than 0K they behave like conductors. Pure germanium and silicon are called intrinsic semiconductors. By adding a small quantity of either group-III, group-V element atoms as impurity into pure Ge or Si, the electrical conductivity of the materials increases. This impure semiconductor is called extrinsic semiconductors.

Figure9.1 (a) Energy band diagram of a semiconductor (b) Fermi level at T=0K & Thermal excitation of electrons at T>0K

9.1 Intrinsic Semiconductor Pure germanium or silicon called an intrinsic semiconductor. Each atom possesses four valence electrons in outer most orbits. At T = 0K a 2-D representation of the crystal of silicon & band diagram is shown in the figure 2.

Figure9.2: Intrinsic silicon crystal at T =0K (a) 2-D representation of silicon crystal (c) Energy band diagram of intrinsic semiconductor

Explanation: At 0K, all the valence electrons of silicon atoms are in covalent bonds and their energies constitute a band of energies called valance band (VB). So at 0K, VB is completely filled & conduction band (CB) is empty. If we rise temperature (T>0K), some of the electrons which are in covalent bonds break the bonds become free and move from VB to CB. The energy required should be

greater than the energy gap of a semiconductor (E>Eg). The electron vacancy or deficiency created in VB is called holes. This is shown in the figure 3 below.

Figure9.3 Silicon crystal at temperature above 0K (a) Due to thermal energy breaking of Covalent bonds take place (b) Energy band representation

9.2 Electron concentration in intrinsic semiconductor in conduction band (n) Definition: The no. of free electrons per unit volume of the conduction band of a given intrinsic semiconductor is called electron concentration, represented by ‘n’.

Figure9.4 (a) Energy band diagram of silicon at T = 0K (b) Energy band diagram of silicon at T > 0K Derivation: Let the no. of free electrons per unit volume of the semiconductor having energies E and E + dE in CB is represented by n (E) dE. It is obtained by multiplying the

density of energy states ZC (E) d (E). [No. of energy states per unit volume] and Fermi –

Dirac distribution function for the Probability of occupation of electrons FC (E)

Therefore n (E) dE = [ZC (E) d (E)] [FC (E)] → (1) Where Z v (E) d (E) = Density of energy states F h (E) = Probability of occupation of electrons given by Fermi – Dirac function The total no. of electrons in CB per unit volume between the energies EC to E ct is given by integrating equation (1) with limits EC to E ct

n = → (2)

But equation (2) can be written as

n = → (3)

n = →(4)

In equation (4) the second term vanishes (disappears).

Since, above E c t electrons do not present. Hence equation (4) becomes

n = →(5)

{Since from equation (1)}

But is Fermi – Dirac distribution function;

→(6)

Here E > EF, >> 1 Hence ‘1’ can be neglected in equation (6)

→ (7) Also the density of electrons

ZC (E) d (E) = ( dE → (8)

Here E > EC .Since EC is the minimum energy state in CB. Hence equation (8) becomes

ZC (E) d (E) = ( dE → (9)

Substituting equations (7) & (9) in equation (5) we get

n = dE

n = dE → (10)

Let d { is constant} . The limits are = 0 to = Hence equation (10) can be written as

n = → (11)

In equation (11) But = → (12)

Substituting (12) in (11) we get

n = 2 → (13)

{Here NC = } Therefore n =

9.3 Hole concentration in the valance band of intrinsic semiconductor(p) Definition: The number of holes per unit volume of the valance band of a given intrinsic semiconductor is called hole concentration, represented by ‘p’. Derivation: Let the number of holes per unit volume of the semiconductor having energies E, E + dE in VB is represented by p (E) dE. It is obtained by multiplying the

density of energy states ZV (E) d (E) [No. of energy states per unit volume] and Fermi – Dirac distribution function for the Probability of occupation of holes F h (E).

Figure9.5 (a) Energy band diagram of silicon at T = 0K (b) Energy band diagram of silicon at T > 0K

Therefore p (E) dE = [Z v (E) d (E)] [F h (E)] → (1) Where Z v (E) d (E) = Density of energy states.

F h (E) = Hole probability given by Fermi – Dirac function The total no. of holes in VB per unit volume between the energies E v b to E v is given by integrating equation (1) with limits E v b to E v

p = →(2)

But equation (2) can be written as

p = → (3)

p = → (4) In equation (4) the first term vanishes (disappears).

Since, below E v b holes do not present. Hence equation (4) becomes

p = → (5) {Since from equation (1)}

But is Fermi – Dirac distribution function; →(6)

Simplifying; =

Divide by we get

= → (6)

Here EF > E, >> 1

Hence ‘1’ can be neglected in equation (6)

→ (7) Also the density of holes

Z v (E) d (E) = ( dE → (8)

Here E < E v .Since E v is the maximum energy state in VB. Hence equation (8) becomes

Z v (E) d (E) = ( dE → (9)

Substituting equations (7) & (9) in equation (5) we get

p = dE

p = dE →(10)

Let d { is constant} . The limits are = = 0 Hence equation (10) can be written as

p = → (11)

In equation (11) But = → (12)

p = 2 → (13)

{Here NV = }

p = 9.4 Fermi energy level in intrinsic semiconductor At temperature T k , the electron concentration ‘n’ is equal to hole concentration ‘p’ in intrinsic semiconductor. i.e. n = p

On simplifying we get

Taking logarithms on both sides we get

= + ] {At T > 0K}

Let T = 0K =

This means EF lies in the middle between ( of the energy gap ‘E g’

As the temperature increases the electrons move from VB to CB. Also the Fermi level

slightly rises upwards towards CB. Hence = + ]. It is shown

in the figure 6 below.

Figure9.6 (a) Fermi level at T = 0K (b) Upward shift of near EC at T> 0k

9.5 Intrinsic carrier concentration (ni) Definition: The no. of free electrons and holes per unit volume of the intrinsic

semiconductor is called intrinsic carrier concentration (n i) remains constant. i .e. n = p = ni n p = (ni) (ni)

= (n p) → (1)

= → (2)

Consider equation (1) = (n p)

4 {Since = E } g

If = , the above equation becomes

Let 2 = C.

Then 9.6. Extrinsic (or) Impure semiconductor Introduction: The conductivity of an intrinsic semiconductor can be increased by adding rd th small amounts of impurity atoms, such as III or V group atoms. The conductivity of silica is increased by 1000 times on adding 10 parts of boron per million part of silicon. The process of adding impurities is called doping and the impurity added is called dopant.

9.6.1 N – Type semiconductor In a pure (intrinsic) semiconductor, when pentavalent an impurity like Phosphorous atom consisting of five valance electrons is doped, and then concentration of electrons increases than holes. Hence the given semiconductor formed is called N – type semiconductor. This is shown in the figure 7a below. By adding donor impurities, the free electrons generated or donated, form an energy level called as “Donor energy level”

i.e. ED is shown in the figure 9.7b below.

Figure9.7 (a) Representation of n- type silicon at T = 0K (b) Energy band diagram at T = 0K

In the figure (b) EF is Fermi energy level is in between EC & ED at T = 0K.

Hence EF = . The donor level‘ ’ is near to EF consisting of free electrons. But CB

is empty.

9.6.2 Variation of Fermi level EF with respect to temperature in N -type

As temperature increases the electrons in the Donor level moves into CB leaving holes. Also The Fermi level slightly shifts upwards towards CB. If further increase of temperature is done, the hole concentration also increases with respect to electron concentration. Hence it reaches again the concentration of n = p i.e. an intrinsic semiconductor. At lasts the Fermi level drops in the middle of the energy gap (or)

Forbidden band gap, indicating a pure semiconductor that is towards ‘Ei’. This happens only for very high temperatures as shown in the figure 8 below.

Fig.8 Variation of Fermi level wrt temperature in n- type semiconductor

9.6.3 Variation of EF with respect to donor concentration As the donor concentration increases the Fermi level decreases (lowers) as in case of

intrinsic semiconductor ‘Ei. This is shown in the figure below

Figure9.9 Variation of Fermi level with temperature for different donor concentrations in an n-type semiconductor

9.6.4. Carrier concentration in N – type semiconductor & Density of electrons in CB

Let ND is the donor concentration (no. of donor atoms per unit volume). Let it be written as

Or written as = exp [

= →(1) The density of electrons in CB in pure semiconductor is given by

= 2 → (2)

At very low temperatures the no. of electrons in CB must be equal to the no. of donor atoms per unit volume. Hence equating equations (1) & (2) we get

Taking logarithms on both sides we get =

{At T>0K}→(3)

Case I: At T= 0K

. That is EF lies between

Case II: At T>0K. As temperature increases the Fermi level slightly shifts upwards towards CB, hence

Density of electrons in CB in extrinsic semiconductor: Here consider equation (2). That is

Substitute the value of EF from equation (3) in equation (2), it becomes

On simplifying We know that exp (a + b) = exp (a) × exp (b) Also exp (log x) = x

9.6.5 P- type semiconductor

P – Type semiconductor is formed by doping with trivalent impurity atoms (acceptor) like III rd group atoms i.e. Aluminum, Gallium, and Indium etc to a pure semiconductor like Ge or Si. As the acceptor trivalent atoms has only three valance electrons & Germanium , Silicon has four valence electrons; holes or vacancy is created for each acceptor dopant atom. Hence holes are majority and electrons are minority. It is shown in

the figure 10a below. Also an acceptor energy level ‘EA’ is formed near VB consisting of holes, as shown in the figure 10b below.

Fig.10 (a) Representation of p- type silicon at T = 0K (b) Energy band diagram at T =0K

As temperature increases (T>0K) the electrons in VB which are in covalent bonds break

the bonds become free and move from VB to acceptor energy level E A.

9.6.6 Variation of Fermi level EF with respect to temperature in P- type semiconductor

As temperature increases the Fermi level EF slightly drops towards VB. For further increase of high temperatures the electron concentration also increases with respect to hole concentration. Hence a condition is reached such that ‘n = p’ i.e. it becomes an intrinsic or pure semiconductor. Hence the Fermi level increases and reaches to intrinsic

level Ei as in case of pure semiconductor. This is shown in the figure 11 below

Figure9.11 Variation of Fermi level wrt temperature in p- type

9.6.7 Variation of Fermi level with respect to acceptor concentration:

Also as the acceptor concentration increases we find that Fermi level EF reaches (increases) towards intrinsic level Ei as in case of pure or intrinsic semiconductor. This is shown in the figure 12 below.

Fig 9.12 Variation of Fermi level with temperature for different acceptor concentrations in a p-type

5.6.8 Carrier concentration of P- type semiconductor & Density of holes in VB

Let N A is the acceptor concentration (no. of acceptor atoms per unit volume). Let it be written as

= exp [

Or written as

= → (1) The density of holes in VB in pure semiconductor is given by

= 2 →(2)

At very low temperatures the no. of holes in VB must be equal to the no. of acceptor atoms per unit volume. Hence equating equations (1) & (2) we get

Taking logarithms on both sides we get

{At T>0K} → (3)

Case I: At T= 0K

.That is EF lies between

Case II: At T>0K. As temperature increases the Fermi level slightly drops towards VB, hence

Density of electrons in CB in extrinsic semiconductor Here consider equation (2). That is

Substitute the value of EF from equation (3) in equation (2), it becomes

On simplifying We know that exp (a + b) = exp (a) × exp (b) Also exp (log x) = x

Fig.13 Energy band diagrams of (a) Intrinsic semiconductor (b) N- type & (c) P- type semiconductors

9.7 Hall Effect

Determination of Hall coefficient RH, Hall voltage VH & Applications of Hall Effect Definition: If a piece of semiconductor carrying a current ‘I’ is placed in a transverse or

perpendicular magnetic field, then an Electric field ‘EH’ is generated in perpendicular to both ‘I’ and ‘B’. Hence production of Hall electric field ‘EH’ and generation of Hall voltage VH by applying current ‘I’ & ‘B’ in perpendicular directions is called Hall Effect.

Derivation of RH, VH, & EH: Assume an N – type semiconductor. Let ‘I’ be the current passed along X- axis. Let ‘B’ the magnetic field applied along ‘Z’ axis. Due to current ‘I’, due to the flow of electrons, the force acting on electrons is in opposite direction to the direction of conventional current. Due to the magnetic field ‘B’ there is a down ward force act on each electron given by Bev. This makes the electrons deviated in a down ward direction along Z – axis. This causes a negative charge to accumulate on the bottom face.

Hence FB = Bev Where ‘e’ = charge of electron, v = velocity of electron

Fig. 9.14 A semiconductor applied to current and magnetic field perpendicularly in Hall Effect

A potential difference is formed from top to bottom of the specimen. The potential

difference causes a field EH to flow along negative Y- axis. Due to EH along negative Y- direction a force of eEH acts in upward direction along positive Y- axis.

Hence FE = e EH Under equilibrium condition

Upward force due to EH = Downward force due to B. FB = FE e EH = Bev v = → (1)

If J is the current density then J = nev v = →(2)

Equating (1) & (2) we get =

EH = ] B J

Let Hall coefficient RH = for electrons

RH = for holes

Hence EH = RH B J → (4)

RH = →(5)

Where n= electron density, P = hole density, EH = Hall electric field, RH = Hall coefficient B = magnetic field, J = current density

Experimental determination of RH Consider equation (5)

Let VH is the Hall voltage across the sample of thickness‘t’ Generally V = Ed

In Hall effect VH = EH × t

EH = →(6)

If ‘b’ is the width of the sample semiconductor, Area ‘A’, thickness‘t’ Area = breadth × thickness A = b × t We know that current density J =

J = → (7)

Substituting equations (6) & (7) in (5) we get

Hall coefficient RH = → (8)

Where VH = Hall voltage, b = breadth of a semiconductor, B = magnetic field I = current due to flow of electron

VH = → (9)

Applications of Hall Effect 1. For determination of type of given semiconductor.

For N-type, Hall coefficient RH = negative; For P-type, Hall coefficient RH = Positive

2. To determine carrier concentration ‘n’ and ‘p’ ; that is n = p=

3. Determination of mobility of charge carriers (

, Where = electrical conductivity

4. For measurement of magnetic flux density ‘B’ & Hall voltage. 5. To determine the sign of charge carriers, whether the conductivity is due to electrons or holes.

9.8. Direct and Indirect band gap semiconductors Direct band gap semiconductor Indirect band gap semiconductor 1.These are impure or Extrinsic or 1.These are pure or intrinsic or elemental compound Semiconductors semiconductors 2. Examples : Ge, Si 2. Examples : InP, GaAs, GaAsP etc 3. 3.

Figure: E-K curve

4. The minimum energy of Conduction band (CB) and maximum energy of valence band (VB) have the different values of wave

vector, i.e. k1 k2

5. Here an electron from CB cannot recombine directly with holes in VB. But can recombine through traps by Figure: E-K curve emitting light without emission of photon or light. 4. The minimum energy of Conduction band (CB) and maximum energy of valence band (VB) have the same value of wave vector , 6. They are used to amplify the signals in i.e. k1 = k2 electronic devices like rectifiers, transistors, amplifiers etc. 5. Here an electron from CB to can recombine with a hole in VB directly by 7. Life time of charge carriers is more. emitting light of photon of energy ‘h ’.

8. No emission of light. It conducts only Electricity. 6. They are used to fabricate LEDs, Laser Diodes etc.

7. Life time (recombination rate) of charge carriers is less.

8. Emission of light has energy gap is

Eg = eV

9.9. Equation of Continuity Definition: The excess carriers in semiconductors undergo the effect like generation, recombination, Drift & Diffusion. “The mathematical relations formed due to these effects of generation, recombination, drift & diffusion are termed as continuity equation of electron and holes”.

Fig. 9.15 A rectangular bar of semiconductor Derivation: Consider an elemental volume as shown in the figure. Consider the combined time rate of change of the excess carrier density as shown in the figure. Let the number of excess electrons in such a volume is .

Let the rate of increase of charge carriers in the volume is

a) Due to generation & recombination rate: The net rate of increase of excess electron concentration due to generation & recombination (G & R) is given by (in an elemental volume ) is

→ (1)

Where &

b) Rate of increase of excess electron concentration due to drift: Let excess electron concentration at plane (1) is The excess electron current in plane is calculated using Current density where I = current, A = area of

the plane

In plane I the flow of current is In plane II the flow of current is The net rate of increase of the number of excess electrons in the volume due to drift is

→ (2)

But is written as → (3)

Substitute in we get

→ (4)

(C) Rate of increase of excess electron concentration due to diffusion: The rate of diffusion of electron out of plane (1) of volume is

→ (5)

Where = diffusion coefficient of electrons,

Similarly the rate of diffusion of electrons out of plane of the volume is

= → (6)

The rate of increase of excess electrons in the volume due to diffusion is

→ (7)

But → (8)

→ (9)

Adding equations (1), (4) & (9) and simplifying we get The net rate of increase excess electrons in volume due to g &R, diffusion & drift is given by For Electrons the Continuity equation is

Similarly for Holes the Continuity equation is

9.10. Introduction to Physics of Semiconductor Devices

 The conductivity of semi conductors is small as compared to the metallic conductors. Conductivity of silicon is about 1011 times lesser than that of copper and is about 1013 times higher than that of fused quartz.  Another distinguishing feature of semiconductor is that its conductivity increases as the temperature is increased.  Devices made of semiconductor are small with less weight, more reliable and in expensive. Hence semiconductor device have replaced almost all vacuum tube services.  Silicon has almost no free electrons at room temperature as compared with germanium. Hence silicon is preferably used in the fabrication of semiconductor devices in this chapter we study how PN junction diodes are formed, what are its characteristics and how are they used.

9.11. Formation of PN Junction  When a layer of P - type semiconducting material is placed with a layer of N- type semiconducting material in such a way that the atoms of P-type combine with the atoms of N-type across the surface of contact, such a surface junction where combination has occurred is known as PN junction.  Combined p-type and n-type semiconductors with PN junction formation is known a junction diode or PN diode.  In practice a PN junction diode may be obtained in any of the following three methods. 1) Grown junction type 2) Fused (or alloyed) junction type and 3) Diffused junction type

1) Grown junction type: When an extrinsic semiconductor is grown from melt, during the middle of the growth process impurities of opposite kind are added to the melt so that the opposite type of crystal grows further for example: when an n type Si crystal containing phosphorous atoms as pentavalent impurity is grown, the growth process is stopped temporarily and

efficient boron atoms are added to the melt as trivalent impurity. This results in growth of p-type thereafter.

2) Fused (or Alloyed) junction type: In this type p-type and n-type materials are kept in contact and fused together by proper heat treatment to form the junction. For example, a small of dot of indium (p-type material) is pressed on thin water of (n-type) germanium. During few minutes of heat treatment, indium atoms fuse in to the surface of germanium and produce p-region inside germanium. This p-type region with n-type germanium water forms a p-n junction.

3) Diffused Junction type: Diffused junctions are formed by impurity diffusion technique. The diffusion process employs either gas diffusion method or solid diffusion method for example in gas diffusion method a wafer of n-type silicon is heated at about 10000 c in a gaseous atmosphere of high concentration gradient the boron atoms. At the temperature due to concentration gradient the boron atom diffuse in to silicon forming p-n junction in solid diffusion process a p-type impurity(say indium) is painted on a n-type substrate and both are heated . Now impurity atoms diffuse into n-type substrate for a short distance and form p-n junction.

Fig. 9.16 (a) Newly made p n junction (b) Formation of P N Junction with potential barrier  The formation PN junction is represented in a figure. Let us consider the formation of a sharp junction when two separate semiconductors of p and n type are brought together.  The p-type has holes as majority carriers and electrons as minority carriers while n- type has electrons as majority carries and holes as minority carriers.

 When they are joined, in the region of contact the free electrons diffuse from n-region and combine with holes in p-region. This leaves n-region near the boundary

positively charged and p-region negatively as a result, electric field EB appears in a small region W on either side of the junction O. This region is called “Depletion region “. The thickness of this region is in the older of 6 ×10-6 cm.

 Due to electric field E B potential difference appears across the depletion region and this potential VB is called “Contact Potential or Barrier Potential or Junction Barrier”.

9.12 Diffusion Current and Drift Current

The diffusion of electrons and holes across the junction depend on the concentration gradient of electrons and holes across the function. If we consider the direction of current flow as x – direction then diffusion current density for electrons is given by

Jn (diffusion) = e Dn →(1)

Where concentration gradient of electron along x—direction

Dn is the electron diffusion constant

Similarly the diffusion current density of holes is given by

Jp (diffusion) = e Dp → (2)

Where concentration gradient of holes along x—direction

Dp is the holes diffusion constant The diffusion of electrons from n-type to p-type leaves the region near the junction on n-type side positively ionized forming positive space charge region. Similarly the region near the junction on p- type side becomes negatively ionized forming negative space

charge region. Hence an electric field EB is produced in a direction opposite to that of diffusion current. Hence electrons, the minority charge carriers of p-region drift to n- region and holes, minority carries of n- region drift to p- region. The drift velocity is a propositional to the electric field strength E developed across the barrier i.e. V = µE Where µ is a constant known as the mobility of charge carriers The current density due to electron Drift is given by

Jn (drift) = (e µn n) E → (3) Where µn is the mobility of the electrons and n is the electron concentration. The current density due to hole Drift is given by

JP (drift) = (e µp p)E →(4) Where µp is the mobility of the holes and p is the holes concentration. Since diffusion and drift process occur simultaneously the total electron and hole current densities are given by

Jn (total) = Jn(diffusion) + Jn(drift)

= e Dn + e µnnE → (5)

Jp (total) = Jp (diffusion)+Jp (drift)

= e Dn + e µp pE → (6)

At equilibrium, the diffusion and drift current cancel each other for both types of charge carriers and hence

Jn (drift) + Jn (diffusion) = 0 → (7) Jp (drift) + Jp (diffusion) = 0 → (8)

9.13 Energy Band Diagram of PN Diode:

Fig. 9.17 Energy level diagram of PN junction diode

The contact potential vB across the junction is due to the potentials of depletion region

on n- side denoted by vn and p – side denoted by vp

ie : VB = Vn - Vp →(7) The energy levels of valance band, conduction band and Fermi level of both p- type and n- type semi conductors . When Pn junction is formed the Fermi levels become common for both the types formation of potential barrier is represented in fig. The contact potential separates the energy bands in p—type and n- type crystals. Since there is no net current flow at equilibrium i.e under unbiased condition , it should have common Fermi level. Hence in PN junction diode the valance and conduction band

energy levels Evp and Ecp of p- type and at higher level compared to the valance and

conduction band energy levels Evn and Ecn of n-type . The electric field EB across the junction is given by

EB = Evp - Evn = Ecp - Ecn = e VB →(8)

9.14. Biased PN Junction

When the PN junction is unbiased, it is in equilibrium and contact potential VB (= Vn- Vp) appears across the depletion region.

Fig. 9.18: Energy level diagram of biased PN junction (a) Unbiased (b) Forward biased (c) Reverse biased  If now an external voltage is applied across depletion region such that positive of the source is connected to the p-type side and negative of the source is connected

to the n-type then junction is said to be “Forward biased”. This voltage Vf appears across the depletion region. As the contact potential difference VB acts from n to p the electrostatic potential barrier is lowered and is given by VB -Vf.

 If the external voltage Vr is applied across the depletion region such that the negative of the source is connected to the p- type side and positive of source is connected to the n- type side then the junction is said to be “Reverse biased :.

Since this potential acts along VB , the electrostatic potential barrier increases. And is given by VB+Vr. thus on application of forward bias , the electric field in the transition region reduces and the application of reverse bias increases the electric field in the transition region.

 Since the width of the transition region is propositional to the square root of the electrostatic potential barrier , the width of the transition region decreases under forward bias and increases under reverse bias.

9.15. Diode Equation

The concentration of electrons in the conduction band of a semiconductor is given by E  E 2m * KT 3 ( F C ) n = 2( e ) 2 exp KT h 2

E  E ( F C ) i.e nc = Nc exp KT →(9)

3 2me * KT 2 where Nc = 2( ) h 2 Similarly the concentration of holes in the valance band of semiconductor is given by E  E ( v F ) P = NV exp KT → (10)

3 2mh * KT 2 Where NV = 2 ( ) h 2 Equations (9) and (10) are valid both for intrinsic and extrinsic semiconductors under thermal

equilibrium condition. For n- type material if we represent electron density by nn , Ec by

Ecn and EF by Efn then, equ. (9) can be written as E  E ( fn cn ) nn = Nc exp KT →(11)

For p- type material if we represent electron density by np , Ec, by Ecp and Ef by Efp then, equ. (9) can be written as

E  E ( fp cp ) Np = Nc exp KT →(12)

Similarly for an n-type material if we represent the hole density by Pn, Ef by Efn and Ev by Evn , the eqn (10) can be written as

Evn E fn Pn = Nv exp ( ) →(13) KT

For p- type materials, If we represent hole density by Pp, Ef by Efp , Ev by Evp then eq(10) can be written as

Evp  E fp Pp = Nv exp ( ) →(14) KT From equation 11 and 12, we get

Since the Fermi levels are equal, Efn = Efp

Hence = exp( Ecp - Ecn)/KT

As Ecp - Ecn= e VB where VB is the potential barrier.

= exp( e VB/KT) →(15)

Similarly from equ (13) and (14), we get

= exp ( -e VB/KT) →(16)

From equation (15):

np = nn exp( -e VB/KT) →(17)

This equation relates the electron density in p- side with electron density in n- side , when no external voltage is applied to the junction. When the junction is forward biased with voltage v, the electron density in the p- region becomes

np + ∆np = nn exp (-e (VB-V))/KT

= nn exp . exp

= np exp

Hence increase in electron density in the p- region is given by

∆np = np exp ( →(18)

during forward bias , the potential barrier decreases and hence more electrons have sufficient energy to cross the barrier and move from n – region to p—region increasing

the electron density in p- region by ∆np .diffusion electron current ie: due to forward bias is therefore given by

i.e C1 ∆np = C1np exp ( → (19)

Where c is constant depending on the property of the semiconductor and its area of the junction. Similarly the hole density in a n- region under un biased condition is given by

pn = pp exp( -e VB/KT) →(20) Increase the hole density in the n – region can be shown to be

∆pn = pn exp ( →(21)

Diffusion hole current in due to forward bias is therefore given by

Ih = c2∆pn = c2pn exp ( → (22)

Hence total current due to forward bias

I = ie + ih = (C1np+C2pn) exp ( → (23)

When the junction is reverse biased V =V1 then reverse current

I = (C1np+C2pn) exp (

At working temperature T, Voltage V.

exp ( , Hence I = ( C1np+C2pn)

If this constant steady current is denoted by I0 Then I = →(24)

This is known as diode equation. Partially it is found that I =

Where β is a numerical constant depending on the material of the diode for Ge, β = 1 & for Si,β = 2.

9.16. I-V Characteristics of PN Junction

Fig. 9.19 V-I curve of p-n diode

1) Unbiased Circuit When the function is not connected to any voltage source, it is said to be unbiased and due to barrier potential across the junction , there is no flow of change carriers and hence there is no current flow through the junction.

Fig.9.20 Unbiased p-n junction

2) Forward Biased Circuit  When positive terminal of the battery is connected to p- type and negative terminal to n- type, the function is said to be forward biased.  Since the potential barrier height is very small (0.2v), when the applied voltage exceeds that value.  The junction resistance becomes almost zero and low resistance path is established. Hence even for small increase of applied voltage, we observe large increase in circuit current such a circuit is called forward bias circuit and the current is called forward current.  When the applied voltage is above the barrier potential the forward current (in (MA) is found to increase linearly with the applied voltage. under forward bias , the d. c. Resistance is about 100Ω for Ge diode.

Fig. 9.21 Forward Biased Circuit

3) Reverse Biased Circuit  When reverse biased, a slight reverse current (in µ A) flows in the circuit. Even for large increase in bias voltage (say 20 volts), there is negligible in reverse current.  When the applied reverse voltage is high enough to break the covalent bonds of the crystal, the current rises suddenly in the reverse direction.  The breakdown reverse bias is called breakdown voltage or zener voltage. For Ge diode reverse resistance is around 1 MΩ.

Fig. 9.22 Reverse Biased Circuit

9.17 PN Diode as a Rectifier  For a PN diode forward resistance is for the order of 100Ω while reverse resistance is of the order of 1 MΩ. Hence an a.c. power source is connected across a PN junction, during positive half of the a.c. cycle diode is forward biased during negative half of the a.c cycle diode is reverse biased and hence almost no current flows.  Thus the diode allows the current to flow in only one direction during forward bias and no current during reverse bias. This unidirectional current flow character is utilized in using PN junction diode as a rectifier.  A rectifier is a device which converts alternating current (a.c) to direct or unidirectional current (d.c). A semiconductor p-n junction diode can be used as a

half-wave as well as a full-wave rectifier as it conducts in one direction only when forward biased.

9.17.1. Half Wave Rectifier  In this as only one p-n junction diode is used, rectification takes place for positive half of a.c wave form.  In circuit diagram of a half-wave rectifier is given. It consists of a transformer whose primary is connected to a.c input. AB is the secondary of the transforms.

Fig. 9.23 Half-wave rectifier (a) input a.c signal (b) the circuit and (c) output d.c signal.  The end A is connected to the p-type while the end B is connected to the n-type of the

diode through the load resistance RL .

 The d.c output is taken across RL . When the input voltage increases, the output voltage also increases and when the input drops to zero, the output also drops to zero.  During the negative half of the input cycle, the diode is reverse bias and the hence there is no current flow through the junction and load resistance. Thus the output is unidirectional and varying and exists only for half the cycle.

9.17.2. Full Wave Rectifier  In full wave rectifier two PN – junction diodes are used. Two half-wave rectifiers are combined such that the output current flows during the entire cycle of the input voltage.  The circuit diagram of a central taped full-wave rectifier across the secondary AB of

the transformer two p-n junction diodes D1 and D2 are connected.

 A load resistor RL is connected between the centre tap point c of the secondary of the transformer and the line connecting the n-sides of the two diodes.  During the first half of the input cycle, if A is positive with respect to centre tap point

c, then diode D1 is forward biased and during that period B will be negative with respect to C and hence the diode D2 will be reverse biased.

 Hence during the half of the cycle current flows through the closed circuit D1 , R1

point C and D1 . During next half of the input cycle B is positive with respect to C and hence D2 is forward biased and D1 is reverse biased. During this half of the period of

the cycle current flows through the closed circuit D2, RL, point C and D2.  Thus we get d.c output during the full cycle. The output is unidirectional but amplitude varying using proper filter circuits, the amplitude of the output could be made constant.

Fig. 9.24 Full-wave rectifier 9.18. Comparison of a Half Wave Rectifier and Full Wave Rectifier:

Average value of current or I(dc) The average value of current I(dc) over one complete cycle is that direct current which sends the same charge through the circuit as sent by the varying current.

Effective value of current for Irms: The effective value of the current or root mean square value of the current is square root of the average of the square of the instantaneous values taken over the whole cycle. Efficiency: The efficiency of a rectifier is defined as the ratio of d.c. output power p(d.c) to the a.c .

Input power P(a.c) supplied to the rectifier. Efficiency =

s.no Quantity Half-wave rectifier Full-wave rectifier 1 No. of diodes 1 2

2 Peak alternating current I0 = I0 =

3 Average value of current (or) I(d.c)

Effective value of current (or) I(r.m.s)

Efficiency ( )

9.19. P - N Junction as a light source - Light Emitting Diode:  Emission of radiation from a solid when it is supplied with some form of energy is known as LUMINESCENCE. Whatever the form of energy input to the luminescent material, the final stage in the process is an electronic transition between two energy

levels, E2 and E1 (E2>E1). With the emission of radiation of wavelength given by

= E2-E1 = Eg

i.e) = → (25)

Usually E1 and E2 are two groups of energy levels and hence instead of a single wave length emission, a band of wavelength is emitted.  PN junction diode operated under forward bias gives rise to injection luminescence. Under forward bias, majority carriers from both sides of the junction across the internal potential barrier and enter the other side of the junction where there are minority carriers. This process is called minority carrier injection. These excess minority carriers while diffusing away from the junction undergo radiative recombination with majority carriers emitting photons.  In reverse bias, no carrier junction takes place and consequently no light is emitted.  The number of radiative recombination’s is proportional to the carrier junction rate and hence to the total current flowing. The current voltage (I-V) relationship for a diode can be written as

I = I0 →(26)

Where I0 is the saturation current and varies between 1 and 2 depending on the

Semiconductor and temperature. 263

Fig. 9.25 Injection of minority carriers and subsequent radiative recombination with the majority

carriers in a forward biased p-n junction.

 Actually, all semiconductors diodes produce radiation when electrons from the conduction band recombine with holes in the valance band and the emitted radiation is absorbed by the surrounding semiconductor material.  In an LED, the semiconductor as a high energy gap and the junction is constructed such that the radiation from the junction can escape.

LED Materials

The main requirements for suitable LED materials are

1) It must have an energy gap of appropriate width (the energy gaps greater than or equal to

about 2ev are required to obtain visible radiation).

2) Both p and n types must exist with low resistivities.

3) Efficient radiative pathways must be present.

Silicon carbide (sic) can be doped as both n and p type doping with B, Al, Sc, and Be gives rise to yellow, blue, green and red emission respectively. But since SIC is an indirect band gap material radiative transition efficiency is low. Ga, As, Ga P and the

ternary alloy GaAs1- xPx are most important III -IV components used.

Gallium arsenide (GaAs)

This is a direct band gap semiconductor with Eg = 1.443ev ( g = 860 nm); p-n junctions are readily formed , when GaAs is doped with silicon , based on growing condition , either silicon can replace Ga or As. On Replacing Ga it acts as a donor and on As replacement it acts as an acceptor. In silicon doped diodes the emission lies between

910 and 1020 nm. This has high quantum efficiency. These diodes have become standard near – IR emitting devices.

Gallium Phosphide (GaP)

This is an indirect band gap semiconductor with Eg = 2.26 ev ( g = 549 nm). Group 5 elements such as n and Bi are commonly used as do pants. Since they have the same valency as the atoms which they replace (p), they do not form normal donor accepts states. As the impurity atom differs considerably from the atom it replaces both in the size and its tendency to acquire negative charge, it acts as a trapping centre called ISOELECTRONIC TRAPS. Once a carrier is trapped, the resulting potential attracts a carrier of the opposite charge and emits radiation. The nitrogen isoelectronic traps result in peak emission at about 565 nm i.e. green radiation. At high nitrogen concentrations the temperature emission shifts to 590nm (i.e. yellow). Using Zn, O for double doping , emission at 690nm(i.e. red) is also possible from GaP.

Gallium arsenide phosphide

This material changes from being a direct band gap material (when x<0.45). to being an

indirect band gap material (when x>0.45).at changeover point , the Eg 2.1v. from

GaAs0.6 P0.6 ; red emission is obtained From GaAs0.35 P0.85; N, orange and yellow emission are obtained respectively.

Gallium Aluminum arsenide (Gax Al1-x As)

In hetero junction formed between n- type Ga0.3 Al0.7 As an p-type Ga0.6 Al0.4 As(with zinc as dopant). Radiative recombination takes place and emission with higher efficiency occurs at 650nm. When Si is used as dopant to form both donor and acceptor states, by varying Al concentration, band gap can be varied and emission between 870 and 890 nm can be obtained.

Construction of LED

 The typical construction of LED is shown in the figure. LED has to be designed and constructed in such a way that most of the radiative recombinations takes place from the side of the junction nearest to the surface so that loss due to reabsorption is minimized

Fig. 9.26 Construction of typical LED

 For this p-n junction layer has to be parallel and closer to the surface layer. The LED is constructed on a GaP n- doped substrate. A thin epitaxial p- doped GaP layer is grown on the top of this substrate.  Electrical contacts are made leaving as much of the upper surface of the p- material uncovered.  The recombination takes place and radiation is generated between p and n layers i.e. p-n junction. Since Gap layer is transparent, the radiation escapes through the top layer.  To the bottom electrode a reflective layer is added to improve the efficiency.  Although the internal quantum efficiencies of some LED materials can approach 100%, the external efficiencies are much lower. The main reason is that most of the emitted radiation strikes the material interface at angles greater than critical angle and so lost due to total internal reflection. Since the III-IV materials have high refractive indices, the critical angles are small.  For example in GaP and air interface, since the critical is 170, the rays’ incident at an angle greater than this are lost.  There are two ways of improving the situations; the first is to ensure that most rays strike surface at angles less than critical angle. This is achieved shaping in the semiconductor/ air interface in to a hemisphere; in this the p- material is made in to a hemisphere dome. More radiation then strikes the semiconductor/ air interface at less than the critical angle and hence are not lost. This technique is too difficult and more expensive.  The second technique is to encapsulate the junction in a transparent plastic medium of high refractive index. Moulding the plastic in to an approximately hemispherical shape, loss at the plastic / air interface is minimized.  Intensity of the emission is an important optical parameter of an LED. The relative intensity increases with LED current. LED output is not monochromatic but contains considerable band width ( 30 to 40 nm) around the central wavelength.

 Output intensity is temperature dependant. If the temperature increases by 250 C, the intensity decreases by 25%.  Another very important characteristic of an LED is its fast response time, typically 90ns for yellow and red and 500ns for green. This response time makes LEDs suitable as a source in communication links.

Fig. 9.27 Two methods used to reduce reflection looses in LEDs.

(a) Hemispherical dome made of p material and (b) Plastic mould in hemispherical shape.

Advantages of LEDs in electronic Display

1) Output is bright and the intensity can be controlled easily by varying current.

2) They can be operated over a wide range of temperature 0 to 700 C

3) Very fast response time in the order of ns and hence very useful as source for optical Communication.

4) Available in different colors.

5) Very small in size and hence can be closely packed for high density display.

6) As long life (about 105 hours) and high degree reliability.

7) The viewing angle is not limited.

8) Very rugged and hence suitable for any environment.

Disadvantages of LEDs in Electronic Display

1) Consume more energy (i.e. high power) when compared to LCDs.

2) Relatively costlier.

9.20 Liquid Crystal Display(LCD)s

 Liquid crystal displays come under passive types of display since no light generation is required.  We use them as display elements in digital watches and pocket calculators.  They consume the least power of all common display devices. There are two basic types of LCDs available. These are (a) Reflective type, required front illumination and (b) Transmissive type required rear illumination.  LCD device consists of cell formed between two glass plates each with an optically transparent but effectively conductive coating.  The cell has a thickness of about 10 µm or less than it is filled with a liquid crystal.

Characteristics of liquid crystal Materials:

Fig. 9.28 (a) Nematic ordering (b) Cholesteric ordering in liquid crystals.

 The liquid crystal state is a phase of matter exhibited by a large number of organic materials over a restricted temperature range.  The material becomes a crystalline solid at the lower end of the temperature range, while it changes in to a clear liquid at the upper end. Within this range it combines some of the optical properties of solids with the fluidity of liquids.  When they are in liquid crystal phase, the rod like shape of their molecules can take up certain orientations relative both to each other and to the liquid crystal surface.

 In any small volume the time – averaged preferred orientation of the molecules is described in terms of a unit vector called a director.  There are three basic types of ordering in liquid crystals namely Nematic, Cholestric and Smectic. The first two types are used in display devices.  In nematic ordering, the molecules are aligned parallel to each other. But they are free to above relative to each other so that the phase has liquid properties.  A nematic liquid crystal molecule usually consists of two Benzene rings linked with a central group. A typical example is 4- methoxylenzylidene-4-butylanaline (MBBA). This has the chemical formula

 MBBA exhibits liquid crystalline property over the temperature range 200c to 470c. In the cholestric phase the molecules arrange in a large number of planes.  In each plane all molecules are arranged along the same direction but from plane to plane they show a progressive change in a director direction is called the PITCH P.  In twisted nematic cell, the opposite walls of the cell are treated to produce a homogenous arrangement but the molecular alignment direction at the walls is at right angle to each other. Thus the molecules undergo a 900 rotation across the cell.

 With electric field (E >> Ec) across the cell the molecular alignment becomes parallel to the applied field.  When a beam of polarized light passes through the cell in the absence of electric field the plane of polarization turns through 900.  But when the plane of polarized beam passes through the cell with electric field applied there is no rotation of the plane of polarization.

Fig.9.29 Behavior of the molecules in a liquid crystal cell with

(a) No applied voltage (E) (b) with a voltage applied such that E > Ec.

Action of LCD Display Device:

 When used as display device, the twisted nematic cell is sandwiched between two pieces of Polaroid’s.

 The polarizing direction of these polarized correspond to the molecular ordering direction of the cell walls i.e. the polarizing directions of the two Polaroid’s are perpendicular to each other.  Polaroid1 is in front of the LCD cell which Polaroid 2 is at its back. Beyond Polaroid 2 a reflector is placed in reflective mode.  After passing through Polaroid 1, plane polarized light rotates through 900 as it transverses the cell, when no field is applied ; then passes through the second polarizer which is held perpendicular to the first and then reflects back at the mirror and the rays retrace the path. Thus with no field applied the device reflects incident radiation and appear bright.  When a field is applied, the direction of polarization of the light travelling the cell is not rotated and hence stopped by the second polarizer. No, light is reflected from the device. And hence it appears dark. This means that in the absence of the field the incident light is reflected while it is blocked when the field is applied.  The amount of light reflected from LCD as a function of applied voltage is shown schematically. The reflectance, initially constant, falls rapidly beyond a critical

voltage Vc and again becomes constant beyond a voltage Vsat A typical value of Vsat is 3v. Since d.c. operation tends to shorten the operating lifetime of the device, a.c. Waveforms are invariably used. A square waveform of frequency between 25H and 1 KH is often used. Transmission LCD displays do not have the reflector. They need rear Illumination for operation.

Fig.9.30 Amount of light reflected from a liquid crystal displays as a function of applied voltage V.

Advantages of LCDs

1) Consume much lesser energy when compared to LED’s.

2) Utilizes the light available outside and no generation of light.

3) Since very thin layer of liquid crystal is used. More suitable to act as display elements ( in digital watches , pocket calculators, etc. ).

4) Since reflectivity is highly sensitive to temperature measuring sensor. 270

5) Very cheap.

Disadvantages:

1) Angle of viewing is very limited.

2) External light is a must for display.

3) Since not generating its own light and makes use of external light for display, contrast is poor.

4) Cannot be used under wide range of temperature.

9.21. P - N junction as a detector – Junction Photodiode

 The function of the photo diode junction is the opposite an LED function. In an LED, photons are released in response to the current flow through the junction. In a photo diode, the photons are absorbed resulting in the generation of the carriers that manifest as current through the junction.  In p-type material holes are majority carriers and in n-type material electrons are minority carriers on formation of a junction, on either side of the junction they diffuse, combine and neutralize one another. Hence a region called DEPLETION REGION is formed on the either side of the junction when there are no free carriers.

Fig. 9.31 The generation & subsequent separation of an electron – hole pair by photon absorption within the depletion region of a p –n junction

 Only ionized atoms of opposite polarity remain i.e in p- type material negatively ionized atoms and in n- type material positively ionized atoms remain. This action generates a barrier voltage that opposes further diffusion of the free carriers in to the junction.

 When an electron – hole pair is generated by photon absorption with this region, the internal field causes the electron and hole to separate. This charge separation can be detected in two ways. If a device is left an open circuit, an externally measurable potential appear between p and n regions. This is known as the PHOTO VOLTAIC

MODE of operation. Instead, if we short circuit the device externally, an external current flows between the p and n regions. This is known as the PHOTOCONDUCTIVE MODE of operation. Under photoconductive mode it is usually operated under reverse bias conditions.  When forward bias is applied to the junction, it opposes the barrier voltage, reduces the width of the deflection region and increases the junction capacitance. When the bias reaches the barrier voltage, the depletion region is eliminated and the junction becomes conductive.  When reverse bias is applied, the depletion region is widened, the junction capacitance reduced, and the junction says non-conductive. The reverse bias junction can conduct current when free carriers are generated in the junction by radiation of sufficient energy.

Advantages of Photoconductive mode:

The following are the advantages of photo conductive mode:

1) Linear response

2) Faster response

3) Better stability and

4) Greater dynamic range.

Construction of a Junction Photodiode

 The most common semiconductor material used for photo diodes is silicon. It has an energy gap of 1.14ev. It provides excellent photodiodes with quantum efficiencies up to 80% at wave length between 0.8-0.9µm. It shows a typical silicon photodiode structure for photoconductive operation.  A junction is formed between heavily doped p- type material ( p+ ) and fairly lightly doped n-type material so that the depletion region extends well into the n- material . The

p+ layer is made fairly thin. Metallic contracts can be made directly to the p+ material but to obtain an ohmic contact to the n – material an intermediate n+ layer is formed.

Fig.9.32Typical silicon photo diode structure for photoconductive operation

 The penetration of photons through the layer depends on the wave length of radiation shorter wave length ( ultra violets ) are absorbed at the surface, while the longer wave length (infrared) can penetrate deep in to the structure .  A wide response photo diode should therefore have a thin p- layer and a thick depletion layer so that most of the photons are absorbed in the depletion layer. Hence the response time of the diode is mainly determined by the thickness of the depletion layer.  The thickness of the layer can be controlled by the doping level of the n- layer lower doping level increase the thickness of the depletion layer and reduce the diode capacitance . This technique is used in the PIN type photo diode, where a thick layer of low doped and highly resistive n- type material is inserted between the p and n- layers. This middle layer is called the intrinsic or I- layer, hence the name PIN diode.  The PIN diode has a relatively thick depletion layer which, with modest reverse bias of 5v , can be extended to the bottom of the intrinsic layer . The result is a photo diode with a faster response time and an extended spectral response. The spectral response of the junction photo diodes ranges from ultraviolet to infrared. It depends mainly on the semiconductor material, and also on the junction design and window material.  The silicon photo diode is one of the most popular of all radiation detectors in the wavelength range 0.4 to 1 µm. it has the following features.

(1) High quantum efficiency

(2) Small size

(3) Good linearity of response.

(4) Large band width

(5) Simple biasing requirements and

(6) Relatively low cost. 9.22 Solar Cell 9.22.1 Definition

Solar cell is a P – N junction semiconducting diode which converts light energy into

electrical energy. The energy of light is transmitted by photons, small packets or quantums of light. Electrical energy is stored in electromagnetic fields, which in turn can make a current of electrons flow. Thus a solar cell converts light, a flow of photons, to electric current, a flow of electrons.

Construction

There are several technologies to construct a solar cell. One of them is Quantum Dot (QD) Solar Cells. These are built up of a semiconductor (silicon) coated with a very thin layer of Quantum dots. Quantum dots is just a fancy name of crystals in the size range typically a few nanometers in diameter. These crystals are mixed into a solution and placed on a piece of silicon which is rotated really fast. The crystals are then spread out due to the centrifugal force. The reason these quantum dots are given so much attention is that normally one photon will excite one electron creating one electron-hole pair. The energy loss is the original energy of the photon minus the energy needed to excite the electron(also called the band gap). However, when a photon hits a quantum dot made of the same material, there may be several electron-hole pairs created, typically 2-3, but 7 has been observed.

Working principle

The solar cell is basically a p – n junction diode that converts sunlight directly to electricity. The working principle of solar cells is based on the photovoltaic effect. In general, the photovoltaic effect means the generation of a potential difference at the junction of two different materials in response to visible or other radiation.

The basic processes behind the photovoltaic effect are

1. Generation of the charge carriers due to the absorption of photons in the materials that form a junction,

2. Subsequent separation of the photo-generated charge carriers in the junction,

2. Collection of the photo-generated charge carriers at the terminals of the junction.

The solar cell works in three steps

1. Photons in sunlight hit the solar panel and are absorbed by semiconducting materials, such as silicon. 2. Electrons (negatively charged) are knocked loose from their atoms, causing an electric potential difference. Current starts flowing through the material to cancel the potential and this electricity is captured. Due to the special composition of solar cells, the electrons are only allowed to move in a single direction. 3. An array of solar cells converts solar energy into a usable amount of direct current (DC) electricity.

Fig. 9.33 A typical solar cell consists of a glass or plastic cover, an antireflective coating, a front contact to allow electrons to enter a circuit, a back contact to allow them to complete the circuit, and the semiconductor layers where the electrons begin and complete their journey.

Explanation

The action of the solar cell is explained as follows:

 When a p – n junction diode is exposed to light, the photons are absorbed and electron pairs are generated in both in both the p – side and n – side of the junction, as shown in the figure below. The electrons and holes that are produced over a small distance from the junction reach the space charge region X by diffusion.  The electron – hole pairs are then separated by the strong barrier field that exists across the region X. The electrons in the p – side slide down the barrier potential

to move to the n – side while the holes in the n – side while the holes in the n – side move towards the p – side.  When the p – n junction diode is open circuited, the accumulation of electrons and

holes on the two sides of the junction gives rise to an open – circuit voltage V0. If a load resistance is connected across the diode, a current will flow in the circuit. The maximum current, called the short – circuit current is obtained when an electric short is connected across the diode terminals. Note that the current flows as long as the diode is exposed to sunlight and the magnitude of the current is proportional to the light intensity.  Solar cells are used extensively in satellites and space vehicles as most important long duration power supply. Solar cells are constructed with Silicon, Germanium, Gallium arsenide, Cadmium sulphide and with many other semiconductors, and in various device configurations.

Fig. 9.34 (a) p-n junction solar cell with load resistance,RL (b) Diffusion of electrons & Holes (c) Energy

band diagram of solar cell (d) Formation of the open circuit voltage Vo

9.22.2 Efficiency of a Solar Cell (η):

 Efficiency is the ratio of the electrical power output Pout, compared to the solar power

input, Pin, into the PV cell. Pout can be taken to be PMAX since the solar cell can be

operated up to its maximum power output to get the maximum efficiency.

Pin is taken as the product of the irradiance of the incident light, measured in W/m2 or in suns (1000 W/m2), with the surface area of the solar cell [m2]. The

maximum efficiency (ηMAX) found from a light test is not only an indication of the performance of the device under test, but, like all of the I-V parameters, can also be affected by ambient conditions such as temperature and the intensity and spectrum of the incident light. For this reason, it is recommended to test and compare PV cells using similar lighting and temperature conditions.

 The efficiency of a solar cell may be broken down into Reflectance efficiency, Thermodynamic efficiency, Charge carrier separation efficiency & Conductive efficiency.  The overall efficiency is the product of each of these individual efficiencies.  A solar cell usually has a voltage dependent efficiency curve, temperature coefficients, and shadow angles.  Due to the difficulty in measuring these parameters directly, other parameters are measured instead: thermodynamic efficiency, quantum efficiency, integrated

Quantum efficiency, VOC ratio, and fill factor. Reflectance losses are a portion of the quantum efficiency under "external quantum efficiency".

Recombination losses make up a portion of the quantum efficiency, VOC ratio, and fill factor. Resistive losses are predominantly categorized under fill factor,

but also make up minor portions of the quantum efficiency, VOC ratio.  Another way to increase the efficiency is to use several layers solar cells with different band gaps in a stack. Each layer will utilise light with different wavelengths, and in this way we can get cells with a higher efficiency.

9.22.3. Different Parameters of a Solar Cell

1) Fill Factor (FF):

 The Fill Factor (FF) is essentially a measure of quality of the solar cell. It is

calculated by comparing the maximum power to the theoretical power (PT) that would be output at both the open circuit voltage and short circuit current together. FF can also be interpreted graphically as the ratio of the rectangular areas depicted in Figure below.

Fig. 9.35 Getting the Fill Factor from the I-V Sweep

 A larger fill factor is desirable, and corresponds to an I-V sweep that is more square-like. Typical fill factors range from 0.5 to 0.82. Fill factor is also often represented as a percentage.  The fill factor is also defined as the ratio of the actual maximum obtainable power to the product of the open circuit voltage and short circuit current. This is a key parameter in evaluating the performance of solar cells. Typical commercial solar cells have a fill factor > 0.70. Grade B cells have a fill factor usually between 0.4 to 0.7. Cells with a high fill factor have a low equivalent series resistance and a high equivalent shunt resistance, so less of the current produced by the cell is dissipated in internal losses. This is the available power at the maximum power

point (Pm) divided by the open circuit voltage

(VOC) and the short circuit current (ISC):

 The fill factor is directly affected by the values of the cell's series and shunt

resistances. Increasing the shunt resistance (Rsh) and decreasing the series

resistance (Rs) lead to a higher fill factor, thus resulting in greater efficiency, and bringing the cell's output power closer to its theoretical maximum.

2) Short Circuit Current (ISC)

The short circuit current ISC corresponds to the short circuit condition when the impedance is low and is calculated when the voltage equals 0.

I (at V=0) = ISC

ISC occurs at the beginning of the forward-bias sweep and is the maximum current value in the power quadrant. For an ideal cell, this maximum current value is the total current

produced in the solar cell by photon excitation.

ISC = IMAX = Iℓ for forward-bias power quadrant

3) Open Circuit Voltage (VOC)

The open circuit voltage (VOC) occurs when there is no current passing through the cell.

V (at I=0) = VOC

VOC is also the maximum voltage difference across the cell for a forward-bias sweep in the power quadrant. VOC= VMAX for forward-bias power quadrant.

4) Maximum Power (PMAX), Current at PMAX (IMP), Voltage at PMAX (VMP)

The power produced by the cell in Watts can be easily calculated along the I-V sweep by the

equation P=IV. At the ISC and VOC points, the power will be zero and the maximum value

for power will occur between the two. The voltage and current at this maximum power point

are denoted as VMP and IMP respectively.

Fig. 9.36 Maximum Power for an I-V Sweep

9.22.4. Theory of I-V Characterization of a Solar Cell:

 PV cells can be modeled as a current source in parallel with a diode. When there is no light present to generate any current, the PV cell behaves like a diode. As the intensity of incident light increases, current is generated by the PV cell, as illustrated in figure below.

Fig. 9.37 I-V Curve of PV Cell and Associated Electrical Diagram

 In an ideal cell, the total current I is equal to the current Iℓ generated by the

photoelectric effect minus the diode current ID, according to the equation

-19 where I0 is the saturation current of the diode, q is the elementary charge 1.6x10

Coulombs, k is a constant of value 1.38x10-23J/K, T is the cell temperature in Kelvin, and

V is the measured cell voltage that is either produced (power quadrant) or applied

(voltage bias).

 A more accurate model will include two diode terms, however, will concentrate on a

single diode model. Expanding the equation gives the simplified circuit model shown

below and the following associated equation, where n is the diode ideality factor

(typically between 1 and 2), and RS and RSH represents the series and shunt

resistances.

Fig. 9.38 Simplified Equivalent Circuit Model for a Photovoltaic Cell

 The I-V curve of an illuminated PV cell has the shape shown in Figure below as

the voltage across the measuring load is swept from zero to VOC, and many performance parameters for the cell can be determined .

Fig. 9.39 Illuminated I-V Sweep Curve

9.23 APPENDIX

9.23.1. Other parameters of a solar cell 281

1) Solar cell efficiency is also defined as the ratio of the electrical output of a solar cell to the incident energy in the form of sunlight. The energy conversion efficiency (η) of a solar cell is the percentage of the solar energy to which the cell is exposed that is converted into electrical energy. This is calculated by dividing a

cell's power output (in watts) at its maximum power point (Pm) by the input light 2 2 (E, in W/m ) and the surface area of the solar cell (Ac in m ).

2) Photovoltaic Cell Performance

Photovoltaic (PV), or solar cells use the energy in sunlight to produce electricity. The amount of electricity produced depends on the quality of the light available and the performance of the PV cell. The measurements of conversion efficiency and quantum efficiency are required to characterize the performance of PV cells. Based on these results, the redesign aspects of the cell e.g., material compositions or thicknesses of layers to improve performance.

3) Photovoltaic Cell Quantum Efficiency

Quantum efficiency (QE) is the ratio of the number of charge carriers collected by a

photovoltaic (PV) cell to the number of photons—or packets of light—of a given energy shining on the solar cell. Quantum efficiency therefore relates to the response of a solar cell to the various wavelengths in the spectrum of light shining on the cell. The QE is given as a function of either wavelength or energy. If all the photons of a certain wavelength are absorbed and the resulting minority carriers (for example, electrons in a p-type material) are collected, then the QE at that particular wavelength has a value of one. The QE for photons with energy below the band gap is zero.

4) Two types of quantum efficiency of a solar cell are often considered

 External QE includes the effect of optical losses such as transmission through the cell and reflection of light away for the cell. However, it is often useful to look at the QE of the light that remains after the reflected and transmitted light has been lost.

 Internal QE refers to the efficiency with which light not transmitted through or reflected away from the cell can generate charge carriers—specifically electrons and holes—that can generate current. By measuring the transmission and reflection of a solar device, the external QE curve can be corrected to obtain the internal QE curve.

9.23.2. Response Time of Photodiodes

Three main factors limit the speed of response of photodiodes; they are:

1) Diffusion time for carriers to the depletion region

Carrier’s diffusion is a relatively slow press. The time taken for carriers to diffuse a distance‘d’ may be written as

Tdiff =

where dc is the minority carrier diffusion coefficient. At wave lengths near the band gap limit and at fairly reverse bias levels the speed of detection is limited by diffusion. For a narrow square optical pulse signal the response of a p-n junction detector. The carriers generated within the depletion region respond rapidly, while those generated outside gives rise to slow tail. By using PIN structure or increasing the reverse bias applied to photo diode, the relative importance of the tail can be decreased.

2) Drift Time of Carriers through the Depletion Region

Normally for the field applied in practice, the drift velocities of carriers in semiconductors tend to saturate. Hence we may assume that the carrier move with a

constant velocity Vsat . Then for a carrier generated at one edge of the depletion region, to transverse the full depletion layer width W, the transit time taken is

Tdrift =

3) Junction Capacitance Effects

A diode under reverse bias exhibits a voltage – dependent capacitance caused by the variation in stored change at the junction. A junction diode has a capacitance given by

1/2 Cj = ( )

Where A is the light sensitive area of the photo diode and Nd is the doping concentration.

The electrical band width, ∆fel is defined as the frequency range over which the output is

above of its maximum value. Usually Rsh >> RL and Rs << RL under this condition.

∆fel =

By reducing the junction capacitance, the band width can be increased. By reducing the

diode area A, reducing the doping level Nd or increasing the reverse bias voltage v, the junction capacitance Cj can be reduced. But there are difficulties associated with each of

these diode areas ‘A’ cannot be made very small. Reduction in Nd and increase in V cause the depletion layer width to increase silicon photodiodes with response time’s less than 1 ns (i.e) band widths greater than 1 GH are readily available

4) Effect of Temperature

According to diode equation, I = I0 exp ( 1

Hence with increase of temperature, the circuit current has to increase. For both Ge and Si the increase is about 7%, this means that for rise of 100c the current nearly doubles.

5) D.C. Resistance of a PN Junction:

The static or d.c. resistance Rdc pn junction diode is defined as the ratio of the applied voltage to the total diode current.

Rdc = = 6)

A.C. Resistance of a PN Junction The dynamic or a.c. resistance of diode is defined as the ratio of a small change in voltage to a small change in current.

ra.c = = =

9.24 Introduction to Acoustics of Buildings and Acoustic Quieting The branch of physics that deals with the process of generation, propagation and hearing (reception) of sound in a room, be it a small room or an auditorium is called acoustics. Architectural acoustics, also called acoustics of buildings, deals with the behavior of sound waves in a closed space. It deals with the design and construction of acoustically good buildings, music halls, recording rooms and movie theatres, where the audience receives the best sound quality.

9.25 Reverberation and Time of Reverberation A sound produced inside a hall will propagate in all directions. Sound waves incident on the surfaces of walls, floor, ceiling and furniture inside a hall, will be multiply reflected. As the source of sound is turned off, the listener hears the sound with gradually

reducing intensity for some time due to the persistence of sound by multiple reflections at different places in the room.

A listener inside the hall will receive the sound waves directly from the source, as well as the reflected waves. The persistence of audible sound even after the source of sound is turned off is called Reverberation. The time taken by the sound intensity to fall to one millionth of its Initial intensity i.e., the intensity just before the source of sound is

turned off is called Reverberation time.

9.25.1. Jaeger’s method for derivation of Sabine’s formula According to Sabine’s law, the reverberation time T is seconds is expressed as, T=

Where, V is the volume of the hall in m3 and ∑aS is given by,

∑aS=a1S1+a2S2+…………………………………+anSn Here, a1, a2 ……………an are the absorption coefficients of the materials in the hall whose surface areas exposed to sound are S1, S2……………….Sn respectively, measured in m2. The average value of absorption coefficient a is given by, a= =∑

or ∑ aS=aS → (1) By statistical method, Jaeger showed that the sound travels a distance of (4V/S) on an average, between two successive reflections. This is known as mean free path. Here, V is the volume of the room and S is the total surface area of all the surfaces exposes to that sound in the room. Time taken between two successive reflections = , where is the velocity of sound.

Number of reflections per second =

The average number of reflections in time t =

Let, a be the fraction of sound absorbed at a single reflection. Then, fraction of sound 2 reflected is (1-a). After two reflections, the fraction of sound reflected is, (1 a)

.Similarly, after a time (S t/4V), the fraction of sound reflected will be, (1 a)(Svt/4V)

Let be the initial velocity of sound and It the intensity after time t, then (Svt/4V) It=Io (1 a) → (2)

When, t=T (reverberation time), It/Io= Substituting the above values in equation (3), = = (1-a)(SVT/4V) → (3)

Taking natural logarithm on both sides, ( ) = (1 a)

T = ( )

(or) T= 2.3026 )

= where, =330 m/s

= ( ) - a)

= → (4)

Substituting the value of a S from equation (1),

This is Sabine’s formula for reverberation time. Reverberation time is, i. Directly proportional to the volume of the auditorium. ii. Inversely proportional to the areas of sound absorbing surfaces such as ceiling, walls, floor and other materials present inside the hall and iii. Inversely proportional to the total absorption.

9.26 Basic requirements of acoustically good hall

The basic requirements of an acoustically good hall are, 1) The volume of the auditorium is decided by the type of programme to be conducted there and also the number of seats to be accommodated. A musical hall requires a large volume where as a lecture hall requires a smaller volume. In deciding the volume of the hall, its height plays an imp role than its length and breadth. The ratio between the ceiling height and breadth should be 2:3. In deciding the volume of the hall, the following guidelines may be followed. i) 3.74-4.2 m3 per seat in cinema theatres.

ii) 2.8-3.7m3 per seat in lecture halls.

iii) 4.2-5.6m3 per seat in musical halls. 2) The shape of the wall and ceiling should be so as to provide uniform distribution of sound throughout the hall. The design of a hall requires smooth decay and growth of sound. To insure these factors, the hall should have scattering objects,

walls should have irregular surface and walls must be fixed with absorptive materials. 3) The reverberation of sound in an auditorium is mainly due to multiple reflections at various surfaces inside. The reverberation should be optimum i.e., neither too large nor too small. The reverberation time should be 1-2 seconds for music and 0.5-1 sec for speech. To control the reverberation, the sound absorbing materials are to be chosen carefully. 4) The sound heard must be sufficiently loud in every part of the hall and no echoes should be present. 5) The total quality of the speech and music must be unchanged i.e., the relative intensities of the several components of a complex sound must be maintained. 6) For the sake of clarity, the successive syllables spoken must be clear and distinct i.e., there must be no confusion due to overlapping of syllables. 7) There should be no concentration of sound in any part of the hall. 8) The boundaries should be sufficiently sound proof to avoid noise from outside. 9) There should be no echelon effect. 10) There should be no resonance within the building. 11) The hall must be full of audience.

9.27 Absorption Co-efficient  The co-efficient of absorption of a material is defined as the ratio of the sound energy absorbed by the surface to that of the total sound energy incident on the surface i.e., Absorption co-efficient, a=

 An open window is considered as an ideal absorber of sound. The unit of sound absorption is open window unit or Sabine. A 1m2 Sabine is equal to the amount of sound energy that is absorbed or passed through one square meter area of open window.  Absorption co-efficient of a surface is also defined as the reciprocal of its area which absorbs the same sound energy, as absorbed by unit area of open window.

9.27.1. Determination of Absorption Co-efficient Method-1 The first method is based on the determination of standard times of reverberation in the room, without and with the sample of the material inside the room.

If T1 is the reverberation time without the sample inside the room, then applying Sabine’s formula, = =

Time is measured with the sample inside the room. =

Where, is the absorption coefficient of the area Now, =

= → (1)

Hence, knowing the terms on the right hand side of the equation (1), the absorption

coefficient of the given material can be calculated.

Method-2 This method consists of finding times of decay of the steady energy density, to the bear

audibility for two sources of power outputs and respectively. From the equation of the decay of energy density,

E= Em Where, = and =

- Velocity of sound A - ∑ ads, the total absorption of all the surfaces on which sound falls. E – Energy density in the room, t seconds after the source is cut off.

Let and be the respective times of decay of energy density to the base audible limit for sources of power outputs and respectively. Then,

= and =

Dividing the two equations,

 A=

(or) aS=

Where, ‘a’ is the average coefficient of absorption,

‘a’ can be calculated knowing the quantities on the right hand side of the equation.

9.28 Factors affecting the Architectural acoustics and their remedies Following factors affect the architectural acoustics. 1) Reverberation

 In a hall, when reverberation is large, there is overlapping of successive sounds which results in loss of clarity in hearing. On the other hand, if the reverberation is very small, the loudness is inadequate. Thus, the reverberation time for a hall should neither to be too large nor too small. The preferred value of reverberation time is called the Optimum reverberation time.  Experimentally it is observed that the time of reverberation depends upon the size of the hall, loudness of sound and on the kind of the music for which the hall is used.  For a frequency of 512 Hz, the best time of reverberation lies between 1 and 1.5 sec for small halls and for large ones, it is up to 2-3 seconds.

Remedy: The reverberation can be controlled by the following factors. i. By providing windows and ventilators which can be opened and closed to make the value of time of reverberation, optimum ii. Decorating the walls by pictures and maps. iii. Using heavy curtains with folds. iv. By lining the walls with absorbent materials such as felt, fiber board etc. v. Having full capacity of audience. vi. By covering the floor with carpets. vii. By providing acoustic tiles.

2) Loudness With large absorption, the time of reverberation will be smaller and the intensity of sound may go below the level of hearing. Sufficient loudness at every point in the hall is an important factor for satisfactory hearing.

Remedy: The loudness may be increased by, i. Using large sounding boards behind the speakers and facing the audience. ii. Low ceilings are of great help to reflect the sound energy towards the audience. iii. Providing additional sound energy with the help of equipments like loud speakers. For uniform distribution of intensity throughout the hall, the loudspeakers should be polished carefully.

3) Focusing If there are focusing surfaces such as concave, spherical, cylindrical or parabolic ones on the walls or ceiling of the hall, they produce concentration of sound in particular regions, while in some other parts, no sound reaches at all. In this way, there will be regions of silence.

Remedy: For uniform distribution of sound energy in the hall, i. There should be no curved surfaces. If such surfaces are present, they should be covered with absorbent material.

ii. Ceiling should be low. iii. A paraboloidal reflected surface, with the speaker at the focus is also helpful in sending a uniform reflected beam of sound in the hall.

4) Echoes An echo is heard when direct sound waves coming from the source, and it’s reflected wave, reach the listener with a time internal of about 1/7 second. The reflected sound arriving earlier helps in raising the loudness while those arriving later produce echoes and confusion. Remedy: Echoes may be avoided by covering the long distant walls and high ceiling with absorbent material.

5) Echelon effect A musical note produced due to the combination of echoes, having regular phase difference is known as Echelon effect. The reflected sound waves from regularly spaced reflecting surfaces such as equally spaced stair cases or a set of railings produce musical note due to the regular succession of echoes of the original sound to the listener. This makes the original sound confused or unintelligible.

Remedy: Echelon effect can be avoided by forming the staircases with unusual spacing between them and covering them with sound absorbing materials like carpet.

6) Resonance Sometimes, window panes loosely fitted wooden portions, wall separators and hollows, start vibrating by absorbing the sound produced in the hall. These may create sound. Certain tones of the original music and the created sound combine to produce interference such that the original sound gets disturbed.

Remedy: Resonance can be suppressed by hanging a large number of curtains in the hall.

7) Noise Generally, there are three types of noise. They are (a) Air-borne noise (b) Structure borne noise (c) Inside noise.

(a) Air-borne noise: The noise that enters the hall from outside through open windows, doors and ventilators is known as air-borne noise. Remedy: i. By using heavy glass doors, windows or ventilators. ii. By using double wall-doors and windows with insulating material in between them. iii. Forming double wall construction.

iv. By fixing doors and windows at proper places. v. Air conditioning the hall and sealing the openings perfectly.

(b) Structure-borne noise: The noise that reaches through the Structures of buildings is known as Structural noise. The activity around the building may cause a structural vibration of the building. Ex: footsteps, operating machinery, street traffic etc; Remedy: i. By using double walls with air space in between them. ii. By using anti-vibration mounts. iii. By properly insulating the equipments such as refrigerators, lifts, fans etc., iv. By using carpets on the floor.

(c) Inside noise: The noise produced inside big halls or offices due to equipment such as air conditioners, type writers and fans is called inside noise. This noise may be minimized as follows. Remedy: i. Placing the machinery on sound absorbent pads. ii. Using noise-free air conditioners. iii. Covering the floor with carpets, walls, ceilings with sound absorbing materials.

9.29. Acoustic Quieting

Acoustic quieting is the process of silencing the noise from machinery i.e., making machinery quieter by damping vibrations. When machinery vibrates, it generates sound waves in air, hydro acoustic waves in water and mechanical stresses in solid matter. By absorbing the vibrational energy or minimizing the source of the vibration, acoustic quieting is achieved.

9.29.1Aspects of Acoustic quieting

To achieve acoustic quieting, a number of different aspects might be considered with an aim to minimize the noise heard by the observer. They are, 1. Reducing the noise generation at its source. 2. Acoustic decoupling to reduce sympathetic vibrations. 3. Increasing acoustic damping or changing the size of the resonator to avoid resonance.

4. Reducing transmission using many methods, depending whether the transmission is through air, liquid or solid. 5. By limiting the reflection by using acoustic absorption materials, trapping the sound, opening windows to let sound out etc;

9.29.2 Methods of quieting

Mechanical acoustic quieting 1) Noise isolation: Noise isolation is the method of isolating noise by using barriers like deadening materials to trap sound and vibrational energy. For example, in home and office construction, sound control barriers such as fiber glass or synthetic rubber are placed in the walls to stop the transmission of noise through them. 2) Noise absorption: The unwanted sounds or noises that are produced inside a room can be made to be absorbed by suitable materials instead of being reflected towards the listener. Thus, the listener receives only the direct sound but not echo reflections. For example, sound proofing rooms are constructed using acoustic tiles for recording studios.

Fig. 9.40 A sound proof room, showing acoustic damping tiles used for noise absorption

3) Acoustic damping: Damping mounts are used to suppress the vibrations. The damping materials prevent the vibrations from being transferred from one material to the other. Motors and rotating shafts are fitted with damping mounts. 4) Acoustic decoupling: Certain parts of a machine such as frame, chasis or external shaft are decoupled from receiving unwanted vibrations from a moving part. 5) Preventing stalls: Whenever a machine undergoes an aero dynamic stall, it abruptly vibrates and hence this has to be avoided. 6) Preventing cavitations: The sounds of gas bubbles imploding is the source of noise, called Cavitation (Rapid formation and vanishing of bubbles produce noise). When a machine is in contact with a fluid, it may be susceptible to cavitation. Ships and submarines have screws that cavitate and this may facilitate their detection through the sonar. 7) Preventing water hammer: In hydraulics and plumbing, water hammer is a known cause for the failure of piping systems. A valve that abruptly opens or shuts is the most common cause for water hammer. It also generates considerable noise. 8) Shock absorption: The shock absorbers present in vehicles, prevent mechanical shocks from reaching the passengers. They quieten the shocks. 9) Reduction of resonance: Every object vibrates with its natural frequency. Many parts of a machine vibrate and produce waves. These waves resonate and form noise.

10) Material selection: The transmission of sound and vibrations can be minimized by using non-metallic components (ex: flexible plastic pipe fittings) in place of metal components. This reduces noise. In some cases, air can be removed from the machine and sealed. The vacuum inside the machine arrests the noise transmission.

9.29.3 Quieting for specific observers

1. Underwater acoustics: Underwater acoustics is mainly applicable to submarines. Submarine may employ a tactic that prevents sounds from reaching a listener at a particular ocean depth. Below certain depth called sound channel axis, the speed of the sound is lowest. Hence, operating below this depth, a submarine can prevent detection by surface ships. 2. Sound refraction: Sound refraction can be used to prevent certain observers from hearing the noise. For example, an outdoor observer, close to the ground, will receive sound waves refracted towards him when the ground is cooler then ambient air, but the sound waves will reflect away when the ground is hotter than ambient air. 3. Sound redirection: The observer placed out of the path of sound of the highest amplitude will receive much quieter sounds. For example, the sound is loudest in line with a jets exhaust. Perpendicular to the exhaust, the sound is significantly quieter. 4. Hearing protection: In highly noise-polluted regions such as open air firing region or airport region, an observer wearing ear plugs may receive less sound. 5. Electronic quieting: Noise can be controlled using electronic methods. Few devices used for electronic quieting are, a) Electronic vibration control: Electronic sensors, computers are used to reduce vibration. Using high speed logic, vibration can be damped quickly, effectively by encountering the motion before it exceeds a certain level. b) Electronic noise control: The noise can be reduced by using phase cancellation, employing electronics, sensors and computers. This method is used in active sound generating devices such as loud speakers. c) Noise reduction: In sound and video equipment, noise reduction is the process of removing noise from the signal. This is strictly for electronic noise. d) Noise canceling: If both the noise and the signals are received by an electronic medium, noise can be filtered from the signal electronically, retransmitted without the noise. Helicopter pilots rely on this technology to speak on the radio.

9.30 Muffler 293

A muffler is nothing but a silencer. It is a device for reducing the amount of noise emitted by a machine. Mufflers are installed along the exhaust pipe as part of the exhaust system to reduce its exhaust noise.

Mini-mufflers

A mini-muffler (also known as ‘hot dog’ in the auto industry), are put in place of a normal muffler in a car exhaust system. By using a specific muffler or mini-muffler or catalytic converter combination, the sound of a car’s exhaust system can be changed considerably. One has to use the mufflers so as to keep the resulting noise below government approved noise levels. Types and positions of mufflers

Different machines are attached with different types of mufflers at different positions. 1) With cars, it is attached length wise underneath, blowing backwards at the rear.

Fig. 9.41 Dual exhaust pipes attached to a car’s muffler

2) With large diesel powered trucks, . Mounted vertically behind the cab . Cross wise under the front of the cab, blowing sideways.

Fig. 9.42 A muffler of a large diesel-powered truck

3) With motor cycles, . Usually beside the engine and rear wheel blowing backwards. . In more modern motor cycles, under the seat blowing backwards from under the back of the seat.

Fig. 9.43 A muffler in a motor cycle

9.30 Sound Proofing

The process of reducing sound pressure with respect to sound source and the receptor is known as sound proofing. The different ways of reducing the intensity of sound are, 1. Distance: The dissipation of sound is proportional to distance. Hence, the intensity of sound felt by an observer decreases progressively as the receiver moves away from the source of sound. If any intervening objects are present, they absorb part of the sound energy and vibrate. The dissipation of sound energy also depends on the weather and reflection from the soil. 2. Damping: Due to damping, the sound energy is converted into heat. This is achieved in several ways. One way is to add a layer of material, such as lead or neoprene which are both heavy and soft. These can be used as a sound deadening layer in such areas as wall, floor and ceiling construction in sound studios. Lead and neoprene suppress high frequency vibrations. Making a sound wave pass through different layers of material with different densities i.e., compound layers assists in noise damping. Open – celled foam is a good sound damper inside a wall. 3. Noise barriers as exterior sound proofing: Along major highways, noise barriers are used to protect adjacent residents from road way noise. The noise barriers used may be constructed with wood, earth or some combination, thereof. 4. Noise cancellation: This is a modern technique. In this, a microphone receives the sound and is analyzed by a computer. Then, sound waves with opposite phase are produced through a speaker. This produces destructive interference and suppresses a lot of noise. 5. Residential sound proofing: This sound proofing is a done mainly to eliminate the effects of exterior noise. Usually, curtains are used to damp sound that enters through doors and windows. Air chambers known as honey combs are also used. Double pane windows give better damping than single pane windows. 295

6. Room within a room: In houses, vibrations pass directly through the brick, wood work and other solid structural elements. The use of acoustic foam and other absorbent materials have no impact on transmitted vibration. Ideal de-coupling eliminates vibrational transfer both in solid material and in air. One method of eliminating the transmittal of sound is to construct a room with in a room.

Chapter – 10 NANOTECHNOLOGY

10.0 Introduction

 Nanotechnology is the study of the control of matter on an atomic scale. Generally, Nanotechnology deals with structures of the size 100 nanometers or smaller, and involves developing materials or devices within that size. These are known as Nanostructures.  Nanotechnology deals with the design, manufacturing and applications of nanostructures in useful nanoscale devices such as electronic circuits and mechanical devices at the molecular level.  In Nanotechnology, the fundamental properties of materials and machines depend on their size.  Nanotechnology deals with the fabrication of devices or materials which lie in the sizes of 1 to 100nm. Nanotechnology produces materials that are built up atom- by-atom. Ex: With carbon atoms, tubular molecules called Carbon nanotubes are made. Spherical molecules with 60 carbon atoms having a diameter slightly less than 1nm, called Bucky balls or Fullerenes are also made with carbon.

10.1 Nano scale  Nano scale means 10-9. A nanometer (nm) is one thousand millionth of a meter (10-9).  The diameter of a single atom can vary from 0.1 to 0.5nm, depending on the type of the element.  For example, one carbon atom is approximately 0.15nm in diameter. A red blood cell is approximately 7000nm wide and a water molecule is almost 0.3nm across. The diameter of a human hair is about 60,000 to 80,000 nms. 1nm would approximately be equal to the length of ten hydrogen atoms in a line.  A material is called a nanomaterial if the size of the material is below 100nm. This is because, many physical and chemical properties of a material start changing significantly when the size of the grains is below 100nm. The length that is of interest in nanoscience is from 100nm down to the atom scale i.e., 0.2nm.

10.2 Origin of Nanotechnology

 The existence of nanostructures and nanodevices is not new, while the word nanotechnology is relatively new.  Richard Feynman, in 1960, presented a visionary lecture at a meeting of the American physical society entitled “There is plenty of room at the bottom.” In his lecture, he speculated on the potential and possibility of nanosized materials.  He envisioned etching lines of few atoms wide with beams of electrons-today’s electron beam lithography, used for making silicon chips. He envisioned building circuits of nanometer scale for use in computer-today’s nanoelectronics. He recognized the existence of nanostructures in biological systems- biological nanoparticles.  He proposed manipulating individual atoms to make new small structures having very different properties.  Ralph landaner, from IBM, proposed nanoscale electronics in 1957. Magnetic fluids consisting of nanosized magnetic particles were developed in 1960s. In the early 1970s, Bell laboratories and IBM fabricated the first 2 dimensional quantum wells. Laser vapour deposition method resulted in development of metal nanoclusters in 1981. Later, this method was used to synthesize the fullerene.  The inventions of Scanning Tunneling Microscope (STM) in 1981 and Atomic Force Microscope (AFM) in 1986 provided new important tools for viewing, characterizing and atomic manipulation of new structures.  Ijima made Carbon nanotubes in 1991. Super conductivity and Ferro magnetism

were found in C60 structures.  In 1994, stable gold nanoparticles were made in solution.  In 1996, scientists at IBM succeeded in moving and precisely positioning individual molecules at room temperature. In 1999, molecular logic gates were created.  From 2000, quite a lot of research and development activities are in progress.

10.3 Nanomaterials  All materials are composed of grains, which in turn comprise of many atoms. Conventional materials have grains of size varying from hundreds of microns to centimeters.  Nanomaterials could be defined as those materials which have structural components with size less than 100nm at least in one dimension.  Materials that are Nano scale in one dimension (and are extended in the other two dimensions) are layers, such as thin films or surface coatings.  Materials that are Nano scale in two dimensions (and are extended in one dimension) include nanowires and nanotubes.

 Materials that are nanoscale in three dimensions are particles, for example, precipitates, colloids and quantum dots (tiny particles of semi conductor devices)

10.4 Factors that differentiate Nano materials from Bulk materials

 The factors that differentiate Nano materials from bulk materials are the increase in surface area to volume ratio and quantum confinement effects.

(i) Surface area to Volume ratio  This value is very large for nanomaterials. Consider a spherical material of radius ‘r.’ Then,

 As the size of the sphere decreases, the above ratio increases. Alternatively, if the material is cubic, as it is divided into small cubes, then also, the surface to volume ratio increases. Hence, the nanomaterials possess large value of surface area to volume ratio as compared to the bulk materials.

(ii) Quantum Confinement Effects  According to band theory, solid materials have energy bands and isolated atoms possess discrete energy levels. Nanomaterials are in intermediate to the above two cases.  For nanomaterials, if the dimensions of potential wells or potential boxes are of the order of the de Broglie wavelength of electrons (mean free path of electrons), then the energy levels of electrons change, and the electron will remain confined to a small region of the material. This is called quantum confinement. Energy is discrete, not continuous in a quantum system.  The electronic properties of solids are altered with the reductions in the particle size. This effect does not come into play by going from macro to micro dimensions. However, it becomes dominant when the nanometer size range is reached.  Materials reduced to nanoscale can suddenly show very different properties compared to what they exhibit on a macro scale. Ex: Opaque substances become transparent (Copper) Inert materials become catalysts (Platinum) Stable materials become combustible (Aluminium) Solids turn into liquids at room temperature (Gold) Insulators become conductors (Si)

10.5 Classification of Nanomaterials

Materials referred to as nanomaterials generally fall into two categories. 1) Fullerenes 2) Nanoparticles

1) Fullerene: A Fullerene is a molecule composed entirely of carbon, in the form of a hollow sphere, ellipsoid or tube. Spherical fullerenes are also called Bucky balls and cylindrical ones are called as Carbon nanotubes or Bucky tubes. The first

Fullerene discovered was Buckminster fullerene C60 in 1985.

Fig. 10.1 Fullerene

(ii)Carbon nanotubes: Carbon nanotubes (CNTs) are allotropes of carbon with a cylindrical nanostructure. Nanotubes are categorized as Single Walled Carbon Nanotubes (SWCNTs) and Multi Walled Carbon Nanotubes (MWCNTs)

Fig. 10.2 Single Walled and Multi Walled Carbon nanotubes

2) Nanoparticles: Nanoparticles are available in different forms such as clusters, metal nanoparticles, colloids, nanoshells, quantum dot etc. Nanoparticles have a very high surface area to volume ratio resulting in a high reactivity. Because of their smaller size, nanoparticles have the tendency to merge into one another.  Hence, such particles are to be stabilized during synthesis. Nanoparticles can be stabilized by ligands in solution, in two basic ways, sterically and electro statically.

Ex: Gold nanoparticles are stabilized sterically by coating with a layer of polymer or protein. By attaching ligands with a negatively charged end group, a repulsive force between the particles can be created, thus avoiding the particles to merge.

Fig. 10.3 Stabilized Gold nanoparticles

10.6 Properties of Nanomaterials

10.6.1 Physical Properties

At the macro scale, the physical and chemical properties are not dependent on the size of the material, but at the nanoscale, everything, including colour, melting point and chemical properties will change, compared to what they exhibit on a macro scale. The various physical properties are explained below.

(i)Geometric structure

 Large nanoparticles have the same crystal structure as that of the bulk material, but different lattice parameters.  In nanomaterials, the surface area to volume ratio increases. Similarly, the interatomic distance decreases by reducing the size of nanoparticles.

(ii) Optical properties

 Different sized nanoparticles scatter different wavelengths of light incident on it and hence, they appear with different colours.

For ex., nanoscale gold particles can be orange, purple, red or greenish in colour, depending on their grain size. The bulk copper is opaque whereas nanoparticle copper is transparent. Porous silicon exhibits red photoluminescence, but bulk silicon does not show this effect.

(iii)Thermal properties

 The melting point of nanogold decreases from 1200k to 800k as the size of the particles decreases from 300Ao to 200Ao.  The Debye temperature and ferro electric phase transition temperature are lower for Nano materials.  Super plasticity of nanomaterial occurs at lower temperatures by reducing the grain size.  Stable Aluminium becomes combustible in nano phase. Solid Gold changes to liquid as it goes from bulk to nanomaterial at room temperature.

(iv) Magnetic properties

The magnetic properties of nanomaterials are different from that of bulk materials.  In nanomaterials single domains are used, unlike large no of domains in bulk materials. The co-ercivity of single domain is very large.  Fe, Co, Ni and Gd are ferro magnetic in bulk but they exhibit super paramagnetism in the nanophase. Na, K and Rh are Paramagnetic in bulk but in nanophase, they are ferromagnetic.  Small clusters (containing less than 80 atoms) of non-magnetic substances show spontaneous magnetic moment where as in the case of magnetic nanoparticles, the magnetic moment is reduced.

 Clusters of non-magnetic element, supported on metal substrates also show magnetism. This shows that small particles possess more magnetism than the bulk material. Nanoparticles of even non-magnetic solids are found to be magnetic.

(v)Electronic properties

 The electrical conductivity and energy band width of some materials change when they pass from bulk phase to nanoparticle phase. For Ex., bulk silicon is an insulator; it becomes a conductor in the nano phase. Nanomaterial semiconductors such as Ge and Si do not show semi-conducting property.  In nanoceramics and in nanomagnetic composites, electrical conductivity increases with reducing particle size. In metals, electrical conductivity decreases with reducing particle size.  By reducing the size of metal particles from bulk top nano, the energy bands become narrower and hence the ionization potential energy increases.

(vi)Mechanical properties

 Mechanical properties such as hardness, toughness, elastic modulus, scratch resistance, fatigue strength, crack initiation and propagation are significantly different for nanostructures than bulk materials.  In metals and alloys, the hardness and toughness are increased by reducing the size of nanoparticles. In ceramics, the ductility and super plasticity are increased on reducing particle size. Hardness increases 4 to 6 times as one goes from coarse grain Cu to nanocrystalline Cu and it is 7 to 8 times for Ni. The hardness increases 2 to 7 times by decreasing the size of nanocrystalline metals from 1 m to 10 m.  Materials with smaller grain size are stronger because crack propagation can be delayed or reduced in nanostructures than in bulk materials. Brittle materials (ceramics, inter metallic) can become ductile by reducing their grain size.  The transition from bulk to nano phase reduces elastic strength and increases plastic behavior. Creep involves atomic transport along grain boundaries. This leads to super plasticity. The creep rate can be increased by 6 to 8 times by reducing the grain size from microns to nanometers.

10.6.2 Chemical Properties

 The nanostructures in chemistry are colloids and these are formed in a condensed phase having sizes in the range of 1-100nm.

 The ionization potential increases as the cluster size drops below the bulk limit and it has limited applicability.  Clusters of Platinum, Iridium or Osmium reduced to the size of 1nm, and supported on Aluminium or silica exhibits electronic properties similar to those found for large crystallites of the metal.  The reactivity of a cluster depends on the cluster size. This is useful in preparation of catalytic agents.  Some chemically inert bulk materials become good chemical catalysts in the nano phase. Ex: Gold, Platinum.

Hydrogen storage in metals

 Most of the metals do not absorb hydrogen; those that absorb have a metal to hydrogen atom ratio of 1. The small positively charged clusters of Ni, Pd and Pt, containing atoms in between 2 to 60, can absorb hydrogen atoms up to 8 per metal atom.  The hydrogen absorption increases with the decrease in the cluster size. Thus, small particles may be very useful in hydrogen storage in metals.

10.7 Production of Nanomaterials

 Nanomaterials can be produced by a number of ways. Nanomaterials can be synthesized by “top down” techniques, producing very small structures from large pieces of the material. One way of doing this is, mechanically crushing of the solid into fine nanopowder (ball milling).  Nanomaterials may also be synthesized by “bottom up” techniques, atom by atom or molecule by molecule. One way of doing this is to allow the atoms or molecules, arrange themselves into a structure, due to their natural properties. Ex: crystal growth.  There are many known methods to produce nanomaterials. They are:

10.7.1 Plasma Arcing

 Plasma is an ionized gas. To produce plasma, potential difference is applied across two electrodes. Plasma can be produced by the discharge of the gas between the two electrodes so that the gas dissociates into electrons and positive ions (i.e., the gas is ionized)  Using plasma arcing, very thin films of the order of atomic dimensions can be deposited on the surface of an electrode (cathode).

 The deposition is carried in vacuum or in an inert gas. An arc passes from one electrode to the other.  The first electrode (anode) vaporizes. So, positively charged ions are produced. These ions deposit on the cathode.  By using carbon electrodes, carbon nanotubes can be formed on the surface of the cathode. A mixture of conducting and non-conducting materials is also used in the electrodes to form thin layers of the materials on the cathode.

Fig. 10.4 Schematic of Arc-Discharge method

10.7.2 Sol-Gel method  In solutions, nanosized molecules are dispersed randomly whereas in colloids, the molecules have diameters in the range of 20µm-100µm and are suspended in the solvent. So, the colloid appears cloudy.  A colloid that is suspended in a liquid is called a Sol. The gelation of the sol in the liquid to form a network is called gel. Gel is the suspension that keeps its shape.  Sol-gel formation occurs in different stages. (a) Hydrolysis (b) Condensation and polymerization of monomers to form particles. (c) Agglomeration of particles. This is followed by the formation of networks which extends throughout the liquid medium and forms a gel.  Using sol-gel method, silica gels, zirconia and yttrium gels and aluminosilicate gels are formed. Nanostructured surfaces are formed using the sol-gel method.

Fig. 10.5 Schematic of Sol-Gel method

10.7.3 Chemical Vapour Deposition (CVD)

 In this method, the material is heated to gaseous phase and allowed to condense on a solid surface in vacuum.  Nanomaterials of metallic oxides or metallic carbides can be formed by heating metal and carbon or metal and oxygen, in a vacuum chamber to gaseous phase, and allowed to deposit on the surface of a solid.  Pure metal powders (nanoparticles) are formed by this method. The metal is melted, excited with microwave frequency and vaporized to produce plasma at 1500oc. By cooling this plasma with water in a reaction column, nanoparticles are produced.  The grain size of the nanoparticles depends on the concentration of the metal vapour, its rate of flow in the reaction column and temperature. This method can also be used to grow surfaces.

Fig.10.6 Schematic of Chemical Vapour Deposition

10.7.4 Physical Vapour Deposition (PVD)

PVD describes a variety of methods of depositing thin films by condensation of a vaporized form of the material onto various surfaces under high vacuum

condition.

Fig. 10.7 Schematic of Physical vapour deposition

Various methods of PVD

(a) Evaporative deposition: The material to be deposited on the mounted substrate within the vacuum chamber is heated and vaporized by electrically resistive heating. Condensation of the vaporized material takes place and by proper control, atomic layer deposition can be obtained. (b) Electron beam physical vapour deposition: The material to be deposited is heated to a high vapour pressure by electron bombardment in high vacuum. (c) Sputter deposition: The material to be deposited is vaporized by bombarding the material with a glow plasma discharge. The plasma is usually confined and localized around the target, by a magnet.

Fig. 10.8 Schematic of Sputter Deposition

(d) Cathodic arc deposition: In this a high power arc is directed at the target material, that blasts away some of the material into vapour. (e) Pulsed laser deposition: In this high power pulsed laser ablates the material from the target into a vapour. In all these methods, the material is converted into vapour phase, for further deposition on the substrate.

Fig. 10.9 Schematic of Pulsed Laser Deposition

10.7.5 Ball Milling

 This is a top-down technique, producing very small structures from larger pieces of the material.  In this method, small balls of the material are made to rotate inside a drum and drop under the influence of gravity, onto a solid present in the drum. The solid is broken into nanocrystallites.  This is also known as mechanical crushing. This method is used for a large number of elements and nano oxides. For example, Iron nanoparticles of 13-30 nm can be formed.

Fig. 10.10 Schematic of Ball milling 308

10.7.6 Electro deposition

 Electro deposition technique is used to electroplate a material. When current is passed through two electrodes immersed inside the electrolyte, certain mass of the substance liberated at one electrode gets deposited on the surface of the other. By controlling the current and the other parameters, it is possible to deposit even a single layer of atoms.  Nanostructured films of Au, Cu, Pt, Ni, polymers, oxides and semi-conductors can be deposited. These films are robust, flat, uniform and shiny.  The deposited plates are used in batteries, solar cells, fuel cells, sensors, photonic devices, field emitters, magnetic read heads etc. 

Fig. 10.11 Schematic of Electro deposition of Nanostructured film of Cu.

10.8 Characterization of Nanomaterials

The characterization technique that is employed for the characterization of Nanomaterial depends on the type and application of the nanomaterial. For all Nanomaterials, particle size determination is essential and this can be done using X-ray diffraction and Electron microscopy.

10.8.1 X-ray Diffraction Method

 Phase identification using X-ray diffraction lies mainly on the position of the peaks in the diffraction profile and to some extent on the relative intensities of these peaks.  The shape, particularly the width of the peak is a measure of the amplitude of thermal vibrations of the atoms at their regular lattice sites.

 It can also be a measure of any deviations from the normal structure. Ex: Plastic deformation, impurity doping, second phase addition to the host phase. When X-ray powder pattern is taken on poly crystalline powders, the crystallite size causes peak broadening. The crystallite size is easily calculated as a function of peak width, specified as Full Width at Half Maximum (FWHM), peak position and wavelength. The average crystallite size ‘L’ is given by,

Where, θB is the Bragg’s angle B1/2 is FWHM of the diffraction peak; K is a constant and λ is the wavelength of X-rays used.

Fig. 10.12 X-ray diffraction of Ag nanoparticles

10.8.2 Electron Microscopy

Electron microscopes are a beam of highly energetic electrons to examine objects on a very fine scale. The examination gives information about, a) Topography: The surface features of the object or “how it looks,” its texture, relation between them and the material properties (hardness, reflectivity etc.,) b) Morphology: The shape and size of the particles making up the object, direct relation between these structures and material properties (ductility, strength, reactivity etc.,) c) Composition: The elements and compounds that the object is composed of, direct relation between composition and material properties (melting point, reactivity, hardness etc.,) d) Crystallographic information: Atomic arrangement in the object, relation between them and the material properties (conductivity, electrical properties, strength etc.,)

Fig.10.13 Electron Microscope

10.8.3 Scanning Electron Microscopy (SEM)

The image in scanning electron microscope is produced by scanning the sample with a focused electron beam and detecting the secondary and back scattered electrons. Electrons and photons are emitted at each beam location and subsequently detected.

Fig. 10.14 Specimen interactions in SEM

Schematic representation of SEM

1) The electron gun produces a stream of monochromatic electrons. 2) The electron stream is condensed by the first condenser lens. It works in conjunction with the condenser aperture to eliminate the high angle electrons from the beam. 3) The second condenser lens forms the electrons into a thin, light coherent beam.

4) Objective aperture further eliminates high angle electrons from the beam. 5) A set of coils acting as electrostatic lens scans and sweeps the beam in a grid fashion (as in television) 6) The objective lens focuses the scanning beam onto the part of the specimen. 7) When the beam strikes the sample, interaction occurs. The intensity of display is determined by the interaction number. More interactions give a brighter pixel. 8) This process is repeated until the grid scan is finished and then repeated. The entire pattern can be scanned 30 times per second.

Fig. 10.15 Schematic representation of SEM

SEM gives useful information on

1) Topography: The surface features of an object or “how it looks,” its texture, detectable features, limited to a few nanometers. 2) Morphology: The shape, size and arrangement of particles making up the object that are lying on the surface of the sample or have been exposed by grinding or chemical etching, detectable features limited to a few nanometers. 3) Composition: The elements and compounds the sample is composed of and their relative ratios, in areas approximately 1 micrometer in diameter. 4) Crystallographic Information: The arrangement of atoms in the specimen and their degree of order, only useful on single-crystal particles >20 µms.

10.8.4 Transmission Electron Microscopy (TEM)

 When transmitted electrons are utilized for imaging, it results in Transmission Electron Microscopy (TEM).  In the TEM analysis, a thin or diluted sample is bombarded under high vacuum with a focused beam of electrons. Electrons that are transmitted through the material, form contrast patterns that reproduce the image of the sample. This pattern arises from the scattering of the electrons. In addition, diffracted electrons give information about the lattice structure of the material. The shape of the sample can also be determined from the image.

Schematic representation of TEM

1. The electron gun produces a stream of monochromatic electrons. 2. This stream is focused to a small coherent beam by the first and second condenser lenses. 3. The condenser aperture knocks off high angle electrons. 4. The beam strikes the specimen. 5. The transmitted portion is focused by the objective lens into an image. 6. Objective aperture enhances the contract by blocking out high angle diffracted electrons. 7. Selected area aperture enables to examine the periodic diffraction of electrons by an ordered arrangement of atoms in the sample. 8. Intermediate and projector lenses enlarge the image.

The beam strikes the phosphor screen and image is formed on the screen. The darker areas of the image represent thicker sample areas since these areas transmit lesser electrons. The brighter areas of the image represent thinner sample areas since these areas transmit more electrons.

TEM gives the information about

1) Morphology: The size, Shape and arrangement of particles as well as their relationship to one another on the scale of atomic diameters. 2) Crystallographic information: The arrangement of atoms in the specimen and their degree of order, detection of atomic scale defects, a few nanometers in diameter. 3) Compositional information: The elements and compounds, the sample is composed of and their relative ratios.

Fig. 10.16 Schematic representation of TEM.

10.9 Carbon Nanotubes (CNTs)

 The allotropes of carbon are diamond, graphite, fullerenes, amorphous carbon and carbon Nanotubes.

Fig. 10.17 Allotropes of Carbon

 CNTs were first observed by sumio Iijima in 1991.  The molecular structure of graphite is one atom thick, a planar network of interconnected hexagonal rings of carbon atoms. In graphite, the sheets of carbon

are stacked on top of one another. They can easily slide over each other. That is why; graphite can be used as a lubricant.  When graphite sheets (one single sheet called as graphene) are rolled into a cylinder and their ends closed with caps containing pentagonal rings, they form carbon Nanotubes i.e., a rolled graphene sheet is called a carbon Nanotube (CNT). The length of CNTs may vary from several micrometers to millimeters and the diameter will vary from 1-20nm.

10.9.1 Types of CNTs

 CNT may contain one cylindrical wall of graphene, called a Single Walled Carbon Nanotube (SWCNT) or a number of concentric, identical, cylindrical graphene sheets, differing in diameter, called Multi Walled Carbon Nanotubes (MWCNTs)  SWCNTs are of three different types, depending upon the ways of rolling the graphene sheet. They are, a) Arm Chair carbon nanotubes b) Zig-Zag carbon nanotubes c) Chiral carbon nanotubes  Depending on the different ways of rolling, CNTs could be a good or poor conductor.  The simplest sequence of a CNT is when concentric layers are identical but differs in diameter. However, mixed variants are possible, consisting of two or more types of concentric CNTs arranged in different orders. These can have either regular layering or random layering.  The structure of CNT influences its properties- including electrical and thermal conductivity, density and lattice structure. Both, the type of CNT and diameter are important.

Fig.10.18 Types of Carbon Nanotubes

10.9.2 Properties of CNTs

CNTs have assumed an important role because of their novel physical and chemical properties. They are, 1) They are mechanically very strong. Their Young’s modulus is over 1 terapascal, making CNTs as stiff as diamond. 2) They are flexible about their axis and can conduct electricity extremely well. The electrical conductivity is a function of diameter and chirality, the degree of twist. 3) CNTs are about 6 times lighter, 10 times stiffer and 20 times stronger than steel. 4) The helicity of the graphene sheet determines whether the CNT is a semiconductor or a metal. As a metal, its electrical conductivity is 1000 times more than that of copper. 5) CNTs have very high thermal conductivity. The strong in plane graphitic carbon- carbon bonds make them exceptionally strong and stiff. 6) The large surface area and high absorbency of CNTs make them ideal candidates for use in air, gas and water filtration.

Because of these remarkable properties, CNTs have a wide range of potential applications in reinforced composites, sensors, nanoelectronics display devices etc.

10.9.3 Production of CNTs

Few methods adopted for the production of CNTs are,

1) Arc Discharge method

 This method creates CNTs through Arc vaporization of two carbon rods placed end to end, separated approximately by 1mm, in an enclosure that is usually filled with inert gas at low pressure.  It is also possible to create CNTs with the Arc method in liquid nitrogen.  In this method, evaporation of one electrode (anode) takes place and the particles are deposited on the surface of the other electrode (cathode). The deposition on the cathode is nanotubes. Normally, MWCNTs are formed from plasma arcing.  SWCNTs are formed if the electrodes are bored out and cobalt or other metals are included as catalysts.

Fig.10.19 Experimental setup for an Arc discharge apparatus for synthesis of CNTs

2) Laser method

 CNTs, specifically SWCNTs in large quantities, were prepared by laser vaporization of graphite rods with a 50:50 catalyst mixture of cobalt and Nickel at 1200oc in flowing argon.  The initial laser vaporization pulse was followed by a second pulse, to vaporize the target more uniformly.  The use of two successive laser pulses minimizes the amount of carbon deposited as soot. The second laser pulse breaks up the larger particles ablated by the first one, and feeds them into the growing nanotube structure. The CNTs produced by this method are 10-20nm in diameter and up to 100µm or more in length.  By varying the growth temperature, the catalyst composition and other process parameters, the average nanotube diameter and size distribution can be varied.

 Fig.10.20 Schematic diagram of Laser Ablation apparatus 3) Chemical Vapour deposition method (CVD)

 Large amounts of CNTs can be formed by catalytic CVD of acetylene over cobalt and iron catalysts supported on Silica and Zeolite. Bundles of SWCNTs were also found among the MWCNTs produced on the carbon or Zeolite catalyst.  CNTs can be formed from ethylene. Supported catalysts such as iron, cobalt and nickel, containing either a single metal or a mixture of metals seem to induce the growth of isolated or bundles of SWCNTs in the ethylene atmosphere.  Production of SWCNTs as well as doubled walled CNTs, on molybdenum and Mo-Fe alloy catalysts has also been demonstrated.

Fig.10.21 (a) schematic of Plasma CVD apparatus Fig.10.21(b) Schematic of Thermal CVD

10.10 Applications of Carbon Nanotubes

1) Electronics:  The SWCNT can act as a transistor.  Pairs of nanotubes or crossed nanotubes show as logic structures.  If a line of hexagons form a helix, then the tube acts as a semiconductor.  A single nanotube with a natural junction acts as a rectifying diode. In flat panel displays, a nanotube works due to its field emission property.  Nanotubes are used in batteries.

2) Hydrogen storage: Nanotubes can store hydrogen and also helium, oxides and metals like copper. 3) Mechanical machines: Nanotubes can act as axles in nanomachines. Building gear teeth on nanotube is favoured to translate different rotational motions. 4) Space elevators:  Long filaments of nanotubes are used in fiber reinforced plastics, which have less weight. So, they are used in aero planes, space ships and land vehicles.  CNTs, like graphite, withstand high temperatures. So, they are used for thermal protection of space craft during their re-entry into the atmosphere.  CNTs have high Young’s modulus. So, they withstand aeronautical strains. 5) Medical: CNTs are thin. So, they can penetrate the body without pain. They can be used for targeted drug delivery.

10.11 Applications of Nanotechnology

1) Electronics industry  Reduction in size of electronic components leads to faster switching times.  Nanotechnology includes fabrication of nanowires used in semiconductors.  Porous silicon emits visible light. So, it finds application in optoelectronics.  Quantum dot laser emits good quality laser beam compared to semiconductor laser diodes. They are cheaper and the emitted wavelength depends on the diameter of the dot.  Materials like nanocrystalline selenide, zinc sulphide, lead telluride and cadmium sulphide synthesized by sol-gel method improve the resolution of the image in television or in a monitor. The resolution depends on the size of the pixel made of phosphors. The resolution increases as the size of the pixel reduces.  Batteries made of nanocrystalline nickel and metal hydrides are long lasting and require minimal recharging.

2) Industry  Nanocrystalline materials such as tungsten carbide, titanium carbide are harder than conventional materials. They are more wear resistant and erosion resistant. So, they can be used in cutting tools and drilling bits.  Nanocrystalline silicon nitride and silicon carbide are used in the manufacturing of high strength springs, ball bearings and valve lifters.  Nanocrystalline ceramics such as zirconia are easily pressed and sintered into various shapes at lower temperatures, because they are softer than conventional materials.

3) High power magnets  The magnetic strength of a material is directly proportional to the surface area per unit volume. 319

 The magnetic nanocrystalline yttrium-samarium-cobalt possesses very high magnetic properties due to large surface area.  The high power rare earth magnets are used in submarines, generators, electric motors, automobile alternators, magnetic resonance imaging (MRI) instruments and in ultrasensitive analytical instruments.

4) Motor vehicles and air craft The wastage of thermal energy generated in engines is reduced by coating the cylinders with nanocrystalline ceramics such as zirconia and alumina.  In air craft, by decreasing the grain size of the materials, the fatigue strength increases as much as 300 %. The components made of nanomaterials are stronger and operate at higher temperatures. So, air craft can fly faster using the same amount of aviation fuel.

5) Aero gels  Aero gels are nanocrystalline and porous. Hence, air is trapped at the interstices. These materials can be used in offices and homes to reduce cooling and heating bills by saving power. These are also used for smart windows so that the materials become darkened when the sun is too bright and lightened when the sun is not shining brightly.

6) Energy efficiency  Energy consumption can be greatly reduced by using quantum edged atoms or LEDs instead of ordinary filament bulbs.  Nanostructures have increased solar energy conversion efficiency as compared to ordinary semiconductor solar cells.  In hydrogen fuel cells, nanostructured catalyst material is used on carbon supported noble metal particles, with diameters between 1 & 5nm.  For hydrogen storage, CNTs, zeolites etc., are used.

7) Medical field  Nanomaterials are used in the development of diagnostic devices, drug delivery vehicles, and analytical tools and in physical therapy applications.  The genetic sequence of a sample can be detected by tagging gold nanoparticles with short segments of DNA.  Magnetic nanoparticles are used to label molecules, structures or micro organisms of an antibody.  Iron nanoparticles are used in cancer treatment. Nanotechnology is used to repair or to reproduce damaged tissue.

8) Textile industry

Clothes made of nanofibers are water and stain repellent and wrinkle free. They can be washed less frequently at low temperatures.

9) Information technology Nanoscale fabricated magnetic materials are used in data storage. Quantum computers use fast quantum algorithms and have quantum bit memory space (qubit). So, it involves several computations at the same time.

10) Cosmetics  Sunscreens based on mineral nanoparticles such as titanium-di-oxide offer several advantages. They have high UV protection compared to the bulk material. In addition to these, there are quantum dots, suntan lotion, nanotubes, and protective coatings.

 Fullerenes (C60) are used as lubricants (molecular ball bearings), diamond seeding, diamond protection, xerographic materials, photo chronic goggles etc.

PREVIOUS UNIVERSITY QUESTIONS

1) a) what is Hall Effect? Derive an expression for Hall efficient for p-type semiconductor b) Write a short note on LED c) For semiconductor, the Hall efficient is -68×10-5m3/coulomb & electrical conductivity is 250m-1Ω-1 calculate the density &mobility of charge carriers. (May/June -2013) 2) a) Derive an expression for density of holes in insect semiconductor. b) Discuss V-I characteristics of a p- n junction on diode c) For intense semiconductor of band gap of 0.78ev, find campier concentrations at 370c [given that the effective mass of electron=effective mass of hole=rest mass of electron] [may/june-2012] 3) a) Derive an expression for campier concentrations of n-type. b) Explain the variation of EF wrt temperature in N - type c) If effective mass of holes is 5 times that of elements at what temperature would the EF be shifted by 15% from middle of forbidden gap( Eg). Given Eg=1.2ev [June -2011] 4) a) Density of electron in N-type –Derivation b) Explain Hall Effect &its significance c) Calculate interstices carrier concentration for Ge at 270c [for Ge atomic weight =72.6, Density=5400kg/m3 Band gap Eg=0.7ev] (june-2011) 5 )a) Draw & Explain the Energy Band diagram of a p-n junction diode in an unbiased Conduction. b) Write a short note on photodiodes (june-20110)

6) a) Derive an expression for density of holes in intrinsic semiconductor b) Write notes on direct &indirect Band gap. 7) a) Density of electrons-Derivation b) Explain the variation of EF wrt Donor &accepter concentration c) Explain the variation of Ef wrt temperature in ‘N & P’ type. 8) a) Draw the energy band diagram of An Intrinsic N-Type, P- Type, EF,ED,EA levels. b) Define Hall effect derive RH,VH &applications of Hall effect

9) a) Show that the application of FB voltage access p-n junction causes an exponential increase of no of charge carrier in opposite regions. b) Write notes on LCD. c) The current in P-N junction at 270c is 0.18µA when a reverse bias voltage is applied. Calculate the current when FB of 0.98 V is applied. (June-2010) 10) a) Distinguish between intrinsic &extrinsic semiconductor b)Derive diode equation. c) Applications of Semiconductors 11) Describe the different methods of acoustic quieting. 12) Discuss various techniques to achieve sound proofing. (June 2010, regular) 13) Define the term Reverberation. What is Reverberation time? 14) Describe any two methods by which the sound absorption co-efficient of a material can be determined. 15) A hall of volume 85000 m3 is found to have a Reverberation time of 2.2 sec. If the area of the sound absorbing surface is 7500 m2, calculate the average sound absorption co-efficient. (Dec, Jan 2011-2012 set 2 ) 16) Define (i) Reverberation (ii) Reverberation time (iii) sound absorption co-efficient of a material. 17) Explain how the Reverberation time of a hall is affected by (i) the size of the hall (ii) nature of wall surfaces and (iii) Audience. 18) Write an essay on Acoustics of buildings. (Dec-Jan 2011-2012 set4 ) 19) Describe an experimental method to determine the sound absorption co-efficient of a material. 20) Discuss the factors which affect the architectural acoustics and suggest your remedy. (June, 2011 set3 ) 21) A hall has a volume of 1500 m3. Its total absorption is equivalent to 100 m2 of open window. What will be the effect on the Reverberation time if the absorption is

increased by 100 m2 of open window by filling the hall with audience? ( June,2011 set3) 22) Write a detailed note on Nanoscience and Nanotechnology. 23) Write the important applications of Nanomaterials in medicine. (June 2009 regular set1) 24) What are Nanomaterials? Explain. 25) Describe the various types of carbon Nanotubes. 26) Explain the fabrication of carbon Nanotubes. (June 2009 regular set 2) 27) Write a detailed note on Nano science. 28) Why Nanomaterials exhibit different properties? Explain. (June 2009 set 3, set4) 29) Describe any three processes by which Nanomaterials are fabricated. 30) Describe the important applications of nanotechnology. (June 2010 set 1) 31) What is the meaning of nanotechnology? Explain. 32) Describe the process of sol-gel and precipitation in the fabrication of nanostructures. 33) Write the applications of nanotechnology in the electronic industry. (June 2010 set2) 34) Describe the top-down methods by which Nanomaterials are fabricated. 35) Explain how x-ray diffraction can be used to characterize nanoparticles. (June 2010,set4) 36) Describe any three processes by which Nanomaterials are fabricated. 37) Describe the important applications of nanotechnology. (Dec-Jan 2011-2012 set1) 38) Write about (i) origin of nanotechnology (ii) nanoscale. 39) Discuss quantum confinement effect on nanoparticles. 40) Explain how TEM can be used to characterize nanoparticles. (Dec-Jan 2011-2012, set3) 41) Write notes on (i) origin of nanotechnology (ii) nanoscale. 42) Describe the combustion method and PVD method for fabrication of nanostructures. 43) Give the applications of nanotechnology in the electronic industry.(June 2011, set1) 44) Give three methods of fabrication of Nanomaterials. 45) Give the important applications of nanotechnology. (June 2011, set2) 46) Describe the top-down methods of fabrication of Nanomaterials.

47) Explain X-ray diffraction method for characterization of Nanomaterials. (June 2011, set4) 48) Write notes on (i) nanoscale (ii) origin of nanotechnology 49) Explain the process of PVD and Pulsed laser vapour deposition method for the fabrication Nanomaterials. (June 2012) 50) How the physical and chemical properties of Nanomaterials vary with their size? (June 2013)

SOLVED PROBLEMS

1) The intrinsic carrier density at room temperature in Ge is 2.37×1019/m3. If electron & hole nobilities are 0.38 &0.18 m2v-1s-1.calculate the resistivity 19 3 Given data: ni = 2.37×10 /m 2 -1 -1 µn = 0.38 m v s . 2 -1 -1 µp = 0.18 m v s . P=? σi = (n e µn + p e µp ) n = p = ni σi = nie (µn + µp)

= nie (µn+µp) {since σi = }

0.471Ω

* * 2) For silicon Eg=1.12ev. Find EF at 300k if me = 0.12mo & mp = 0.28 mo Given data: Eg=1.12ev T=300k * * me = 0.12mo & mp = 0.28 mo

kB = ev [Joules to eV]

= + T ln ( )

= (OR) - =

= (OR)

= + ln ( )

=0.576ev.

3) For intrinsic semiconductor with Band gap of =0.78ev calculate intrinsic 0 * * carrier at 37 c (300k) if me = mp = m0=rest mass of electron.

Given data 0.78×1.6× Joules. T= 370c = 300k

= 9.1× kg

= 1.38× J/K

Formula =2 exp (

= 2 × exp ( )

4) If the effective mass of holes is 5 times that of electrons, at what temperature would the Fermi level be shifted by 15% from the middle of Eg. [Given ] 5) For a semiconductor, the Hall coefficient is -6.85 / & electrical conductivity is 250 calculate the coulomb density & mobility of charge carriers.

Given data : =250 Density of elements ‘n’ =no of atoms/unit volume=? Mobility We know Hall coefficient

n = n = 9.124×1022atoms/m3

Conductivity = neµ

Mobility µ =

= 17.125× m2v-1s-1

6) The RH is 33.66× its resistivity is 8.93× m. find µ & n

Given data: RH=3.66×

Formula:-

Density’ n’ =

n=1.708 atoms/m3 Mobility µ=? But σ=neµ µ = =

= = 0.041m2v-1s-1

8) The volume of a hall is 475m3. The area of wall is 200m2, area of floor and ceiling each is 100m2. If absorption coefficient of the wall, ceiling and floor are 0.025, 0.02 and 0.55 respectively, calculate the reverberation time for the hall.

Solution:

Reverberation time T =

∑aS = (200x0.025+100x0.02+100x0.55) = 5+2+55= 62

Therefore T = = 1.264 s (Answer)

8) A hall has a volume of 12500m3 and reverberation time of 1.5 sec. if 200 cushioned chairs are additionally placed in the hall, what will be the new reverberation time of the hall. The absorption each chair is 1.0 O.W.U. Solution:

Formula: T =

Before adding the cushioned chairs

Reverberation time T1 = i.e., ∑aS =

= 1375s After adding 200 cushioned chairs having absorption of 1 O.W.U. each

T2 = =

= 1.31 s (Answer)

9) A cinema hall has a volume of 7500m3. What should be the total absorption in the hall if the reverberation time of 1.5 seconds is to be maintained? Solution:

Formula: T =

Hence total absorption ∑aS =

= 825 OWU (Answer)

10) The volume of as auditorium is 9500cubic meters. The period of reverberation is found to be 1.5 sec. calculate the total absorption in the auditorium. If the floor of the auditorium is now covered with carpets whereby the total absorption is found to have increased by 100 sabines, calculate the new period of reverberation. Solution: ∑aS = = = 145 OWU (Answer)

With carpet T = = 1.369 second (Answer)

11) A hall has dimensions 20*15*5 m3.the reverberation time is 3.5 sec. calculate the total absorption of its surfaces and the average absorption coefficient. Solution: ∑aS = = = 70.714 OWU

Total absorption = 70.714 OWU (Answer) Average absorption coefficient = =

= 0.074sabine/m2 (Answer) 12) A concert hall has volume of 2265 m3 and its total absorption is equivalent to 92.9m2. How many persons should be seated in the hall so that the reverberation time becomes 2 seconds? Given that the absorption area of one person is equivalent to 18.6m2 of open window. Calculate the reverberation time of to empty hall also. Solution:

T = = = 4.023s

Reverberation time of empty hall = 4.023s (Answer) ∑aS + X = = 186.86

Increase in absorption = 186.86-92.9 = 93.96sabine or OWU Absorption due to one person = 18.6 m2 Hence number of persons to be seated = = 5 (Answer).

OBJECTIVE QUESTIONS Semiconductor physics and devices: 1. In intrinsic semiconductor the carrier concentration varies as T3/2. 2. At 0K pure silicon is insulator. 3. The majority charge carriers in p-type Ge are holes . 4. The electric current in a pure semiconductor when voltage is applied is due to holes and electrons. 5. In an intrinsic semiconductor .The resistivity of pure si at 00c is 300 m. The intrinsic carrier concentration per cubic metre is =0.14 and =0.05m2 v-1 s-1 is 1.095 x 1017 7. If the drift velocity of electrons under a field gradient of 1000vm-1 is 4ms-1, their mobility is 4x10-3 8. In an intrinsic semiconductor at 0k, the fermi level lies at centre of top of the valence band and bottom of conduction band. 9. The density of charge carriers in a pure semiconductor is proportional to exp(-Eg/2KT). 10.The ratio of the diffusion constant for hole(DP) to the mobility of hole ( ) is proportional to T. 11. The einstein relationship between the diffusion constant D and mobility for electron is Dn/ is KT/e . * * 12. Since mh > me in the energy gap with increse of temoerature the fermi level is just above the middle and rises slightly. 13. Mobilities of electrons and holes are proportional to T-3/2. 14. To prepare a p-type semiconductor the element to be added with Ge is aluminium. 15. To prepare a n-type semiconductor,the element to be added with si is Arsenic . 16. In n-type semiconductors,density of electrons in the conduction band is given by 1/2 * 2 3/4 n=(2Nd) (2 me kT/h ) exp(Ed-Ec/2kT). 17. In an intrinsic semiconductor the density of electrons is given by * 2 3/2 n=2 ( me kT/h ) exp(Ef-Ec/2kT). 18. In an intrinsic semiconductor the density of holes is given by * 2 3/2 n=2 (2 π mh kT/h ) exp(Ev-Ef/2kT)

2 3/2 * * 3/4 19. Intrinsic carrier concentration is given by ni=2(2π KT/h ) (me mh ) exp-( Eg/2kT) 20. In p-type semiconductor,density of holes in the valence band is given by 1/2 * 2 3/4 p= ( 2Na) (2 π mh kT/h ) exp(Ev-Ea/2kT). 21. At moderately low temperatures the density of electrons in the conduction band of a n-type semiconductor is proportional to square root of the donor concentration. 22. At moderately low temperatures the density of holes in the valence band of a p-type semiconductor is proportional to square root of acceptor concentration. 23. Hall co.efficient is used to calculate mobility of charge carrier. 24. Concentration of charge carrier can be calculated by Hall co.efficient. 25. For an n-type semiconductor the hall co.efficient RH is related to carrier concentration by RH=-1/ne. 26. If b is the width of the sample,Vh the hall voltage developed when placed in magnetic field of induction B,and current I flows through it,then hall co.efficient is given by RH=VHt/IB . 27. The ratio of diffusion co.efficient to mobility of carriers is proportional to T.

28. Type of semiconductor can be identified using Hall co.efficient. 29. In an intrinsic semiconductor, the electrical conductivity is proportional to T1/2. 30. Minority carrier life time is the time taken for the excess charge carriers to reduce to 1/e of its initial value. 31. Electrical conductivity of a semiconductor at absolute zero of temperature is zero. 32. In intrinsic semiconductor the electron concentration is equal to hole concentration. 33. For p-type semi conductor trivalent typr of atoms are added to si. 34. For n-type semiconductor pentavalent type of atoms are added to Ge. 35. Diffusion coefficients and mobility relation is given by Einstein relation. 36. If the effective mass of a hole is greater than the effective mass of an electron,then the position of fermi level in intrinsic semiconductor is just above the centre of energy gap. 37. The excess concentration of holes injected into a semiconductor decreases exponentially with time. 38. If the Hall coefficient is negative then the semiconductor is n-type. 39. Solids are classified into metals,semiconductors,insulators.

40. In metals the valence band and the conduction band overlap with each other. 41. Semiconductors have finite forbidden energy gap between the valence band and the CB. 42. Insulators have much larger band gaps. 43. Ge and Si are the important semiconductors which are widely used in manufacturng of diodes and transistors. 44. At room temperature the thermal energy is sufficient to break covalent bonds.When a covalent bond is broken a free electron-hole pair is generated. 45. Pentavalent impurities contribute through donor electrons. 46. Trivalent impurities contribute through acceptor holes. 47. The conductivity of semiconductor is dependent on the number of charge carriers, their mobilities and electron charge. 48. Recombination centre are the energy levels at about the centre of the energy gap. 49. During recombination process one hole and one electron disappear and the time taken for this process is called the minority carrier life time. 50. Minority carrier life time can be defined as the time taken for the excess charge carriers to reduce 1/e times its initial values, once the source generating these excess charge carriers is out off. 51. Under equillibrium conditions,the rate of generation of charge carriers to the rate of loss due to recombination. 52. The equation relation these generation and loss rates are known as equation of conservation of charges of the continuity equation. 53. Mean lifetime of a charge carrier is the time taken for the injected concentration to fall to 1/e of its initial value. 54. Diffusion length for charge carrier is the distance into the semiconductor at which the injected carrier concentration falls to 1/e of its value at the surface. 55. Diffusion length can also be defined as the average distance which an injected charge carrier travels before lost due to recombination. 56. An electric field produced inside the conductor ina direction normal to both electric and magnetic fields is known as Hall effect. 57. The voltage generated in the phenomenon of Hall Effect is called as hall voltage. 329

58. Combined p-type and n-type semiconductors with PN junction formation is known as junction diode or PN diode. 59. PN junction can be obtained by growth junction type method. 60. Diffused junction type is used to obtain pn junction diode. 61. When an electric field appears on either side of pn junction that particular region is called as depletion region. 62. The potential difference which appears across the depletion region of pn junction is known as junction barrier. 63. Contact potential and barrier potential are the other names for junction barrier. 64. When pn junction is formed the Fermi levels of both n-side and p-side become common. 65. Under unbiased condition there is no current flow across the junction. 66. In pn junction the barrier potential is of the order of 0.2ev. 67. In forward bias condition the function resistance becomes almost zero and low resistance path is established. 68. Under forward bias, d.c. resistance is around 100 ohms for Ge diode and forward current is in mA. 69. During reverse bias, the junction resistance is very high and reverse current is in the order of micro ampere. 70. For Ge diode reverse resistance is around 1 mega ohm. 71. When applied reverse voltage is high and it raises the current suddenly in reverse direction , this break down reverse bias is called breakdown voltage . 72. The breakdown voltage is also called as zener voltage. 73. When an a.c. power source is connected across a pn junction during positive half of the a.c. cycle diode is forward biased. 74. During negative half of the a.c. cycle diode is reverse biased and no current flows. 75. The current flows in one direction during forward bias and no current during reverse bias 76.The unidirectional current flow character is utilized in using pn junction diode as a rectifier. 77. A rectifier is a device which converts alternating current to direct or unidirectional current. 78. Rectification takes place for positive half of a.c waveform, when diode is forward biased in rectifier circuit containing single PN junction diode.

79. One PN junction diode is used in rectifier circuit, during negative half of input cycle, the diode is reverse biased and no current flows through the junction. 80. There are two types of rectifiers Half wave rectifier and full wave rectifier. 81. The average value of the current I(dc) over one complete cycle is that direct current which sends the same charge through the circuit as sent by the varying current. 82. The effective value of the current is the square root of the average of squares of the instantaneous values taken over the whole cycle. 83. The effective value of the current is also called as the root mean square value of the current. 84. The efficiency of the rectifier is defined as the ratio of the d.c. output power P(d.c.) to the a.c. input power P(a.c.) supplied to the rectifier . 85. The number of diodes in half wave rectifier is one. 86. The peak alternating current in half wave rectifier is IO= E0/(r + RL). 330

87. The average value of current in half wave rectifier is I0/π. 88. The effective value of current in a half wave rectifier is given by Io/2. 2 89. The efficiency of a half wave rectifier is given by 4/(π) ( RL/(r+ RL) ) . 90. The peak alternating current in full wave rectifier is IO= E0/(r + RL). 91. The average value of current in full wave rectifier is 2I0/π. 1/2 92. The effective value of current in a full wave rectifier is given by Io/ (2) . 2 93. The efficiency of a half wave rectifier is given by 8/ (π) ( RL/(r+ RL) ) . 94. The number of diodes in full wave rectifier is two. 95. When a PN junction is forward biased, electrons diffuse from n-type side and undergo radiative recombination with holes in p-type side emitting photons.such PN junction diodes are called light emitting diodes. 96. In order to construct a suitable LED the material required must have energy gap of appropriate width. 97. Both p and n types must exist with low resistivities in order to form a suitable LED. 98. GaAs,GaP,Gallium arsenide phosphide are the important materials required for LED. 99. LED is very useful as a source for optical communication. 100. Full form of LCD is liquid crystal display.

101. In LCD’s the concept used is rotation of plane of polarization of light. Acoustics: 102. Optimum reverberation time for music is 1 to 2 sec. 103. Sabine’s formula is T = 0.165V / ∑aS. 104. Two square meter of a fiber board absorbs sound energy as that of unit area of an open window. The absorption coefficient of fibre board is 0.5. 105. The walls of a hall built for music concerts should absorb sound. 106. A cinema hall has a volume of 7500 m3. To have reverberation time of 1.5 sec total absorption in the hall should be 825 OUW. 107. Human body has maximum absorption coefficient. 108. Glass has minimum absorption coefficient. 109. The reverberation time is equal to 0.165V/A. 110. In a good acoustic hall sound should be distributed uniformly. 111. The reverberation time should be optimum. 112. The walls of a concert hall should absorb sound. 113. To keep optimum reverberation time either windows should be provided or sound absorbing materials should be arranged. 114. The echo effect is produced due to reaching of direct sound before reflected sound.

115. If reflected sound reaches the audience earlier than direct sound loudness increases. 116. Carpets are the sound absorbing material. 117. Acoustic quieting deals with the reduction of machinery sound & vibration, redirection of sound. 118. Lingering of sound is called reverberation. 119. An open window is a perfect absorber of sound. 120. To avoid interference effect there is no concave surface. 121. Echelon effect refers to the combination of echoes having regular phase difference. 122. The prolongation of sound inside a room or hall even after source producing the sound is turned off is called reverberation. 123. The time required for the intensity of sound to drop to one millionth (10-6) of its initial value is called reverberation time. 124. The reverberation time sound should be 1 to 2 sec for a music and 0.5 to 1 sec for speech.

125. The absorption coefficient of a material is given by a = 0.165V/S1(1/T1-1/T2). 126. Acoustic quieting is the process of silencing the noise from machinery, damping vibrations. 127. Muffler is also a silencer for reducing the amount of noise emitted by a machine. 128. Sound proofing is reducing the sound pressure with respect to a specified sound source and receptor. 129. Sound proofing effects sound in two different ways, Noise reduction and noise absorption. 130. A Room within a Room (RWAR) is one method of isolating sound and stopping it from transmitting to the outside worlds where it may be desirable. 131. The user is required to break the connection between the room that contains the noise source and the outside world is called acoustic quieting. 132. Electronic quieting is electronic vibration control, noise control, noise reduction and noise cancelling.

133. The reverberation time is the time required for the intensity to drop by 60 decibels. 134. The reverberation of sound is mainly due to multiple reflections at various surfaces inside. 135. The volume and the shape of the auditorium and absorption of sound influence the behavior of sound. 136. A musical hall requires a large volume while a lecture requires a smaller volume. 137. The ratio between ceiling height and breadth should be 2:3. 138. Echelon effect, echoes, resonance should be reduced. 139. Sound proofing materials should be fixed and hall must be full of audience for good acoustical hall. 140. Absorption coefficient a = sound energy absorbed by the surface/ Total sound energy incident on the surface. 141. Open window is a perfect absorber of sound and absorption coefficient of all substances is measured in terms of Open Window Unit (OWU). 142. Absorption coefficient is defined as reciprocal of its area which absorbs the same sound energy as absorbed by unit area of an open window. 143. The absorption coefficient of a given material depends on the frequency of the sound and it is higher at higher frequencies. 144. Sabine’s formula for reverberation time is directly proportional to volume of the auditorium, inversely proportional to absorbing surface area and total absorption. 145. The time of reverberation is optimum by providing windows, ventilators, heavy curtains, carpets, acoustic tiles and having full capacity of audience. 146. The time of reverberation is optimum by lining the walls with absorbent materials like felt, celotex, fiber board , glass wool etc. 147. Loudness may be increased by loud speakers at required places, low ceilings help in reflecting sound energy for uniform distribution of intensity throughout the hall. 148. Focusing surfaces such as concaves, spherical, cylindrical, parabolic on the walls should be minimized. 149. The prevention of transmission of noise inside or outside the hall is called sound insulation or sound proofing.

150. Air – borne noise, structure borne noise, inside noise should be minimized. 151. Automotive shock absorbers prevent mechanical shocks from reaching the passengers in a vehicle. 152. Electronic quieting is done for reduction of vibrations, noises produced by electronic equipments. 153. Mufflers are installed along exhaust pipe which is a chamber that causes destructive interference where opposite sound waves cancel each other out. 154. Specific mufflers, mini mufflers, catalytic converters are used to reduce noises. 155. Open - celled foam is a good sound damper inside a wall which reduces sonic vibrations. 156. To have a good sound effect inside a hall, the reverberation time has to be optimum

Nanotechnology 157. Diameter of one carbon atom is 0.15nm. 158. In Nano materials with decrease of size the interatomic spacing decreases. 159. In Nano materials with decrease of size the melting point decreases. 160. In Nano materials the ionization potential is higher when compared with that of the bulk. 161. Gold Nano spheres of 100nm appear in orange in color while 50nm appear green in color. 162. With decrease of coordination number, the magnetic moment increases. 163. Young’s modulus value of single walled Nanotubes is about 1 Terapascal. 164. Nanotechnology is the engineering of functional systems at the molecular scale. 165. Nanomaterials are having domain size about 100nm. 166. Properties of Nano materials differ from bulk materials due to the presence of less no. of atoms. 167. When a metal particle having bulk properties, is reduced in size , the density of states decreases. 168. Nano sized particles are chemically very active because the no. of surface molecules of atoms is more.

169. In the fabrication of nanoparticles, macro crystalline structures are broken down to Nano crystalline structures in ball milling. 170. The advantages of sol-gel technique in the fabrication a low temperature process, the product can be obtained in any form and polished to optical quality. 171. An electrochromic device is display device which displays information by changing color when a voltage is applied. 172. Fullerene is carbon molecule with carbon atoms arranged in a spherical shape. 173. Carbon Nanotubes are sheets of graphite rolled into a tube. 174. The approach used to construct Nanomaterials by merging smaller components into more complex assemblies is bottom-up approach. 175. The approach used to construct Nanomaterials by deforming large complex molecules in to small ones is top-down approach. 176. Lithographic method of preparing the Nanomaterials falls under top-down approach. 177. Epitaxial growth falls under top-down approach. 178. STM, AFM and SFM fall under top-down approach. 179. A nanometer is one thousand of a meter (i.e. 10-9) . 180. Photographic film is an emulsion consisting of silver nanoparticles in gelatin. 181. Nano materials are defined as those materials which have structured components with size less than 100nm at least in dimension. 182. Materials that are nanoscale in one dimension are layers, such as thin films or surface coatings. 183. Materials that are nanoscale in two dimensions are nanotubes and nanowires. 184. Materials that are nanoscale in three dimensions are particles like precipitates, colloids, quantum dots, grains. 185. Nanoscience is the study of phenomena and manipulation of materials at atomic, molecular and macromolecular scales where properties differ significantly from those at a larger scale . 186. Nanotechnology can be defined as the design, characterization, production and application of structures, devices and systems by controlling shape and size at the

nanometer scale.

187. A fullerene is a molecule entirely of carbon, in the form of hallow sphere, called bucky balls. 188. Cylindrical fullerenes are called nanotubes and bucky tubes 189. Fullerenes are similar to graphite containing pentagonal rings.

190. Fullerene was discovered by Buckminsterfullerene C60. 191. CNTS are two types SWNTs and MWNTs. 192. Properties of Nano materials changes due to Increase in surface area to volume ratio and Quantum confinement effects. 193. When the dimensions of potential wells, potential boxes are of the order of de Broglie wavelength of electrons, the energy levels of electrons changes. This effect is called Quantum Confinement. 194. The physical, electronic, magnetic, mechanical, chemical, thermal, optical properties of materials depend on size. 195. The interatomic spacing and melting point decreases with size. 196. Nano particles may emit or absorb specific wave lengths of light by controlling size. 197. Ionization potential and mechanical properties are higher in Nano materials. 198. CNTs are rolled graphite sheets which are mechanically strong and flexible and first observed by Sumio Iijima. 199. Nano particles are more magnetic than bulk materials. 200. Plasma arcing is used produce CNTs. 201. Evaporative deposition is the method where the material is heated and vaporized by electrically resistive heating under high vacuum. 202. Electron beam physical vapor deposition is the method where the material to be deposited is heated to a high vapor pressure by electron bombardment in high vacuum. 203. Sputter deposition is the method where the material to be deposited is vaporized by bombarding the material with aglow plasma discharge. 204. Cathodic arc deposition is a high power arc directed at the target material. 205. Pulsed laser deposition is a high power pulsed laser ablates the material from the target into a vapor.

206. Chemical vapor deposition is the method where the material is heated to form a gas and allowed to deposit on a solid surface under vacuum condition by chemical reaction. 207. Nano powders of oxides and carbides of metals are produced by chemical vapor deposition. It is also used to grow surfaces. 208. Sol-gels are suspensions of colloids in liquids that keep their shape. 209. Sol-gel process is done by three stages i.e. 1. Hydrolysis 2.condensation and polymerization of monomers to form particles 3. Agglomeration of particles followed by the formation of networks which extends throughout the liquid medium and forms a gel. 210. Electro deposition is used to electroplate a material. 211. Nano structured films of copper, platinum, nickel, gold etc are produced by Electro deposition.

212. Ball milling is used for preparing metal oxides like CeO2 and ZnO. 213. CNTs are prepared by Arc method, Laser method and Chemical vapor deposition. 214. CNTs have high electrical conductivity, high strength and elasticity, high thermal conductivity and highly absorbent due to large surface area. 215. Hydrogen absorbing capacity increases by decreases size of Nano particles of Ni, Pd, Pt used in hydrogen storage devices. 216. Nano Bio-ceramics are used in implantation of heart valves, interact with bone cells because of low weight, high strength. 217. Nano magnetic materials are used in data storage such fuel cells, solar cells and batteries. 218. Nano Light emitting phosphors are used in T.V. screens, computer monitors, flat panel displays. 219. Nano-Engineered membranes are used in energy efficient water purification processes. 220. Nano fabrics are breathable, water proof and stain resistant. 221. Nano coatings are wear and scratch –resistant. 222. Nano paints change color in response to change in temperature . 223. Carbon nanoparticles are used fillers in a matrix to reinforce car tyres, bumpers.

224. Nano crystalline magnets are used in MRI instruments. 225. Characterization of Nano particles is done by XRD and TEM techniques. 226. Scanning electron microscope (SEM) is produced by scanning the sample with a focused electron beam and detecting the secondary and back scattered electrons. 227. Transmission electron microscope (TEM) is produced by analyzing the electrons which are elastically scattered, inelastically scattered and unscattered. 228. Electron microscope has high magnification of 1,000,000X times and high resolution of 0.1nm than optical microscope. 229. Electron microscope consists of electron gun, electromagnetic fields as lenses maintained under high vacuum. 230. SEM and TEM gives information on topography, morphology, composition and crystallographic information. 231. Nano sized ball bearings are used as lubricants to reduce friction. 232. Nano particles in cosmetics are harmful to humans. 233. Nano materials in food and drinks for package materials, storages like additives are harmful to humans.