IA TEST -3 SCHEME & SOLUTION TEST ON : 10-05-2016 Marks: 40 Subject & Code: ENGG.PHYSICS & 15PHY22 Sec : A, B, C, D & E Name of faculty: Dr.Gajanan v Honnavar, Dr. M. Faisal & Mohana Lakshmi Time : 11.30 AM to 1.00 PM

Q. no. Scheme Marks P Definition of A a i) Mean free path ii) Relaxation time iii) Drift velocity 3 x 01 R 1 b Definition of Fermi factor. 02 T Variation of Fermi factor with temperature 3 x 01 a Success of QFET over CFET 3 x 02 1 2 b Equation , Substitution & Answer 02 P a Statement and explanation of law of mass action 02 + 02 A 3 b Equation , Substitution & Answer 04 R Use of equations of ohms law , current density, electrical T a current , mobility of charge carriers and arriving at the final 04 expression electrical conductivity of an ISC 2 4 b Equation , Substitution & Answer 04 P Temperature dependence of resistivity in metals 02 A 5 a Temperature dependence of resistivity in superconductors 02 R Definition of Superconductivity 01 T b Explanation of BCS theory of superconductivity 03 6 a Description of Type I and Type II superconductors 2 x 03 3 b Explanation of maglevs 02 P a Construction and working of shock tube 02 A 7 Diagram, Principle, Working 03 R b Any three applications of shock wave 03 T 8 a Explanation of ultrasonic and supersonic waves 01 +01 4 B Three conservation laws, Normal shock relationships 03 + 03

P 9 Principle, construction, working and applications of SEM 08 A R a Synthesis of carbon nanotubes by the Arc-discharge method 04 T b Graphical representation of the variation of

10 in 0D, 1D, 2D and 3D quantum structures 04 5

SOLUTION PART 1 1. a) Define the terms i ) Mean free path ii) Relaxation time iii) Drift velocity (3M)

Mean free path (λ): It is the average distance travelled by the free between successive collisions with the lattice ions.

Relaxation time (흉r): When the electric field applied across a conductor is switched off, the drift velocity decreases exponentially to zero. The time during which the drift velocity reduces to 1/e times the initial value is called relaxation time.

Drift velocity (vd): When an electric field is applied, some of the free electrons acquire a constant, terminal velocity in a direction opposite to that of the field. This steady and very small velocity is called drift velocity.

1. b) Explain Fermi factor. Discuss the probability of occupation of various energy states by electrons at T=0oK, and T>0oK, on the basis of Fermi factor. (5M)

Fermi factor: Under thermal equilibrium the free electrons occupy various energy levels in accordance with a statistical rule known as Fermi – Dirac statistics. Fermi – Dirac statistics enables the evaluation of probability of finding in energy levels over a certain range of energy values. The evaluation is done with the help of a quantity called Fermi factor f (E) given by:

1 f E  EEF  kT e 1 Effect of temperature on Fermi function

Case 1: Probability of occupation for E< EF at T= 0k.

Substituting this condition in F-D equation, we get

1 푓(퐸) = 1 + 푒−∞ 1 푓(퐸) = 1 + 0 푓(퐸) = 1

The distribution 푓(퐸) = 1 means that at T = 0k, all the energy levels below the Fermi level are fully occupied by electrons leaving the upper levels vacant.

I.e. there is 100 % probability of finding an electron below the Fermi .

Case 2: Probability of occupation for E >EF at T= 0k.

Substituting this condition in F-D equation, we get

1 푓(퐸) = 1 + 푒∞ 1 푓(퐸) = 1 + ∞ 1 푓(퐸) = = 0 ∞ This indicates that at T= 0k, the energy levels above the Fermi level is not occupied by electrons, they are vacant.

I.e. there is 0% probability of finding the electron above the Fermi level at absolute zero K

Case 3: Probability of occupation at ordinary temperatures.

At temperatures above 0 K and E = EF .

Substituting this condition in F-D equation, we get

1 푓(퐸) = 1 + 푒0 1 푓(퐸) = 1 + 1 1 푓(퐸) = = 0.5 2 At temperatures above 0K, there is only 50 % probability for an electron to occupy Fermi level.

2. a) Describe how quantum free electron theory was successful in overcoming the failures of classical free electron theory. (6M) Merits of Quantum Free Electron Theory:

1. Specific heat of free electrons:

From quantum theory of free electrons, the specific heat of free electrons is given by Cv = (2k/EF) RT For a typical value of EF = 5 eV, we get Cv = 10−4RT. Hence gives the correct molar specific heat at constant volume for free electrons in metals. 2. Temperature dependence of conductivity of metals:

The quantum free electron theory gives the correct temperature dependence of conductivity. From the equation for conductivity according to Q.F.E.T, 흈 ∝ 흀 ퟏ But, 흀 ∝ 흅풓ퟐ Where ‘r’ is the amplitude of lattice vibrations which is directly proportional to temperature i.e. 푟2 ∝ 푇 1 ∴ 휆 ∝ 푇 Since, 흈 ∝ 흀 ퟏ ∴ 흈 ∝ 푻 3. Dependence of electrical conductivity on electron concentration:

ne2 The electrical conductivity in metals is given by  * , m vF It is clear from this equation that the electrical conductivity depends on the electron

*. concentration, the ratio of λ/vF and m Now by taking above expression, if we calculate σ, we get the observed conductivities of the metals. - 2. b) Find the relaxation time of conduction electrons in a metal of resistivity 1.54 x 10 8 28 3 ohm-m, if the metal has 5.8 x 10 conduction electrons per m . (2M) σ = n e2휏/m 흉 = ퟑ. ퟗퟕퟒ × ퟏퟎ−ퟏퟒ 풔 PART 2 3. a) State and explain the law of mass action for a . (4M)

Law of Mass Action: The law of mass action states that the product of electron concentration (ne) in the conduction band and the hole concentration (nh) in the valence band is equal to the square of the intrinsic carrier concentration (ni).

We have the equation for electron concentration and hole concentration as

Concentration of Free Electrons in the Conduction Band ne:

E E  4 2 3 F g n  m kT 2 e kT e h3 e Concentration of Holes in the Valence Band nh:

3  E  4 2(m kT) 2  F  n  h e  kT  h h3

In both the equations, we see the dependence of both ne and nh on EF

Let us consider the product of ne and nh

3  E  2  g  32(kT) * * 3 2  kT  n n  m m  e e h h6 e h

The above equation shows that the product of nenh does not depend on EF, but remains a constant at a given temperature ne nh = a constant

This condition is applicable for both extrinsic and intrinsic . The above equation is called the law of mass action for semiconductors. 2 In case of intrinsic semiconductor , ne nh = ni

3. b) The Fermi level for a metal is 3.1eV. Calculate the energies for which the probabilities of occupancy at 300 K are 98% and 50 %. (4M)

1 f E  EEF  e kT 1  1  E  EF  kT ln 1  f E  EF = 3.1 eV i) For f (E) = 98% = 0.98 E = 2.999 eV ii) For f (E) = 50% = 0.50 E = 3.1 eV

4. a) Derive the expression for electrical conductivity of an intrinsic semi-conductor The current in a semiconductor is contributed by both electrons and holes, moving in opposite directions. Consider a semiconductor with area of cross-section, A, free electron

concentration ne and number density of holes, nh. The current due to free electrons is: Ie = ne A e ve, Where e is electronic charge and ve is the drift velocity of electrons. And, the current due to holes is:

Ih = nh A e vh,

Where vh is the drift velocity of holes.

Hence the total current, I = Ie + Ih = ne A e ve + nh A e vh = A e (ne ve + nh vh)  Current density J = I/A = e (ne ve + nh vh)

But, current density, J = 휎 E, where 휎 is conductivity and E, the field across the semiconductor.  휎 E = e (ne ve + nh vh)  휎 = e (ne ve/E + nh vh/E) Or, 흈 = e (ne 흁 e + nh 흁 h)

Where, ve/E = e is the mobility of electrons and vh/E = h is the mobility of holes.

For intrinsic semiconductors ne = nh = ni, where ni is the density of intrinsic charge carriers.

Hence for intrinsic semiconductors,  = e ni (e + h)

For n-type semiconductors, ne >> nh and therefore,  = e ne e

For p-type semiconductors, np >> ne and therefore,  = e nh h

4. b) The effective mass of the electron in silicon is 0.31 m0, where m0 is the free electron mass. Find the electron concentration for silicon at 300 K assuming that the Fermi level lies exactly in the middle of the energy gap, given that the energy gap of silicon is 1.11 eV. (4M)

Sol: T= 300K ; Eg= 1.11 eV; EF= Eg/2

E E 3  F g  4 2 kT n  m kT 2 e e h3 e

15 3 ne= 2.07 x 10 /m

PART 3

5. a. Explain the temperature dependence of resistance in metals and superconductors. (4M) In case of metals the resistance decreases witht temperature and reaches a minimum value at T=0 K. Even at ‘0K’, metals offer finite resistance, called residual resistance which is due to the scattering of electrons by impurities and crystal defects present in the material. In addition to being scattered by thermal vibrations of the lattice ions, the electrons are also scattered by the impurities and this impurity scattering is more or less independent of temperature. As a result there will be certain residual resitivity which remains at the lowest temperature. In case of superconductors, the resistance decreases with decrease in temperature when the material is in non-superconducting state, as in case of a normal metal up to a

particular temperature Tc. At Tc, the resistance abruptly drops to zero and signifies a transition from normal state to the superconducting state.

5. b. Define superconductivity and describe how BCS theory explains superconductivity. (4M) Superconductivity: Superconductivity is a state of materials in which their electrical resistance is zero. For certain materials, when they are cooled, the resistivity abruptly falls to zero at a temperature characteristic. This phenomenon is known as superconductivity. The temperature at which transition takes place from normal conducting to

superconducting state is the transition temperature or critical temperature TC.

BCS Theory:

Explanation to superconductivity was given by John Bardeen, Leon N Cooper and John Robert Schrieffer on the basis of quantum theory. The theory is named after the three scientists as BCS theory. When an electron approaches a positive ion in the lattice, there is a Coulomb attraction between them, which produces a distortion in the lattice. Now if another electron passes through the distorted lattice, interaction between the distorted lattice and the electron occurs, which results in lowering of energy of the electron. This in effect is an interaction between two electrons via the lattice distortion or the phonon (quantum of energy of lattice vibration), which results in the lowering of energy for the two electrons. This lowering of energy implies that the force between the two electrons is attractive. This type of interaction is called the electron-phonon-electron interaction. This interaction is strongest if the two electrons have equal and opposite momentums and spin. Such pairs of electrons are called Cooper pairs. At very low temperatures the density of Cooper pairs is large and they collectively move through the lattice with small velocity. Low speed reduces collisions and thereby resistivity decreases, which explains superconductivity. The attraction between the electrons in the Cooper pair is very weak. Therefore they can be separated by small increase in temperature, which results in a transition back to the normal state.

6. a) Describe Type I and Type II superconductors (6M)

Types of Superconductors:

Based on the magnetic behavior, superconductors are classified into two categories, viz. Type I and Type II superconductors.

Type I Superconductors (Soft Superconductors):

In type I superconductors, the transition from superconducting to normal state, in the presence of , occurs sharply at the critical field HC. Below the critical field the magnetization in the specimen is proportional the applied field and it behaves like a diamagnet. For H > HC, magnetization of the specimen is negligible and the specimen becomes a normal conductor. This kind of behavior is exhibited by pure elements. The critical field is relatively low for these superconductors. Examples are aluminium, lead and indium.

Type II Superconductors (Hard Superconductors): Type II superconductors are characterized by two critical fields HC1 and HC2. Up to HC1, the lower critical field, the specimen is perfectly diamagnetic and hence is

superconducting. From HC1 to HC2, Meissner effect is incomplete but the specimen exhibits superconductivity. This state is called the mixed or vortex state. Above the

upper critical field HC2, the specimen becomes an ordinary conductor. HC2 can be as high as 20 to 50 T, and retention of superconductivity at such high magnetic fields makes them useful in creating very high magnetic fields. Examples of type II superconductors are transition metals and alloys containing niobium, silicon and vanadium.

6. b) What are Maglevs? (2M)

Magnetic Levitation:

When a magnet is brought near a superconducting coil, a current is induced in the coil. This induced current produces a magnetic field in the coil, which results in a repulsive force between the coil and the magnet (Lenz’s law). The magnet therefore floats in air above the superconducting coil. This is called magnetic levitation (Maglev). This phenomenon has been used to develop a prototype magnetic levitation train in Japan, which reached high speeds of the order of 500 km/h, as there is no friction between the wheels and the rails. The height to which the vehicle is levitated is about 10 to 15 cm. PART 4 7. a) Explain the construction and working of shock tube.(5M)

Shock Tube: It is a device used to create and study shock waves in the laboratory. Reddy Shock Tube: Reddy shock tube is a hand operated shock tube. It consists of a cylindrical steel tube of about 30 mm diameter and length about 1 m. It is divided into two sections of length 50 cm each. One half is the driver tube and the other half is the driven tube, and are separated by thin aluminum or paper diaphragm, about 0.1 mm thick. The end of the driver section is fitted with a piston and the far end of the driven section is closed. A digital pressure gauge is mounted in the driver section close to the diaphragm. Piezoelectric sensors S1 and S2, separated by a distance 70 mm, are mounted near the closed end. The driver section is filled with a gas at a high pressure with the help of the piston.The driver section is filled with a gas (driver gas) at high pressure and the driven section is filled with a gas (driven gas) at a low pressure.

The driver gas is compressed using the piston until the diaphragm ruptures. Due to this the gas rushes into the driven section and pushes the driven gas towards the far downstream end of the tube. This generates a moving shock wave which moves towards the end of the driven section. The shock wave raises the temperature and pressure of the driven gas. The time over which the extreme temperature and pressure are sustained is of the order of milliseconds. However the actual duration depends on the properties of the gases and the dimensions of the shock tube.

8. b). Mention the applications of shock waves.(3M)

Applications of Shock Waves: 1. Shock waves are used for the treatment of kidney stones. Shock waves break kidney stones into small fragments which pass out of the body through urinary tracts. 2. Using mild shock waves DNA can be pushed into biological cells. 3. Shock waves are used in pencil industry to soften wood. 4. Using shock waves drugs can be injected into the body without using needles.

8. a) What are ultrasonic and supersonic waves?(2M) Ultrasonic waves: Ultrasonic Waves are pressure waves having frequencies beyond 20,000 Hz.

Supersonic waves: Mechanical waves which move with speeds greater than that of sound are called Supersonic waves. Supersonic conditions occur for Mach numbers greater than one. i.e. 1

8. b) Based on the conservation laws, give an account of the normal shock relationships. (6M) The conservation of mass, momentum and energy are the fundamental principles of physics.

According to the principle of conservation of mass the total mass of a closed system remains unchanged irrespective of whether any physical or chemical changes occur within the system.

The law of conservation of momentum states that the total momentum of a closed system is a constant.

The law of conservation of energy states that the total energy of a closed system remains constant and is independent of the changes taking place within the system.

Starting from the basic conservation equations, and assuming that the gas is perfect, the jumps in pressure, temperature and density can be calculated.

Along with the three equations of conservation laws and also considering the equation of state i.e. A perfect gas obeys the equation of state Pv =RT , where R is the gas constant.

The normal shock relations called Rankine-Hugoniot Equations can be derived.

The pressure ratio in terms of Mach No. is

Temperature ratio

Density ratio

PART 5

9. a) Explain the principle, construction, working and applications of SEM.(8M)

Principle: The surface of the sample under study is scanned using a high energy beam of electrons. This results in secondary electrons, back-scattered electrons and X-rays. Of these, the secondary electrons give the topographical information about the surface, the back- scattered electrons give information on the chemical composition and X-rays give the elemental composition.

Construction: The electron gun produces high energy electrons which are converged using magnetic condenser lenses to form a fine beam. A scanning coil is used to deflect the electron beam to scan the specimen. The objective lens is used to focus the scanning beam on the desired spot on the sample. The intensities of the secondary electrons, the back- scattered electrons and X-rays are recorded using detectors. The outputs from the detectors are fed to the TV monitor, which displays the images.

Working: When the beam of high energy electrons hit the target some electrons are elastically scattered (back-scattered electrons). Some electrons are knocked off from the surface of the sample (secondary electrons). And some electrons penetrate deep into the inner shells of the sample atoms and knock off inner electrons. This results in the production of characteristic X- rays. These are detected using detectors and the signals are amplified and displayed on a TV monitor.

Applications of SEM  SEM is used to study biological specimens like pollen grains  SEM is used to study corroded layers on metal surfaces.  SEM images are used to study properties of the sample like hardness, reflectivity, melting point etc.

10. a). Describe the synthesis of carbon nanotubes by the Arc - discharge method.(4M)

The arc discharge technique involves the use of two high purity graphite electrodes as the anode and the cathode. The carbon rods are placed end to end separated by approximately 1mm in an enclosure filled with inert gas at low pressure. Helium is circulated in the chamber at a pressure of 500 torr. A direct current of 50 to 100 A, driven by a potential difference of approximately 20 V, creates a high temperature discharge b/w the two electrodes.The discharge vaporizes the surface of the anode and some part of the evaporated carbon deposits on cathode tip layer by layer. Thus a carbon rod is built up at the cathode.

10. b). Give the graphical representation of the variation of density of states in 0D, 1D, 2D and 3D quantum structures.(4M)

Density of states for different Nanosystems:

The density of states g (E) for various quantum structures as function of energy is shown in the figure.

The DOS increases with energy and the curve for a 3 –D material is parabolic in shape. From the equation we can conclude that the DOS in a bulk material is proportional to E1/2 and hence increases with energy.

If the electrons are confined in one or more directions by reducing the dimensions of the material in those directions, the DOS changes due to quantization of energy.

For a quantum well, the DOS for a given will be constant. For different quantum states, it has different constant values and the curve varies as a step function.

For a , the DOS for a given quantum state will decrease with increasing energy. For different quantum states, it has variation as shown in the figure. For a , only some discrete energy states are allowed for the electrons. As only certain energy states are allowed, the density of energy levels appear as discrete lines.