H K B K COLLEGE OF ENGINEERING BENGALURU Department Of Engineering Physics
Course Material
Subject: Engineering Physics Subject Code: 18PHY12/22 Semester: I/II Sem B.E.
MODULE 5 MATERIAL SCIENCE Engineering Physics
MODULE - 5 MATERIAL SCIENCE
The properties of metals such as electrical conduction, thermal conduction, specific heat etc are due to the free electrons or conduction electrons in metals. The first theory to explain the electrical conductivity of metals is ‘Free electron theory’ and it was proposed by Drude in the year 1900 and later developed and refined by Lorentz. Hence the classical free electron theory is known as Drude- Lorentz’s theory. It failed to account the facts such as temperature dependence of conductivity and dependence of electrical conductivity on electron concentration.
Some important Definitions:
• Thermal Velocity (vth): The velocity with which the free electrons keep moving 푘푇 due to thermal agitation is called Thermal velocity. vth = √ 푚 • Drift velocity (vd): The velocity of electrons in the steady state in an applied −푒퐸휏 electric field is called drift velocity. vd = 푚 • Mean free path (): The average distance travelled by the conduction electrons between successive collisions with lattice ions. • Mean collision time (τ): The average time that elapses between two consecutive collisions of an electron with the lattice points is called mean collision time.
• Relaxation time (r): From the instant of sudden disappearance of an electric field across a metal, the average velocity of the conduction electrons decays exponentially to zero, and the time required in this process for the average velocity to reduce to (1/e) times its value is known as Relaxation time. ne 2λ • Electrical conductivity σ = (according to classical free electron theory) 3kTm
Failures of Classical Free Electron Theory: 1. Temperature dependence of electrical conductivity: According to classical free electron theory ne 2λ σ = 3kTm . 1 The above equation shows that, σ α . But, experimentally electrical conductivity σ is T 1 found to be inversely proportional to the temperature T. i.e. σ exp α . It is clear that the T experimental value is not agreeing with the theory.
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2. Dependence of electrical conductivity on electron concentration:
According to classical free electron theory
ne 2λ σ = therefore, σ α n 3 k Tm
Experimentally determined values of some metals are given below
Metal Electron Concentration (n) Electrical Conductivity (σ) 1028/ m3 107 /Ωm Copper 8.45 5.88 Aluminum 18.06 3.65
From the table it is clear that, though electron concentration in copper is 2.13 times less than that of Aluminum, its electrical conductivity is much greater than that of Al. Hence, the classical free electron theory fails to explain the dependence of σ on electron concentration.
From the above discussions, it is clear that classical free electron theory fails to explain many of the experimentally observed facts.
Quantum Free electron theory of metals:
In order to rectify the draw backs of the classical free electron theory, Sommerfeld proposed the quantum free electron theory in 1928.This theory is based on quantum concepts.
Assumptions of Quantum free electron theory:
• The energy values of the conduction electrons are quantized. • The distribution of energy among the free electrons is according to Fermi-Dirac statistics. • The distribution of electrons in the various allowed energy levels occur as per Pauli’s exclusion principle. • The electrons travel with a constant potential inside the metal but confined within its boundaries. Density of states:
There are large numbers of allowed energy levels for electrons in solid materials. A group of energy levels close to each other is called as energy band. Each energy band is spread over a few electron-volt energy ranges. In 1mm3 volume of the material, there will be large number of permitted energy levels in an energy range of few electron-volts. Because of this, the energy values appear to be virtually continuous over a band spread. To represent it technically it is stated as density of energy levels. The dependence of
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density of energy levels on the energy is denoted by g(E). It is called density of states function. The graph shows variation of g(E) versus E.
‘It is the number of allowed energy levels per unit energy interval in the band associated with material of unit volume’. The density of states in range E and (E+dE) is denoted by g(E)dE. 8√2휋푚3/2 i.e. g(E)dE = E1/2 dE ℎ3 It is clear g(E) is proportional to √퐸 in the interval dE. Fermi – Dirac statistics: Fermi Dirac statistics explains the distribution electrons in the permitted energy states. At T=0K, all the energy states in all the energy bands below the valence band are completely filled. The Valence band is partially occupied by the electronsand the energy states above te fermi level are completely vacant. At T>0K, it was expected that there will not be any orderliness in the occupation of energy states by the electrons as large number of energy levels will be avaiable.But contrary to the expectation, there exist a orderliness in the occupation of higher energy levels. It follows the orderliness or statistical rule called Fermi- Dirac statistics when there is a thermal equilibrium. Fermi – Dirac statistics is applicable to the assembly of particles which are identical, indistinguishable and having ½ integral spin. The distribution of the particle in various energy levels will be according to Pauli’s exclusion principle. The electrons satisfies these conditions.
Fermi energy and Fermi factor
Fermi energy: In a metal having N atoms, there are N allowed energy levels in each band. In the energy band the energy levels are separated by energy differences. It is characteristic of the material. According to Pauli’s exclusion principle, each energy level can accommodate a maximum of two electrons, one with spin up and the other with spin down. The filling up of energy levels occurs from the lowest level. The next pair of electrons occupies the next energy level and so on till all the electrons in the metal are accommodated. Still number of allowed energy levels, are left vacant. Department of Engineering Physics, HKBKCE Page 3
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‘The energy of the highest occupied level at absolute zero temperature (0K) is called the Fermi energy and the energy level is called Fermi level’. It is denoted by Ef.
Fermi factor: When the temperature is greater than zero kelvin, the material will receive thermal energy from its surroundings at room temperature. The available thermal energy is small and therefore the electrons occupying energy levels much below the Fermi level cannot absorb the thermal energy due to non-availability of higher energy levels. But this energy can be absorbed by the electrons occupying energy levels which are just below the Fermi level. Because there are a large number of unoccupied energy levels just above the Fermi level to which electrons get excited. Though the excitations are random, the distributions of electrons in various energy levels will be systematically governed by a statistical function at the steady state.
‘Fermi factor is defined as the probability of occupancy of a given energy state for a material in thermal equilibrium’.
The probability f (E) that a given energy state with energy E is occupied at a steady temperature T is given by 1 f(E) = (E−EF ) e kT +1 f (E) is called the Fermi factor.
Dependence of Fermi factor with temperature and energy:
The dependence of Fermi factor on temperature and energy is as shown in the figure.
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• Case I : Probability of occupancy for E When T=0K and E 1 1 푓(퐸) = = = 1 푒−∞+1 0+1 f (E)=1 for E Therefore at T=0K all the energy levels below the Fermi level are occupied. • Case II: Probability of occupancy for E>Ef at T=0K: When T=0K and E >Ef 1 1 푓(퐸) = = = 0 푒∞ + 1 ∞ . . . f (E)=0 for E >Ef ... At T=0K, all the energy levels above fermi levels are unoccupied • Case III: The probability of occupancy at T˃0K (퐸−퐸 )⁄푘푇 0 At E=Ef, 푒 푓 = 푒 = 1 . 1 1 1 . . 푓(퐸) = (퐸−퐸 )⁄푘푇 = = 푒 푓 +1 1+1 2 Hence, the Fermi energy is the most probable or the average energy of the electrons across which the energy transitions occur at temperature above zero degree absolute. Derivation of the expression for Fermi Energy: Fermi energy ‘EF’ at 0K is denoted EF0. The number of electrons/unit volume which possess energy only in the range E and E+dE is given by N(E)dE = [number of available states in the energy range E and (E+dE) ] ×[probability of the occupation of those energy levels by the electron] Department of Engineering Physics, HKBKCE Page 5 Engineering Physics If g(E) is the density of state function, then the number of energy states in the range E and (E+dE) is g(E) and the probability of occupation of any given energy state by the electron is given by the Fermi factor f(E). ... N(E)dE = g(E)dE × f(E) The number of electrons per unit volume of the material ‘n’ can be evaluated by integrating the above expression from E=0 to E=Emax , where Emax is the maximum energy possessed by the electrons. 퐸 푛 = 푚푎푥 푁(퐸)푑퐸 ∫퐸=0 퐸 or 푛 = 푚푎푥 푔(퐸)푓(퐸)푑퐸 ∫퐸=0 But f(E)=1 at T=0K 퐸 ... 푛 = 푚푎푥 푔(퐸)푑퐸 × 1 ∫퐸=0 8 2휋푚3⁄2 Where 푔(퐸)푑퐸 = √ 퐸1⁄2푑퐸 ℎ3 Where ‘m’ is the mass of the electron and ‘h’ is the Planck’s constant. 8√2휋푚3⁄2 퐸 푛 = ∫ 푚푎푥 퐸1⁄2 푑퐸 ℎ3 퐸=0 3⁄2 퐸 8√2휋푚 2 3⁄2 푚푎푥 푛 = 3 [ 퐸 ] ℎ 3 0 But at T=0K, the maximum energy that any electron of the material can have is EFo. Hence Emax= EFo. 8 2휋푚3⁄2 2 푛 = √ (퐸 )3⁄2 ℎ3 3 퐹표 8×23⁄2푚3⁄2 휋 푛 = [ ] (퐸 )3⁄2 ℎ3 3 퐹표 3 3 ( 8 2 = 888 = 82 8 =8 2 2 2 =8 2 2 ) ℎ3 3푛 (퐸 )3⁄2 = [ ] 퐹표 (8푚)3⁄2 휋 풉ퟐ ퟑ풏 ퟐ⁄ퟑ 푬 = ( ) 푭풐 ퟖ풎 흅 Fermi Temperature: It is the temperature at which a metal is to be heated such that the free electrons acquire energy equal to Fermi energy. Department of Engineering Physics, HKBKCE Page 6 Engineering Physics 2E T = F F 3k Fermi Velocity: It is the velocity of the electrons occupying the Fermi level. 2E V = F F m Success of Quantum Free Electron Theory: Quantum free electron theory has successfully explained following observed experimental facts where as the classical free electron theory failed. a) Temperature depends on electrical conductivity. 1 1 Electrical conductivity is proportional to but not which is as follows: 푇 √푇 ne 2 τ Electrical conductivity σ= m* Where m* is called effective mass of an electron. λ According to quantum free electron theory is τ = vF ne 2λ σ= m* v F Conduction electrons are scattered by the vibrating ions of the lattice. The vibration occurs such that the displacement of ions takes place equally in all directions. If ‘r’ is the amplitude of vibrations then the ions can be considered to present effectively a circular cross section of area πr2 that blocks the path of the electron irrespective of direction of approach. Since the vibrations cover larger area of cross section should scatter more number of electrons, it results in the reduction of mean free path of the electron. 1 λ α 2 π r Considering the facts The energy of vibrating body is proportional to the square of amplitude. The energy of ions is due to the thermal energy. The thermal energy is proportional to the temperature ‘T’. Department of Engineering Physics, HKBKCE Page 7 Engineering Physics We can write r2 T 1 λ T 1 σ T 1 σ is correctly explained by quantum free electron theory. T b) Electrical conductivity and electron concentration: Aluminum which has three free electrons per atom, have lower electrical conductivity than that of copper which has only one free electron per atom. As per quantum free electron the electrical conductivity is ne 2λ σ= * m vF The value of n for aluminum is 2.13 times higher than that of copper. But the value of λ/vf for copper is about 3.73 times higher than that of aluminum. SEMICONDUCTOR PHYSICS Conductivity of Semiconducting materials • The valence band and the conduction band are separated by a narrow forbidden gap of about 1ev at 0 K. • At 0 K, the conduction band is completely empty and the valence band is completely filled. • When a small amount of energy is supplied, the electrons easily move from valence band to the conduction band. • Resistivity of semiconductor lies in the range of 10-6 to 108Ωm. Fig: Energy level diagram for a metal, semiconductor and an insulator. Department of Engineering Physics, HKBKCE Page 8 Engineering Physics Concentration of electrons and holes in intrinsic semiconductors: Semiconductors in the purest form are known as intrinsic semiconductors. Ge and Si are common examples. Consider Ge crystal lattice, at absolute zero, electrons are bonded together and are not available for electrical conduction. As a result, Ge behaves like a perfect insulator. In an intrinsic semiconductor for every electron freed from the bond, there will be one hole created in the crystal. It means that “the number of conduction electrons is equal to the number of holes at any given temperature”. i.e., at constant temperature, ni = pi = constant In an extrinsic semiconductor, the carriers are generated by two different ways. They might be from the impurity atoms (electrons for n-type and holes for p-type) or from the broken covalent bonds (equal number of holes and electrons). Majority charge carriers (electrons for n-type/holes for p-type) are more in numbers than minority charge carriers (holes for n-type/electrons for p-type), as minority charge carriers are only from broken covalent bond. Expression for concentration in conduction band: Electron density in conduction band is given by 3 E −E 2m*kt 2 − C F n = 2 e e kT e h2 Expression for hole concentration in valence band: Hole density in valence band may be obtained from the result 3 E −E 2m*kT 2 − F V n = 2 h e kT h h2 Fermi level in an intrinsic semiconductor: At T=0K, all the energy levels in the valence band are completely filled where as all the energy levels in the conduction band are completely empty. The difference in the energy of the bottom of conduction band and the valence band is energy gap denoted by Eg .At room temperature some of the electrons from the top of the valence band jump to the bottom of the conduction band due to thermal excitation. After sometime the electrons get de-excited and hence the process of excitation and de-excitation continues and the electrons become conduction electrons. The average energy of the electron 1 taking part in conduction will be almost equal to Eg. Therefore Fermi level lies in the 2 middle of the forbidden gap for an intrinsic semiconductor. Department of Engineering Physics, HKBKCE Page 9 Engineering Physics Valence Band ------EF Conduction Band For an intrinsic semiconductor, ne= nh 3 3 E −E E −E 2m*kt 2 − C F 2m*kT 2 − F V 2 e e kT = 2 h e kT 2 2 h h 3 m* 2 −E f + Ev + Ec −E f e = e kT * mh 3 m* − 2E + E + E ln e = f v c * 2 mh kT E + E 3 m* E = v c − kTln e f * 2 4 mh Conductivity in a Semiconductor: In a semiconductor, the current is due to free electrons as well as holes. The current due to electrons can be written as Ie = ne e A ve where ve = Drift velocity of electrons; ne = Number density of electrons Similarly the current due to holes is Ih = nh e A vh where vh = Drift velocity of holes; nh = Number density of holes The total current is, I = Ie + Ih I = e A (neve + nhvh) I The current density J = A = e (neve + nhvh) J σ = E v e v h σ = e n e + n h E E Department of Engineering Physics, HKBKCE Page 10 Engineering Physics v v e = μ , the mobility of electrons; h = μ , the mobility of holes E e E h σ = e(n eμ e + n h μ h ) For intrinsic semiconductors, ne = nh = ni where ni is called the density of intrinsic charge carriers. σ = n e μ + μ i ( e h ) Hall Effect : When magnetic field is applied perpendicular to direction of current in a conductor, a potential difference develops along on axis perpendicular to both current and magnetic field. This effect is known as Halleffect and the potential difference developed is known as Hall Voltage. Probe Y + + + + + + + + + + + + + + + + + + + + + + + + + + + + + n- type material w d I I ------ X B Z Probe Consider a rectangular slab of an n-type semiconductor material in which a current I is flowing in the positive X-direction, magnetic field B is applied along Z-axis and Hall voltage develops along Y-axis. Under the influence of magnetic field, the electrons experience the Lorentz force FL given by FL = -Bev ------(1) Hall voltage across the upper and lower surfaces of the semiconductor material establishes an electric field EH, called the Hall field exerts an upward force FH on the electrons given by FH = -eEH ------(2) Under equilibrium conditions, force on charge carriers due to magnetic field will be balanced by the force on them due to Hall field FL = FH -Bev = -eEH Bv = EH------(3) Department of Engineering Physics, HKBKCE Page 11 Engineering Physics VH EH = , where d is the distance between the upper and lower surfaces of the slab, then 푑 VH = EH d = Bv d------(4) Current , I= n (Av) e I v = n e A If w is the thickness of the semiconductor slab, A = w d I substituting v = in equation (4) newd I B d I B VH = = n e w d n e w 1 The quantity is the reciprocal of charge density and is defined as the Hall co-efficient ne RH 1 RH = ne I B VH = R H w V w R = H H I B Knowing the Hall coefficient, the concentration of charge carriers can be determined using the relation 1 R H = ne Thus, Hall effect can be used to determine i) Whether Semiconductor is n-type or p-type ii) Density of charge carriers iii) Mobility of charge carriers Department of Engineering Physics, HKBKCE Page 12 Engineering Physics DIELECTRIC MATERIALS Dielectric materials have the property to get polarized when kept in electric field. These are the materials which do not conduct electricity. But each molecules of the dielectric material exbihits modified motion of the electrons, when the electric field is applied. As the result of reorientation of electrons and their shift in the direction of field, the molecules behave like an electric doublet or called Dipoles. Dipole: Two equal and opposite charges separated by a small distance constitutes a dipole. +q l -q Polarisation : The separation of effective centre of positive and negative charges in a substance by the application of electric field is known as polarization. Atoms in a dielectric in the absence of electric field E Atom in a dielectric in presence of electric field Dipole moment is the product of charge and the separation distance. = ql Atomic polarisability p = E where is the polarization constant. Polarisation P= Dipole Moment per unit volume= 0 E( r −1) = NEi Types of Dielectric Materials: Polar dielectrics: These possess permanent dipole moment .They are permanently polarized in nature. Ex: Water, KCl, NH3, NaCl. Non Polar dielectrics: These are the materials which do not possess permanent dipole moment. Department of Engineering Physics, HKBKCE Page 13 Engineering Physics They get polarized only in the presence of external electric field. Ex: O2, N2, CO2, H2 Expression for static dielectric constant: Consider a parallel plate capacitor of area A, Charge density and total charge Q. From Gauss’s law, Electric intensity between the plates E = o r Potential difference V = E . d = . d o r (a)In the absence of a dielectric Q A A Capacitance C without dielectic = = = V d o d o o r A (a) In the presence of a dielectric Cwith dielectric = d Cwithdielectric Static dielectric constant r = Cwithoutdielectric Dielectric constant is defined as the ratio of capacitance of a capacitor with a dielectric to its capacitance in the absence of a dielectric. Different polarization mechanisms: When a dielectric material is subjected to external electric field, the electrical polarization of the material occurs by any one of the following mechanisms.There are 4 mechanisms namely, 1. Electronic polarization 2. Ionic polarization 3. Orientation polarization Department of Engineering Physics, HKBKCE Page 14 Engineering Physics 4. Space charge polarization Electronic polarization: These are generally seen in the case of covalent compounds. When a covalent compound is placed in electric field, displacement of electron cloud takes place relative to the nucleus. This displacement creates a dipole which develops dipole moment. o ( r −1) Electronic polarisability αe = N N is number of dipoles per unit volume It is independent of temperature. Electronic polarization set over a period of time of 10-14 to 10-15 s Ionic polarization: This is exhibited by ionic compounds like NaCl, KCl. When ionic compounds are kept in an electric field, displacement of positive and negative ions occurs developing a dipole moment. o ( r −1) Ionic polarisability αi= N i It is also independent of temperature. The ionic compounds experience Electronic polarisation in addition to it. The ionic polarisation occurs for a period of 10-11 to 10-14 s. Orientation polarization: This type of polarization occurs in polar substances in which there are permanent molecular dipoles but are orientated randomly due to thermal agitation.When polar molecules are kept in an electric field, the already existing dipoles tend to align in the direction of applied electric field. This increases the dipole moment. 2 Orientation polarization αo = kT The build up time is in the order of 10-10s. The orientational polarization is strongly temperature dependent. Higher the temperature, lower will be the polarization. Space charge polarization: This polarization exists in materials possessing different phases due to difference in temperatures. In such materials charge carriers drift and accommodate in certain regions of Department of Engineering Physics, HKBKCE Page 15 Engineering Physics higher conductivity (electrodes) causing dipole moment. It occurs in ferrites and semiconductors. Its magnitude is very small compared to other mechanisms. RELATION BETWEEN POLARIZATION AND DIECLECTRIC CONSTANT: Consider a parallel plate capacitor with plates between which an electric field E0 exists. If σ is the charge per unit area on the plates, then from Guass law, we have E0= σ / Ԑ0. (1) Let the given dielectric slab be placed between the two plates. Due to polarization, charges appear at the faces of the dielectricslab and establish another field E’ within the dielectric which is opposite to the applied field’s direction.The resultant field is given by E= E0-E’ (2) Let σp be the charge per unit area on the surface of the dieectric, then E’ = σp/ Ԑ0. (3) + + + + + + + + + + + + + + + + ------E0 ili E’ + + + + + + + + + + + + + + + + + + + + ------ Therefore eqn (2) becomes E = σ / Ԑ0 - σp/ Ԑ0 E Ԑ0 = σ - σp (4) Since, Polarization=Induced charges per unit area, P= σp and by Gauss law D= σ where D is called electric flux density. Therefore, E Ԑ0 = D - P P = D - E Ԑ0 From Electrostatics, D = Ԑ0 Ԑr P = Ԑ0 Ԑr E - Ԑ0 E Department of Engineering Physics, HKBKCE Page 16 Engineering Physics P = Ԑ0 (Ԑr -1)E P = Ԑ0 ᵡ E where ᵡ is the dielectric susceptability of the material. INTERNAL FIELDS IN A DIELECTRIC: Internal field or local field is the resultant of the applied field and the field produced due to all the dipoles. Consider a dielectric material kept in an electric field Ea. Ea Eθ B1 A1 O A2 B2 The dipole is assumed to be in a one dimensional array and are oriented in the same direction. In that array, let us consider an atom O which s at a distance ‘d’ from A1 and ‘2d’ from B1 and so on. The electric field at O due to dipole A1 is given by cos sin EA1= + = as θ = 0 2d 3 4d 3 2d 3 The electric field at O due to A2 EA2 = as all dipoles are oriented in the same direction. Field at O due to A1 and A2 is E A1 ,A2 = = E1 d 3 Similarly field at O due to B1 and B2 is = 3 = E2 (2d ) The resultant field due n dipoles is given by ER = E1 + E2 + E 3+ E 4+ E5 + E6+ ……………. Department of Engineering Physics, HKBKCE Page 17 Engineering Physics = + 3 +…………….. d 3 (2d ) 1 1.2 1 = 1.2 = 3 3 3 3 d n=1 n d n=1 n 1.2 The internal electric field is Ei = Ea + d 3 1.2 Ei = Ea + = E d 3 a In three dimensional case , (1/d3) could be replaced by N, the number of atoms per unit volume and (1.2/Π) by a constant γ which depends on the crystal structure. N Hence Ei = Ea + = E + P N = P 0 CLAUSIUS – MOSOTTI RELATION: This expression relates dielectric constant of an insulator (ε) to the polarization of individual atoms(α) comprising it. −1 N r = r + 2 3 0 where N is the number of atoms per unit volume α is the polrisability of the atom εr is the relative permittivity of the medium εo is the permittivity of free space. Proof: If there are N atoms per unit volume ,the electric dipole moment per unit volume –known as polarization is given by P = NαE By the definition of polarization P, it can be shown that Department of Engineering Physics, HKBKCE Page 18 Engineering Physics P = 0 E( r −1) = NEi 0 r E − 0 E = NEi NEi …………………..(1) r = 1+ 0 E The internal field at an atom in a cubic structure(γ =1/3) is of the form p NEi Ei = E + = E + 3 0 3 0 E 1 i = E N 1− 3 0 E Substituting for i in equation (1) E N N0 2 N 0 1− + 1+ N 1 3 0 0 3 0 r = 1+ = = N N 1 N 0 1− 0 1− 1− 3 0 3 0 3 0 N 1+ (2 / 3) 0 −1 N 1− (1/ 3) r −1 0 N = = + 2 N 3 r 1− (2 / 3) 0 0 + 2 N 1− (1/ 3) 0 Solid, Liquid and Gaseous Dielectrics: Dielectric materials are broadly classified as Solid , liquid and gaseous dielectrics.the main application of the dielectrics is to provide electrical insulation and for the storage of charges. During the manufacturing process, the insulating materials are subjected to various stresses like thermal, electrical and mechanical stresses to study its durability. Department of Engineering Physics, HKBKCE Page 19 Engineering Physics Solid Dielectrics: Mica, Porcelain , glass and synthetic materials like plastic are inorganic whereas cloth, rubber and paper are Organic insulating materials.Paper are hygroscopic. Hence it is dried and impregnated with mineral oil or vegetable oils. High density papers are preferred in d.c and energy storage capacitors. Liquid Dielectrics: Liquid dielectrics are used mainly in transformers, switches and circuit breakers. Liquid insulators are used to transfer the heat that is developed due to both eddy currents and Joules heating effect in the windings of the electrical devices to the surroundings. Examples are transformer oils, silicon fluids, Chlorinated hydrocarbons, viscous vaseline, fluoro organic fluids etc. Gaseous Dielectrics: Air, Nitrogen, CO2, Inert gases, hydrogen are some of the examples for Gaseous dielectrics. They are also used as heat transfering media. Pressure of the gases affects the dielectric strength of the gases. Higher pressure reduces its ability to insulate. Applications of Dielectrics in transformers A transformer consist of two insulated conducting coils wound on a core. The core is also insulated. In case of high voltage transformers, further insulation is required to be provided between individual windings in the coils and also between the core and the coils. The insulation is provided with paper, cloth or mica. The paper is impregnated with varnish or wax to fill the air gaps. If there is a air gap, then it will damage the insulation since the permittivity of the air is less, ionization of air occurs at high voltage which leads to excessive heating of coil. This effect is called corona. Mica is used to guard against corona but upto 3kV.If the operatinfg voltage exceeds 3 kV, mineral oils are used for guarding. Apart from guarding against corona upto 100kV, the oil keep the transformer cool. It remains stable even at high temperatures. Solved Problems 1. Calculate the fermi energy in eV for silver at 0K, given that its density is 10500 kg/m3, atomic weight is 107.9 and it has only one conduction electron per atom. From the free electron model, the fermi energy EF is given by Department of Engineering Physics, HKBKCE Page 20 Engineering Physics ℎ2 3푛 2/3 퐸 = ( ) ( ) 퐹 2푚 8휋 h = 6.63×10-34 J.s m = 9.1×10-31 kg The number density of conduction electrons is obtained using the Avogadro number. 107.9 kg of silver contains 6.025×1026 atoms. As each atom contributes one electron, 107.9 kg of silver contains 6.025×1026 electrons. 10500 kg i.e. 1 m3 contains 6.025×1026 푛 = × 10500 푒푙푒푐푡푟표푛푠/푚3 107.9 n = 5.863×1028 per m3 2 2/3 (6.63×10−34) 3×5.863×1028 퐸 = × ( ) 퐹 2×9.1×10−31 8휋 = 8.837×10-19 J EF= 5.52 eV 2. Find the temperature at which there is 1.0% probability that a state with an energy 0.5 eV above fermi energy will be occupied. The fermi distribution function is 1 푓(퐸) = 1+푒(퐸−퐸퐹)/퐾푇 f(E) = 1.0% = 0.01 -19 E-EF = 0.5 eV = 0.5×1.6×10 J -20 E-EF = 8×10 J k = 1.38×10-23 J/K 1 0.01 = 8×10−20 ( −23 ) 1+푒 1.38×10 푇 5797.1 ( ) 1 + 푒 푇 = 100 5797.1 ( ) 푒 푇 = 99 5797.1 = 4.595 푇 T = 1261.6 K 3. Calculate the probability of an electron occupying an energy level 0.02 eV above the fermi level at 200K and 400K in a material. The fermi distribution function is 1 푓(퐸) = 1+푒(퐸−퐸퐹)/퐾푇 E-EF = 0.02 eV k = 1.38×10-23 J/K = 8.625×10-5 eV/K For T = 200K, Department of Engineering Physics, HKBKCE Page 21 Engineering Physics 1 푓(퐸) = 0.02 ( ) 1+푒 8.625×10−5×200 f(E) = 0.239 For T = 400K 1 푓(퐸) = 0.02 ( ) 1+푒 8.625×10−5×400 f(E) = 0.359 4. Calculate the probability of an electron occupying an energy level 0.02 eV above the fermi level and 0.02 eV below the fermi level at 200K. For 0.02 eV below the Fermi level at 200K E-EF = -0.02 eV k = 8.625×10-5 eV/K T = 200K 1 푓(퐸) = 1+푒(퐸−퐸퐹)/퐾푇 1 푓(퐸) = −0.02 ( ) 1+푒 8.625×10−5×200 f(E) = 0.761 For 0.02 eV above the Fermi level at 200K E-EF = 0.02 eV 1 푓(퐸) = 0.02 ( ) 1+푒 8.625×10−5×200 f(E) = 0.238 5. Calculate the fermi velocity and the mean free path for the conduction electrons in silver, given that its fermi energy is 5.5 eV and relaxation time for electrons is 3.97×10-14 s. 1 퐸 = 푚 푣2 퐹 2 퐹 2퐸 푣 = √ 퐹 퐹 푚 -19 -19 EF = 5.5 eV = 5.5×1.6×10 J = 8.8×10 J m = 9.1×10-31 kg 2×8.8×10−19 푣 = √ 퐹 9.1×10−31 6 vF = 1.3910 m/s λ = vF τ τ = 3.97×10-14 s Department of Engineering Physics, HKBKCE Page 22 Engineering Physics λ = 1.39×106×3.97×10-14 λ = 5.52×10-8 m 6. Calculate the conductivity of pure silicon at room temperature when the concentration of carriers is 1.6×1010 per c3 2 2 Take μe = 1500 cm /V-s and μh = 500 cm /V-s at room temperature. σi = ni e ( μe+μh) 10 3 ni = 1.6×10 per cm e = 1.6×10-19 C 2 μe = 1500 cm /V-s 2 μh = 500 cm /V-s 10 -19 σi = 1.6×10 ×1.6×10 (1500+500) -6 σi = 5.12×10 mho/cm 7. Find the polarization produced in a dielectric medium of relative permittivity 15 when it is subjected to an electric field of 500V/m. Given : Ԑr = 15, E = 500V/m P = Ԑ0 (Ԑr – 1) E = 8.85x10-12x (15-1) x500 = 6.19x10-8 C/m2 = 6.19 nC/m2 8. The polarisability of Neon atom is 3.5x10-41 Fm2. Find the dielectric constant of Neon gas at 0ºC and normal pressure. -41 2 Given : αe = 3.5x10 Fm N αe = Ԑ0 (Ԑr – 1) No. Of atoms per unit vol. At STP is given by 퐴푣푎𝑔푎푑푟표 푛표. N= 푀표푙푎푟 푉표푙 = 6.023x1023 / 22.4x10-3 = 2.69x1025 atoms/m3 N αe. Ԑr = +1 Ԑ0 2.69푥1025푥 3.5푥10−41 = +1 8.85푥10−12 =1.063x10-4+ 1 Ԑr = 1.0001 Department of Engineering Physics, HKBKCE Page 23 Engineering Physics 9. An elemental solid containing 2x1028 atoms per m3 shows an electonic polarizibility of 2x10-40 Fm2. Assuming the Lorentz field to be operative, calculate the dielectric constant of the material. 28 3 -40 2 Given : N = 2x10 atoms/ m , αe = 2x10 Fm Ԑ푟−1 푁훼푒 = Ԑ푟+2 3Ԑ0 2푥1028푥 2푥10−40 Ԑr -1 = (Ԑr +2) 3푥8.85푥10−12 = 0.150 (Ԑr +2) = 0.15 Ԑr + 0.3 Ԑr – 0.15 Ԑr = 0.3+1 (1-0.15) Ԑr = 1.3 1.3 Ԑr = 0.85 Ԑr = 1.529 10. An elemental dielectric material has Ԑr =12 and it contains 5x1028 atoms/m3. Calaulate its electronic polarisability assuming Lorentz field. Given : Ԑr = 12, N = 5x1028 atoms/m3 Ԑ푟−1 푁훼푒 = Ԑ푟+2 3Ԑ0 (12−1) 푋3푋8.85푋10−12 훼 = 푒 (12+2)푋5푋1028 11푋3푋8.85푋10−12 = 14푋5푋1028 -40 2 훼푒 = 4.17X10 Fm 11. The centers of two identical atoms of polarisability α = 2x 10-40Fm2 are seperated by a distance 5x10-10m. A homogeneous electric field E is applied in a direction to the line joining the centers of the two atoms. Calculate the ratio between internal field Ei and applied field E. -40 2, -10 Given : 훼푒 = 2X10 Fm d = 5x10 m 1.2휇 Ei = E + 3 휋푑 Ԑ0 1.2훼푒 Ei/E = 1 + 3 휋푑 Ԑ0 1.2푋2푋10−40 = 1+ 3.14푋8.85푋10−12푋(5푋10−10) 3 = 1+0.069 Ei/E =1.069 Department of Engineering Physics, HKBKCE Page 24 Engineering Physics 12. Three identical atoms in a string are subjected to a homogenous electric field E V/m along the line joining the centers. If the polarisability is 2.5x10-40Fm2 and the center spacing is 0.3nm, find the Ei/E at the position of center of the atom. -40 2, -9 Given : 훼푒 = 2.5X10 Fm d = 0.3x10 m 1.2휇 Ei = E + 3 휋푑 Ԑ0 1.2훼푒 Ei/E = 1 + 3 휋푑 Ԑ0 1.2푋2.5푋10−40 = 1+ 3.14푋8.85푋10−12푋(3푋10−9) 3 = 1+ 0.399 Ei/E =1.399 13. A solid dielectric material has 5x1028 identical atoms per m3. If the polarisability is 3.6x10-40 Fm2, calculate the Lorentz field normalized with respect to the external field. -40 2, 28 3 Given : 훼푒 = 3.6X10 Fm N = 5x10 atoms /m 푃 Ei = E + 3Ԑ0 푁훼푒퐸 Ei/E = 1 + 3퐸Ԑ0 5푋1028푋3.6푋10−40 = 1+ 3푋8.85푋10−12 = 1+ 0.6779 Ei/E =1.6779 14. A solid containig 8x1028 atoms/m3 shows an internal field which is 1.5 times the applied field. If the solid has an internal field of Lorentz field type, calculate the polarization associated with each atom. 28 3 Given : N = 5x10 atoms /m , Ei = 1.5E 푃 Ei = E + 3Ԑ0 푁훼푒퐸 Ei/E = 1 + 3퐸Ԑ0 푁훼푒 Ei/E = 1 + 3Ԑ0 8푋1028훼 (1.5 -1) = 1 + 푒 3푋8.85푋10−12 0.5푋3.6푋10−40 αe = 8푋1028 -40 2 αe = 1.659X10 Fm Department of Engineering Physics, HKBKCE Page 25 Engineering Physics 15. The dielectric constant of sulphur is 3.4. assuming a cubic lattice for its structure, calculate the electronic polarisability of sulphur. Given density = 2.07gm/cc and atomic weight = 32.07 3 3 23 Given : Ԑr = 3.4, ρ = 2.07X10 Kg/m , M = 32.07, NA = 6.023X10 gmol Ԑ푟−1 푁훼푒 = Ԑ푟+2 3Ԑ0 Ԑ푟−1 3Ԑ0 훼푒 = ( ) Ԑ푟+2 푁 3.4−1 3푋8.85푋10−12푋32.07 훼 = ( ) 푒 3.4+2 2.07푋6.023푋1026 = 30.35x10-41 Fm2 Department of Engineering Physics, HKBKCE Page 26