Final m5 march 2019

Module 5 : Material Science
(Quantum Free Electron Theory of Metals, Physics of Semiconductor & Dielectric Materials)
CONTENTS
I. Quantum Free Electron Theory of Metals
1. Review of Classical Free Electron Theory (CFET)
(Free Electron Concept / Drude-Lorentz’s Theory)
2. Failures of CFET
3. Assumptions of Quantum Free Electron Theory (QFET)
4. Density of States (Only Expression)
5. Fermi Dirac Statistics (Qualitative)
6. Fermi Factor & Fermi Level
7. Expression for Fermi Energy
8. Success of QFET
9. Numerical Problems
II. Physics of Semiconductor
1. Fermi Level in Intrinsic Semiconductors
2. Concentration of Electrons in Conduction Band (Only Expression)
3. Concentration of Holes in Valence Band (Only Expression)
4. Expression for Fermi energy and Energy Gap in an Intrinsic Semiconductor
5. Expression for Conductivity of Semiconductors
6. Hall Effect
7. Expression for Hall Coefficient
III. Dielectric Materials
1. Polar & Non Polar Dielectrics
2. Internal Fields in a Solids
3. Clausius-Mossotti Equation
4. Solid, Liquid & Gaseous Dielectrics (Explain with an Example)
5. Applications of Dielectrics in Transformers
IV. Question Bank

MODULE - 4 QUANTUM MECHANICS & LASER
Dr. DIVAKARA S, PROF. & HEAD, DEPT. OF PHYSICS, VVCE, MYSURU Page 2
Books Referred:
1. Electrical Properties of Materials By Walsh
2. Engineering Physics By S P Basavaraju
3. Solid State Physics By Saxena-Gupta-Saxena
4. Solid State Physics By S O Pillai
5. Engineering Physics By Wiley
6. Principles of Electronics By V K Mehta
7. Introduction to Solid State Physics By C Kittel
8. Solid State Physics By Dekker
9. Solid State Physics By A V Azroff
10. Engineering Physics By M N Avadhanulu & P G Kshirsagar
11. Engineering Physics By R K Gaur & S L Gupta
12. Problems on Solid State Physics By M A Wahab

I. Quantum Free Electron Theory of Metals (QFET)
1. Review of Classical Free Electron Theory
(Free Electron Concept -Drude-Lorentz Theory)
 A copper atom consists of 29 electrons out of which 28 electrons fill the first three
shells and form the core part. The remaining electron of the atom is the valence
electron of the copper and which is loosely bound in an atom.
 When a large number of copper atoms join to form a metal, the boundary
overlapping occurs. Due to overlapping, the valence electron finds continuity
from atom to atom and then can move easily throughout the metal.
 The free movement of the electrons means that none of them belongs to any atom
in particular. Thus, each such electron is called free electrons.
 The free electron is responsible for electrical conductivity in a solid, they are also
called conduction electron.
On Basis of Drude-Lorentz Theory, expression for electrical conductivity (σ) is,
Core
Valence
Electrons

2. Failure of Classical free Electron Theory
a) Specific heat capacity
Experimentally it is observed that, contribution of electrons to specific heat of the
metal is,
;
According to Classical free electron theory, the electrons behaves like a gas molecule
and obeys the laws of kinetic theory of gases,
Total K.E. (Internal Energy) for a mole of electrons,
The Specific heat capacity at constant volume,
( )
The value of specific heat capacity is very high because all the conduction electrons
behaves like a gas molecule and all the conduction electrons will participate in
absorbing heat energy. Therefore, CFET could not explain the results obtained by
experiment.
b) Temperature dependence of electrical conductivity
From experimental observations,
From Classical free electron theory,
√

√
√
√
√
The prediction of CFET is not agreeing with the experimental observations.
c) Electron concentration dependence of electrical conductivity
From Classical free electron theory,
From Experimental,
Element
Ωm)
Cu 5.88 8.45
Ag 6.30 5.85
Zn 1.09 13.1
Cd 0.15 9.28
The electrical conductivity of Zn and Cd are much less than that of Cu and Ag whereas
the electron concentration of Zn and Cd are much higher than that of Cu and Ag. The
prediction of CFET ( ) does not holds good.
3. Assumptions of quantum Free Electron Theory
 The energy levels of the conduction electron are quantized.
 The distribution of the electron in the energy levels occurs as per Pauli’s Exclusion
Principle.
 The electron travel in a constant potential inside the metal but stays confined within
its boundaries.
 The attraction between the electrons & the lattice ions and the repulsion between the
electrons are ignored.

4. Density of States [g(E)dE]
 It is defined as the number of available energy states per unit volume of a metal in an
energy interval E and E+dE.
 Unit : Number of states/m3
/J or Number of states/m3
/eV
 The density of states does not depends on the shape or size of the solid.
 Expression for the density of states for three dimensional solid of unit volume,
) [
√
]
 The graph shows the variation of density of states of a metal as a function of energy
Energy (E)
DensityofStatesg(E)
g (E) α E
½

5. Fermi Dirac Statistics (Qualitative)
a) Fermi Energy
 The energy of highest occupied level at zero degree absolute is called Fermi
energy and energy level is called Fermi level.
 When there is no external energy supply for the electrons (Thermal or electrical
energy), the electrons are free and thus settle in the lowest allowed energy states
available.
 At T = 0K, All energy level lying above the Fermi level are vacant and those
lying below are filled energy state.
b) Fermi-Dirac Statistics
 During thermal excitation (i.e. T>0K), the electrons which absorb the thermal
energy move into higher energy levels, which were unoccupied at zero degree
absolute (T=0K).
 Though such excitation seems to be random, the occupation of various energy
levels obeys a statistical distribution (i.e. probability of occupation of a particular
energy level) called Fermi-Dirac distribution.
 Fermi-Dirac statistics applicable to the assembly of particle which obeys Pauli’s
Exclusion Principle and also for an identical particle of spin ½.
Filled Energy Level
Vacant Energy Level
Energy(E)→
𝑬 𝑭

c) Fermi Factor [f(E)]
 Fermi factor is statistical distribution function which gives the probability of
occupation of a given energy state for a material in thermal equilibrium in terms
of Fermi energy, Temperature and Boltzmann’s Constant.
 The probability f(E) that a given energy state with energy E is occupied at a
steady temperature is given by,
)
(
;
)
 Dependence of Fermi Factor on temperature and effect on occupancy of energy
levels is as shown in the figure.
 Let us consider the different cases of distribution as follows,
)
)
(
;
)
)
(
;
)
:
; : :
At T = 0K, all the energy levels below Fermi level are occupied.
)
)
(
;
)
)
( )
:
: :
At T = 0K, all the energy levels above Fermi level are unoccupied.
𝐾
𝐾

)
At ordinary temperature, the value of probability starts reducing from 1 for
values of E close to but less than EF (see the graph at T > 0K). i.e. the value
of f(E) becomes ½ at E = EF,
)
(
;
)
)
( )
:
: :
Further for E>EF, the probability value falls off to zero rapidly.
6. Expression for Fermi Energy at 0K ( )
The number of electrons per unit volume in the energy range E and E+dE [N(E)dE] is the
product of density of states and the Fermi factor.
) ) )
The number of electrons per unit volume of the material from E = 0 to E = Emax is given
by,
∫ )
∫ ) )
At T = 0K, f(E) = 1 and Emax =
∫ )
Where ) *
8√
+
∫ [
8√
]
[
8√
] ∫

[
8√
] [ ]
[
8√
] ( )
[
8 √
] ( )
( ) [
8 8
] ( )
( ) [
8
] ( )
8 ; 8
Note :
Fermi energy at T > 0K, * ( ) +
Fermi Temperature,
Fermi Velocity, √
𝑬 𝑭 𝟎
*
𝒉 𝟐
𝟖𝒎
+ (
𝟑𝒏
𝝅
)
𝟐
𝟑

7. Success (Merits) of Quantum Free Electron Theory
a) Specific heat capacity
Experimentally it is observed that, contribution of electrons to specific heat of the
metal is,
;
According to quantum free electron theory, it is only those electrons which occupy
energy level close to Fermi energy (EF) are capable of absorbing heat energy to get
excited to higher energy levels. As a result, the specific heat capacity value becomes
very small for the conduction electrons.
Total internal energy (U),
* +
( )
For a mole of electron,
( )
( )
The Specific heat capacity at a constant volume,
( )
( ) )
( )
;
;

b) Temperature dependence of electrical conductivity
From experimental observations,
From quantum free electron theory, the electrical conductivity,
( )
The amplitude of vibration of the lattice ions (r) can be shown that
And the energy of vibration ion α (amplitude)2
And the thermal energy is proportional to T,
Therefore,
The prediction of QFET is agreeing with the experimental observations.
c) Electron concentration dependence on electrical conductivity
From Quantum free electron theory,
( )
 From the above equation, it is clear that, the value of σ depends on both n and ( )
 If we compare the cases of copper and aluminum: The value of n for Al is 2.13
times the value of n for Cu. But the value of ( ) for Cu is about 3.73 times higher
than that of Al. Thus the conductivity of copper exceeds that of aluminum.

Formulas at a glance


 ) *
8√
+
 )
(
;
)
:
 *
8
+ ( ) 8 ; 8
 & √
01 Calculate the thermal energy and thermal velocity of electron in a metal at a temperature of
300K. Jan 2007
02 Calculate the Fermi velocity and the mean free path for the conduction electron in silver (Given;
EF=5.5eV and τ =3.97 x 10-14
s). Jan 2013
03 Calculate the probability of an electron occupying an energy level 0.02eV above the Fermi level
at 200K. Jan 2010
𝑣𝑡 ?
𝐸𝑡 ?
Sol :
T = 300K
𝑎)𝐸𝑡 𝑘𝑇 8 𝑋 ;
𝑋 ) 𝟔 𝟐𝟏 𝑿 𝟏𝟎;𝟐𝟏
𝑱
𝑏)𝐸𝑡 𝑚𝑣𝑡 𝑣𝑡 √
𝐸𝑡
𝑚
√
𝑋 6 𝑋 ;
9 𝑋 ;
𝟏 𝟏𝟕 𝑿 𝟏𝟎 𝟓
𝒎 𝒔
𝑣 𝐹 ?
𝜆 ?
𝐸 𝐹 𝑒𝑉
𝜏 97 𝑋 ;
𝑠
Sol :
T = 300K
𝑎)𝐸 𝐹 𝑚𝑣 𝐹 𝑣 𝐹 √
𝐸 𝐹
𝑚
√
𝑋 𝑋 6 𝑋 ; 9
9 𝑋 ;
𝟏 𝟑𝟗 𝑿 𝟏𝟎;𝟔
𝒎 𝒔
𝑏)𝜆 𝑣 𝐹 𝑋 𝜏 9 𝑋 ;6
𝑋 97𝑋 ;
𝟓 𝟓𝟕 𝑿 𝟏𝟎;𝟖
𝒎
𝑓 𝐸) ?
𝐸 𝐸 𝐹 𝑒𝑉
Sol :
T = 200K
Above 𝐸 𝐹,
𝒇 𝑬)
𝟏
𝒆
(
𝑬;𝑬 𝑭
𝒌𝑻
)
𝟏
=
𝟏
𝒆
(
𝟎 𝟎𝟐𝑿𝟏 𝟔𝟎𝟐;𝟏𝟗
𝟏 𝟑𝟖𝑿𝟏𝟎;𝟐𝟏 𝑿𝟐𝟎𝟎
)
:𝟏
𝟎 𝟐𝟒

04 Calculate the probability of an electron occupying an energy level 0.02eV below the Fermi level
at 300K. Jan 2018
05 Find the temperature at which there is 1% probability that a state with an energy 0.5eV above
Fermi energy is occupied. Jan 2019, Jan 2014
06 The Fermi level in silver is 5.5eV. What is the energy for which the probability of occupying at
300K is 0.99. July 2014
07 Potassium is a BCC crystal with lattice constant 5.25Å and the metal has one free electron per
atom. Calculate the Fermi energy in eV for the metal. Also evaluate its Fermi factor at a
temperature of 600K for an energy value 0.075eV higher than EF.
𝑓 𝐸) ?
Sol :
T = 300K
Below 𝐸 𝐹,
𝒇 𝑬)
𝟏
𝒆
(
𝑬;𝑬 𝑭
𝒌𝑻
)
:𝟏
=
𝟏
𝐞
(
;𝟎 𝟎𝟐𝐗𝟏 𝟔𝟎𝟐;𝟏𝟗
𝟏 𝟑𝟖𝐗𝟏𝟎;𝟐𝟏 𝐗𝟑𝟎𝟎
)
:𝟏
𝟎 𝟔𝟖
𝑓 𝐸)
Sol :
T = ?
Above 𝐸 𝐹, 𝑓 𝐸) 𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
𝑓 𝐸)
𝑙𝑛 𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
𝑙𝑛
𝑓 𝐸)
𝑓 𝐸)
𝑓 𝐸)
𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
:
(
𝐸;𝐸 𝐹
𝑘𝑇
) 𝑙𝑛 *
;𝑓 𝐸)
𝑓 𝐸)
+ 𝑇
𝑘𝑙𝑛*
;𝑓 𝐸)
𝑓 𝐸)
+
𝐸;𝐸 𝐹
=
8 𝑋 ; 𝑙𝑛*
;
+
5 𝑋 6 𝑋 6 ; 9
= 1263K
𝑓 𝐸) 99
𝐸 𝐹 𝑒𝑉
𝐸 ?
Sol :
T = 300
(
𝐸 𝐸 𝐹
𝑘𝑇
) 𝑙𝑛
𝑓 𝐸)
𝑓 𝐸)
𝐸 𝐸 𝐹 𝑘𝑇𝑙𝑛
𝑓 𝐸)
𝑓 𝐸)
𝑓 𝐸)
𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
:
𝑓 𝐸) 𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
* 𝑓 𝐸)
+ 𝑙𝑛 𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
𝑙𝑛 *
;𝑓 𝐸)
𝑓 𝐸)
+
𝑋 6 𝑋 ; 9
8 𝑋 ;
*
; 99
99
+ 8.61 X 𝟏𝟎;𝟏𝟗
𝑱 𝟓 𝟑𝟕𝟒𝒆𝑽
𝐵𝐶𝐶 𝑁
Sol :
a=5.25Å
No. of electrons per
atom=1
T=600K
E-𝐸 𝐹 7 𝑒𝑉
𝐸 𝐹 ? & f(E)=?
𝑁𝑜 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 𝑛
𝑁
𝑎 𝑋 ; )
𝟏 𝟑𝟖𝟐 𝑿 𝟏𝟎 𝟐𝟖
𝒎 𝟑
𝑎)𝐸 𝐹 *
8𝑚
+ (
𝑛
𝜋
) 8 𝑋 ; 8
𝑛 8 𝑋 ; 8
8 𝑋 8)
𝟑 𝟑𝟔𝟗 𝑿𝟏𝟎;𝟏𝟗
J = 2.1eV
𝑏)𝑓 𝐸)
𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
:
=
𝑒
(
75 𝑋 6 𝑋 ; 9
8 𝑋 ; 𝑋 6
)
:
𝟎 𝟏𝟖𝟗𝟖
𝑋 6 𝑋 ; 9
8 𝑋 ;
*
; 99
99
+ 𝟏𝟎;𝟏𝟗
𝑱 𝟓 𝟑𝟕𝟒𝒆𝑽

08 Show that occupation probability at E=EF+ΔE is equal to the non-occupation probability at
E=EF-ΔE. June 2008
09 Show that sum of the probability of occupation of a level ∆E above the Fermi level and a level ∆E
below the Fermi level is 1.
10 The Fermi level in silver is 5.5ev at 0K. Calculate the number of free electrons per unit volume
and the probability of occupation for electrons with energy 5.6eV in silver at same temperature.
𝑓 𝐸) 𝐸 𝐸 𝐹 ∆𝐸
𝑒
(
∆𝐸
𝑘𝑇
)
𝑓 𝐸) 𝐸 𝐸 𝐹 ∆𝐸
𝑒
(
;∆𝐸
𝑘𝑇
)
Sol :
Occupation Probability at
𝐸 𝐸 𝐹 ∆𝐸
Non occupation Probability
at 𝐸 𝐸 𝐹 ∆𝐸
𝑓 𝐸) 𝐸<𝐸 𝐹;∆𝐸 𝑓 𝐸) 𝐸<𝐸 𝐹:∆𝐸
𝑓 𝐸)
From the theory of probability,
𝑁𝑜𝑛 𝑜𝑐𝑐𝑢𝑝𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑂𝑐𝑐𝑢𝑝𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦
𝑒
(
;∆𝐸
𝑘𝑇
)
:
𝑒
(
∆𝐸
𝑘𝑇
)
:
:𝑒
(
∆𝐸
𝑘𝑇
)
𝑒
(
∆𝐸
𝑘𝑇
)
𝑒
(
∆𝐸
𝑘𝑇
)
:𝑒
(
∆𝐸
𝑘𝑇
)
:𝑒
(
∆𝐸
𝑘𝑇
)
= 𝒇 𝑬) 𝑬<𝑬 𝑭:∆𝑬
𝐸 𝐸 𝐹 𝑏𝑦 ∆𝐸 𝐸 𝐸 𝐹 ∆𝐸
𝑓 𝐸) 𝑎𝑏𝑜𝑣𝑒
𝑒
(
∆𝐸
𝑘𝑇
)
𝐸 𝐸 𝐹 𝑏𝑦 ∆𝐸 𝐸 𝐸 𝐹 ∆𝐸
𝑓 𝐸) 𝑏𝑒𝑙𝑜𝑤
𝑒
(
;∆𝐸
𝑘𝑇
)
Sol :
𝑒
(
∆𝐸
𝑘𝑇
)
𝑒
(
;∆𝐸
𝑘𝑇
)
𝑓 𝐸)
𝑒
𝐸;𝐸
𝑠𝑢𝑚 𝑜𝑓 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑓 𝐸) 𝑎𝑏𝑜𝑣𝑒+𝑓 𝐸) 𝑏𝑒𝑙𝑜𝑤
𝑒
(
∆𝐸
𝑘𝑇
)
:
𝑒
(
∆𝐸
𝑘𝑇
)
:
𝑒
(
∆𝐸
𝑘𝑇
)
: :𝑒
(
∆𝐸
𝑘𝑇
)
𝑒
(
∆𝐸
𝑘𝑇
)
𝑒
(
∆𝐸
𝑘𝑇
)
:
𝑒
(
∆𝐸
𝑘𝑇
)
:𝑒
(
∆𝐸
𝑘𝑇
)
:𝑒
(
∆𝐸
𝑘𝑇
)
:𝑒
(
∆𝐸
𝑘𝑇
)
= 1
𝐸 𝐹 𝑒𝑉
𝑛 ?
𝐸 ?
𝑓 𝐸) ?
Sol :
𝐸 𝐹 *
8𝑚
+ (
𝑛
𝜋
) 8 𝑋 ; 8
𝑛
𝑛
𝐹𝐹
8 𝑋 ; 8
𝟓 𝟖𝟒𝟓 𝑿 𝟏𝟎 𝟐𝟖
𝒎 𝟑
𝐹𝑟𝑜𝑚 𝑑𝑎𝑡𝑎 𝑎𝑡 𝑇 𝐾 𝐸 6𝑒𝑉 𝐸 𝐹 𝑒𝑉
𝐸 𝐸 𝐹 𝑓 𝐸)

11 Calculate the Fermi energy in eV for a metal at 0K, whose density is 10500kg/m3
, atomic weight
is 107.9 and it has one free electron per atom.
𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
𝑛
𝑛𝑜 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑝𝑒𝑟 𝑎𝑡𝑜𝑚 𝑋 𝑁𝐴 𝑋 𝐷
𝑊
𝑋 6 𝑋 6
𝑋
7 9
𝟓 𝟖𝟔 𝑿 𝟏𝟎 𝟐𝟖
𝒎 𝟑
𝐸 𝐹 *
8𝑚
+ (
𝑛
𝜋
) 8 𝑋 ; 8
𝑛 8 𝑋 ; 8
86 𝑋 8)
𝟖 𝟑𝟑 𝑿𝟏𝟎;𝟏𝟗
J = 5.509eV
𝑏)𝑓 𝐸)
𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
:
=
𝑒
(
75 𝑋 6 𝑋 ; 9
)
:
𝟎 𝟏𝟖𝟗𝟖
𝐸 𝐹 ?
𝐷 𝑘𝑔 𝑚
𝑊 7 9
𝑁𝑜 𝑜𝑓 𝑓𝑟𝑒𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑝𝑒𝑟
𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒
Sol :

II. Physics of Semiconductor
1. Fermi Level in an Intrinsic Semiconductors
 At T = 0K, the conduction band is completely empty and the valence band is
completely filled energy state.
C Conduction Band
Energy Gap (Eg)
V Valence Band
 When the intrinsic semiconductor is forward bias with a large current. The
electrons from the valence band raises to conduction band but this is an unstable
state and with a short time (10-13
s), electrons in the conduction band drops to the
lower level in that band. The lowest level of conduction band is filled with
electrons and the top of valence band is filled is full of holes.
 Due to this, average energy of the conduction electrons will be almost equal to ½
Eg. Thus the Fermi level lies in the mid-part of the forbidden gap or Energy gap
for an intrinsic semiconductor.
← Electrons
← Holes

2. Concentration of Electrons in Conduction Band ( )
The expression for electron concentration is given by,
√
)
(
;
)
Where, is the effective mass of electron
k is the Boltzmann’s constant
T is the temperature in absolute scale
EF is the Fermi energy
Eg is the Energy gap
h is the Planck’s Constant
3. Concentration of holes in Valence Band ( )
The expression for hole concentration is given by,
√
) (
;
)
Where, is the effective mass of hole
k is the Boltzmann’s constant
T is the temperature in absolute scale
EF is the Fermi energy
h is the Planck’s Constant

4. Expression for Fermi energy and Energy Gap in an Intrinsic Semiconductor
For an intrinsic semiconductor,
√
)
(
;
) √
) (
;
)
)
(
;
)
) (
;
)
( )
(
;
)
(
;
)
( )
(
;
)
Taking the natural logarithm on both sides,
( ) ( )
( ) ( )
Under practical considerations,
Thus, the Fermi level is in the middle of the band gap for an intrinsic semiconductor.
𝑬 𝑭
𝑬 𝒈
𝟐
𝟑
𝟒
kT ln(
𝒎 𝒉
𝒎 𝒆
)

5. Expression for Electrical Conductivity of Semiconductors
If Ne is the number of electrons per unit volume, e is the charge of electron, A is the
area of cross section and v is the velocity of electron, then the charge flow per
second (I),
Therefore, the current density is,
The electron mobility ( ) is given by,
⁄
Therefore, the current density is,
From ohm’s law,
If is the conductivity due to electrons in the semiconductor material,
)
Similarly, is the conductivity due to holes in the semiconductor material,
)
Where
The total conductivity for a semiconductor is given by,
)
In case of intrinsic semiconductor,
Therefore,
𝝇𝒊 𝒏𝒊 𝒆 𝝁 𝒆 𝝁 𝒉)

6. Hall Effect
 If a piece of conductor (Metal or semiconductor) carrying a current is placed in a
transverse magnetic field, an electric field is produced inside the conductor in a
direction normal to both the current and the magnetic field. This phenomenon is
known as Hall Effect and the generated voltage is called Hall voltage.
 Consider a rectangular slab of a n-type semiconductor in which current I is
flowing in the positive X direction.
 Let a magnetic field B is applied along the Z-direction. Under the influence of
the magnetic field, the electron experience the Lorentz force (FL),
)
 Applying Fleming’s left hand rule, we see that the force is exerted on the
electron in the negative Y direction. The electron are therefore deflected
downwards. As a results, the density of electrons increases in the lower end of
the material, due to which surface 1 become negatively charged and surface 2
become positively charged (loss of electron).
 Hence a potential VH (Hall voltage) appears between the upper and lower
surfaces of the slab which establishes an electric field EH (Hall field)
Therefore, the force acting on the electron in the upward direction is,
)
 Now, as the deflection of electron continues in the downward direction due to
Lorentz force FL, it also contributes to the growth of hall field. As a result, the
force FH which acts on the electron in the upward direction also increases.

Opposing force reach an equilibrium,
) ),
=
(3)
)
Where d is the distance between surface 1 and surface 2.
The current density, J = I/A = I/wd (5)
Where w is the thickness of the slab along Z-axis
But we know that, (6)
W
) 6),
=
= (7)
Substituting equ. (7) in equ. (4),
( )
By measuring VH, I, w and by knowing B, the charge density can be calculated
7. Expression for Hall Coefficient (RH)
For a given semiconductor, the Hall field (EH) depends upon the current density (J)
and the applied field (B).
We know that, EH = Bv & J = , then
𝝆
𝑩𝑰
𝑽 𝑯 𝒘
𝑹 𝑯
𝟏
𝒏𝒆
𝒐𝒓
𝟏
𝝆

Note :
 The force which is exerted by a magnetic field on a moving electric charge is
known as Lorentz Force.
 Fleming’s Left Hand Rule states that, if we stretch the thumb, the center
finger and the middle finger of our left hand such that they are mutually
perpendicular to each other. If the center finger gives the direction and the
middle finger points in the direction of the magnetic field then the thumb
points towards the direction of the force or motion of the conductor.

√
) (
;
)

√
) (
;
)
 )

 kT ln( )

 ( )
 ;
01 The following data are given for intrinsic Germanium at 300K. Given; ni = 2.4 x 1019
/m3
, µe =
0.39 m2
V-1
s-1
, µh = 0.19 m2
V-1
s-1
. Calculate the resistivity of the sample. Jan 2018
𝜌𝑖 ?
Sol :
ni = 2.4 x 1019/m3
µe = 0.39 m2V-1s-1
µh = 0.19 m2V-1s-1
𝜍𝑖 𝑛𝑖 𝑒 𝜇 𝑒 𝜇 ) 𝜍𝑖
𝜌𝑖
𝜌𝑖
𝑛𝑖 𝑒 𝜇 𝑒 𝜇 )
𝑋 9 𝑋 6 𝑋 ; 9 9 9)
𝟐 𝟑𝟕 𝑿 𝟏𝟎;𝟏𝟗
𝒎 𝟑

02 The resistivity of intrinsic Germanium at 27o
C is equal to 0.47Ω-m.(Given; µe = 0.38 m2
V-1
s-1
, µh
= 0.18 m2
V-1
s-1
) Calculate the intrinsic carrier density. Jan 2019
03 Pure Germanium at 300K has a density of charge carrier 2.5 x 1019
/m3
. A specimen of pure
Germanium is doped with donor impurity atoms at the rate of impurity atoms for every 106
atoms of Ge. Find the resistivity of the doped germanium. (Given; µe = 0.36 m2
V-1
s-1
, µh = 0.18
m2
V-1
s-1
and number of silicon atoms per unit volume, N = 4.2 x 1028
atoms/m3
).
04 The effective mass for the electron in Germanium is 0.55mo. Where mo is the free electron mass.
Find the electron concentration in Germanium at 300K, assuming that, the Fermi energy is half
of energy gap and energy gap for Germanium is 0.66eV.
𝜌𝑖 7𝛺𝑚
Sol :
ni = ?
µe = 0.38m2V-1s-1
µh = 0.198m2V-1s-1
𝜍𝑖 𝑛𝑖 𝑒 𝜇 𝑒 𝜇 ) 𝜍𝑖
𝜌𝑖
𝜌𝑖
𝑛𝑖 𝑒 𝜇 𝑒 𝜇 )
𝑛𝑖
𝜌𝑖 𝑒 𝜇 𝑒 𝜇 )
𝑛𝑖
7 𝑋 6 𝑋 ; 9 8 8)
𝟐 𝟑𝟕 𝑿 𝟏𝟎 𝟏𝟗
𝒎 𝟑
Sol :
ni = 2.5 x 1019/m3
µe = 0.36 m2V-1s-1
µh = 0.18 m2V-1s-1
N = 4.2 X 1028/m3
Ratio of impurity atoms to Ge
atom(x)=
6
;6
ρ ?
𝑁𝑜 𝑜𝑓 𝑓𝑟𝑒𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑖𝑚𝑝𝑢𝑟𝑖𝑡𝑦 𝑝𝑒𝑟 𝑚
𝑁𝑒 𝑥 𝑋 𝑁 ;6
𝑋 𝑋 8
𝟒 𝟐𝑿𝟏𝟎 𝟐𝟔
𝒎 𝟑
𝑁𝑒 𝑁 𝑛𝑖 𝑁
𝑛𝑖
𝑁𝑒
𝑋 9)
𝑋 6
𝟏 𝟒𝟖𝟖𝑿𝟏𝟎 𝟏𝟔
𝒎 𝟑
𝜍 𝑁𝑒 𝑒𝜇 𝑒 𝑁 𝑒𝜇 𝜌
𝑁𝑒 𝑒𝜇 𝑒 𝑁 𝑒𝜇
𝑋 6 6 𝑋 ; 9 𝑋 6) ( 88𝑋 6
𝑋 6 𝑋 ; 9 𝑋 8)
𝟒 𝟏𝟑𝟑 𝑿 𝟏𝟎;𝟒
𝜴𝒎
𝑚 𝑒 𝑚
𝑁𝑒 ?
𝐸𝑔 66𝑒𝑉
𝐸 𝐹
𝐸𝑔
𝑒𝑉
Sol :
T=300K
𝑁𝑒
√
𝜋𝑚 𝑒 𝑘𝑇) 𝑒
(
𝐸 𝐹;𝐸 𝑔
𝑘𝑇
)
𝟐 𝟗𝟒 𝑿𝟏𝟎 𝟏𝟗
𝒎 𝟑
= √
6 6 5 𝑋 ; 4)
𝜋𝑋 𝑋9 𝑋 ;
𝑋 8𝑋 ;
𝑋 ) 𝑒
6 𝑋 ; 9 ; 66)
8 𝑋 ; 𝑋

05 The electric conductivity of intrinsic semiconductor increases from 19.96/Ω-m to79.44/Ω-m.
When the temperature rises from 600
C to 1000
C. Find the energy gap.
06 An intrinsic semiconductor has an energy gap of 0.4eV. Calculate the probability of occupation
of the lowest level in conduction band at 500
C.
07 The effective masses of hole and electron in GaAs semiconductor are 0.48mo and 0.067mo
respectively, its band gap energy is 1.43eV. Calculate the Fermi energy in eV at 300k.
08 Calculate the position of the Fermi level for pure Silicon at 300K. If the electron concentration is
2 x 1015
/ m3
. (Given; energy gap= 1.1eV, =0.31mo)
𝜍 9 96 𝛺𝑚
𝜍 79 𝛺𝑚
𝑇 𝐾
𝑇 7 𝐾
𝐸𝑔 ?
Sol : 𝜍
𝜍
𝑒
;
𝐸 𝑔
𝑘
*
𝑇
;
𝑇
+
𝐸𝑔
𝑘𝑙𝑛 (
𝜍
𝜍
)
* 𝑇 𝑇
+
𝑋 8 𝑋 ;
(
9 96
79 )
* 7
+
𝟏 𝟏𝟖 𝑿 𝟏𝟎;𝟏𝟗
𝑱 𝟎 𝟕𝟒𝒆𝑽
𝑙𝑛 (
𝜍
𝜍
) 𝑙𝑛𝑒
;
𝐸 𝑔
𝑘
*
𝑇
;
𝑇
+
𝑙𝑛 (
𝜍
𝜍
)
𝐸 𝑔
𝑘
*
𝑇 𝑇
+ 𝑙𝑛𝑒
𝐸𝑔 𝑒𝑉
Sol :
f(E)=?
T=323K
𝑓 𝐸)
𝑒
(
𝐸;𝐸 𝐹
𝑘𝑇
)
𝑬 𝑬 𝒈 𝒂𝒏𝒅 𝑬 𝑭
𝑬 𝒈
𝟐
𝑓 𝐸)
𝑒
(
𝐸 𝑔
𝑘𝑇
)
𝑒
(
𝑋 6 𝑋 ; 9
𝑋 8𝑋 ; 𝑋
)
𝟕 𝟓𝟓 𝑿 𝟏𝟎 𝟒
𝑚 8𝑚
𝑚 𝑒 67𝑚
𝑁𝑒 ?
𝐸𝑔 𝑒𝑉
𝐸 𝐹 ?
Sol :
T=300K
𝐸 𝐹
𝐸 𝑔
kT ln(
𝑚
𝑚 𝑒
)
𝑋 6 𝑋 9
)
X1.38 X ;
𝑋 ln(
8 𝑚
67𝑚
)
𝟏 𝟐𝟎𝟓 𝑿 𝟏𝟎 𝟏𝟗
𝟎 𝟕𝟓𝒆𝑽
𝑁𝑒
√
𝜋𝑚 𝑒 𝑘𝑇) 𝑒
(
𝐸 𝐹;𝐸 𝑔
𝑘𝑇
)
𝑁𝑒
√ 𝜋𝑚 𝑒 𝑘𝑇) 𝑒
𝑒
(
𝐸 𝐹;𝐸 𝑔
𝑘𝑇
)
𝒍𝒏 [
𝑵 𝒆 𝒉 𝟑
𝟒√𝟐 𝝅𝒎 𝒆 𝒌𝑻)
𝟑
𝟐
]
⬚
𝑙𝑛𝑒
(
𝐸 𝐹;𝐸 𝑔
𝑘𝑇
) 𝑬 𝑭 𝑬 𝒈
𝒌𝑻
𝐸 𝐹 𝐸𝑔 𝑘𝑇𝒍𝒏 *
𝑵 𝒆 𝒉 𝟑
𝟒√𝟐 𝝅𝒎 𝒆 𝒌𝑻)
𝟑
𝟐
+
⬚
=8.722 X 𝟏𝟎;𝟐𝟎
𝑱 𝟎 𝟓𝟒𝒆𝑽
𝑚 𝑒 𝑚
𝑁𝑒 𝑋 5
𝑚
𝐸𝑔 𝑒𝑉
𝐸 𝐹 ?
Sol :
T=300K

09 In an intrinsic semiconductor the energy gap is 1.2eV. What is the ratio between its conductivity
at 600K and that at 300K.
10 For intrinsic GaAs, the electrical conductivity at room temperature is 10-6
/ Ω-m. The electron
and hole mobility are respectively 0.85 m2
/V-s and 0.04 m2
/V-s. Calculate the intrinsic carrier
concentration at room temperature
11 Calculate the concentration at which the acceptor atoms must be added to a Germanium sample
to get a P-type semiconductor with conductivity 0.15 / Ω-m.(µholes = 0.17m2
/Vs)
12 The Hall Coefficient of a specimen of a doped silicon is found to be 3.66 X 10-4
m3
/coulomb.
(Given: Resistivity = 8.93 X 10-3
Ωm). Find the mobility and density of charge carriers.
𝜍
𝜍
𝐶𝑒
𝐸 𝑔
𝑘𝑇
𝐶𝑒
𝐸 𝑔
𝑘𝑇
𝑒
𝐸 𝑔
𝑘 𝑇 𝑇 𝑒
( 𝑋 6 𝑋 9
)
𝑋 8𝑋 6
𝟏 𝟎𝟗𝑿𝟏𝟎 𝟓
𝜍 𝐶𝑒
;
𝐸 𝑔
𝑘𝑇 𝑎𝑛𝑑 𝜍 𝐶𝑒
;
𝐸 𝑔
𝑘𝑇
𝜍𝑖 𝑛𝑖 𝑒 𝜇 𝑒 𝜇 )
𝑛𝑖
𝜍𝑖
𝑒 𝜇 𝑒 𝜇 )
𝑛𝑖
;6
6 𝑋 ; 9 8 )
𝟕 𝟎𝟐 𝑿 𝟏𝟎 𝟏𝟐
𝒎 𝟑
𝜍 𝑁 𝑒𝜇
𝑁
𝜍
𝑒𝜇 6 𝑋 9
𝑋 7
𝟓 𝟓𝟏 𝑿 𝟏𝟎 𝟏𝟖
𝒎 𝟑
𝑅 𝐻 66 𝑋 ;
𝑚 𝐶
𝜌 8 9 𝑋 ;
𝛺𝑚
𝑁 ?
Sol :
µh = ?
𝑅 𝐻
𝜌 𝑛𝑒 𝑁 𝑒
𝑁
𝑅 𝑒 66 𝑋 𝑋 6 𝑋 9
𝟏 𝟕𝟎𝟓 𝑿 𝟏𝟎 𝟐𝟐
𝒎 𝟑
𝜍 𝑁 𝑒𝜇 𝜇
𝜍
𝑒𝑁 𝜌 𝑒𝑁
𝜇
8 9 𝑋 𝑋 6 𝑋 9
𝑋 7 𝑋
𝟎 𝟎𝟒𝟏𝒎 𝟐
𝒗𝒔
𝜍 𝛺𝑚
Sol :
nh =?
µh = 0.17m2V-1s-1
𝑇 6 𝐾
𝑇 𝐾
𝐸𝑔 𝑒𝑉
𝜍
𝜍
?
Sol :
𝑇 6 𝐾
𝑇 𝐾
𝐸𝑔 𝑒𝑉
𝜍
𝜍
?
Sol :

13 A rectangular plane sheet of a semiconductor material has dimensions 2cm along y-axis and
1mm along z-axis. Hall probes are attached on its two surfaces parallel to x-z axis and a
magnetic field of flux density 1weber/m2
is applied alone z-axis. A current 3mA is flowing in it in
the x-axis. Calculate Hall voltage (RH = 3.66 X 10-4
m3
/C). Also calculate the charge carrier
concentration.
14 The conductivity and Hall coefficient of n-type semiconductor are 112 /Ωm and 1.25 X 10-3
m3
/C
respectively. Calculate the charge carrier concentration and electron mobility.
𝑉𝐻 𝐸 𝐻 𝑋 𝑑
𝑉𝐻 𝑅 𝐻 𝐽𝐵) 𝑑
𝑉𝐻 𝑅 𝐻 (
𝐼
𝐴
) 𝐵 𝑑 66 𝑋 𝑋
𝑋
𝑋
𝑋 𝑋 𝑋 𝟏 𝟏𝒎𝑽
𝑅 𝐻
𝑛𝑒
𝑛
𝑅 𝐻 𝑒 66𝑋 ; 𝑋 6 𝑋 ; 9
𝟏 𝟕𝟏𝑿𝟏𝟎 𝟐𝟐
𝒎 𝟑
𝑅 𝐻 66 𝑋 ;
𝑚 𝐶
Sol :
Area, A = 2cm X 1mm
= 2 X 10-5
m2
I =3mA
d=2cm
w=1mm
B=1w/m2
n=?,
VH=?
𝑅 𝐻 𝑋 ;
𝑚 𝐶
Sol :
σ = 112/Ωm
n = ?
μe=?
𝑅 𝐻
𝜌 𝑛𝑒 𝑁𝑒 𝑒
𝑁𝑒
𝑅 𝑒 𝑋 𝑋 6 𝑋 9
𝟒 𝟗𝟗 𝑿 𝟏𝟎 𝟐𝟏
𝒎 𝟑
𝜍𝑒 𝑁𝑒 𝑒𝜇 𝑒 𝜇 𝑒
𝜍 𝑒
𝑒𝑁 𝑒
𝜇
6 𝑋 9
𝑋 99 𝑋
𝟎 𝟏𝟒 𝒎 𝟐
𝒗𝒔

III. Dielectric Materials
Introduction
 All dielectrics are insulators, the distinction between a dielectric material and
insulator lies in the application to which one is employed. The insulating
materials are used to resist the flow of current through it. On the other hand,
dielectric materials are used to store electrical energy.
 Dielectric materials are those materials which have the ability to get
electrically polarized and in which electric field can exist. They do not allow
current to flow through it when subjected to ordinary voltages.
 A dielectric is an electrically non-conducting material such as wood, rubber,
porcelain etc. This provides electrical insulation between two media.
 Dielectric Constant,
 Dipole moment (μ): The dipole moment is the product of magnitude of the
charge and distance of separation between the charges. i.e. μ = qr (Cm)
 Polarization (P): The displacement of charges in the molecule of dielectric
material under the action of applied electric field leading to the development
of dipole moment is called polarization. (C/m2
)
 Polarizability (α), (Fm2
)
 Relation between Polarization and Dielectric constant
⃗⃗ )⃗⃗

1. Types of Polarization
There are four different mechanisms through which electrical polarization can
occur in dielectric material when they are subjected to an external electric field.
a) Electronic polarization
b) Ionic polarization
c) Orientational polarization
d) Space charge polarization
a) Electronic polarization
 The electronic polarization occurs due to displacement of the positive and
negative charges in a dielectric material owing to the application of an
external electric field. This leading to development of dipole moment.
 The electronic polarizability for a rare gas atom is,
)
b) Ionic polarization
 Ionic polarization occurs in those dielectric materials which possess ionic
bonds (NaCl, KCl, KBr).
 When ionic solids are subjected to an external electric field, the adjacent ions
of opposite sign undergo displacement. The displacement causes an increases
or decrease in the distance of separation between the atoms.
Charge distribution in
the absence of the field
Charge displacement
due to the applied field
Ion placement in the
absence of the field
Ion displacement due
to the applied field

c) Orientaional polarization
 Orientational polarization occurs in those dielectric materials which possess
permanent dipole moment in molecule. i.e. The orientation of these
molecules will be random (Net dipole moment is zero).
 But under the influence of an applied field, each of the dipole undergoes
rotation along the direction of the field.
 The orientational polarization is strongly temperature dependent and
decreases with increase of temperature. Then,
d) Space charge polarization
 Space charge polarization occurs in multiphase dielectric materials in which
there is a change of resistivity between different phases.
 When such material is subjected to an electric field, especially at high
temperature, the charges get accumulated at the interface because sudden
change in conductivity. It leads to development of dipole moment within the
low resistivity phase.
Dipole orientation in
the absence of the field
Dipole alignment due
to the applies field
Charge distribution within
the low resistivity phase
Charge accumulation at the
interface due to applied field

2. Polar and Non-polar Dielectrics
a) Polar Dielectrics
 The center of gravity of positive charges do not coincides with the center of
gravity of negative charges even in absence of an electric field are called Polar
Dielectrics.
 Examples: NH3, HCl, H2O, CO2, etc.
 They have net dipole moment
 (a) In the absence of electric field, dipole moment of these polar molecules
point in the random direction and arrange themselves in the closed chain.
(b) When the external field is applied, the chains are broken and those
molecules again themselves parallel to the direction of applied field.
b) Non-polar Dielectrics
 The center of gravity of positive charges coincides with the center of gravity
of negative charges. Thus, neutralizing each other effects are called Non polar
dielectrics.
 Examples: N2, H2, O2, methane, benzene, etc.
 They do not have permanent dipole moment.
 (a) The electric field is applied across the non-polar molecules, the center of
gravity of negative and positive charges suffers a displacement and the
molecule acquires a temporary character by induction.
(b)When the applied field is removed, the induced dipole moment disappears
and molecule again becomes a non-polar.

3. Internal Fields in a Solids
 When a dielectric material either a solid or liquid is subjected to an external
electric field, each of the atoms develops a dipole moment and acts as electric
dipole.
 Hence, the resultant internal field will be sum of the applied electric field and the
electric field due to surrounding dipoles. This resultant local field is called the
internal field.

 Expression for internal field in case of liquid or solid dielectrics in one
dimensional,
 Expression for internal field in case of liquid or solid dielectrics in three
dimensional,
( )
( )

4. Clausius-Mossotti Equation
Consider a dielectric material of dielectric constant , N is the number of atoms per
unit volume of the material and μ is the dipole moment,
The field experienced by the atom in the internal filed ( ). Hence, if is the
electronic polarizability of the atoms then,
Therefore, Dipole moment per unit volume = N
i.e. Polarization, P = N
)
We know that Polarization, )
)
)
We have equation for internal filed in solids (3 dimensional),
( )
Considering the internal field to be Lorentz field, i.e.
( )
)
( )
)
)
*
:
;
+
𝜺 𝒓 𝟏
𝜺 𝒓 𝟐
𝐍𝜶 𝒆
𝟑𝜺 𝟎

5. Solid, Liquid & Gaseous Dielectrics (Explain with an example)
 Solid dielectric materials are mica, porcelain, glass, synthetic materials
(Plastic), cloth, rubber, paper etc. High density papers are used in DC and energy
storage capacitors.
 Liquids dielectric material are transformer oil, silicon fluids, askarels
(chlorinated hydrocarbon), Viscous vaseline, Fluoro-organic fluids etc. They are
used in transformers, switches and circuit breakers.
 Gaseous dielectric material are air, nitrogen, inert gases, hydrogen, CO2, etc.
They used as heat transferring media.
6. Applications of Dielectrics in Transformers
 A transformer consists of two insulated conducting coil (Primary coil &
Secondary coil) wound on a core. The core is also insulated.
 In case of high voltage transformers, further insulation is required to be provided
between individual winding in the coil and also between the core and the coils.
Hence their size grows. The size of the transformer increases also with
operational AC frequency. The insulation provided by using paper, mica and
cloth.
 The paper is soak with polish or wax to fill air gaps. If there are air gap then
ionization of air occurs high voltage leading to excessive heating which damages
the insulation. This effect is called Corona. Mica is used to guard against corona.
However, when the operating voltage cross 3kV and upto 100kV, a kind of oil
called transformer oil (mineral oil) is used to guard against Corona. The oil helps
to keep the transformer cool. It remains stable at high temperature.



; )



;
 ( )
 )
 *
;
:
+
 P = Nμ = N
 μ =

01 A parallel plate capacitor has an area of 6.45 X 10-4
m2
and plates are separated by a distance of
2mm across which a potential of 10V is applied. If a material with dielectric constant 6 is
introduced between the plates. Find the capacitance, charge stored on each plate, the
polarization and also find the dielectric displacement. July 2018
02 What is the polarization produced in NaCl by an electric field of 600v/mm. If it has a dielectric
constant of 6. Jan 2007
𝐴 6 𝑋 ;
𝑚
𝜀 𝑟 6
Sol :
d=2mm
V=10v
C=?, Q=?,
P=?, D=?
𝐶
𝜀 𝜀 𝑟 𝐴
𝑑
8 8 𝑋 ;
𝑋 6 𝑋 6 𝑋 ;
𝑋 ;
𝟏 𝟕𝟏 𝑿 𝟏𝟎;𝟏𝟏
𝑭
𝑄 𝐶𝑉 7 𝑋 ;
𝑋 𝟏 𝟕𝟏 𝑿 𝟏𝟎;𝟏𝟎
𝑪
𝐷 𝜀 𝑜 𝜀 𝑟 𝐸 8 8 𝑋 ;
𝑋6𝑋
𝑋 ;
𝟐 𝟔𝟕 𝑿 𝟏𝟎;𝟓
𝑪 𝒎 𝟐
𝑃 𝜀 𝜀 𝑟 )𝐸 𝜀 𝜀 𝑟 )
𝐸
𝑑
8 85 𝑋 ; 𝑋 6; )𝑋
𝑋 ; 𝟐 𝟐𝟏 𝑿 𝟏𝟎;𝟕
𝑪 𝒎 𝟐
𝑃 ?
𝜀 𝑟 6
Sol :
E=600V/mm
𝑃 𝜀 𝜀 𝑟 ) 𝐸
𝑃 8 8 𝑋 ;
𝑋 6 ) 𝑋6 𝑋 𝟐 𝟔𝟓 𝑿 𝟏𝟎;𝟓
𝑪 𝒎 𝟐

03 Find the electric field required to produce a polarization of 6.20 X 10-12
C/m2
in a dielectric
medium of dielectric constant 12.
04 If NaCl crystal is subjected to an electric field of 500v/m the resulting polarization is 2.3 X 10-8
C/m2
. Calculate the relative permittivity of NaCl.
05 The dielectric constant of Helium at 0o
C is 1.000074. The density of atom is 2.7 X 1025
/m3
.
Calculate the dipole moment induced in each atom when the gas is in an electric field of 3 X
104
v/m
06 The atomic weight and density of sulphur are 32 and 2.08 X 103
Kg/m3
respectively. The
dielectric polarizability of an atom is 3.28 X 10-40
Fm2
. If sulphur solid has cubic structure.
Calculate its dielectric constant. July 2011
07 The lattice parameter of KCl is 0.629nm. it crystallizes like the NaCl crystal structure. The
electronic polarizability of K+
is 1.264 X 10-40
Fm2
and that of Cl-
is 3.408 X 10-40
Fm2
. Find its
dielectric constant.
𝑃 6 𝑋 ;
𝐶 𝑚
𝜀 𝑟
Sol :
E=?
𝑃 𝜀 𝜀 𝑟 ) 𝐸 𝐸
𝑃
𝜀 𝜀 𝑟 )
6 𝑋
8 8 𝑋 ; )
𝑬 𝟎 𝟎𝟔𝟒 𝑽 𝒎
𝑃 𝑋 ;8
𝐶 𝑚
𝜀 𝑟 ?
Sol :
E=500v/m
𝑃 𝜀 𝜀 𝑟 ) 𝐸
𝑃
𝜀 𝐸
𝜀 𝑟 )
𝜀 𝑟
𝑃
𝜀 𝐸
𝑋
8 8 𝑋 ; 𝑋
𝟔 𝟏𝟗
𝜀 𝑟 7
𝑁 7 𝑋 5
𝑚
Sol :
E=3 X 104
v/m
μ=?
𝜇 𝛼 𝑒 𝐸 *
𝜀 𝜀 𝑟 )
𝑁
+ 𝐸
𝜇 *
8 8 𝑋 ;
𝑋 7 )
7 𝑋
+ 𝑋 𝟕 𝟐𝟖 𝑿 𝟏𝟎;𝟑𝟕
𝑪 𝒎
𝛼 𝑒 8 𝑋 𝐹𝑚
𝜀 𝑟 ?
Sol :
W=32
D=2.08 X 103
Kg/m3
𝑁
𝑁𝐴 𝑋 𝐷
𝑊
6 𝑋 6
𝑋 8 𝑋
𝟑 𝟗𝟐 𝑿 𝟏𝟎 𝟐𝟖
𝒎 𝟑
𝛼 𝑒 *
𝜀 𝜀 𝑟 )
𝑁
+ 𝜀 𝑟
𝑁𝛼 𝑒
𝜀
9 𝑋 8
𝑋 8𝑋 ;
8 8 𝑋 ;
𝜇 *
8 8 𝑋 ;
𝑋 7 )
+ 𝑋 𝟕 𝟐𝟖 𝑿 𝟏𝟎 𝑪 𝒎
𝟐 𝟒𝟓𝟑
𝛼 𝑒 𝑓𝑜𝑟 𝐾𝐶𝑙
𝛼 𝑒 𝑓𝑜𝑟 𝐾𝐶𝑙 𝛼 𝑒 𝑓𝑜𝑟 𝐾𝐶𝑙
67 𝑋 ;
𝐹𝑚
𝜀 𝑟 ?
𝛼
Sol :
a=0.629nm
𝑁𝑜 𝑜𝑓 𝑑𝑖𝑝𝑜𝑙𝑒𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 𝑁
𝑛
𝑎
𝐹𝐶𝐶)
6 9𝑋 ;9)
𝟏 𝟔𝟏 𝑿 𝟏𝟎 𝟐𝟖
𝒎 𝟑
𝛼 𝑒 *
𝜀 𝜀 𝑟 )
𝑁
+ 𝜀 𝑟
𝑁𝛼 𝑒
𝜀
6 𝑋 8
𝑋 67 𝑋 ;
8 8 𝑋 ;
𝟏 𝟖𝟓
8 8 𝑋 𝑋 7 )

08 The dielectric constant of the gas at NTP is 1.0000684. Calculate the electronic polarizability of
the atoms. If the gas contains 2.7 X 1025
atom/m3
and hence calculate the radius of helium atom.
Dec 2010
09 An electric field of 105
v/m is applied on a sample of Neon at NTP. Calculate the dipole moment
induced in each atom ( 1.00014). Also find the electronic polarizability of Neon gas.
10 A solid contains 5 X 1028
atoms/m3
, each with a polarizability of 2 X 10-40
Fm2
. Assuming that the
internal filed is given by the Lorentz relation. Calculate the ratio of internal field to the applied
filed.
11 The dielectric constant for sulphur is 3.4. Assuming a cubic lattice for its structure. Calculate
electronic polarizability of sulphur (Given: density = 2.07g/cc, W=32.07) Jan 2008
𝜀 𝑟 68
𝛼 𝑒 ?
𝑁 7 𝑋 𝑚
𝑅 ?
Sol :
𝛼 𝑒 *
𝜀 𝜀 𝑟 )
𝑁
+
8 8 𝑋 ;
68 )
7𝑋 5
𝟐 𝟐𝟒𝟐𝑿𝟏𝟎 𝟒𝟏
𝑭𝒎 𝟐
𝛼 𝑒 𝜋𝜀 𝑅 𝑅 (
𝛼 𝑒
𝜋𝜀
) (
𝑋 ;
𝜋𝑋 8 8 𝑋
) 𝟎 𝟓𝟗Å
𝜀 𝑟
𝛼 𝑒 ?
𝑁
6 𝑋 6
7 𝑋 5
𝑚
Sol :
E=105
V/m
μ=?
𝑃 𝜀 𝜀 𝑟 )𝐸 𝑁𝜇 𝜇
𝜀 𝜀 𝑟 )𝐸
𝑁
8 8 𝑋 )
7 𝑋 5
𝟒 𝟓𝟕𝑿𝟏𝟎;𝟑𝟕
𝑪𝒎
𝛼 𝑒
𝜇
𝐸
7 𝑋 ; 7
5
𝟒 𝟓𝟕 𝑿 𝟏𝟎;𝟒𝟏
𝑭𝒎 𝟐
𝑁 𝑋 8
𝑚
𝛼 𝑒 𝑋 ;
𝐹𝑚
𝐸𝑖
𝐸
?
Sol : 𝐸𝑖 𝐸 (
𝑃
𝜀
) 𝐸 (
𝑁𝜇
𝜀
) 𝐸 (
𝑁𝛼 𝑒 𝐸𝑖
𝜀
)
𝐸𝑖 (
𝑁𝛼 𝑒 𝐸𝑖
𝜀
) 𝐸
𝐸𝑖
𝐸 * (
𝑁𝛼 𝑒
𝜀
)+ (
𝑋 8 𝑋 𝑋 ;
𝑋8 8 𝑋 ; )
= 1.604
𝜀 𝑟
𝛼 𝑒 ?
𝐷 7 𝑋 𝐾𝑔 𝑚
Sol :
W=32.07
𝑁
𝑁𝐴 𝑋 𝐷
𝑊
6 𝑋 6
𝑋 7 𝑋
7
𝟑 𝟑𝟕 𝑿 𝟏𝟎 𝟐𝟖
𝒎 𝟑
𝜀 𝑟
𝜀 𝑟
𝛼 𝑒
𝜀
𝛼 𝑒
𝜀
𝑁
𝜀 𝑟
𝜀 𝑟
𝛼 𝑒
𝑋8 8 𝑋
7𝑋 8
𝟓 𝑿 𝟏𝟎 𝟒𝟎
𝑭𝒎 𝟐
For Cubic Lattice,

12 An element solid dielectric material has polarizability 7 X 10-40
Fm2
. Assuming the internal filed
to be Lorentz field. Calculate dielectric constant if the material has 3 X 1028
atoms/m3
. Jan 2012
13 A parallel plate capacitor of area 650mm2
and a plate separation of 4mm has a charge of 2 X
10-10
C on it. What is the resultant voltage across the capacitor when a material of dielectric
constant 3.5 is introduced between the plates.
14 An air-filled parallel plate capacitor has a capacitance of 1.5pF. If the separation between the
plates is doubled and wax is inserted between them, the capacitance increases to 3pF. Compute
the dielectric constant of wax.
𝜀 𝑟 ?
𝛼 𝑒 7 𝑋 ;
𝐹𝑚
𝑁 𝑋 8
𝑚
Sol : 𝜀 𝑟
𝜀 𝑟
𝛼 𝑒
𝜀
𝜺 𝒓 𝟏)
𝛼 𝑒
𝜀
𝜀 𝑟 )
𝐍𝜶 𝒆 𝜺 𝒓
𝟑𝜺 𝟎
𝟐𝐍𝜶 𝒆
𝟑𝜺 𝟎
(𝜺 𝒓
𝐍𝜶 𝒆 𝜺 𝒓
𝟑𝜺 𝟎
) 𝟏
𝟐𝐍𝜶 𝒆
𝟑𝜺 𝟎
𝜺 𝒓
𝟏:
𝟐𝐍𝜶 𝒆
𝟑𝜺 𝟎
(𝟏;
𝐍𝜶 𝒆
𝟑𝜺 𝟎
)
𝟏𝟐 𝟑𝟑
𝐴 6 𝑋 ;6
𝑚
𝜀 𝑟
Sol :
d=4mm
Q=2 X 10-10
C
V=?
𝑄 𝐶𝑉 𝑉
𝑄
𝐶
𝑄
𝜀 𝜀 𝑟 𝐴
𝑑
𝑋 ;
8 8 𝑋 ; 𝑋 𝑋6 𝑋 ;6
𝑋 ;
𝟑𝟗 𝟕𝟑𝑽
𝐶 𝑋 ;
𝐶
𝜀 𝑟 ?
𝜀 𝑟 𝑓𝑜𝑟 𝑎𝑖𝑟
Sol :
𝐶 𝑋 ;
𝐶
𝐹𝑜𝑟 𝐶 𝐴𝑖𝑟) 𝐶
𝜀 𝜀 𝑟 𝐴
𝑑
𝜀 𝐴
𝑑
𝐶
𝐶
𝜀 𝐴
𝑑
𝜀 𝜀 𝑟 𝐴
𝑑
𝜀 𝑟
𝜀 𝑟
𝐶
𝐶
𝑋 𝑋
𝑋
𝟒
𝑊 𝑒𝑛 𝑑 𝑖𝑠 𝑑𝑜𝑢𝑏𝑙𝑒𝑑 𝑎𝑛𝑑 𝑤𝑎𝑥 𝑖𝑠 𝑖𝑛𝑠𝑒𝑟𝑡𝑒𝑑 𝐶
𝜀 𝜀 𝑟 𝐴
𝑑

IV. Question Bank
Module 5: Material Science
Q. No. Question Bank
01
Explain the salient features of Drude – Lorentz theory (Free electron Concept)
Or
Explain the review of classical free electron theory.
02 Explain the failure of classical free electron theory.
03 Mention the assumptions of quantum free electron theory.
04 Explain the Fermi-Dirac statistics.
05 Explain Fermi energy and Fermi factor.
06 Discuss the dependence of Fermi factor on temperature and energy.
07 Write a note on density of states.
08 Derive an expression for Fermi energy at zero Kelvin.
09 Explain the success / merits of quantum free electron.
10 Problems on quantum free electron theory.
11 Explain the Fermi energy in an intrinsic semiconductor.
12
Derive an expression for the Fermi energy of an intrinsic semiconductor.
Or
Derive the relation between Fermi energy and energy gap of a semiconductor.
13 Derive an expression for the electrical conductivity of an intrinsic semiconductor.
14 What is Hall effect and derive an expression for Hall Coefficient.
15 Problems on physics of semiconductor
16 Mention the relation between polarization and dielectric constant
17 Describe polar and non-polar dielectrics.
18 Describe the different polarization mechanisms or types of polarization.
19 Explain the term internal field.
20 Derive Clausius-Mossotti equation.
21 Describe solid, liquid and gaseous dielectrics with an example.
22 Explain the application of dielectrics in transformer.
23 Problems on Dielectric materials

Final m5 march 2019

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