UNIT I - ELECTRICAL PROPERTIES OF MATERIALS

 Electrical Conductivity, Thermal conductivity – Wiedemann Franz law

 Fermi Distribution Function - Particle in a 3-Dimensional box
 Density of energy states - Classification of Solids
 Tight binding approximation
 Effective mass of an electron – Concept of Hole

1. Electrical conductivity, Thermal conductivity – Wiedemann-Franz law

Electrical Conductivity :
Definition: Coefficient of electrical conductivity (σ) which is defined as the quantity
of electricity flowing per unit area per unit time maintained at unit potential gradient.
Unit: Ω – 1 m-1.
Based on Drude and Lorentz theory, when we apply potential difference across the
metal, the current density ‘J’ (which is defined as the current flowing through unit area) is
given by
J  nev d ………(1)
where ‘n’ is the free electron density and ‘e’ is the charge of an electron. The electrical
forces acting on a free electron is given by F  eE
By Newton’s second law of motion F  ma
where ‘a’ is the acceleration gained by the electron in the presence of the applied electric
field.
Therefore,
eE
(or) a
m
But acceleration = drift velocity / relaxation time =

eE
Therefore vd  a 
m
Substituting the value of vd in eqn.(1) we get,
neeE ne 2 E
J  ….. (2)
m m
From the microscopic form of Ohm’s law,
J  E …… (3)
Comparing the equations (2) and (3)
..... (4)
Thus the electrical conductivity of a metal depends on n and σ.

Thermal Conductivity
Definition: Thermal conductivity of the material is defined as the amount of heat
conducted per unit area per unit time maintained at unit temperature gradient.
Unit:
The amount of heat (Q) conducted by the rod from the end
A to B of length 2 λ is given by

1
 (or)
KA(T1  T2 )t
Q (1)
2

where K → Coefficient of thermal conductivity

A → Area of cross section of the rod
(T1 - T2) → Temperature difference between the ends A and B
t → Time for conduction
2λ → Length of rod (A to B)
From equation (1) we can write
Coefficient of thermal conductivity per unit area per unit time

(2)

Let us assume that there is equal probability for the electrons to y

move in all the six directions as shown in figure. Since each -z
electron travels with thermal velocity v, if ‘n’ is the free
electron density, then on an average (1/6) nv electrons will -x x
travel in any one direction.
Therefore, the number of electrons crossing unit area per unit
time at z
-y
(3)

According to kinetic theory of gas, (Since free electrons are assumed to be gas molecules
which are freely moving)
The average kinetic energy of an electron at hot end ‘A’ of temperature (T1)

The average kinetic energy of an electron at cold end ‘B’ of temperature (T2)

Where KB → Boltzmann constant = 1.380 x 10-23 J/K

The heat energy transferred per unit area per unit time from end A to B across C =
Number of electrons x Average K.E. of electron moving from A to B

1
 nvK B T1 (4)
4

Similarly, the heat energy transferred per unit area per unit time from end B to A across C

2
 1
 nvK B T2 (5)
4
Therefore, the net heat energy transferred from end A to B per unit area per unit time
across ‘C’ can be got by subtracting Eqn. (5) from Eqn. (4)

(6)
Substituting Eqn. (6) in Eqn.(2) we have
Thermal conductivity

(7)
We know for metals Relaxation time (τ) = Collision time (τc)

τv = λ (8)
Substituting Eqn.(8) in Eqn.(7) we have

(or) Thermal Conductivity

(9)

Eqn. (9) is the classical expression for thermal conductivity.

Wiedemann-Franz law
The ratio between the thermal conductivity and electrical conductivity of a metal is
directly proportional to the absolute temperature of the metal.

where L is a constant called as Lorentz number; it's value is 2.44 x 10-8 WΩK-2
(Quantum mechanical value) at temperature T = 293K.
Proof:
We know electrical conductivity (from classical theory)

3
 Thermal conductivity (from classical theory)

Therefore

Therefore

……….(1)

We know kinetic energy of an electron =

Therefore substituting this in equation (1) we can write

Where

Substituting the value of Boltzmann constant KB = 1.38 x 10-23 JK-1 and the charge of electron
e = 1.6021 x 10-19 Joules, we get Lorentz Number

(or) L = 1.12 x 10 -8 W Ω K -2

It is found that the classical value of Lorentz number, is only one half of the experimental
value (i.e.,) 2.44 x 10-8 W Ω K-2. This Discrepancy in the experimental and theoretical value
of ‘L’ is the failure of classical theory. This discrepancy can be rectified by quantum theory.

4
2. Fermi Distribution Function
Definition :
Fermi function F(E) represents the probability of an electron occupying a given
energy state. To find out the energy states actually occupied by the free electron at any
temperature (T), we can apply the Fermi-Dirac statistics. The Fermi - Dirac statistics
deals with the particles (electrons) having half integral spin, named as Fermions. Thus we can
write the Fermi distribution function (i.e.) the probability of an electron occupying a
given energy state as
1
F (E)  ( E  E F ) / K BT
… (1)
1 e

Where EF is the Fermi energy, KB is the Boltzmann constant.

Effect of temperature on fermi function
The effect of temperature on Fermi function F(E) can be discussed with respect to eqn (1).

(i) At 0 Kelvin
At 0 Kelvin, the electrons can be filled only upto a maximum energy level called Fermi
energy , above all the energy levels will be empty. It can be proved from the
following conditions.
i. When E < EF, equation (1) becomes
F(E) = = =1
[i.e. 100 % chance for the electron to be filled within the Fermi energy level]

ii. When E > EF, equation (1) becomes

F(E) = = =0
[ (i.e.) 0 % chance for the electron not to be filled within the Fermi energy level]

iii. When E = EF, equation (1) becomes

F (E) = = = 0.5
[(i.e) 50 % chance for an electron to be filled and not to be filled within the Fermi
energy level]
This clearly shows that at 0 Kelvin all the energy states
below are filled and all those above it are empty.
The Fermi function at 0 kelvin ( ) can also be
represented graphically as shown in figure. It is seen from the
figure that the curve has step-like character at 0 kelvin.

(i.e) F(E) = 1 below and F(E) = 0 above

(ii) At any temperature T Kelvin

When the temperature is raised slowly from absolute zero, the Fermi distribution
function smoothly decreases to zero as shown in figure.
Explanation:
Due to the supply of thermal energy the electrons within the range of K BT below the
Fermi level ( ) alone takes the energy = KBT and goes to higher energy state. Hence at any
temperature (T), empty states will also be available below .

5
 Electrons lying very deep (far below) the Fermi
energy level ( ) will not go to excited state.

Also under classical limit

When KBT << EF ; F(E) is said to be degenerate
function.
KBT >> EF ; F(E) is said to be non-degenerate
function.

3. Density of energy states

Definition:
Density of states Z(E) dE is defined as the number of available electron states per unit
volume in an energy interval (dE).
Explanation:
In order to fill the electrons in an energy state we have to first find the number of available
energy states within a given energy interval. We know that a number of available energy
levels can be obtained for various combinations of quantum numbers nx, ny and nz
((i.e) n2 = + + )
Therefore, let us construct a three dimensional space of points which represents the
quantum numbers nx, ny and nz as shown in figure. In this space each point represents an
energy level.
Number of energy levels in a cubical metal piece
To find the energy levels in a cubical metal piece
and to find the number of electrons that can be filled in
the given energy level, let us construct a sphere of
radius ‘n’ in the space.

The sphere is further divided into many shells

and each of this shell represents a particular
combination of quantum numbers (nx, ny and nz) and
therefore represents a particular energy value.
Let us consider two energy values E and E + dE.
The number of energy states between E and E + dE can
be found by finding the number of energy states
between the shells of radius n and n + n, from the origin.

The number of energy states within the sphere of radius n = π n3

Since nx, ny and nz will have only positive values, we have to take only one octant of
the sphere (i.e.) 1/8th of the sphere volume.
 The number of available energy states within the sphere of radius ‘ n ‘ is

‘n’ =

Similarly, the number of available energy states within the sphere of radius ‘ n + dn’ is
n + dn =

6
 The number of available energy states between the shells of radius n and n + dn
(or) between the energy levels.
E and E + dE =

(i.e) the number of available energy states between the energy interval dE is
Z (E) dE = [ n3 + dn3 + 3n2 dn + 3dn2 n-n3]

Since the higher powers of dn is very small, dn2 and dn3 terms can be neglected.
𝜋
 Z(E) dE = 3n2 dn
6

Z(E) dE = n2 dn ….(1)
We know the energy of the electron in a cubical metal piece of sides 'l'
n2h2
E
8ml 2

n2 = …(2)

…(3)

Differentiating equation (2) we get

2ndn

ndn = dE …(4)

Equation (1) can be written as

Z(E) dE = n(ndn)

Substituting equations (3) and (4) in the above equation we have

1/ 2
  8ml 2 E   8ml 2 
Z ( E )dE     2 dE 
2  h2   2h 
1/ 2
 1  8ml 2 E   8ml 2 
Z ( E )dE     2 dE 
2 2  h2   h 
1/ 2
 1  8ml 2   8ml 2  1 / 2
Z ( E )dE     2  E dE
2 2  h2   h 
3/ 2
 1  8ml 2 
Z ( E )dE    E 1 / 2 dE
2 2  h2 

7
  1  8m 
3/ 2

Z ( E )dE  l 3 E 1 / 2 dE
2 2  h 2 

Here l 3 represents the volume of the metal piece.

If l 3  1 , then we can write that

The number of available energy states per unit volume (i.e). Density of States

  8m 
3/ 2

Z ( E )dE  E 1 / 2 dE (5)
4  h 2 

Since each energy level provides 2 electron states one with spin up and another with
spin down (pauli’s exclusion principle), we have

  8m 
3/ 2

Density of states Z ( E )dE  2. E 1 / 2 dE

4  h 2 

  8m 
3/ 2

 Z ( E )dE  E 1 / 2 dE (6)
2  h 2 
Carrier Concentration in metals
Let N(E) dE represents the number of filled enrgy states between the interval of energy dE. The
probability of filling of electrons in a given energy state is given by Fermi function F(E).
3
1
N(E) dE = Z(E) dE . F(E) 𝜋 8𝑚 2
𝑁(𝐸)𝑑(𝐸) = [ ] 𝐸 2 𝑑𝐸. 𝐹(𝐸)
2 ℎ2
Fermi energy at 0 K
At 0 K the maximum energy level that can be occupied by the electron is known as Fermi
energy level (𝐸0 ).
At 0 K for E < EF ; F(E) = 1

Integrating above equation within the limits 0 to 𝐸𝐹0 we can get the number of energy states
electrons (N) within the Fermi energy 𝐸𝐹0
3
𝑁
𝜋 8𝑚 2 𝐸𝐹0 1
∫ 𝑁(𝐸)𝑑(𝐸) = [ ] ∫ 𝐸2 𝑑𝐸
0 2 ℎ2 0

3
3
2
𝜋 8𝑚 2 𝐸𝐹0
𝑁= [ 2] 3
2 ℎ
2

Number of filled energy states at 0 K

3
𝜋 8𝑚 2 32
𝑁= [ ] 𝐸𝐹
3 ℎ2 0

2
3𝑁 3 ℎ2
(or) Fermi Energy 𝐸 𝐹0 = [ 𝜋 ] 8𝑚

8
4. Tight binding approximation
In solid, ionic cores at fixed lattice locations and free electron gas enveloping these
ionic cores. In other words, it is assumed that the solid already exists. The ionic cores are
'tightly bound' to their lattice locations. The electrons are free to run through the extend of the
solid. This is called the 'Free electron approximation'.
There is another approach to modeling materials which starts from opposite position.
In this approach, the atoms are independent to begin with and they are brought together to
build the solid. The electrons are bound to their respective individual atoms to begin with.
In this case the atoms are free to begin with while the electrons are tightly bound to the atom.
• In view of the electronic properties of the materials, this approach is referred to as the
'Tight binding approximation' - highlighting the status of the electrons at the start of the
model.
• Figure shows how the tight binding approximation builds the band structure of the
solid.

 When the atoms are far apart, all the bound electrons associated with each atom, have
fixed energy levels.
• Assuming that building the solid starts using atoms of the same element. Thus, the
energy levels occupied by the respective electrons in each atom will be identical.
• As we bring the atoms closed to each other to form the solid, the electrons will still
maintain their original energy levels as long as the interatomic separation is large.
• When the atoms get close enough, the outer shell electrons begin to overlap with each
other.
• The energy levels of these outer shell electrons are forced to split into energy levels
above and below the energy level of these electrons when they belonged to individual
atoms.
• The splitting of energy levels occurs because electrons obey the Pauli's exclusion
principle.
• Initially only the outer shell electrons overlap, therefore only their levels split. But inner
shell electrons still maintain their energy levels like individual atom.
• If the interatomic separation keeps decreasing even further, progressively more of the
inner shell electron levels will overlap and hence also split.

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 • At each energy level, the level will split to enough new energy levels (band) so as to
accommodate the electrons of all the atoms in the solid taken together.
• So, for example, if a hundred atoms come together, and there is one electron in the
outer shell, the solid will split the energy level to a hundred energy levels. Thus the
hundred outer shell electrons are filled corresponding to the combined solid.
• In view of the starting points, the free electron approximation lends itself more easily to
the treatment of metallic system. The tight binding approximation is typically more
consistent with the state of the material in the case of insulators, so it is better suited for
modeling insulators.

5. Effective mass of electron

Effective mass of electron

The mass acquired by an electron when it is accelerated in a periodic potential is
called effective mass of an electron. It is denoted by m*.
Explanation:
When an electron is accelerated by an electrical or magnetic field in a periodic
potential, the mass of the electron is not a constant. But, it varies with respect to the field
applied. This varying mass is called effective mass (m*).
Derivation of effective mass of electron
Consider a crystal subjected to an electric field of intensity 'E'. Due to this applied
field, the electron gains a velocity which can be described by a wave vector k.
According to wave mechanics, a particle moving with a velocity v is equivalent to a
wave packet moving with a group velocity vg
Group velocity with which the electron can travel
d
vg  -------(1)
dk
K - wave vector,  - angular velocity

We know that   2

E
 2   --------(2)
h
Substitute equation (2) in (1)

2  dE 
vg   
h  dk 

1  dE 
vg    ----------(3)
  dk 
d (v g )
The acceleration a 
dt

d  1  dE  ------(4)
  
dt    dk 

10
 Multiply and divide by dk
1 d 2 E dk
a=
 dk 2 dt

h
Momentum P  (From de-Broglie)

Multiply and divide by 2π

h 2

2 
2
P  k -------------(5) Since k 

Differentiating the equation (5) with respect to 't'

…… (6)

Substituting equation (6) in (4)

 

 
 
 2 
F .a …..(7)
d 2E 
 2 
  dk  

When an electrical field is applied, acceleration of the electron due to this field

eE F
a  [ F  eE ]
m* m*

(or) F  m*a ...(8)

Comparing the equations (7) and (8), we have

11
  
 2 
  a
m*a 
 d 2 E  
  2  
  dk  

2
m* 
 d 2E 
 2 
 dk 
The above equation indicates that the effective mass of an electron is not a constant,

but depends on the value of

Concept of hole (or) effective or negative mass of electron

Definition: Electron with negative effective mass is called hole (or) the electrons in the upper
band behaves as a positive charged particle is called hole.

Let us take E-k curve of a single electron in a periodic potential and this E-k curve can be
directed into 2 bands viz. upper and lower band w.r.to point of inflection (P).

The electron is found to exist with positive effective mass in lower band and negative effective
mass (hole) in the upper band.

The advantage of hole is for a nearly filled band with ‘n’ number of empty states (holes) arises.

Concept of hole used to study Thomson and Hall effect.

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6. Particle in a three dimensional box
The solution of one-dimensional potential well can be
extended for a three- dimensional potential box. In a three-
dimensional potential box, the particle can move in any
direction in space. The particle can be represented by three
quantum numbers nx , ny , nz corresponding to x, y, z axes
respectively.
Consider a particle enclosed in a three-dimensional
potential box of length a, b, c along X, Y, Z axes
respectively. The potential energy of the electron inside
the box is constant and is taken as zero and the potential
energy outside the box is infinite (∞)
The boundary conditions are
V(x,y, z) = 0 when 0 < x < a & V(x,y,z) = ∞ 0 ≥ x ≥ a
V(x,y,z) = 0 when 0 < y < b & V(x,y,z) = ∞ 0 ≥ y ≥ b
V(x,y,z) = 0 when 0 < z < c & V(x,y,z) = ∞ 0 ≥ z ≥ c
The three dimensional Schroedinger time independent wave equation
is  2  2  2 2m
   2 E  V   0 …………..(1)
x 2 y 2 z 2 
 2  2  2 2m
   2 E  0 Since V= 0 ………(2)
x 2 y 2 z 2 
The solution for equation (2) can be written as
 x, y, z   X ( x)Y ( y )Z ( z ) ………….(3)
 is a function of x,y and z is equal to product of three functions X,Y and Z.
The solution can be written as
  XYZ ………….(4)
Differentiating equation (4) partially with respect to ‘x’ twice we get
 2 d2X
 YZ ……………….(5)
x 2 dx 2
 2 d 2Y
 XZ ………………(6)
y 2 dy 2
 2 d 2Z
 XY …………..(7)
z 2 dz 2
Substituting equations (4), (5),(6) and (7) in (2)
d2X d 2Y d 2 Z 2m ………………..(8)
YZ  XZ  XY  EXYZ  0
dx 2 dy 2 dz 2  2
Dividing by XYZ in both sides we get
1 d 2 X 1 d 2Y 1 d 2 Z 2mE …………………….(9)
   
X dx 2 Y dy 2 Z dz 2 2

13
 1 d 2 X 1 d 2Y 1 d 2 Z
   (k x  k y k z ) …………….(10)
2 2 2
2 2 2
X dx Y dy Z dz
2mE
 (k x  k y k z ) …………………………(11)
2 2 2
2

In equation (10) the L.H.S is independent of each other and is equal to constants in R.H.S
By equating
1 d2X d2X
 k x  k x X
2 2
2
or 2
X dx dx
d2X
 k x X  0 ……………(12)
2
2
dx
Similarly we get
d 2Y
 k y Y  0 …………….(13)
2
2
dy
d 2Z
 k z Z  0 …………….(14)
2

dz 2
Equations (12), (13) and (14) represents the differential equations in x,y,z coordinates.
The solution of (12) can be written as
X(x) = Axsinkxx + Bx coskxx …………….(15)
Where Ax and Bx are arbitrary constants
Applying the boundary conditions
(i) When x = 0; X = 0
Eqn (15) becomes 0 = 0+Bx ; Bx = 0 ……………..(16)
(ii) When x = a ; X = 0
Axsinkxa = 0; Bx = 0 ; Ax≠ 0
Therefore sin kxa = 0 …….(a)
We know that sin nxπ = 0 …….(b)
Comparing (a) & (b) kxa = nxπ
n
k x  x …………..(17)
a
Eqn (16) & (17) in eqn (15)
n x
X ( x )  Ax sin x ………..(18)
a
Eqn (18) is unnormalized function
NORMALIZATION
Equation (18) can be normalized by integrating it within the limits (i.e) boundary condition
0 to a
a

 X ( x) dx  1
2

n xx
a

A dx  1
2
(or) x sin 2
0
a

14
 2
Ax a
Solving we get 1
2
2
Ax  ………………(19)
a
2 n x
Eqn (19) in (18) X ( x )  sin x ………….(20)
a a
Similarly by solving eqn (13) & (14) with boundary conditions 0 to b and 0 to c respectively,
we can write
2 n yy
Y ( y)  sin ……………………(21)
b b
2 n z
Z ( z)  sin z ……………………..(22)
c c
Eigen Functions
The complete wave function for eqn (2) can be written as
 x, y, z   X ( x)Y ( y )Z ( z ) …………………(23)
Substituting eqn (20), (21) & (22) in eqn (23) will get
2 n x 2 n y y 2 n z
 nx n y nz  sin x sin sin z
a a b b c c
2 2 n x n yy n z
n n n  sin x sin sin z …………….(24)
x y z
abc a b c
Eigen Values
From eqn (11)
2mE
 (k x  k y k z )
2 2 2
2

2
E (k x  k y k z )
2 2 2

2m ………………..(25)
From eqn (17) we can write
nx  2
2
kx 
2

a2
ny  2 nz  2
2 2
Similarly we get k y 2  kz 
2

; b2 c2
Substituting these values in eqn (25) we get
2 nx  2 n y  nz  2
2 2 2 2

E ( 2   )
2m a b2 c2
h 2  2 nx
2 2 2
ny nz
E (  2  2)
4 2 2m a 2 b c
2 2 2
h 2 nx ny n
E ( 2  2  z2 )
8m a b c ……………..(26)
Equation (26) represents the energy Eigen value of an electron in a rectangular box.

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If the box is cubical i.e the sides a = b = c
h2
 (n x  n y  n z )
2 2 2
We get En x n y n y 2
8ma ………………(27)
And
8 n xx n yy n z
n n n  3
sin sin sin z
x y z
a a a a ………………(28)
This shows that different combination of three quantum number (nx, ny, nz) leads to different
Eigen value and Eigen function.
DEGENERATE STATES:
For several combinations of quantum numbers, Eigen values are same but Eigen functions
are different. Such a states are called Degenerate states.
Example:
nx =1 ny =1 nz = 2 nx2 + ny2 + nz2 = 6
nx =1 ny = 2 nz = 1 nx2 + ny2 + nz2 = 6
nx = 2 ny =1 nz = 1 nx2 + ny2 + nz2 = 6
Eigen value,
6h 2
E 112E 121E 211 
8ma 2
Corresponding wave function
8 x y 2z
 112  3
sin sin sin
a a a a
8 x 2y z
 121  3
sin sin sin
a a a a
8 2x y z
 211  3
sin sin sin
a a a a
NON-DEGENERATE STATES:
When only one wave function corresponds to energy Eigen value, such a states are called
Non-Degenerate states.
Example:
12h 2 8 2x 2y 2z
E 222  2
 222  3
sin sin sin
Eigen value 8ma ; Eigen function - a a a a

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Classification of solids:
On the basis of energy band materials are classified as Conductors, Semiconductors and
Insulators.
According to band theory of solids, it consists of a set of closely spaced energy levels are
called energy bands, at which electrons of different energies are going to be occupied.
Types of energy bands
a. Valence band-consists of a group of states containing outer most electrons (valence
electrons).
b. Conduction band-lies above valence band with higher permitted energy bands
(conduction electrons).
c. Forbidden energy band - valence band and conduction band are separated by a not
allowed energy gap levels are called Forbidden energy gap.
Conductors
 Substances like copper, aluminium, silver which allow the passage of current through
them are conductors.
 The valence band of these substances overlaps the conduction band as shown in fig (a).
 Due to this overlapping, a large number of free electrons are available for conduction.
This is the reason, why a slight potential difference applied across them causes a heavy
flow of current through them.

Fig. a - Conductors Fig. b – Semi Conductors Fig. c - Insulators

Semiconductors
 Substances like carbon, silicon, germanium whose electrical conductivity lies in between
the conductors and insulators are known as semiconductors.
 The valence band of these substances is almost filled, but the conduction band is almost
empty.
 The forbidden energy gap between valence and conduction band is very small (1eV) as
shown in fig (b).
 Therefore comparatively a smaller electric field is required to push the valence electrons
to the conduction band. This is the reason, why such materials under ordinary conditions
do not conduct current and behaves as an insulator.
 Even at room temperature, when some heat energy is imparted to the valence electrons, a
few of them cross over to the conduction band imparting minor conductivity to the

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 semiconductors. As the temperature is increased, more valence electrons cross over to the
conduction band and the conductivity of the material increases.
 Thus these materials have negative temperature co-efficient of resistance.
Insulators
 Substances like wood, glass, which do not allow the passage of current through them, are
known as insulators.
 The valence band of these substances is full whereas the conduction band is completely
empty.
 The forbidden energy gap between valence band and conduction band is very large (8eV)
as shown in the fig (c).
 Therefore a large amount of energy, i.e. a very high electric field is required to push the
valence electrons to the conduction band. This is the reason, why such materials under
ordinary conditions do not conduct at all and are designated as insulators.

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