The chapter gives brief knowledge about formation of bands in solids. What are free electrons how they contribute for conductivity in conductors, but can be extended to semiconductors also.

1. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 1 of 11
Electric properties of Materials:
Formation Bands in Solids:
 The discrete energy levels of individual atom
convert into energy bands when atoms come
nearer to the distance equals effective atomic
distance (distance between atoms at which
they bind to form crystals).
 If n-atoms come nearer then each energy level
split into n- sub energy level as shown in
below figure
 Hence each energy level of individual atom
converts into group of energy levels called
energy band.
 The outermost energy band in a solid that is
completely filled, called the Valance band.
 The band above the outermost completely
filled band (valance band) is called as
Conduction band.
 The conduction band is either partially filled
or completely empty at 0k.
 One must have electrons in conduction band
to get electric conductivity.
 In case of metals topmost energy band that
contains electrons is partially filled ( Sodium,
Copper, Silver, etc.,) or completely filled
valance band overlap with empty conduction
together they act as partially filled band
(Manessium, Aluminium etc.)
Review of classical free electron theory:
Assumptions of Classical Free electron Theory
1) The free electrons in metal move in a constant
potential field of fixed ionic core.
2) The force of repulsion between the electrons
and force of attraction between electrons and
lattice ions is neglected.
3) The motion of electrons in the metal is random
and similar to the motion of gas molecules in a
vessel and it obeys the classical kinetic theory of
gases.
4) As the motion is random between successive
collisions, kinetic energy of electrons is,
2
2
1
2
3
thmvkT  ; where, vth is the average thermal
velocity.
Expression for electrical conductivity -
Consider a cylindrical conductor carrying current
(I) per unit area’, then Current density (J) is given
by,
Volume of wire, which contains free electrons
Axl
No of electrons in volume V, N =nx Axl
Charge of electrons is Q = Nxe = nx Axl xe
Current through conductor, I =Q/t, t-time
= (nx Axl xe) / t, but l/t =vd,i.e. drift velocity,

2. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 2 of 11
because this is velocity acquired by electron after
applying electric field. Also J = I/A
Therefore substituting both we have.
dnevJ  ---- (1)
Where n- no. of electrons per unit volume
(electron concentration), e – charge of electron
and vd is drift velocity of electron due to applied
electric field (E), and is called as drift velocity
and is given by

m
eE
vd  ---------- 2
 - mean collision time or relaxation time
Put eqn. (2) in eqn. (1)
Therefore, 
m
eE
neJ 
Therefore, 
m
Ee
nJ
2
 ------------ (6)
Electric conductivity
The electric conductivity is a physical property
that determines ability of metallic conductor to
conduct electricity,
If an electric field E applied across the two ends
of conductor then electric conductivity (σ) is
given by charge density (J) per applied electric
field.
Therefore
E
J
 ----------------- (7)
Put eqn. (6) in eqn. (7), hence
mE
Ene 

2

Or
m
ne 

2
 ohm-1 meter -1
This is the classical expression for electrical
conductivity
The unit of conductivity is ohmm-1 meter-1.
The reciprocal of conductivity is called as
resistivity.
Therefore,

 2
ne
m
 ohm meter
The electric conductivity or Resistivity is also
written in terms of – Mobility of electron (μ) is
defined as drift velocity (vd) acquired by electrons
per unit electric field (E)
Therefore
E
vd

Substituting value of 
m
eE
vd  ,
τ – mean collision time or relaxation time (average
time taken between two successive collision of electrons
with lattice ions)
We have,
m
e
  V-1 s-1
Substitute for μ in
m
ne 

2
 we get,
 ne or


ne
1

Thermal velocity of electrons in conductor is
average velocity with which an electron, travels
between two successive scatterings in the absence
of electric field, net displacement of electron is
zero hence current, I=0. It is of the order of 106
m/sec.
When external electric field ‘E’ applied to
conductor, electron attains resultant
displacement along with random motion per unit
time, called as ‘Drift velocity’. It is of the order
of 10-2 to 10-3m/sec.
l
I
E
A
E

3. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 3 of 11
Failures of classical Free Electron theory
Classical free electron theory explained some
concepts conductivity rightly, it fails to account
many concepts some example are given below.
1) Specific heat capacity
According to classical free electron theory (it
follows kinetic theory of gases) the specific heat
capacity if a free electron is given by.
KkmolekJR
dT
dE
Cv //5.12
2
3
 ;
Where, R is universal gas constant. It is
independent of temperature of solid
However, experimental results shown that,
Cv is dependent on the temperature (Cv α T)
i.e. specific heat due to electrons is dependent on
temperature.
2) Temperature dependence of electrical
conductivity
According to Classical free electron theory
electrical conductivity of the metals is inversely
proportional to square root of its temperature (
T
1
 ), but experimental results suggest us
that electrical conductivity is inversely
proportional to its temperature
T
1
 . Classical
free electron theory failed to account for this
difference.
3) Dependence of electrical onductivity on
free electron concentration
According to classical free theory electrical
conductivity ( ) is directly proportional to free
electron concentration of the metal (n).
For instance, value of ‘n’ for silver and copper are
5.85 x 1028 / m3 and 8.45 x 1028 / m3 respectively.
According to classical free electron theory,
conductivity of silver must be less than that of
copper. However, it is not so, there is no solution
for this problem with the classical free electron
theory.
Quantum free electron theory
Assumptions:
1. The free electrons in metal move in a constant
potential field of fixed ionic core.
2. The force of repulsion between the electrons
and force of attraction between electrons and
lattice ions is neglected.
3. Electrons in a metal possess discrete energy
values. Hence, there energy Eigen values is
quantized.
4. The distribution of electron among various
allowed energy levels occurs according to
Pauli’s exclusion principle.
Fermi level, Fermi energy and Fermi velocity:
According to the assumptions of quantum free
electron theory, conduction / valance band of the
conductor have quantized energy states and the
distribution of electrons among these states
follows Pauli’s exclusion principle. Figure
indicates the conceptual representation of
quantization of conduction band of a metal.
Fermi level: The top most energy level occupied
by electrons at absolute zero temperature is called
Fermi level.
Fermi energy (EF): Energy associated with the
top most energy level (Fermi level) occupied by
the electron is called Fermi energy.
At absolute temperature (T = 0 K)
)(1065.3
8
3
2
3
2
19
3
2
2
0
eVnX
n
m
h
EF
















4. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 4 of 11
The value of Fermi energy at T > 0 is given by,

















2
2
0
0
12
1
F
FF
E
kT
EE

Note: at normal temperatures, 0FF EE  .
Fermi velocity (vF): Velocity possessed by an
electron occupied at the Fermi level, called as
Fermi velocity.
K.E. of electron on Fermi energy level is
2
2
1
FF mvE 
Hence Fermi velocity is,
m
E
v F
F
2

Fermi Temperature: The temperature of the
electrons corresponding to the Fermi level is
called as Fermi temperature.
It is purely quantum mechanical concept.
EF = kbTF or TF = EF /kb
Fermi Dirac Distribution:
At T = 0K, electrons distributed regularly up to
Fermi level and above we get empty states.
When T > 0, electrons absorb energy excite to
higher states, the distribution electrons in higher
energy states, is not random, but according to a
systematic statistical approach, this statistical rule
is called as Fermi-Dirac Statistics.
It permits the evaluation of probability of
occupation of electrons in energy levels of a given
energy range.
The probability of occupation of states is also
called as Fermi-Dirac Distribution function or
Fermi factor.
Fermi factor f (E): It is the probability of
occupation of an electron in the given energy state
for a material in thermal equilibrium and it is
expressed as
   
1
1

 
kT
EE F
e
Ef
Where E energy of an electron at temperature T,
EF - Fermi energy, k - Boltzmann constant.
Temperature dependence of f(E):
(i) At absolute zero temperature (T= 0):
(a) For the energy states below Fermi
energy i.e., E < EF
f(E) =1, All the energy levels below Fermi level
(E< EF) are occupied at T=0k.
(b) For the energy state is above Fermi energy
i.e., E > EF
f(E) =0 All the energy levels above Fermi level (E
> EF) are unoccupied at absolute zero
temperature.
EF (Fermi
Energy)
Fermi
Level
filled
energy
states
Empty
energy
states

5. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 5 of 11
The graph of f(E) verse E at T=0 is as shown in
the figure.
(ii) For temperatures T > 0,
at E = EF, then,
 
1

kT
EE F
e ,
Hence the value of f (E) = 21 .
At higher temperature, some of the electrons
occupied the energy states near to the Fermi
Energy gets thermal energy and makes transition
to the higher energy states.
The graph below gives variation of f(E)->E
Density of states g(E)dE:
The density of states is defined as the number of
allowed energy states in an energy interval per
unit volume of a material.
Therefore 𝐷( 𝐸) =
𝑑𝑁/𝑑𝐸
𝑉
------------- 1
𝑁 = 2×
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑘 − 𝑠𝑝𝑎𝑐𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑢𝑛𝑖𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑐𝑒𝑙𝑙
𝑁 = 2 ×
𝑉
𝑉𝑘
= 2
4
3
𝜋𝑘3
(
2𝜋
𝐿
)
3
=
𝑘3
𝐿3
3𝜋2
, put above eqn.
Therefore
----2
Differentiate eqn 2 w.r.t. E and take V = L3,
where L- is length of potential well.
Therefore, equation 1 becomes
This implies, density of states is directly
proportional to E and it is independent of
absolute temperature.
Occupation of density of states N(E)dE: The
number of energy states occupied by the electrons
in an energy interval under the thermal
equilibrium is given by N(E)dE = g(E)dE×f(E).
Temperature dependance of N(E)dE:
Occupation of density of states depends upon two
factors on the density of states g(E)dE and Fermi
factor f(E).
Even though density of states (g(E)dE) is
temperature independent the occupation of
density of states (N(E)dE) becomes temperature
dependent, because f(E) is temperature
dependent.
2
22
2
2 
 mE
k
m
k
E 
3 23 3 3
2 2 2
2
3 3
/
k L L mE
N
 
 
   
 h
3/2
3 2 2
1 1 2
( )
2
dN m
D E E
L dE 
 
   
 h

6. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 6 of 11
Case 1) At T = 0 K
Since g(E)dE is a temperature independent
parameter, it remains unaffected by the
temperature. The probability of occupation f(E)
changes with temeprature’T‘.
at T =0, f(E) =1, at E < EF,
so, N(E)dE = f(E)g(E)dE) = g(E)d(E),
Case 2) At T > 0 K
The elcetrons nearer (below) fermi level absorb
thermal energy and occupy the levels above fermi
level, according to pauli exclusion principle.
Hence, f(E) below the Fermi level (E < EF) falls
below one and f (E) value above fermi level
increases by same magnitude.
hence the value of N(E)dE reduces at energy, E <
EF and increases at energy (E > EF) . due
Case 3) At T > > 0 K
With the further increase in temperature the value
of f(E) below the Fermi level (E < EF) decreases
further and hence more decrease in the value of
N(E)dE occurs, whereas N(E)dE above Fermi
level (E > EF) increases further, due to further
increase of f(E).
Expression for Electric conductivity by
Quantum Free electron theory.
According to assumptions of quantum free
electron theory the energy of electron is quantized
and they distribute in various energy states
according to pauli’s exclusion principle. At zero
Kelvin electrons, all the energy levels are filled
below Fermi level and energy levels above Fermi
level are empty.
If temperature T>0k, under thermal equilibrium
equal quantity of energy absorbed by all electrons
and this energy is very small, hence the electron
very near to Fermi level can cross Fermi level,
and participate in the conduction, the momentum
of such electrons is Ћk.
Under the application of electric field,
𝑑𝑘 = −
𝑒𝐸
ℏ
𝑑𝑡
The electric field change the ‘k’ value of the
electrons near Fermi level in time t = τF
If n- electron density per unit volume, vF -
velocity of electron, near Fermi surface
then current density,
= σE ------------2
from equations (1) and (2) we have,
, therefore, we have
Therefore the electric conductivity is of metal is
large depends on the electrons near the Fermi
surface whose mean collision time is τF.

7. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 7 of 11
.The τF is large for less carrier concentration and
decreases for large carrier concentration. This
greately affects of electric conductivity.
The mean collision time τF for such electron is
τF =
F
F
v

,
Above equation can also be written as,
Where are mean free path and Fermi
velocity of electrons near Fermi level
respectively.
As the electron is in potential field of lattice ions
of crystal, the concept to effective mass of
electron (i.e. m =m*) is used here.
Merits of Quantum free electrontheory:
The quantum free electron theory explained
successfully number of experimental facts those
could not explained by the classical theory.
Those are as follows.
1. Electronic Specific heat of solid:
It observed experimentally that CV the specific
heat of conduction electron very low. This is
because of, those electrons very near to Fermi
energy (EF) can absorb small amount of heat and
get excited into higher energy states. Hence, a
small percentage of the conduction electrons
contribute for the specific heat.
It is observed that
The typical value of EF ~ 5 eV
RTCV
4
10
 , which is very small
2. Temperature Dependence of Electrical
conductivity
From quantum free electron theory,
F
F
vm
ne 

*
2

Out of amount of energy available, part energy
used to cross Fermi energy and Fermi velocity of
electron (K.E.) and these two are independent of
temperature.
However, a part of thermal energy also
absorbed by lattice ion and set the lattice
vibrations in all directions.
Let ‘r’ is the amplitude of lattice vibrations and
hence area around the ion πr2 blocks the motion
Hence mean free path of electron gives
λ α 1 / r2
Thermal energy absorbed by the vibrating body is
proportional to square of the amplitude of
vibration, i.e. r2
E α r2
Also thermal energy is directly proportional the
Temperature (T) of the vibrating body,
E α T
From the above explanation we have
λ α 1 / T
3. Electrical conductivity dependent on
electron Concentration:
From classical, free electron theory σ α n
But metals like Aluminum, Iron like tri-valent
(having 3-valence electrons) materials, have less
conductivity compared to mono valent (1-valence
electron) metal like copper.

8. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 8 of 11
It can be explained by the quantum theory that, it
is due to τF i.e. mean collision time of Fermi
electrons,
If number of valence electron increased the value
of τ F =
F
F
v

decreases.
For example, value of n in aluminum is 2.13
times of copper. But τF of copper is 3.73 times of
the aluminum. Thus, conductivity of copper
exceeds that of aluminum.
Effect of Lattice defects and temperature
on resistivity in metals:
The resistivity of a material at a given
temperature is given by,

 2
ne
m

But resistivity of metal varies with temperature.
When T=0k, metal retain some value of
resistivity, is known as residual resistivity (ρo).
This arises is due to lattice defects (fig).
At higher temperatures due to greater lattice
vibrations, mean collision time reduces intern
resistivity increases, hence(𝜌 𝛼
1
𝜏
)
1
𝜏
=
1
𝜏 𝑑𝑒𝑓𝑒𝑐𝑡
+
1
𝜏𝑡ℎ𝑒𝑟𝑚𝑎 𝑙
Gives the total contribution of mean collision
time and corresponding resistivity is sum of ρo
(resistivity due to defect at 0k) and ρthermal -
(resistivity at temp T), is known as Matthiesen’s
rule.
ρ = ρ defect + ρ thermal.
Superconductivity:
The resistivity of certain materials drop to zero at
certain temperature before zero kelvin, this
phenomenon is called as Superconductivity and
materials are superconductors.
An undiminished current (persistence current)
sets up in the material without expense energy
or consumption of power.
MAGLEV (magnetic levitation) vehicles,
SQUIDs (Superconducting Quantum Interference
devices) are working on the principle of
superconductivity.
……..Questions:
1. List the assumptions of Classical free electron
theory of metals.
2. Show that Classical free electron theory is
failed to explain the phenomenon of
conduction taking three examples.
3. Deduce the classical expression Electrical
conductivity of metal. Arrive at an expression
for the resistivity of the metal.
4. What are the assumptions of quantum free
electron theory? Explain the distribution of
electrons and Fermi energy of in metal by use
of Pauli’s exclusion principle.
5. What is Fermi factor? Determine the how the
probability of occupation of energy states
dependent on temperature with graphical
representation.
ρo

9. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 9 of 11
6. What is density of state? Write the expression
for it. What is the condition to get electron
density=density of states and electron density
= half of the density of states.
7. Derive an expression of electric conductivity
by using Quantum free electron theory of
metals.
8. Give the proof for the merits of quantum free
electron taking three examples.
9. Using Mathieson’s rule. Give an account
of dependence of electric conductivity on
temperature and impurities.
NUMERICALS:
1. What is the drift velocity of electrons in a
copper conductor having a cross-sectional area
of 5 x 10-6 m2 if the current is 10A? Assume
that there are 8 x 1028 electrons/m3A.
Suggested solution:
Current, I = 10 A Cross-sectional area, A = 5 x
10-6 m2 Drift velocity, Vd = ?
Number density of free electrons, n = 8 x 1028
electrons/m3
We know that
neA
I
vd 
2. Calculate the drift velocity of electrons in a
silver wire having cross-sectional area of 3.14
x 10-6 m2 and carrying a current of 20
amperes. Given: Avogadro's number = 6.023
x 1023, atomic weight of silver = 108.
Density of silver = 10.5 x 103 kg m-3, e = 1.6 x 10-
19C.
Suggested solution:
Since the valency of silver is 1, therefore each
atom of silver contributes one free electron.
At.wt of silver is108. i.e., 1 kilo mole silver
weighs 108 kg and contains atoms equals to
Avogadro’s number ‘N’(= 6.023 x 1026 atoms).
Number of electrons per unit volume of silver,
Now,
or
3. The 2 ohm resistance measured when a current
flows through the wire of diameter 1mm and
length 2m. If the density and at.wt. of the
material is 4x103kg and 92 respectively calculate
of relaxation time of electrons (each atom
contribute one electron for conduction).
 
m
l
dR
l
RA








7
232
10855.7
42
)101(142.324




1
1.27x106 mho-m-
28
326
1062.2
92
1041002.6



W
dN
n /m3
Where n – electron concentration, W – atomic wt.
d – Density of material , N – Avogadro’s no.
 21928
316
2
2
106.11062.2
101.91027.1





en
m
or
m
ne 



= 1.72x10-15 s
a) If Fermi energy and resistivity of copper are
7eV and 1.7 x10-7-m respectively, find the

10. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 10 of 11
mean free path of electron, if concentration of
valence electrons is equals to 8.5x1028.
a) Fermi energy meaning
31
19
101.9
106.1722



x
xxx
m
E
v
e
F
F
= .57x106m/s
b)
  

xxxx
xxx
x
ne
vm Fe
21928
631
7
2
106.1105.8
1057.1101.9
107.1




 0.386 x 10-8m = 38.6 Å
a) Calculate the Fermi energy of a metal at 0 K
whose density is 10500 kgm-3, and has an
atomic weight of 107.9 and contributes 1
conduction electron per atom.
Electronic concentration is given by,
n = (NAx D)/A
Where n – electron concentration, A – atomic wt.
D – Density of material , NA – Avogadro’s no.
n = 5.863 x 1028
Fermi energy 3/23/22
2
)3(
2
n
m
h
EF 





 
EF = 5.5 eV
b) Find the temperature at which there is 1%
probability that a state with energy of 0.55
eV above the Fermi energy is occupied.
f(E) = 1% = 0.01, E – EF =0.55 eV, T = ?
=
T
8.6376
T= 1387.76 k
1. Determine the drift velocity of electrons in
metallic wire of diameter 0.5mm, carrying a
current of 5A. Electron density of in metal is
2x1028/m3.
2. The relaxation time of electron in a conductor
is 10-15s when an electric field of 20V applied,
determine the drift velocity.
3. The density and atomic weight of a metallic
conductor 8960kg/m3 and 63.54 respectively,
if resistivity of copper is 1.7 x 10-8 ohm m.
Determine the relaxation time and mobility of
electron in copper.
4. What would be the mobility of electrons in
copper if there are 9x1028 valence
electrons/m3 and conductivity of copper is 6x
107mho/m. (et/m).
5. Calculate the Fermi energy in eV for a metal
at 0k, whose density is 10500kg/m3 and
atomic weight 107.9 that contributes one
electron/atom for conduction. Also determine
the Fermi temperature and Fermi velocity.
6. Calculate the probability of an electron
occupying an energy level 0.02eV below the
Fermi level at 200k and above the Fermi level
at 400k in a material. Similarly determine the
probability of un occupation above Fermi
level at 200k.

11. Praveen Vaidya, Department of Physics, SDMCET, Dharwad-02 UNIT - III
Page 11 of 11