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SNS COLLEGE OF ENGINEERING
Kurumbapalayam(Po), Coimbatore – 641 107
Accredited by NAAC-UGC with ‘A’ Grade
Approved by AICTE & Affiliated to Anna University, Chennai
INTERNAL ASSESMENT EXAMINATIONS - I
Answer key- PH6251- Engineering Physics -II
Part-A
1. Drift velocity
The average velocity acquired by the free electron in a particular direction due to the
application of electric field is called as Drift velocity.(2)
2. drawbacks of classical free electron theory

It is a macroscopic theory.

According to this theory, K/sT = L, a constant (Wiedemann-Franz law) for all
temperatures. But this is not true at low temperatures.

The theoretically predicted value of specific heat of a metal does not agree with the
experimentally obtained value.

This theory fails to explain ferromagnetism, superconductivity, photoelectric effect,
Compton Effect and black body radiation.

It is a macroscopic theory.

Dual nature is not explained and atomic fine spectra could not be accounted.(2)
3. Drift velocity
The average velocity acquired by the free electron in a particular direction due to the
application of electric field is called as Drift velocity.(1)
Thermal velocity
It is the velocity of an electron without any exteranl field. This is very high.(1)
4. Mean free path.
The average distance travelled between two successive collisions is called mean free
path.
(i.e) λ =c̅τc, where c̅ is the root mean square velocity of the electron.(2)
5. Formula F (E ) =
1
(
E−EF
)
1+e KBT
F (E ) = 1 at 5.0 eV
F (E ) = 1/2 at 5.2eV (2)
6. Formula
K
σ
= LT
Ans : K = 470.4 Wm/K (2)
7. Formula Electrical conductivity σi = nie(μe+μh)
Ans: 36.256 X 106 (2)
8. Properties of a semiconductor.
(i) They have crystalline structure.
(ii)They have empty conduction band.
(iii)They have filled valence band.
(iv) Atoms are bonded by covalent bond.(2)
9. Law of mass action
(2)
Part-B
1.a.(i). The postulates of classical free electron theory:
In the absence of Electric field:
In the absence of an electric field the electrons move in random directions making collisions
from time to time with positive ions. All the collisions are elastic.
In the presence of Electric field:
When an external field is applied the free electrons are slowly drifting towards the positive
potential. Classical free electrons in the metal obey Maxwell-Boltzmann statistics.
(ii). Electrical conductivity
Coefficient of electrical conductivity which is defined as the quantity of electricity flowing
per unit area per unit time maintained at unit potential gradient. Unit is Ω-1m-1
Comparing equation (7) and (8) we can write
σ =
ne2 τ
m
σE =
nEe2 τ
m
… (9)
Expression for Electrical conductivity:
We know in the absence of external electric field, the motion of electrons in a metal moves
randomly in all directions. When an electric field (i.e.,) potential difference is maintained between the
two ends of a metallic rod, the electrons will move towards the positive field direction.
If ‘n’ is the free electron density and ‘e’ is the charge of electron then the
current density (i.e.,) the current flowing through unit area is given by
… (1)
J = nvd(-e)
The (-ve) sign implies that the charge of the electron is negative and it also indicates that the
conventional direction of current is in the opposite direction to the electron movement.
Due to the electric field applied, the electron gains the acceleration ‘a’
(i.e.,)
Acceleration (a) =
drift velocity(vd)
Relaxation time(τ)
… (2)
vd = aτ
If E is the electric field intensity and ‘m’ is the mass of the electron.
Then, the force experienced by the electron is
… (3)
F = -Ee
From Newton’s Second law of motion,
… (4)
The force on the electron F = ma
Equating equation (3) and equation (4) we have
-Ee = ma
a=
−Ee
m
… (5)
Substituting equation (5) in equation (2) we have
Drift velocity vd =
−Ee
m
… (6)
τ
Substituting equation (6) in equation (1) we have
−Ee
m
Current density J = n
J=
τ(-e)
nEe2 τ
m
… (7)
Here the number of electrons flowing per second through unit area (i.e.,) the current density
depends on the field applied. Thus if the field (E) applied is more, current density (J) will also be
more.
J∞ E
We can write
J = σE
I
A
=σE
… (8)
[J=
If A =1 and E =1, then
I
A
]
σ=I
OR
b. Thermal conductivity:
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 Wm-1K-1
Expression for thermal conductivity :Let us consider a uniform rod AB with the temperature T1
(Hot) at end A and T2 (cold) end at B. Heat flows from hot end ‘A’ to the cold end ‘B’. Let us
consider a cross sectional area ‘C’ which is at a distance equal t the mean free path (λ) of the electron
between the ends A and B of the rod.
The amount of heat (Q) conducted by the rod from the end A to B of length 2λ is given by
Qα
A (T1 − T2 )t
2λ
(or)
Q=
KA (T1 − T2 )t
2λ
… (1)
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
Q. 2 λ
1 − T2 )
…(2)
K = (T
Let us assume that there is equal probability for the electrons to move in all the six directions
as shown in fig. Since each electron travels with thermal velocity v. If ‘n’ is the free electron density,
1
6
then on an average nv electrons will travel in any one direction.
The number of electrons crossing unit area per unit time at
1
C = 6 nv
…(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)
=
3
K T1
2 B
The average kinetic energy of an electron at hot end ‘A’ of temperature (T2)
=
3
K T2
2 B
Where, K B = 1.380 X 10−23 J/K
The heat energy transferred per
unit area per unit time from end = Number of electrons X Average K.E. of electron moving
A to B across from A to B
1
3
= 6nv .2 K BT1
1
4
= nvK B T1
… (4)
Similarly, the heat energy transferred per unit area per unit time from end B to A across C
1
3
= 6nv.2 K BT2
1
4
=
K B T2
nv
…(5)
The net heat energy transferred from end A to B per unit area per unit time across ‘C’ can be got by
subtracting equation (5) from equation (4)
1
1
Q = 4nvK B T1- 4 nv K B T2
(i.e)
1
4
Q = nvK B (T1 − T2 )… (6)
Substituting equation (6) in equation (2) we have
Thermal conductivity K =
1 nvKB (T1 − T2 )2 λ
(T1 − T2 )
4
(or)
K=
nvKB λ
2
… (7)
We know for metals,
Relaxation time(τ) = Collision time(τc) λ = τv
… (8)
Substituting equation (8) in equation (7) we have
K=
nv2 KB τ
…(9)
2
Equation (9) is the classical expression for thermal conductivity.
2.a.Density of states:
DENSITY OF STATES AND CARRIER CONCENTRATION IN METALS
Definition:
Density of States Z (E) dE is defined as the number of available energy 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 = n2x+n2y+n2z)
Therefore, let us construct a three dimensional space of points which represents the quantum
numbers as shown in fig. 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 a 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+, from
the origin.
The number of available energy states within the sphere of radius
4
‘ n ‘ = [ 3πn3]
Since 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
1
4
‘ n ‘ =8 [ 3πn3]
Similarly, the number of available energy states within the sphere of radius
n + dn=
1 4
[
8 3
π (n + dn)3]
Therefore, the number of available energy states between the shells of radius n and n + dn (or)
between the energy levels E and E + d E
1 4
8 3
4
3
= [ π(n+dn)3 - πn3]
(i.e) The number of available energy states between the energy interval d E is
Z(E) d E =
1 4π
(n3
8 3
+ dn3 + 3n2 dn + 3dn2 dn − n3 )
Since the higher powers of dn is very small, dn2 and dn3 terms can be neglected.
π
Z( E ) d E = 6 3n2dn
(or)
π
2
Z( E ) d E = n2dn
… (1)
We know the energy of the electron in a cubical metal piece of sides ‘l’,
n2 h2
E = 8ml2
(or)
n2 =
8ml2 E
h2
n=[
8ml2 E
]1/2
h2
…(2)
…(3)
Differentiating equation (2) we get
2ndn =
ndn =
8ml2 E
2h2
8ml2
h2
dE
dE
…(4)
Equation (1) can be written as
π
Z (E) d E = 2 n(ndn)
Substituting equation (3) and (4) in the above equation we have
π 8ml2 E
]1/2
h2
Z (E ) d E = 2 [
=
π
2
[
8ml2
]
2h2
1 8ml2 E
]1/2
2
h2
. [
π 8ml2
]3/2
h2
= 4[
π 8m
]3/2
4 h2
Z (E) d E = [
[
dE
8ml2
]
2h2
dE
E1/2 d E
E1/2 l3d E
Here l3 represents the volume of the metal piece.
If l3 = 1, then we can write that
The number of available energy states per unit volume (i.e) Density of states.
π 8m
]3/2
h2
Z(E)d E = 4 [
E1/2 dE
… (5)
Since each energy level provides 2 electron states one with spin up and another with spin down
(pauli’s exclusion principle), we have
Density of states
π
Z(E)dE = 2. 4 [
π 8m
]3/2
h2
Z(E)dE = 2 [
8m
]3/2
h2
E1/2 dE
E1/2 dE
…(6)
CARRIER CONCENTRATION IN METALS
Let N (E) d E represents the number of filled energy states between the interval of energy d E.
Normally all the energy states will not be filled. The probability of filling of electrons in a given
energy state is given by Fermi Function F (E).
N (E) dE = Z(E) dE . F(E)
… (7)
Substituting equation (6) in equation (7), we get
Number of filled energy states per unit volume
π 8m
]3/2
h2
N (E) dE = 2 [
E1/2 d E .F(E )
…(8)
N (E) is known as carrier distribution function (or) Carrier concentration in metals.
Fermi Energy at 0 Kelvin:
We know at 0K maximum energy level that can occupied by the electron is called Fermi energy
level (EF0 )
(i.e) at 0 Kelvin for E< EF and
Therefore F (E) = 1
Integrating equation (8) within the limits 0 to EF0 we can get the number of energy states electrons
(N) within the Fermi energy EF0
E
π 8m
]3/2 ∫0 F0
h2
∫ N(E)d E = 2 [
1
E 2 E.dE
3
3
π 8m 2 EF0 2
= 2 ( h2 )
3
2
3
(or) Number of filled energy states at zero =
3
π 8m 2
2 …(9)
(
)
E
F
2
0
3 h
(or)
3
EF 0 2 =
3N h2
( )
π 8m
3/2
EF 0 =
Fermi energy
h2 3N
( )2/3
8m π
…(10)
Average energy of an electron at 0K:
Average energy of an electron
(Eave) =
Total energy of the electrons at 0K (ET )
…(11)
Number of energy states at 0K (N)
Here, the total energy of the electrons at 0K = (Number of energy states at 0K) X
(Energy of the electron)
EF0
(i.e) ET = ∫0
N(E)dE . E
3
1
E
π 8m 2
= 2 ( h2 ) ∫0 F0 E 2 E.dE
5
3
=
[F (E) = 1]
π 8m 2 EF0 2
( ) 5
2 h2
2
π 8m
3
2
5
ET = 5 ( h2 ) EF0 2 …(12)
Substituting equation 9 and 12 in 11 we get
Eave =
3
5
π 8m 2
( 2 ) EF0 2
5 h
3
3
π 8m 2
( ) EF0 2
3 h2
3
5
−3
Eave = 5 EF0 2 EF0 2
3
The average energy of an electron at 0K is Eave = 5 EF0
OR
b.fermi distribution function:
Fermi distribution function F (E) represents the probability of an electron occupying a given energy
state. The Fermi – Dirac statistics deals with the particles (Electrons) having half integral spin, named
as Fermions.
F (E ) =
1
E−EF
(
)
1+e KBT
…(1)
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 equation
(1).
(i) At 0 kelvin
At 0 kelvin, the electrons can be filled only upto a maximum energy level called Fermi energy
(EF0 ), above EF0 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
1+e−∞
1
1
= =1
[ i.e., 100% chance for the electron to be filled within the Fermi energy level]
(ii) When E>EF, equation ( 1 ) becomes,
1
1
F(E ) = 1+e∞ = ∞ = 0
[i.e Zero % chance for the electron not to be filled within the Fermi energy level]
(i)
When E=EF, equation ( 1 ) becomes,
F(E ) =
1
1+1
1
2
= = 0.5
[i.e 50% chance for the electron to be filled and not to be filled within the Fermi energy level]
This clearly shows that 0 kelvin all the energy states belowEF0 are filled and all those above it
are empty.
The Fermi function at 0 kelvin (EF0 ) can also be represented graphically as shown in fig. It is
seen from the figure that the curve has step-like character at 0 kelvin.
(i.e) F (E) = 1 below (EF0 )
and F (E) = above (EF0 )
(i)
At any temperatue T kelvin:
When the temperature is raised slowly from absolute zero, the Fermi distribution function
smoothly decreases to zero as shown in fig.
Explanation:
Due to the supply of thermal energy the electrons within the range of KBT below the Fermi level
(EF0 ) alone takes the energy (=KBT) and goes to higher energy state. Hence at any temperature (T),
empty states will also be available belowEF0 .
3.a.(i).Electron concentration in intrinsic semiconductor:
Density of electrons in conduction band
∞
Density of electrons in conduction band ne =∫𝐸 𝑍(𝐸). 𝐹(𝐸)𝑑𝐸
𝑐
………. (1)
From Fermi-Dirac statistics we can write
1
Z(E)dE =
8𝑚∗ 2 1
2. 4 [ ℎ2𝑒 ] 𝐸 2
𝜋
………. (2)
𝑑𝐸
Considering minimum energy of conduction band as Ec and the maximum energy can go upto
∞ we can write eqn (2) as
3
Z(E)dE =
8𝑚𝑒∗ 2
[
] (𝐸
2
ℎ2
𝜋
1
……….. (3)
− 𝐸𝑐 )2 𝑑𝐸
1
………… (4)
We know Fermi function, F(E)=1+𝑒 (𝐸−𝐸𝐹)/𝑇𝐾𝐵
Sub. Eqn (4) and (3) in eqn (1) we have Density of electrons is conduction band within the
limits Ec to ∞
3
8𝑚𝑒∗ 2
[
]
2
ℎ2
𝜋
ne =
1
∞
(𝐸−𝐸 )2
∫𝐸 1+𝑒 (𝐸−𝐸𝑐𝐹)/𝑇𝐾𝐵
𝑐
. 𝑑𝐸
………. (5)
Since to move an electron from valence band to conduction band the energy required is
greater than 4KB T. (i.e) 𝑒 (𝐸−𝐸𝐹)/𝑇𝐾𝐵 ≫ 1 & 1 + 𝑒 (𝐸−𝐸𝐹)/𝑇𝐾𝐵 = 𝑒 (𝐸−𝐸𝐹)/𝑇𝐾𝐵
Eqn. (5) becomes
3
ne =
8𝑚𝑒∗ 2
[
]
2
ℎ2
𝜋
∞
1
∫𝐸 (𝐸 − 𝐸𝑐 )2 . 𝑒 (𝐸−𝐸𝐹)/𝑇𝐾𝐵 𝑑𝐸
𝑐
………. (6)
Let us assume that E-Ec = xKBT Differentiating we get dE = KBT.dx,
Limits: when E=Ec; x=0, when E=∞; x=∞ Therefore limits are 0 to ∞
By solving Eqn (6) using this limits we can get,
3
Density of electrons in conduction band is ne =2
2𝜋𝑚∗ 𝐾 𝑇 2
[ ℎ𝑒2 𝐵 ] .
𝑒 (𝐸−𝐸𝐹)/𝑇𝐾𝐵 ……….. (7)
(i)Density of holes in valence band
 We know, F(E) represents the probability of filled state.
 As the maximum probability will be 1, the probability of unfilled states will be [1-F(E)]
 Example, if F(E) = 0.8, then 1-F(E) = 0.2
 Let the maximum energy in valence band be Ev and the minimum energy be -∞. So density of
holes in valence band nh is given by
E
v
nh =∫−∞
Z(E). [1 − F(E)]dE
3
We know Z(E)dE =
π 8m∗e 2
[ ] (E
2 h2
…………. (8)
1
………….. (9)
− Ec )2 dE
1-F(E) = e(E−EF )/TKB
………… (10)
Sub eqn (10) and (9) in (8), we get
3
nh =
1
π 8m∗h 2 Ev
2 . e(E−EF )/TKB
(E
−
E
)
[
]
∫
c
2
−∞
2 h
dE
……………. (11)
Let us assume that Ev-E = xKBT Differentiating we get dE = -KBT.dx,
Limits: when E=-∞; we have Ev–(-∞) = x Therefore x= ∞
When E=Ev;
x=0,
Therefore limits are ∞ to 0
Using these limits we can solve eqn (11) and we can get the Density of holes.
Density of holes in valence band is
3
nh =2
2πm∗ K T 2
[ hh2 B ] . e(Ev −EF )/TKB
……………… (12)
OR
b.(i) Electrical Conductivity
The electrical conductivity of an intrinsic semiconductor in terms of mobility of charge
carriers is given by,
The electrical conductivity σi = nie(μe+μh)
……………..(1)
Where
ni is intrinsic carrier concentration
e is charge of electron
μe is mobility of electrons
μh is mobility of holes.
According to intrinsic carrier concentration
3
 2K B T  2


  2e 
 mh . me
2
h



e

Eg
2 K BT
 e
+ h

Here the electrical conductivity depends on the exponential of forbidden energy gap
between valence band and conduction band and on the mobility of charge carriers, both µ e
and µh. But the term (µe + µh) has temperature dependence and it will cancel with the T3/2
term and hence we can write
σi = C𝑒 −𝐸𝑔/2𝐾𝐵 𝑇
…………. (2)
Where, C is a constant.
Taking log on both sides we have
log σi = log C − (Eg /2K B T)
If a plot is made between log σi and 1/T we get a straight line.
Log
σ
Here σi increases with temperature as shown in figure.
(ii)
1/T
Determination of Band gap energy
We know for intrinsic semiconductor σi = C𝑒 −𝐸𝑔/2𝐾𝐵 𝑇
We know resistivity ρi = 1/ Conductivity
1
Ρi =𝐶 𝑒 𝐸𝑔 /2𝐾𝐵 𝑇
….…….. (1)
We know resistivity is resistance per unit area per unit length
𝜌𝑖 =
𝑅𝑖 𝑎
𝑙
………….. (2)
Where 𝜌𝑖 – Resistance,
a – Cross sectional area,
l – Length
Log
Ri
Equating eqns (1) and (2) and Taking log on both sides we get,
log 𝑅𝑖 = log C1 + (Eg /2K B T)
dy
dx
1/T
The resistance can be found using meter bridge or carey
fosters bridge or post office box experiments at various temperatures.
If a graph is plotted between 1/T and log Ri a straight line is obtained as shown in
figure, with a slope dy/dx = Eg/2KB. Therefore by finding the slope of line we can
calculate the energy band gap Eg with the following expression.
(iii)
Eg = 2KB(dy/dx) Joules