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S.K.P. Engineering College, Tiruvannamalai II SEM

Physics Department 1 Engineering Physics II

SKP Engineering College

Tiruvannamalai – 606611

A Course Material

on

Engineering Physics-II

By

Dr.D.Narayanasamy

Professor

Physics Department



Quality Certificate

This is to Certify that the Electronic Study Material

Subject Code:PH6251

Subject Name:Engineering Physics-II

Year/Sem:I/II

Being prepared by me and it meets the knowledge requirement of the University

curriculum.

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Name: Dr.D.Narayanasamy

Designation: Professor

This is to certify that the course material being prepared by Dr.D.Narayanasamy is of

the adequate quality. He has referred more than five books and one among them is

from abroad author.

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Name: Mrs.Bincy Name: Dr.V.Subramania Bharathi

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PH 6251 ENGINEERING PHYSICS II

Lecture : 3 hrs/Week InternalAssessment: 20

Tutorial : 0 hr/week FinalExamination: 80

Practical : - Credits: 3

PREREQUISITE: Basic concepts in electrical circuits, magnetism and electrostatics. COURSE OBJECTIVES:

Students will be well equipped to pursue research and development careers in new and emerging technologies such as properties of new materials.

To gain the students for application of science in the design, construction, analysis of machines, processes or work for practical purposes.

COURSE OUTCOMES: Upon completion of this course the student will be familiar with:

CO1 Ability to conduct electricity

CO2 Understand the property of a semiconductor relies on quantum physics to explain the movement of electrons and holes in a crystal lattice.

CO3 Produce a magnetic field in response to an applied magnetic field. Understands the magnetic behavior of dia, para and ferromagnetic materials.

CO4 Study of dielectric properties concerns storage and dissipation of electric and magnetic energy in materials.

CO5 Development of novel metallic-glass-matrix composite materials and studies of the atomic-scale structure to very practical studies of mechanical behavior, including both deformation and fracture and also behavior of some nanomaterials is well understood.



SYLLABUS

TOTAL : 45 Periods

UNIT – I CONDUCTING MATERIALS 9

Conductors – classical free electron theory of metals – Electrical and thermal conductivity – Wiedemann – Franz law – Lorentz number – Draw backs of classical theory – Quantum theory – Fermi distribution function – Effect of temperature on Fermi Function – Density of energy states – carrier concentration in metals.

UNIT – II SEMICONDUCTING MATERIALS 9

Intrinsic semiconductor – carrier concentration derivation – Fermi level – Variation of Fermi level with temperature – electrical conductivity – band gap determination – extrinsic semiconductors – carrier concentration derivation in n-type and p-type semiconductor – variation of Fermi level with temperature and impurity concentration – compound semiconductors – Hall effect –Determination of Hall coefficient – Applications.

UNIT – III MAGNETIC AND SUPERCONDUCTING MATERIALS 9

Origin of magnetic moment – Bohr magneton – Dia and para magnetism – Ferromagnetism – Domain theory – Hysteresis – soft and hard magnetic materials – anti –ferromagnetic materials – Ferrites – applications – magnetic recording and readout –storage of magnetic data – tapes, floppy and magnetic disc drives. Superconductivity : properties - Types of super conductors – BCS theory of superconductivity(Qualitative) - High Tc superconductors – Applications of superconductors – SQUID, cryotron, magnetic levitation.

UNIT – IV DIELECTRIC MATERIALS 9

Electrical susceptibility – dielectric constant – electronic, ionic, orientational and space charge polarization – frequency and temperature dependence of polarisation – internal field – Claussius – Mosotti relation (derivation) – dielectric loss – dielectric breakdown – uses of dielectric materials (capacitor and transformer) – ferroelectricity and applications.

UNIT V MODERN ENGINEERING MATERIALS 9

Metallic glasses: preparation, properties and applications. Shape memory alloys (SMA): Characteristics, properties of NiTi alloy, application, advantages and disadvantages of SMA Nanomaterials: synthesis –plasma arcing – chemical vapour deposition – sol-gels –electrodeposition – ball milling - properties of nanoparticles and applications. Carbon nanotubes: fabrication – arc method – pulsed laser deposition – chemical vapour deposition - structure – properties and applications.



CONTENT BEYOND SYLLABUS LEARNINGRESOURCES: TEXT BOOKS: 1. Charles Kittel ‘Introduction to Solid State Physics’, John Wiley & sons, 7th edition,

Singapore

(2007)

2. Charles P. Poole and Frank J.Ownen,’Introduction to Nanotechnology’, Wiley

India(2007)

(for Unit V)

REFERENCES:

1. Rajendran, V, and Marikani A, ‘Materials science’Tata McGraw Hill publications,

(2004)

New delhi.

2. Jayakumar, S. ‘Materials science’, R.K. Publishers, Coimbatore, (2008).

3. Palanisamy P.K, ‘Materials science’, Scitech publications(India) Pvt. LTd.,

Chennai,second

Edition(2007)

4. M. Arumugam, ‘Materials Science’ Anuradha publications, Kumbakonam, (2006).

WEB RESOURCES:

1. http://www.hyperphysics.com

2. http://www.sciencejoywagon.com/physicszone

http://www.sciencejoywagon.com/physicszone



CONTENTS

S.No Particulars Page

1 Unit – I 7

2 Unit – II 20

3 Unit – III 55

4 Unit – IV 84

5 Unit – V 109



Unit – I

Conducting Materilas

Part – A

1.Define drift velocity and the mobility of an electron? [CO1- L1- May 2008,Dec

2012]

Drift velocity is defined as the average velocity acquired by the free electrons of a

metal in a particular direction by the application of an electric field.

Mobility is defined as the drift velocity per unit electric field applied.

2.Write the main postulates of the classical free electron theory?[CO1-L1-May/jun

2013]

A metal consists of large number of free electrons, which can move freely

throughout the volume of the metal. These free electrons are fully

responsible for electrical conduction in metal.

The free electrons collide with the positive ions in the lattice and also among themselves. All the collisions are elastic, i.e., there is no loss of energy.

These free electrons are just like gas molecules and they obey the laws of kinetic theory gases.

In the absence of electric field, the random motion of free electron constitutes a zero net current.

In the presence of electric field, electron gains a velocity called drift velocity and moves opposite to the field direction.

3.Give the important assumptions quantum free electron theory. [CO1-L1- May

2009]

The potential energy of an electron is constant within the metal.

The allowed energy levels of an electron are quantized



The electrons are free to move within the crystal and they are restricted from leaving the crystal

When energy is supplied to the system only few electrons can absorb the energy.

Part – B 1. Explain the drawbacks of Classical Free Electron Theory.[CO1-L2-May

2009,Jun 2010,Dec 2012]

Contradiction in the absorption of supplied energy. The classical theory states that all the free electrons will absorb the supplied energy. But the quantum theory states that only a few electrons absorb the supplied energy. Fails to explain the properties of non-metals.

The electrical conductivity and resistivity of the non-metals such as

semiconductors and insulators are not explained by this theory. Fails to explain the concept of photo-electric effect, Compton Effect and

black body radiation.

The concept of photo-electric effect, Compton Effect and black body radiation could not be explained on the basis of this theory.

Specific heat of solids

The molar specific heat of a gas 𝐶𝑣 at constant volume is given by

𝐶𝑣 = 3

2𝑅.



R – Universal gas constant.

But experimentally it was found that the contribution to the specific heat of a metal by its conduction electron is given by

𝐶𝑣 = 10−4 𝑅

Thus the value of specific heat as per the prediction is much higher than the experimentally observed value.

Temperature dependence of electrical conductivity

By experimentally the electrical conductivity 𝜍𝑒𝑥𝑝 of a metal is inversely

proportional to temperature T but theoretically it is inversely proportional to 𝑇 Widemann – Franz Law

The theoretical value of Lorentz No is in agreement with experimental values. 2. Write an expression for the Fermi-energy distribution function F(E) and discuss its behaviour with change in temperature. Plot F(E) versus E for T = OK and T > OK. [CO1-L1-A.U.June 2009] FERMI DISTRIBUTION FUNCTION:

It is the probability function F(E) of an electron occupancy in a given energy level

at absolute temperature. It is given by

𝐹 𝐸 = 1

1+ 𝑒 𝐸−𝐸𝐹 /𝑘𝑇

Where

E - Given Energy level

𝐸𝐹 – Fermi energy level k - Boltzmann’s constant T - Absolute temperature Effect of Temperature on Fermi Function



Case (i) Probability of occupation for E 𝐸𝐹 , 𝐸 − 𝐸𝐹 is +ve, we have

𝐹 𝐸 = 1

1+ 𝑒 +𝑣𝑒/0 =

1

1+ 𝑒∞=

1

1+ ∞=

1

∞= 0 [∵ 𝑒∞ = ∞]

F(E) = 0

Thus, there is 0% chance for the electrons to occupy energy level above the

Fermi energy level i.e., all the energy levels above Fermi energy level are not occupied

by the electrons.

From the above two cases, at T = 0K the variation of F(E) for different energy

values becomes a step function as shown in fig (a).



Case (iii) Probability of occupation at ordinary temperature at E = Ef

At ordinary temperature, the value of probability starts reducing from 1 for values

of E slightly less than EF

With the increase of temperature, i.e. T >0 K, Fermi function F(E) varies with E

as shown in Fig. (b). At E = Ef ,

F(E) = 1

1+ 𝑒0

F(E) = 1

1+ 1 =

1

2 = 0.5 [∵ 𝑒0 = 1]

% of F(E) = 0.5 100

= 50%

Hence, there is 50% chance for the electrons to occupy Fermi energy level. i.e.,

the value of F(E) becomes 1

2 𝑎𝑡 E = 𝐸𝐹

Further, for E >𝐸𝐹 the probability value falls off rapidly to zero fig. (b).

Fermi energy of the metal at any temperature is defined as the energy level in which

the Fermi function is 0.5.



3.Obtain an expression for electrical and thermal conductivities and hence prove

Wiedemann - Franz law? [CO1-L2-A.U.Dec 2009/Jun 2012 ]

Electrical conductivity of a metal:

The electrical conductivity is defined as the quantity of electrical charge (Q) flowing per unit area (A) per unit time (t) maintained at a unit potential gradient (E).

i.e., electrical conductivity σ = 𝑄

𝑡𝐴𝐸

When an electric field (E) is applied to an electron of charge ‘e’ of a metallic rod,

the electron moves in opposite direction to the applied field with a velocity vd. This velocity is known as drift velocity.

Lorentz force acting on the electron F1 = eE (1) This force is known as the driving force of the electron. Due to this force the electron gains acceleration ‘a’ in between the terminals According to Newton’s second law of motion the force F2 = Mass of the electron × acceleration F2 = ma (2)

At equilibrium , F1 = F2 ma = eE



a = 𝑒𝐸

𝑚 (3)

acceleration a = 𝑑𝑟𝑖𝑓𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣𝑑

𝑟𝑒𝑙𝑎𝑥𝑎𝑡 𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 (𝜏)

𝑎 = 𝑣𝑑

𝜏

𝑣𝑑 = 𝑎𝜏 (4)

Substituting (3) in (4)

𝑣𝑑 = 𝑒𝜏

𝑚 𝐸 (5)

The Ohm’s law states that current density (J) is expressed as J = σE

(Or) σ = 𝐽

𝐸

(6) Where σ is the electrical conductivity of the electron. But the current density in terms

drift velocity is given as J = nevd

(7)

Substituting equation (5) in equation (7),

We have 𝐽 = 𝑛𝑒 𝑒𝜏

𝑚 𝐸

𝐽

𝐸=

𝑛𝑒2𝜏

𝑚

(8)

On comparing the equations (6) and (8), we have electrical conductivity σ = 𝑛𝑒2𝜏

𝑚.

Thermal conductivity of a metal:

Thermal conductivity (K) is defined as the amount of heat (Q) flowing per unit time (t) through unit area of cross section (A) of the material maintained at a unit temperature gradient (dT / dx).

𝐾 = 𝑄

𝑑𝑇 𝑑𝑥

Consider two cross section A and B in a uniform metallic rod AB separated by a

distance 𝝀. Let A at a high temperature (T) and B at a low temperature (T- dT). Now heat conduction takes place from A to B by the electrons.



The conduction electron per unit volume is n and average velocity of these electrons is

v. during the movement of electrons in the rod, the collision take place. Hence, the

electrons near A lose their kinetic energy while electrons near B gain kinetic energy.

At A, average kinetic energy of an electron = 3

2𝑘𝑇

(1)

At B, average kinetic energy of an electron = 3

2𝑘(𝑇 − 𝑑𝑇)

(2)

The excess of kinetic energy carried by the electron from A to B = 3

2𝑘𝑇 −

3

2𝑘(𝑇 − 𝑑𝑇)

=3

2𝑘𝑑𝑇 (3)

Number of electrons crossing per unit area per unit time from A to B = 1

6𝑛𝑣

The excess of energy carried from A to B per unit area in unit time = 1

6𝑛𝑣 ×

3

2𝑘𝑑𝑇

= 1

4𝑛𝑣 𝑘𝑑𝑇 (4)

Similarly the energy carried from B to A per unit area per unit time = − 1

4𝑛𝑣 𝑘𝑑𝑇

(5)

Hence the net amount of energy transferred from A to B per unit area per unit time,

𝑄 = 1

4𝑛𝑣 𝑘𝑑𝑇 ( )

(6)



From the basic definition of thermal conductivity, the amount of heat conducted per unit

area per unit time is given by

ie., =

(7)

We known that for the metals relaxation time (τ) = collision time (τc)

ie., τ = τc =

τ (8)

Substituting the equations (8) in (7) we have

This is the expression for thermal conductivity of metals.

Wiedemann – Franz Law:

According to Wiedemann – Franz law, the ratio between thermal conductivity and

the electrical conductivity of a metal is directly proportional to the absolute temperature

of the metal.

Where L is constant called Lorentz number.

We can derive Wiedemann – Franz law using the expression for electrical and

the thermal conductivity of metals.



The kinetic energy of electron is given by

Substituting this in the above equation, we get,

Where L = is a constant known as Lorentz number.

Hence, it is proved that the ratio of thermal conductivity to electrical conductivity of a metal is directly proportional to the absolute temperature of metal. 4.Derive an expression for the density of states, and based on that calculate the

carrier concentration in metals.[CO1-L2- A.U May 2012, Dec 2014 etc.,]

Definition:

Density of states Z(E) dE is defined as the number of available electron states

per unit volume in an energy interval E and E+ dE.

Density of stateZ(E)dE =

Number of energy levels in cubical metal piece:



Let us consider a cubical sample with side ‘a’. A sphere is constructed with three

quantum numbers the number as coordinate

axes in three dimensional spaces as shown in fig.

n is the radius vector of the sphere. All the points on the

surface of the sphere have the same energy E.

Thus denotes the radius of the

sphere with energy E.

The sphere is further divided into many shells and each

of this shell represents a particular combination of quantum

number ( and therefore represents a particular

energy value.

Let us consider two energy values E and E+dE with a small energy difference 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+dn, from the origin.

The number of energy states within a sphere of radius n =

. . (1)

Since can have only positive integer values, we have to consider

first octant of the sphere

Hence number of available energy states within the sphere of radius n =

Similarly, the number of available energy states within a sphere of radius

(n+dn) =

Now, the number of available energy states between E and E +dE is,

N(E)dE = -



=

N(E)dE = = (2)

(Higher powers of dn are very small and terms can be neglected.) The

energy of free electron is the same as the energy of particle in a box. i.e., the energy,

E = (3)

(4)

n = (5)

Also differentiating, eqn (4), we get

2ndn =

ndn= (6)

Substituting (5) and (6) in (2)

N(E)dE =

N(E)dE (7)



Number of energy states in the range from E and E+dE per unit volume is

Z(E)dE = N(E)dE Volume of cubic specimen

Z(E)dE =

Here represents the volume of cubical metal piece.

(8)

From this we write the density of energy states as

Z(E)dE (9)

The Pauli’s exclusion principle permits two electrons in each state. Hence, the

actual number of available states is given by

Z(E)dE

Z(E)dE (10)

This is the expression for density of energy states at low temperature and it can

be used to calculate the charge carrier concentration in metals.



Carrier concentration in metals

Let N(E) be the number of electrons in a system with energy E and Z(E) be the

number of states within energy interval around the energy E. Then

Z(E)dE = F(E) Z(E) dE

Now, the actual number of electrons lying in the energy interval dE is

given by

N(E)dE = Z(E) dE.F(E)

i.e. N(E)dE = (11)

Calculation of density of electrons and Fermi energy at 0 K:

In the case of a material at absolute zero, when T = 0 K, F(E) = 1

From eqn (1), we get the carrier concentration of electrons (n) as,

n =

n =

n =

Number of filled energy states at 0 K IS n =

(12)

Using eqn. (12), the Fermi energy is easily obtained as



(13)



UNIT – II SEMICONDUCTING MATERIALS

Part-A

1.Difference between elemental and compound semi conductor?[ CO2-H1-A.U

May 2012]

Sl.no Elemental semiconductor Compound semiconductor

1 They are made of single IV group elements Eg: Germanium , silicon

They are made of compounds such as II and VI group , III and V group elements. Eg: GaA, GaP

2 They are called indirect band gap semi conductor.

They are called direct band gap semi conductor.

3 During electron – hole recombination heat energy is produced.

During electron – hole recombination photos are emitted.

4 Current amplification is more. Current amplification is less

5 They are used for the manufacturing of diodes and transistors.

They are used for making LED’s, laser diodes.

2.Difference between intrinsic semi conductor and extrinsic semi

conductor.[CO2-H1-May 2012]

Sl.no Intrinsic semiconductor Extrinsic semiconductor

1 Semi conductor in a pure form is called Intrinsic semi conductor.

Semi conductor which are doped with impurities are called extrinsic semi conductor.

2 Here the charge carriers are produced only due to thermal agitations.

Here the charge carriers are produced due to impurities and also due to thermal agitations.

3 They have low electrical conductivity They have high electrical conductivity

4 At 0K Fermi level exactly lies between the conduction band and valence band.

At 0K Fermi level lies closer to conduction band in N-type semi conductor and lies near to the valence band in P– type semi conductor.

5 Eg: Si, Ge.

Eg: Si and Ge doped with Al, In, P, As.



3.What are the properties of semi conductors? [CO2-L1-A.U May 2012]

They are formed by covalent bond

They have empty conduction band

They have almost filled valence band

These materials have comparatively narrow energy gap.

4.What is a Fermi level energy? [CO2-L1-A.U May 2012]

The maximum energy level occupied by the electron at absolute Zero. The energy level in which the Fermi function is 0.5 at any temp.

5.What is meant by donar energy level? [CO2-L1-Dec 2012]

When a pentavalent impurity is doped with an intrinsic semi conductor, it donates

one electron which is placed in the energy gap near Ec. This extra energy level is

called donar energy level.

6.What is meant by acceptor energy level? [CO2-L1-Dec 2012]

When a trivalent impurity is doped with an intrinsic semi conductor, a hole is

developed which is placed in energy gap near Ev. This extra level is called

acceptor energy level.

PART-B

1. Derive an expression for density of electron in the conduction band of an

intrinsic semiconductor. [CO2-L2-A.U. June 2010]

Definition

Number of electrons in the conduction band per unit volume of the semi

conducting materials is called the Density of electrons or concentration of electrons. At

0K, an intrinsic semi conductor behaves as an insulator. When its temperature is

increased, some electrons move from valence band to conduction band as shown in the

fig.i.



The number of electrons per unit volume in the energy range E and E + dE is

given by

dn = Z(E) F (E) dE (1)

The number of electrons in the conduction band for the entire region is calculated

by integrating eqn (1) between energy range EC and +

Fig i. Energy band diagram of an intrinsic semiconductor

𝑑𝑛 = 𝑛 = 𝑍 𝐸 𝐹 𝐸 𝑑𝐸∞

𝐸𝐶 (2)

EC is the energy corresponding to the bottom most level of the conduction band

and +∞ is energy corresponding to the upper most level of the conduction band (Fig.i).



Since the electrons are moving in a periodic potential the mass of the electron m is

replaced by 𝑚𝑒∗

Density of states in the conduction band between the energy range E and E+dE

is given by

𝑍 𝐸 𝑑𝐸 = 4𝜋

3 2𝑚𝑒

∗ 3/2

𝐸1/2𝑑𝐸 (3)

Conduction band energy starts from EC. Thus in eqn (3), E can be written as (E –

EC)

𝑍 𝐸 𝑑𝐸 = 4𝜋

3 2𝑚𝑒

∗ 3/2

(𝐸 − 𝐸𝐶)1/2𝑑𝐸 (4)

The Fermi distribution function

F(E) = 1

1+𝑒 𝐸−𝐸𝐹 /𝑘𝑇 (5)

Substituting the eqns (4) and (5) in (2), we get

n = 4π

h3 2me

∗ 3/2

(E − EC)1/2 ×

1

1+e E−E F /kT

+∞

EC dE

n = 4π

h3 2me

∗ 3/2

(E−EC )

1/2

1+e E−E F /kT dE

+∞

EC (6)

𝐸 − 𝐸𝐹 is greater than kT

1 + 𝑒 𝐸−𝐸𝐹 /𝑘𝑇 ≈ 𝑒 𝐸−𝐸𝐹 /𝑘𝑇

Now the eqn (6) becomes



n = 4𝜋

3 2𝑚𝑒

∗ 3/2

(𝐸−𝐸𝐶 )

1/2

𝑒 𝐸−𝐸𝐹 /𝑘𝑇 𝑑𝐸

∞

𝐸𝐶

(or) n = 4π

h3 2me

∗ 3/2

(E − EC)1/2e− E−EF /kT dE

+∞

EC

n = 4π

h3 2me

∗ 3/2

(E − EC)1/2e EF−E /kT dE

+∞

EC

n = 4π

h3 2me

∗ 3/2

eEF /kT (E − EC)1/2e−E/kT dE

∞

EC (7)

To solve the integral in the eqn (7), let us assume

When when

𝐸 − 𝐸𝐶 = 𝑥𝑘𝑇𝐸 = 𝐸𝐶 + 𝑥𝑘𝑇𝑑𝐸 = 𝑑𝑥𝑘𝑇

𝐸 = 𝐸𝐶

𝐸𝐶 − 𝐸𝐶 = 𝑥∴ 𝑥 = 0

𝑘𝑇 𝐸 = + ∞

+ ∞ − 𝐸𝐶 = 𝑥𝑘𝑇∴ 𝑥 = + ∞

Substituting the above values is eqn (7), we have

n = 4π

h3 2me

∗ 3/2

eEF /kT (xkT)1/2e−(EC +xkT )/kT dxkT∞

0

n = 4π

h3 2me

∗ 3/2

e EF−EC /kT x1/2(kT)3

2 e−xdx∞

0

n = 4π

h3 2me

∗ 3/2

e EF−EC /kT (kT)3

2 x1/2 e−x dx∞

0 (8)

Using the gamma function, it is shown that



x1/2e−xdx∞

0=

π1/2

2 (9)

Substituting the eqn (9) in the eqn (8), we have

n = 4π

h3 2me

∗ 3/2

e EF−EC /kT (kT)3

2 π1/2

2

n = 2π

h3 2me

∗ 3/2

(kT)3/2π1/2e EF−EC /kT

or n = 2 2πme

∗kT

h2

3/2e EF−EC /kT (10)

the equation (10) gives the expression for density of electrons in the conduction band of

an intrinsic semiconductor.

2. Derive an expression for density of holes in the valence band of an intrinsic

semiconductor.[ CO2-L2-A.U. May 2009]

DENSITY OF HOLES IN THE VALENCE BAND OF AN INTRINSIC

SEMICONDUCTOR

Let dp be the number of holes per unit volume in the valence band between the energy

E and E + dE.

dp = Z (E) (1-F(E))dE (1)

Since F(F) is the probability of electron occupancy, 1 – F (E) is the probability of

an unoccupied electron state, i.e., hole.

1 – F (E) = 1 - 1

1+𝑒 𝐸 −𝐸𝐹 /𝑘𝑇

= 1+e E −E F /kT −1

1+e E −E F /kT



1 – F(E) = e E −E F /kT

1+e E −E F /kT (2)

Since E is very small when compared to EF,

(E - EF) is a negative quantity. Therefore 𝑒 𝐸 −𝐸𝐹 /𝑘𝑇 is very small and it is neglected

i.e.,1 + e E −EF /kT ≈ 1

∴ 1 − F E = e E −EF /kT (3)

Density of states in the valence band,

Z E dE = 4π

h3 2mh

∗ 3/2

E1/2dE (4)

Here 𝑚∗ is the effective mass of the hole in the valence band.

𝐸𝑣, 𝑖𝑠 top level in the valence band So the term 𝐸1/2 is replaced by (𝐸𝑣 − 𝐸)in eqn

(4).

∴ 𝑍 𝐸 𝑑𝐸 = 4𝜋

3 2𝑚

∗ 3/2

(𝐸𝑣 − 𝐸)1/2𝑑 (5)

Substituting the eqns (3) and (5) in (1), we get

dp = 4𝜋

3 2𝑚

∗ 3/2

(𝐸𝑣 − 𝐸)1/2

𝑒 𝐸−𝐸𝐹 /𝑘𝑇𝑑 (6)

The number of holes in the valence band for the entire energy range is obtained

by integrating the equation (6) between the limits - ∞ 𝑡𝑜 𝐸𝑣 .

dp = p = 4π

h3 2mh

∗ 3/2

(Ev − E)1/2

e E−EF /kT dEEv

−∞



p = 4π

h3 2mh

∗ 3/2

e −EF /kT Ev − E 1/2

e(E/kT )dEEv

−∞ (7)

To solve the above integral in eqn (7), let us assume,

When When

𝐸𝑣 − 𝐸 = 𝑥𝑘𝑇𝐸 = −𝑥𝑘𝑇 + 𝐸𝑣∴ 𝑑𝐸 = −𝑘𝑇𝑑𝑥

𝐸 = −∞

𝐸𝑣 + ∞ = 𝑥∴ 𝑥 = ∞

𝐸 = 𝐸𝑣

𝑥𝑘𝑇 = 𝐸𝑣 − 𝐸𝑣∴ 𝑥 = 0

Substituting these values in eqn (7), we have

p =4π

h3 2mh

∗ 3/2

e −EF /kT (xkT)1/2

e(−xkT +Ev )/kT (−kTdx)0

∞ (8)

or p = 4π

h3 2mh

∗ 3/2

e Ev−EF /kT x1/2(kT)3/2e−x dx∞

0 (9)

using the gamma function, it is shown that

x1/2e−xdx = π1/2

2

∞

0 (10)


p = 4π

h3 2mh

∗ 3/2

e Ev−EF /kT (kT)3/2 π1/2

2

p = 2π

h3 2mh

∗ 3/2

(kT)3/2π1/2e Ev−EF /kT

or p = 2 2πmh

∗ kT

h2

3/2e Ev−EF /kT (11)



3. Obtain an expression for the intrinsic carrier concentration in an intrinsic

semiconductor.[ CO2-L2-A.U. May 2009]

(or)

Derive the expressions for density of electrons in conduction band and density of

holes in the valence band also derive the following

INTRINSIC CARRIER CONCENTRATION

In an intrinsic semiconductor, the concentration of electrons in the conduction

band is equal to the concentration of the holes in the valence band.

∴ 𝑤𝑒 𝑐𝑎𝑛 𝑤𝑟𝑖𝑡𝑒 𝑛 = 𝑝 = 𝑛𝑖

Where 𝑛𝑖 is intrinsic carrier concentration

np = 𝑛𝑖 × 𝑛𝑖= 𝑛𝑖2

(1)

substituting the expressions for n and p in eqn (1), we have

ni2 = 2

2πme∗kT

h2

3/2e EF−EC /kT × 2

2πmh∗ kT

h2

3/2e Ev−EF /kT

ni2 = 4

2πkT

h2

3 me∗mh

∗ 3/2

e Ev−EC /kT

ni2 = 4

2πkT

h2

3 me∗mh

∗ 3/2

e−Eg /kT (2)

Where 𝐸𝐶 − 𝐸𝑣 = 𝐸𝑔 is the forbidden energy gap.

𝑛𝑖 = 2 2𝜋𝑘𝑇

2

3/2 𝑚𝑒

∗𝑚∗

3/4𝑒−𝐸𝑔/2𝑘𝑇



4. Derive an expression for Fermi level and discuss how it varies with

temperature in intrinsic semiconductor.[CO2-L2-A.U. Jul 2009]

FERMI LEVEL AND ITS VARIATION WITH TEMPERATURE

In an intrinsic semiconductor, the density of electrons in the conduction band is

equal to the density of holes in the valence band

i.e. n = p (1)

Substituting the expressions of n and p in eqn (1), we have

2 2πme

∗kT

h2

3/2e EF−EC /kT = 2

2πmh∗ kT

h2

3/2e Ev−EF /kT

Rearranging, we get

e EF−EC /kT = mh

∗

me∗

3/2e Ev−EF /kT

Or e2EF /kT = mh

∗

me∗

3/2e Ev +EC /kT

Taking log on both sides, we get

𝑙𝑜𝑔𝑒𝑒2𝐸𝐹 /𝑘𝑇 = 𝑙𝑜𝑔𝑒

𝑚∗

𝑚𝑒∗

3/2+ 𝑙𝑜𝑔𝑒𝑒

𝐸𝑣+𝐸𝐶 /𝑘𝑇

2𝐸𝐹

𝑘𝑇=

3

2𝑙𝑜𝑔𝑒

𝑚∗

𝑚𝑒∗ +

𝐸𝑣+ 𝐸𝐶

𝑘𝑇

𝐸𝐹 = 𝑘𝑇

2

3

2𝑙𝑜𝑔𝑒

𝑚∗

𝑚𝑒∗ +

𝐸𝑣+ 𝐸𝐶

𝑘𝑇

𝐸𝐹 = 3𝑘𝑇

4𝑙𝑜𝑔𝑒

𝑚∗

𝑚𝑒∗ +

𝑘𝑇

2 𝐸𝑣+ 𝐸𝐶

𝑘𝑇



𝐸𝐹 = 3𝑘𝑇

4𝑙𝑜𝑔𝑒

𝑚∗

𝑚𝑒∗ +

𝐸𝑣+ 𝐸𝐶

2 (2)

If 𝑚∗ = 𝑚𝑒 ,

∗ then 𝑙𝑜𝑔𝑒 𝑚

∗

𝑚𝑒∗ = 𝑙𝑜𝑔𝑒 1 = 0

Hence the eqn (2) becomes

𝐸𝐹 = 𝐸𝑣+ 𝐸𝐶

2 (3)

Thus Fermi, level is located half way between the top of the valence band and

bottom of the conduction band as shown in fig. Its position is independent of

temperature.

Fig.(1)position of Fermi level in an intrinsic semiconductor at various

temperatures. (a) At T = 0K. Fermi Level is in the middle of the forbidden band.

(b) As the temperature rises, it shifts upward since 𝑚∗ > 𝑚𝑒

∗

In reality 𝑚∗ > 𝑚𝑒

∗thus, Fermi level is just above the middle of energy gap and it

rises slightly with increasing temperature.



5. Derive an expression for electrical conductivity in intrinsic semiconductor.

How does electrical conductivity vary with temperature for an intrinsic

semiconductor?[CO2-L2-A.U. Jun 2009]

Electrical Conductivity:

On the basis of free electron theory of solids, the electrical conductivity (𝜍) of

metal is given by

𝝈 = 𝒏 𝒆 𝝁

(1)

Using the equation (1), the electrical conductivity (𝜍) of a semiconductor due to

electrons in the conduction band is given by

𝝈𝒆 = 𝒏𝒆 𝝁𝒆

(2)

Similarly, the electrical conductivity of a semiconductor due to holes in the

valence band is given by

𝝈𝒉 = 𝒑𝒆 𝝁𝒉

(3)

Therefore, the total electrical conductivity 𝜍𝑖of the semiconductor is the sum of

electrical conductivities due to electrons and holes.

i.e. 𝝈𝒊 = 𝝈𝒆 + 𝝈𝒉 (4)



Substituting the equations (2) and (3) in the equation (4),

We have

𝜍𝑖 = 𝑛𝑒𝜇𝑒 + 𝑝𝑒𝜇

Or 𝝈𝒊 = 𝒆(𝒏𝝁𝒆 + 𝒑𝝁𝒉) (5)

For an intrinsic semiconductor, the number of electrons in the conduction band is

equal to the number of holes in the valence band,

i.e., n = p = 𝑛𝑖

where 𝑛𝑖 is intrinsic carrier concentration.

Hence, the electrical conductivity of an intrinsic semiconductor is given by

𝜍𝑖 = 𝑒 𝑛𝑖𝜇𝑒 + 𝑛𝑖𝜇

Or 𝝈𝒊 = 𝒆𝒏𝒊 𝝁𝒆 + 𝝁𝒉 (6)

Substituting the expression for 𝑛𝑖 in eqn (6), we get

𝝈𝒊 = 𝟐𝒆 𝟐𝝅𝒌𝑻

𝒉𝟐 𝟑/𝟐

𝒎𝒆∗𝒎𝒉

∗ 𝟑/𝟒

𝒆 −𝑬𝒈/𝟐𝒌𝑻 𝝁𝒆 + 𝝁𝒉 (7)

Variation of electrical conductivity with temperature

From the expression (7), we have

𝝈𝒊 = A𝒆 −𝑬𝒈/𝟐𝒌𝑻 (8)

WhereA =2𝑒 2𝜋𝑘𝑇

2

3/2 𝑚𝑒

∗𝑚∗

3/4 𝜇𝑒 + 𝜇

Taking log on both sides of equation (8),



logeσi = logeAe−Eg /2kT

logeσi = logeA+logee−Eg /2kT

logσi = log A − Eg

2kT (9)

A typical graph between 1

𝑇 and log 𝜍𝑖 is as shown in fig. From the graph, it is

noted that the electrical conductivity increases with temperature.

6. Describe a method of determining the band gap of a semiconductor.

[CO2-L2-A.U. Jun 2010]

Electrical conductivity of an intrinsic semi conductor is given by

𝝈𝒊 = A𝒆 −𝑬𝒈/𝟐𝒌𝑻

(1)

Where A =2𝑒 2𝜋𝑘𝑇

2

3/2 𝑚𝑒

∗𝑚∗

3/4 𝜇𝑒 + 𝜇

Resistivity of an intrinsic semi conductor,

𝜌𝑖 = 1

𝜍𝑖=

1

𝐀𝒆 −𝑬𝒈/𝟐𝒌𝑻 =

𝒆 𝑬𝒈/𝟐𝒌𝑻

𝐴

𝑅𝑖𝑎

𝑙=

𝒆 𝑬𝒈/𝟐𝒌𝑻

𝐴

Where 𝜌𝑖 =𝑅𝑖𝑎

𝑙

𝑅𝑖 = resistance, a = area of cross section, l = length.



𝑅𝑖 =𝑙𝒆 𝑬𝒈/𝟐𝒌𝑻

𝑎𝐴

𝑅𝑖 = 𝐶𝒆 𝑬𝒈/𝟐𝒌𝑻

Where C = 𝑙

𝑎𝐴

Taking log an both sides of the above equation

𝑙𝑜𝑔𝑒𝑅𝑖 = 𝑙𝑜𝑔𝑒(𝐶𝒆 𝑬𝒈𝟐𝒌𝑻

)

𝑙𝑜𝑔𝑒𝑅𝑖 = 𝑙𝑜𝑔𝑒𝐶 + 𝑙𝑜𝑔𝑒𝒆 𝑬𝒈𝟐𝒌𝑻

𝑙𝑜𝑔𝑒𝑅𝑖 = 𝑙𝑜𝑔𝑒𝐶 + Eg

2kT

The above equation is similar to the equation of a straight line, y = mx+c.

y= 𝑙𝑜𝑔𝑒𝑅

x=1

T

m= Eg

2k

c=𝑙𝑜𝑔𝑒𝐶

from the equation , we understand that by measuring the resistance of an intrinsic semi

conductor at different temperatures, we can determine the forbidden energy gap 𝑬𝒈

If a graph is drawn between log R and1

T, the value of 𝑬𝒈 is determined from the slope of

a straight line.

Slope = Eg

2k



Eg = 2k × slope

From the graph, slope = 𝑑𝑦

𝑑𝑥

Substituting the above expression we get

Eg = 2k ×𝑑𝑦

𝑑𝑥

Where k is the Boltzmann’s constant.

Experimental determination of band gap:

A semiconductor specimen is connected in series with a battery, rheostat and an

ammeter. It is immersed in an oil bath. A voltmeter is connected in parallel to the

specimen. The temperature is noted.

A known potential difference V is applied across the specimen and current is

noted. The potential difference is kept constant and the experiment is repeated at

different temperatures (by heating the oil) and corresponding currents are noted.

Then , a graph is drawn between log R and 1

𝑇. The slope of the straight line is

determined. The energy gap is determined by using the relation.

𝐸𝑔 = 2𝑘 × 𝑠𝑙𝑜𝑝𝑒



7. Obtain an expression for density of electrons in the conduction band of an n –

type semiconductor.[CO2-L2-A.U. Jun 2010]

The energy band diagram of an n-type semiconductor is shown in figure. In the

n-type semiconductor, the donor level is just below the conduction band. Nd denotes

donor concentration, i.e., the number of donor atoms per unit volume of the material and

Ed represents the energy of the donor level.

Density of electrons per unit volume in the conduction band is given by

n = 2 𝟐𝝅𝒎𝒆

∗𝒌𝑻

𝒉𝟐 𝟑/𝟐

𝒆 𝑬𝑭−𝑬𝑪 /𝒌𝑻

(1)

EC – Energy corresponding to the bottom most level of conduction band.



Fig. Energy band diagram of an-type semiconductor

At equilibrium, the density of electrons in the conduction band is equal to the density of

ionised donors.

Density of ionised donors = Nd 1 − F Ed

= Nd 1 −1

1+𝐞 𝐄𝐝−𝐄𝐅 /𝐤𝐓

= Nd 1+ 𝐞 𝐄𝐝−𝐄𝐅 /𝐤𝐓−𝟏


= Nd 𝐞

𝐄𝐝−𝐄𝐅 /𝐤𝐓

1+𝐞 𝐄𝐝−𝐄𝐅 /𝐤𝐓 =

Nd


𝐞 𝐄𝐝−𝐄𝐅 /𝐤𝐓

= Nd

1

e E d−E F /kT+1



= Nd

e−(E d−E F )/kT +1

Density of ionised donors = 𝑁𝑑

1+𝒆 𝑬𝑭−𝑬𝒅 /𝒌𝑻 (2)

Equating (1) and (2), we have

2 𝟐𝛑𝐦𝐞

∗𝐤𝐓

𝐡𝟐

3/2𝐞 𝐄𝐅−𝐄𝐂 /𝐤𝐓 =

Nd

1+𝐞 𝐄𝐅−𝐄𝐝 /𝐤𝐓 (3)

𝐞 𝐄𝐅−𝐄𝐝 /𝐤𝐓is very large when compared to 1. Hence 1 + 𝒆 𝑬𝑭−𝑬𝒅 /𝒌𝑻

𝒆 𝑬𝑭−𝑬𝒅 /𝒌𝑻

Hence, the eqn (3) becomes

2 𝟐𝛑𝐦𝐞

∗𝐤𝐓

𝐡𝟐

3/2𝐞 𝐄𝐅−𝐄𝐂 /𝐤𝐓 =

Nd

𝐞 𝐄𝐅−𝐄𝐝 /𝐤𝐓

or 2 𝟐𝛑𝐦𝐞

∗𝐤𝐓

𝐡𝟐

3/2𝐞 𝐄𝐅−𝐄𝐂 /𝐤𝐓 = Nd𝐞

𝐄𝐝−𝐄𝐅 /𝐤𝐓 (4)

Taking log on both sides, we get

loge 2 2πme

∗kT

h2

3/2 + loge𝐞

𝐄𝐅−𝐄𝐂 /𝐤𝐓 = logeNd𝐞 𝐄𝐝−𝐄𝐅 /𝐤𝐓

𝑙𝑜𝑔𝑒 2 𝟐𝝅𝒎𝒆

∗𝒌𝑻

𝒉𝟐

3/2 +

𝐸𝐹−𝐸𝐶

𝑘𝑇= 𝑙𝑜𝑔𝑒𝑁𝑑 +

𝐸𝑑−𝐸𝐹

𝑘𝑇 (5)

(

Rearranging, we have



EF−EC−Ed +EF

kT = logeNd − log 2

𝟐𝛑𝐦𝐞∗𝐤𝐓

𝐡𝟐

3/2

2EF−Ed−EC

kT= loge

Nd

2 𝟐𝛑𝐦𝐞

∗𝐤𝐓

𝐡𝟐 𝟑/𝟐

Or 2EF = Ed + EC + kT loge Nd

2 𝟐𝛑𝐦𝐞

∗𝐤𝐓

𝐡𝟐 𝟑/𝟐

EF = Ed +EC

2+

kT

2loge

Nd

2 𝟐𝛑𝐦𝐞

∗𝐤𝐓

𝐡𝟐 𝟑/𝟐 (6)

To find the number of free electrons in the conduction band of an n- type semi

conductor the above 𝐸𝐹 is substituted in the equation (1).

n = 2 2𝜋𝑚𝑒

∗𝑘𝑇

2

3/2 𝑒𝑥𝑝

𝐸𝑑+𝐸𝐶

2+𝑘𝑇

2𝑙𝑜𝑔𝑒

𝑁𝑑

2 2𝜋𝑚𝑒

∗𝑘𝑇

2

3/2 −𝐸𝐶

𝑘𝑇

(7)

n = 2 2𝜋𝑚𝑒

∗𝑘𝑇

2

3/2 𝑒𝑥𝑝

𝐸𝐶+𝐸𝑑−2𝐸𝐶

2𝑘𝑇+

1

2𝑙𝑜𝑔𝑒

𝑁𝑑

2 2𝜋𝑚𝑒

∗𝑘𝑇

2 3/2

n = 2 2𝜋𝑚𝑒

∗𝑘𝑇

2

3/2 𝑒𝑥𝑝

𝐸𝑑−𝐸𝐶

2𝑘𝑇+ 𝑙𝑜𝑔𝑒

𝑁𝑑1/2

2 2𝜋𝑚𝑒

∗𝑘𝑇

2

3/2 1/2



n = 2 2𝜋𝑚𝑒

∗𝑘𝑇

2

3/2 𝑁𝑑

2

1/2

2𝜋𝑚𝑒

∗𝑘𝑇

2

3/4 𝑒 𝐸𝑑−𝐸𝐶 /2𝑘𝑇 (8)

Rearranging the expression (8), we get

n = 21/2𝑁𝑑1/2

2𝜋𝑚𝑒

∗𝑘𝑇

2

3/4𝑒 𝐸𝑑−𝐸𝐶 /2𝑘𝑇 (9)

This gives the number of electrons present in the conduction band of an n – type semi

conductor.

n = 2𝑁𝑑 1/2

2𝜋𝑚𝑒∗𝑘𝑇

2

3/4𝑒−∆𝐸/2𝑘𝑇 (10)

Where ∆𝐸 = 𝐸𝐶 − 𝐸𝑑 is the ionization energy of the donor atoms.

VARIATION OF FERMI LEVEL WITH TEMPERATURE AND IMPURITY

CONCENTRATION

The expression for the Fermi level in n – type semi conductor is given by

𝐸𝐹 = 𝐸𝑑 + 𝐸𝐶

2+𝑘𝑇

2𝑙𝑜𝑔𝑒

𝑁𝑑

2 𝟐𝝅𝒎𝒆∗𝒌𝑻

𝒉𝟐 𝟑𝟐

At T = 0 K, the above expression reduces to

𝐸𝐹 = 𝐸𝑑+𝐸𝐶

2



At 0 K, the Fermi level lies exactly in the middle of the donor level (Ed) and bottom of the

conduction band.

As the temperature increases more and more donor atoms are ionized and Fermi

level EF shifts down wards.

At very high temperature Fermi level EF shifts to intrinsic Fermi level and n – type

semi conductor behaves like an intrinsic semi conductor.

When the concentration of donor atoms increases, extrinsic behaviour increases

and Fermi level reaches the intrinsic Fermi level only at very high temperature.



8. Obtain an expression for density of holes in the valence band of p – type

extrinsic semiconductor. [CO2-L2-A.U. Jun 2009]

(OR)

Concentration of Holes in the Valence Band of P– Type Semiconductor

[Derivation]

In P−𝑡𝑦𝑝𝑒 semiconductor, acceptor energy level is just above the valence band

(fig.). 𝐸𝑎 represents the energy of the acceptor level and 𝑁𝑎 denotes the number of

acceptor atoms per unit volume.

Density of holes per unit volume in the valence band is given by

𝒑 = 𝟐 𝟐𝝅𝒎𝒉

∗𝒌𝑻

𝒉𝟐 𝟑/𝟐

𝒆 𝑬𝒗−𝑬𝑭 /𝒌𝑻 (1)

At equilibrium,

Density of holes = density of

In valence band ionised acceptor (2)

Density of ionised acceptors = 𝑁𝑎𝐹(𝐸𝑎 )



= 𝑁𝑎

1+𝒆 𝑬𝒂−𝑬𝑭 /𝒌𝑻 …… (3)

Since 𝐸𝑎 − 𝐸𝐹 is very large when compared to kT, 𝑒 𝐸𝑎−𝐸𝐹 /𝑘𝑇 is a large quantity

and thus ‘1’ from the denominator of R.H.S. of the equation (3) is neglected.

Now, the eqn (3) is modified as,

𝑁𝑎𝐹(𝐸𝑎 ) = 𝑁𝑎

𝒆 𝑬𝒂−𝑬𝑭 /𝒌𝑻

Density of ionized acceptors = 𝑁𝑎𝒆 𝑬𝑭−𝑬𝒂 /𝒌𝑻 (4)

Substituting(4)and(1) in(2)

𝟐 𝟐𝝅𝒎𝒉

∗𝒌𝑻

𝒉𝟐 𝟑/𝟐

𝒆 𝑬𝒗−𝑬𝑭 /𝒌𝑻 = 𝑁𝑎𝒆 𝑬𝒂−𝑬𝑭 /𝒌𝑻 (5)

Taking log on both sides of the equation (5), we have

𝑙𝑜𝑔𝑒 2 𝟐𝝅𝒎𝒉

∗𝒌𝑻

𝒉𝟐

3/2 𝒆 𝑬𝒗−𝑬𝑭 /𝒌𝑻 = 𝒍𝒐𝒈𝒆𝑁𝑎𝒆

𝑬𝒂−𝑬𝑭 /𝒌𝑻 (6)

𝑙𝑜𝑔𝑒 2 𝟐𝝅𝒎𝒉

∗𝒌𝑻

𝒉𝟐

3/2 +

𝐸𝑣−𝐸𝐹

𝑘𝑇= 𝒍𝒐𝒈𝒆𝑁𝑎 +

𝑬𝑭−𝑬𝒂

𝒌𝑻

Rearranging the expression (6), we have

𝑬𝑭−𝑬𝒂−𝐸𝑣+𝐸𝐹

𝒌𝑻 = − 𝒍𝒐𝒈𝒆𝑁𝑎+𝒍𝒐𝒈𝒆 2

𝟐𝝅𝒎𝒉∗𝒌𝑻

𝒉𝟐

3/2

Or 𝟐𝑬𝑭−(𝑬𝒂−𝐸𝑣)

𝒌𝑻= − 𝒍𝒐𝒈𝒆

𝑵𝒂

2 𝟐𝝅𝒎𝒉

∗ 𝒌𝑻

𝒉𝟐

3/2



Or 𝟐𝑬𝑭 − 𝑬𝒂 − 𝐸𝑣 = −𝑘𝑇𝒍𝒐𝒈𝒆 𝑵𝒂

2 𝟐𝝅𝒎𝒉

∗ 𝒌𝑻

𝒉𝟐 3/2

(7)

𝑬𝑭 =𝑬𝒂+𝐸𝑣

𝟐−

𝒌𝑻

𝟐𝒍𝒐𝒈𝒆

𝑵𝒂

2 𝟐𝝅𝒎𝒉

∗ 𝒌𝑻

𝒉𝟐 3/2

(8)

Substituting the expression of EF from (8) in (1) we get


∗𝒌𝑻

𝒉𝟐 𝟑/𝟐

𝒆𝒙𝒑

𝑬𝒗−

𝑬𝒗+𝑬𝒂𝟐

+𝒌𝑻


𝑵𝒂

2 𝟐𝝅𝒎𝒉

∗𝒌𝑻

𝒉𝟐 3/2

𝒌𝑻


∗𝒌𝑻

𝒉𝟐 𝟑/𝟐

𝒆𝒙𝒑

𝟐𝑬𝒗−𝑬𝒗−𝑬𝒂

𝟐+𝟏


𝑵𝒂

2 𝟐𝝅𝒎𝒉

∗𝒌𝑻

𝒉𝟐 3/2

𝒌𝑻

(9)


∗𝒌𝑻

𝒉𝟐 𝟑/𝟐

𝑵𝒂𝟐 𝟏/𝟐

𝟐𝝅𝒎𝒉

∗ 𝒌𝑻

𝒉𝟐 𝟑/𝟒

𝒆(𝑬𝒗−𝑬𝒂)/𝟐𝒌𝑻

p = 𝟐𝑵𝒂 𝟏/𝟐

𝟐𝝅𝒎𝒉

∗𝒌𝑻

𝒉𝟐 𝟑/𝟒

𝒆(𝑬𝒗−𝑬𝒂)/𝟐𝒌𝑻 (10)



p = 𝟐𝑵𝒂 𝟏/𝟐

𝟐𝝅𝒎𝒉

∗𝒌𝑻

𝒉𝟐 𝟑/𝟒

𝒆−∆𝑬/𝟐𝒌𝑻 (11)

Where ∆𝑬 = 𝑬𝒂 − 𝑬𝒗.

9.Explain with a sketch the variation of Fermi level and carrier concentration with

temperature with the p-type semiconductor. [CO2-L2-A.U. Dec. 2008]

(OR)

Variation of Fermi Level with Temperature and Impurity Concentration in p-type

Semiconductor

The expression for Fermi level in a p-type semiconductor is given by


𝟐−

𝒌𝑻


𝑵𝒂

2 𝟐𝝅𝒎𝒉

∗ 𝒌𝑻

𝒉𝟐 3/2

(1)

Substituting T = 0 K in the above expression (1), we have


𝟐

(2)

Referring figure, we get the following results.

At 0 K, Fermi level lies exactly halfway between acceptor level Ea and top of the

valence band 𝐸𝑣 .

As temperature increases, more and more acceptor atoms are ionised and Fermi

level EF (P-type) shifts upwards. At a particular temperature, when all the

acceptor atoms are ionised, Fermi level crosses the acceptor level.



At very high temperature, Fermi level shifts to intrinsic Fermi level and p-type

semiconductor behaves like an intrinsic semiconductor.

Further when the concentration of acceptors increases, extrinsic behaviour

increases and Fermi level reaches the intrinsic Fermi level only at a very high

temperature.

10.What is Hall Effect? Derive an expression of Hall coefficient. Describe an

experimental set up for the measurement of Hall coefficient. [CO2-L2-A.U.Jun

2011/2012.Dec 2012,jun 2013/2014.May 2015]

Statement:

When a conductor (metal or semiconductor) carrying a current (I) is placed

perpendicular to a magnetic field (B), a potential difference (electric field) is developed

inside the conductor in a direction perpendicular to both current and magnetic field.

This phenomenon is known as Hall effect.



Hall effect in a n-type semiconductor:

Let us consider a n-type semiconducting material in the form of rectangular slab.

In such a material current flows in X-direction and magnetic field B is applied in Z-

direction. As a result, Hall voltage is developed along Y-direction as shown in fig.

The current flow is entirely due to the flow of electrons moving from right to left

along X-direction as shown in fig. When a magnetic field (B) is applied in Z-direction,

the electrons moving with velocity v will experience a downward force.

Downward force experienced by the electrons = Bev (1)

Where e is the charge of an electron

This downward force deflects the electrons in downward direction and therefore, there is

an accumulation of negatively charged electrons on the bottom face of the slab as

shown in fig.



This causes the bottom face to be more negative with respect to the top face.

Therefore, a potential difference is developed between top and bottom of the

specimen. This potential difference causes an electric field EH called Hall field in

negative Y – direction. This electric field develops a force which is acting in the upward

direction on each electron.

Upward force acting on each electron = eEH (2)

At equilibrium, the downward force Bev will balance the upward force eEH

∴ Bev = eEH

Or EH = Bv (3)

The current density (𝐽𝑥) acting along the X – direction is related to the velocity v

as

𝐽𝑥 = −𝑛𝑒𝑣

Where n is the concentration of current carriers (electrons).

v = −Jx

ne (4)




EH =− B Jx

ne (5)

𝐸𝐻 = 𝑅𝐻𝐽𝑥 𝐵 . (6)

Or RH = EH

Jx B

Where 𝑅𝐻 = − 1

𝑛𝑒(for electrons)

𝑅𝐻 is a constant and it is known as Hall coefficient.

Hall coefficient interms of Hall voltage

If t is the thickness of the sample and VH the voltage developed, then

VH = EH t (7)

Where EH is Hall field


𝑉𝐻 = 𝑅𝐻𝐽𝑥 𝐵 𝑡 (8)

If b is the breadth of the sample, then

Area of the sample (A) = Breadth (b) × 𝑇𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑡

Current density, 𝐽𝑥 = 𝐼𝑥

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 (𝐴) =

𝐼𝑥

𝑏𝑡 (9)




𝑉𝐻 = 𝑅𝐻𝐼𝑥𝐵 𝑡

𝑏𝑡

Or 𝑉𝐻 = 𝑅𝐻 𝐼𝑥 𝐵

𝑏

Hall coefficient 𝑹𝑯 = 𝑽𝑯𝒃

𝑰𝒙 𝑩 (10)

Determination of Hall Coefficient

The experimental set up for the measurement of Hall-coefficient is shown in fig.

A semiconducting material is taken in the form of a rectangular slab of thickness t and breadth b.



A suitable current 𝐼𝑥 ampere is passed through this sample along X-axis by

connecting it to a battery.

This sample is placed in between two poles of an electromagnet such that the magnetic field is applied along Z-axis, i.e., perpendicular to the plane of paper. Due to Hall effect, Hall voltage (VH) is developed in the sample. This voltage is measured by fixing two probes at the centres of the bottom and top faces of the sample. By measuring Hall voltage, Hall coefficient is determined from the formula.

𝑅𝐻 =𝑉𝐻 𝑏

𝐼𝑥 𝐵

From Hall coefficient, carrier concentration and mobility can be determined.

10. Describe applications of Hall Effect.[ CO2-L1- May 2009]

(i)Determination of semiconductor type

For a n-type semiconductor, Hall coefficient is negative whereas for a p-type

semiconductor it is positive. Thus, the sign of the Hall coefficient is used to determine

whether a given semiconductor is n-type or p-type.

(ii)Calculation of carrier concentration

By measuring Hall coefficient RH, carrier concentration is obtained from the

relation

𝑛 = 1

𝑒 𝑅𝐻

(iii)Determination of mobility

We know that electrical conductivity, 𝜍𝑒 = 𝑒𝑛 𝜇𝑒

∴ 𝜇𝑒 = 𝜍𝑒

𝑛𝑒

𝜇𝑒 = 𝜍𝑒𝑅𝐻

Thus, by measuring electrical conductivity and Hall coefficient of a sample, the mobility

of charge carries can be calculated.



(iv) Magnetic field meter

The Hall voltage VH for a given current is proportional to B. Hence, VH measures

the magnetic field B.



UNIT-III MAGNETIC MATERIAL AND SUPERCONDUCTORS

Part - A

1. Define magnetic susceptibility (χ). [CO3-L1- Dec.2011]

Magnetic susceptibility (χ) of a material is defined as the ratio of intensity of magnetization induced in the material to the magnetic field strength in which material is placed.

𝜒 =𝐼

H

2. Define magnetic permeability (μ). [CO3-L1-Dec 2013]

Magnetic permeability (μ) is defined as the ratio magnetic flux density (B) in the sample to the applied field intensity (H).

μ =𝐵

H

3. What is D.C. Josephson Effect?[CO3-L1-Dec 2012]

The tunneling of super conducting electrons pairs through Josephson junction leads to the flow of current without a voltage drop. This phenomenon is known as DC Josephson effect.This current persists for a longer time.

4. What is A.C. Josephson Effect? [CO3-L1-Dec 2012]

Consider a dc voltage, V applied to a Josephson device, and then there will be a flow of ac voltage through the device. This phenomenon is known as AC Josephson effect. This current persists for a short time.

5. What is Bohr Magneton? [CO3-L1-A.U. May2009, Dec 2012]

The orbital magnetic moment and the spin magnetic moment of an electron in an atom can be expressed in terms of atomic unit of magnetic moment called Bhor Magneton.

1 Bhor Magneton = 𝑒ħ

2𝑚 = 9.27 × 10-24 Am2.



6. What is curie Weiss law? [CO3-L1- May 2011]

Curie Weiss law is given by

𝜒 = 𝐶

𝑇 − 𝜃

Where C is curie constant, T is absolute temperature and θ is curie temperature. It determines the susceptibility of the magnetic material interms of temperature. If the temperature is less than curie temperature a paramagnetic material becomes dia magnetic and if the temperature is greater than curie temperature a ferro magnetic material becomes para magnetic material.

7. Define cooper pairs?[CO3-L1- May 2013]

Attractive interaction between two electrons mediated by means of phonon exchange dominates the usual repulsive interaction. A pair of such electrons having opposite spins which interact attractively in the phonon field are called cooper pair

8. Define energy product of a magnetic material?[CO3-L1-May 2009/2010]

The product of rentivity and corecivity is known as energy product. It represents the maximum amount of energy stored in the specimen.

9. Define anti ferro magnetism. Mention two materials that exhibit anti ferro magnetism? [CO3-L1-May 2014] In anti ferro magnetism, the electron spin of neighbouring atoms are aligned in

anti parallel direction. Anti ferro magnetic susceptibility is small and positive and

it depends on temperature. Example: MnO, MnS, Cr2O3

10. Define superconductivity.[CO3-L1- May 2009/2010,Dec 2012]

The phenomenon of losing the resistivity absolutely to zero when cooled to sufficiently low temperature is known as superconductivity.

11. Define critical temperature and critical magnetic field.[CO3-L1-Jan2012]

The temperature at which a normal conductor loses its resistivity and becomes a superconductor is known as critical temperature.

The minimum field required to destroy superconducting property and become a normal conductor is known as critical magnetic field.



12. Explain Meissner Effect. [ CO3-L3-May 2010]

When a superconducting material kept in a external magnetic field which is less than the critical field and if the material is cooled below critical temperature, then the magnetic lines of force are ejected from the material. This behavior was first discovered by Meissner and hence this property is known as Meissner effect.

13. Prove that superconductor is a perfect diamagnetic. [CO3-L1-Dec 2009] The magnetic induction B is given by the relation

B = μ0(H+I) Where H is the applied magnetic field and I is the intensity of magnetization For a super conductor, B =0

Thus, 0 = μ0(H+I) since, μ0≠0

therefore H + I = 0 I = -H

𝐼

𝐻= −1

χ=-1 i.e., when B = 0 susceptibility is negative. So super conductor is a perfect

diamagnetic.

14. What are coercivity and retentivity of magnetic materials?[CO3-L1-Dec 2010] Coercivity: The amount of magnetizing field applied in the reverse direction to remove the residual magnetism completely from the material. Retentivity:

The amount of magnetic induction retained in the material after removing the magnetising field.



15. Define hysteresis. [CO3-L1- 2013,Dec 2009]

The intensity of magnetization lagging behind the applied magnetic field is called Hysteresis.

16. What do you understand by the term “magnetic domains” and “domain

walls”? [CO3-L1- Dec 2009,Jan 2012]

Due to the phenomenon of hysteresis, a ferromagnetic material is divided into a

large number of small regions called domains. The walls of small regions or

domains are called domain walls.

17. Mention any properties changes that occur in superconductor[ CO3-L1-

May2012]

They exhibit perfect diamagnetism.

Electrical resistance of superconductor is very less.

When the superconducting materials are subjected to a large value of

magnetic field, it will result in destruction of superconducting property.

The critical temperature superconductor varies with isotropic mass.

18. What are high temperature superconductors? Give example. [CO3-L1-

May2011] Any superconductor, if transition temperature is above 30K is called as high temperature superconductors (Tc).

Eg. YBa2Cu3O7 Tc=97K.



PART-B 1. Briefly explain different types of magnetic materials and their properties?

[ CO3-L2- May2011]

SI.NO DIA MAGNETIC

MATERIALS PARA MAGNETIC

MATERIALS FERRO MAGNETIC

MATERIALS

1

Definitions

It is material in which no permanent magnetic moment

It is material in which there is permanent magnetic moment

It is material in which there is enormous permanent magnetic moment

2

Magnetization

If we apply a magnetic field the magnetic moments are created and aligned opposite to the applied field direction. The net magnetization is zero.

If we apply a magnetic field all the magnetic moments are aligned along the applied field direction. The net magnetization is high.

If we apply a small magnetic field all the magnetic moments aligned along the applied field direction. The net magnetization is very high.

3

Behavior of the material in the presence of magnetic field

The magnetic flux lines are expelled from the material.

The magnetic flux lines are attracted towards the material.

The magnetic flux lines are highly attracted towards the material.



4

Spin alignment

No spin magnetic moment

All spin magnetic moments are randomly oriented

All spin magnetic moments are orderly oriented

5

Susceptibility The magnetic susceptibility is negative (χ = -ve)

The magnetic susceptibility is positive and small (χ = +ve)

The magnetic susceptibility is positive and large (χ = +ve)

6 Permeability Permeability is less

than one μ < 1 Permeability is greater than one μ >1

Permeability is very much greater than one μ >>1

7

Temperature dependence

It is independent of temperature

It is inversely proportional to absolute temperature of the material

It dependence upon the temperature



8

Magnetic phase transition

When the temperature is greater than critical temperature the dia magnetic material becomes normal material

When the temperature is less than curie temperature para magnetic material becomes dia magnetic material

When the temperature is greater than curie temperature the ferro magnetic material becomes para magnetic material

9

Behavior in a varying magnetic field

When placed in a varying magnetic field they have a tendency to move from stronger part to weaker part of the field.

When placed in a varying magnetic field they have a tendency to move from weaker part to stronger part of the field.

When placed in a varying magnetic field they have a tendency to move from weaker part to stronger part of the field.

10 Example Water, Hydrogen, Ag,

Gold Platinum, Aluminium, Chromium

Iron, Nickel, Cobalt



2.Explain the domain structure in ferro magnetic material. [CO3-L1-

May2009,2012]

The magnetic moment are aligned in parallel over a small region in the

ferromagnetic material called domain. A Ferromagnetic material consists of a large

number of small regions in which magnetic moments are aligned in parallel called

domains. These domains are spontaneously magnetized, due to the parallel

alignment of all magnetic dipoles Fig.3.6 (a). But in each domain, the magnetization

is different. But the resultant magnetization of a ferromagnetic material will be nearly

zero.

When an external magnetic field is applied to a material like iron (Fe) there are

two possible ways of magnetization.

1. By the motion of domain walls

2. By the rotation of domains.

Fig. 3.6 (a) Domains, 3.6 (b) Domain wall movement, 3.6 (c) Domain rotation

1. Magnetization by the motion of Domain walls:

When a weak external magnetic field is applied to the specimen of a

ferromagnetic material, the magnetic moment increases and the dipoles are oriented

unfavorably. Thus the volume of the domain changes and the boundaries of the

domain are displaced fig. 3.6 (b).



2. Magnetization by the rotation of domains:

When a strong magnetic field is applied to the specimen of a ferromagnetic

material, the magnetic moment tends to change their direction towards the applied field

direction. Then due to this alignment, rotation of domains occurs Fig. 3.6. (c).

When an external magnetic field is applied to it, due to the increase in magnetic

moment, the volume of the domains changes and the domain walls start moving. This

can be viewed through the microscope.

In this domain growth, four types of energy are involved.

1. Exchange energy

2. Anisotropy energy

3. Domain wall energy or Bloch wall energy.

4. Magnetostrictive energy.

Exchange Energy:

The energy required in assembling the atomic magnets (or) dipoles into a single

domain in a ferromagnetic material is called exchange energy.

This energy is stored as potential energy. It depends upon the inter atomic

distance. It arises due to the interaction of electron spins. This interaction energy

makes the adjacent dipoles to align themselves. The volume of the domain may vary

between 10−2 𝑡𝑜 10−6𝑐𝑚3

Anisotropy energy:

In Ferro magnetic materials / crystals, the energy of magnetization is found to be

a function of crystal orientation.

There are two magnetization directions.

1. Easy direction of magnetization for which weak magnetic field can be applied.

2. Hard direction of magnetization for which strong magnetic field can be

applied.

The excess energy required to magnetize a specimen in a hard direction than the

energy required to magnetize it along the easy direction is called crystalline anisotropy

energy.



Example: Hard and intermediate directions in iron, and nickel:

Fe Ni

BCC FCC

Easy (1 0 0) (1 1 1)

Intermediate (1 1 0) (1 1 0)

Hard (1 1 1) (1 0 0)

Domain wall energy:

Two adjacent domains magnetized in different directions, are separated by a thin

region called domain wall. The thickness of the Block walls is about 200 to 300 lattice

constant (a), the electron spin changes gradually from one domain to another and it

leads to thick wall. Here the energy required for this gradual spin change is called

domain wall energy.

The rotation of magnetic moments through a 180 domain wall

Magnetostrictive Energy:

When a ferromagnetic material placed in a magnetic field, the dimension of the

specimen changes.This is known as magnetostriction. The change in dimension of the

specimen is different in different directions. Also change in dimension is different for

different material. When we applied a magnetic field on a ferro magnetic material the

domains are magnetized along the fields either by expand or shrink i.e a work must be

done against the elastic restoring forces. The work done by the magnetic fields against

these elastic restoring force is called magnetostrictive energy.



3.Explain the hysteresis on the basis of domain theory of ferro magnetism. [CO3-

L1- Dec 2009]

A Ferro magnetic material is placed in a magnetic field ‘H’. Magnetic induction

‘B’ varies with respect to the variation in magnetic field ‘H.”

This variation can be represented by a closed curve called Hysteresis. The

Hysteresis of ferromagnetic materials refers to the lagging of magnetization behind the

applied field. Consider a Ferro magnetic material and magnetic field ‘H’ applied to it.

Fig. 3.8 Hysteresis curve

If the value of ‘H’ is increased, the magnetic induction‘B’ also increase as shown

in the figure as oa.

If ‘H’ is decreased from its maximum value to zero, then ‘B’ will not fall rapidly to zero,

but falls to ‘b’. Thus it possesses the property called retentivity.

This residual magnetism can be removed by reversing the applied field in the

specimen as shown by the value ‘oc’ in the curve. This property is called as coercivity.

Further increased of magnetic field the magnetic induction increases along cd in

the reverse direction as shown in the graph.

The loss of energy in taking a ferromagnetic specimen through a complete cycle

of magnetization is called Hysteresis loss and the enclosed area is called as hysteresis

loop.



Retentivity or Residual magnetism:

Retentivity or residual magnetism is the amount of magnetic imduction retained

in the material after removing the magnetic field.

Coercivity:

Coercivity is the amount of magnetizing field applied in the reverse direction to

remove the residual magnetism from the material.

Explanation of hystresis on the basis of domain theory:

When a ferro magnetic material is subjected to external magnetic field there is an

increase in the value of the resultant magnetic moment due to

i. Motion of domain walls

ii. Rotation of domain walls

When a small magnetic field is applied domain walls are displaced in the direction of

field and the magnetization increases. It represents the curve OA in the fig. when the

applied field is removed the domains return to the original state and

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