Department of Physics Question Bank Engineering Physics … ·  · 2016-08-29Department of Physics Question Bank Engineering Physics II – PH6251 UNIT-I – Conducting Materials

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Engineering Physics II – PH6251 UNIT-I – Conducting Materials

PART – A

1. What are the classifications of conducting materials?

Conducting materials are classified into

Zero resistive materials. Ex.Superconductors

Low resistive materials. Ex. Silver , Aluminium , Alloys

High resistive materials. Ex. Tungsten, platinum ,Nichrome

2. What are the sources of resistance in metals?

The resistance in metals is due to

Impurities present in metals

Temperature of the metal

Number of free electrons

3. Give any three postulates of classical free electron theory.

The free electrons in the metal move freely, similar to the gas molecules moving in a vessel

and it obeys the classical kinetic theory of gases.

These free electrons move in a constant potential field due to the ions fixed in the lattice

When the field is applied the free electrons move in a direction opposite to that of the field

direction

4. Define drift velocity. How is it different from thermal velocity of an electron?

It is the average velocity acquired by the free electron in a particular direction, due to the

application of electric field but thermal velocity is the velocity of an electron without any external

filed. Here the electrons move in random direction.

5. Define the terms relaxation time, collision time and Mean free path of an electron.

Relaxation time: The average time taken by a free electron to reach its equilibrium position from

its disturbed position due to application of an external electric field is called relaxation time.

Collision time: It is the average time taken by a free electron between two successive collisions.

Mean free path: The average distance travelled by a free electron between any two successive

collisions in the presence of an applied field is known as mean free path.

6. The mobility of electron in copper is 3x10-3 m2/Vs. Assuming e = 1.6x10-19C and me= 9.1x10-31

kg, calculate the Mean free time.

Equating (1) and (2) we get,

7. Differentiate between electrical conductivity and thermal conductivity.

S. No Electrical Conductivity Thermal Conductivity

1 Electrical conductivity is based on the no of free electrons

Thermal conductivity is based both on electrons and phonons

2 The quantity of electrical charges flowing per unit time across unit area in the metal for unit applied electric field is called electrical conductivity

It is defined as the amount of heat conducted per unit time through the metal having unit area of cross section maintaining at unit temperature gradient between the two ends

3 Electrical conductivity takes place from higher potential side to lower potential side

Thermal conductivity takes place from hot end to cold end.

4 unit :Ω-1 m-1 unit :W-1 m-1K-1

8. Mention the drawbacks of classical free electron theory of metals.

It is a microscopic theory

Classical theory states that all the free electrons will absorb energy, but quantum theory states

only few electrons will absorb energy.

This theory cannot explain the Compton, photo-electric effect, paramagnetism,

ferromagnetism, etc.

The theoretical and experimental values of specific heat and electronic specific heat are not

matched.

The Lorentz number by classical theory does not have good agreement with the experimental

value and it is rectified by quantum theory.

9. What are the merits of classical free electron theory of metals?

It is used to verify ohm’s law.

The electrical and thermal conductivities of metals can be explained by this theory.

It is used to derive Wiedemann-Franz law.

It is used to explain the optical properties of metals.

10. Find the drift velocity of electrons copper wire whose cross sectional area is 1 mm2. When the

wire carries a current of 10 A. Assume that each copper atom contributes one electron of the

electron gas. Given n = 8.5x1028 /m3.

Drift velocity

11. A conducting rod contains 8.5x1028 electrons per m3. Calculate the electrical conductivity at

room temperature if the collision time for electron is 2 X 10-14 s.

12. State any three assumptions of quantum free electron theory.

The potential energy of an electron is uniform or constant throughout the metal.

The electrons have wave nature.

The free electrons obey Fermi – Dirac statistics.

13. What are the merits and demerits of quantum free electron theory?

Merits:

This theory attracts the electron quantum mechanically rather than classically.

It explains the electrical conductivity, thermal conductivity, specific heat capacity of

metals, photoelectric effect and Compton Effect.

Demerits

Even though it explains most of the physical properties of the metals, it fails to state the

difference between conductor, semiconductor and insulator.

It also fails to explain the positive value of Hall coefficient and some of the transport

properties of metals.

14. Define Fermi level and Fermi energy with its importance.

Fermi level: It is the highest reference energy level of a particle at 0K.

Importance: It is the reference energy level that separates the filled energy levels and vacant

energy levels.

Fermi energy: It is the maximum energy of the quantum state corresponding to Fermi energy

level at 0K.

Importance: It determines the energy of the particle at any temperature.

15. Write down the expression for Fermi-Dirac distribution function and plot it as a function of

energy.

Fermi distribution function is given by F(E) =

16. Calculate the Fermi energy of copper at 0° K if the concentration of electrons is 8.5x1028 m-3

17. Define Density of Energy states.

Density of state Z (E) dE is defined as the number of available states present in a metal

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

18. Define Lorentz number.

The ratio of thermal conductivity (K) of a metal to the product of electrical conductivity

(σ) of a metal and absolute temperature (T) of the metal is a constant. It is called Lorentz number

and it is given by

Where L is Lorentz number whose value is 2.44 x 10-8 W Ω K-2 at 293 K.

19. The thermal conductivity of a metal is 123.92 W/m/k. Find the electrical conductivity and

Lorentz number when the metal possess relaxation time 10-14 seconds and 300 K.(Density of

electrons = 6 x 1028 /m3).

Electrical Conductivity

20. The Fermi temperature of a metal is 24600 K. Calculate the Fermi velocity.

m/s.

PART – B

1. i) Define Electrical conductivity. Derive an expression for electrical conductivity of a metal by

using classical free electron theory.

Electrical conductivity: The quantity of electric charge flows in unit time per unit area of

cross section of the conductor per unit potential gradient.

ohm–1 m–1

Expression for electrical conductivity

ii) Find the mobility of electrons in copper if there are 9x1028 valence electrons/m3 and the

conductivity of copper is 6x107 mho/m.

4.166 X

2. i) Define thermal conductivity Derive an expression for thermal conductivity of a metal.

Thermal Conductivity (K): The amount of heat flowing through an unit area of a material per unit temperature gradient.

K = - W/m/K.

The negative sign indicates that heat flows from hot end to cold end.

Expression for thermal conductivity

ii) Calculate the electrical and thermal conductivities for a metal with a relaxation time 10-14

second at 300 K. Also calculate Lorentz number using the above result. (Density of electrons

=6x1028 m-3).

Thermal conductivity

K = 123.80 W/m/K

3. Deduce mathematical expression for electrical conductivity and thermal conductivity of a

conducting material and hence obtain Wiedemann-Franz law.

Electrical conductivity:

When an electric field of strength (E) is applied to the conductor, the force experienced

by the free electrons is given by F = – e E

From Newton’s 2nd law of motion, the force acquired by the electrons can be written as

F = m a

Equating the above equations – e E = m a

Thermal conductivity:

The average kinetic energy of an electron at hot end ‘A’ of temperature (T) =

The average kinetic energy of an electron at cold end ‘B’ of temperature (T-dT) =

The net heat energy transferred from end A to B per unit area per unit time

Amount of heat transferred per unit area per unit time is given by

We know for metals relaxation time ( collision time

Therefore thermal conductivity

Wiedemann-Franz law: This law states that “the ratio between the thermal conductivity and

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

the metal”.

Proof:

T

4. State and prove Wiedemann-Franz law. Why does the Lorentz number determined

experimentally does not agree with the value calculated from the classical theory.

Wiedemann-Franz law: This law states that “the ratio between the thermal conductivity and the

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

metal”.

Proof:

T 5. Derive an expression for the density of states and based on that calculate the carrier

concentration in metals.

The Fermi function F (E) gives only the probability of filling up of electrons in a given energy

state, it does not gives the information about the number of electrons that can be filled in a

given energy state. To know that we should know the number of available energy states, so

called density of states

Density of state N (E) dE is defined as the number of available states present in a metal per

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

Z(E)dE =

Z(E)dE =

Z(E)dE =

Carrier concentration

6. i) Starting with the density of energy states obtain the expression for the Fermi energy of an

electron at 0 K and hence obtain the expression for the average energy of an electron.

Carrier concentration

Expression for Fermi energy at 0K

Expression for Mean energy at 0K

=

=

=

ii) The Fermi energy of silver is 5.51 eV. What is the average energy of a free electron at 0 K.

7. i) Define mobility.

When electric field (E) is applied to metals, the electrons move in the direction

opposite to the field direction with the velocity ‘v’, then mobility of an electron is defined as the

velocity acquired by the electron per unit electric field (E).

Unit: m2V-1s-1.

ii) What are Fermi particles or Fermions?

Fermi particles or fermions or particles with ½ integral spin values. Eg., electrons.

iii) A uniform silver wire has a resistivity of 1.54x10-8 ohm/m at room temperature. For an

electric field along the wire of 1 Volt/cm, compute the average drift velocity of electron assuming

that there is 5.8x1028 conduction electrons/m3. Also calculate the mobility.

8. 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=0K, and T > 0K.

Fermi distribution function: The probability F (E) of an electron occupying a given energy

level is given by Fermi-Dirac distribution function

At T =0K and E< Ef

At T =0K and E> Ef

At T =0K and E= Ef

At T = any K

Due to the supply of thermal energy electrons in the lower energy level absorb

the supplied energy and move to higher energy level thereby creating vacant energy levels

below the Fermi energy level.

9. i) Use the Fermi distribution function to obtain the value of F (E) for the level just 0.01eV

above the Fermi level at 200 K.

ii) Evaluate the Fermi function of energy KBT above the Fermi energy.

10. i) The density of silver is 10.5x103 kg/m3. The atomic weight of silver is 107.9. Each silver

atom provides one conduction electron. The conductivity of silver at 20°C is 6.8 x107Ω-1m-1.

Calculate the density of electron and also the mobility of electrons in silver.

ii) Calculate the electrical and thermal conductivities of a metal with the relaxation time of

10-14 second at 300 K. The electron density is 6x1026 m-3.

Thermal conductivity K = L

K = 1.241 W/m/K.

iii) Calculate the Fermi energy and Fermi temperature in a metal. The Fermi velocity of

electrons in the metal is 0.86x106 m/s.

Fermi energy

Fermi Temperature

UNIT-II – SEMICONDUCTING MATERIALS

PART A

1. Define semiconductor and mention its properties.

The material which has the electrical conductivity between conductors and insulators are

called semiconductors. It is special class of material very small in size and sensitive to heat,

light and electricity.

The resistivity lies between 10-4 to 0.5 ohm meters.

At 0K, they behave as insulators.

They empty conduction band and almost filled valence band.

The conductivity of a semiconductor increases both due to the temperature and

impurities.

They have negative temperature coefficient of resistance.

In semiconductors both the electron and holes are charge carriers and will take part in

conduction.

2. What is the band gap energy Eg ? Give its value for Germanium & Silicon.

Energy band gap (Eg) is the region of energies which are not occupied by the electrons

and is equal to the energy difference between minimum energy of conduction band (Ec) and

the maximum energy of valence band (Ev).

3. What are the differences between elemental and compound semiconductors? Give examples.

S.No Elemental semiconductors Compound semiconductors

1 They are made up of single element. They are made up of compounds.

2 They are called indirect band gap semiconductors.

They are called as direct band gap semiconductors.

3 Here heat is produced during recombination.

Here the photons are emitted during recombination.

4 Life time of charge carriers is more. Life time of charge carriers is less.

5 Current amplification is more. Current amplification is less.

6 They are used in the manufacture of

diodes and transistors, etc. They are used for making LED’s, laser diodes, IC’s, etc.

7 Examples: Ge, Si, etc. Examples: GaAs, GaP, CdS, MgO, etc.

4. Distinguish between intrinsic and extrinsic semiconductors.

5. Write an expression for electrical conductivity of an intrinsic semiconductor

Electrical conductivity 2 e (me* mh *) 3/4 e-Eg/2KT ( )

Where Intrinsic carrier concentration

Mobility of electron

Mobility of hole.

6. What are the limitations of intrinsic semiconductors?

Electrical conductivity is low. Germanium has a conductivity of 1.67 Ω-1m-1 which is nearly

107 times smaller than copper.

Electrical conductivity is a function of temperature and increases exponentially as temperature

increases.

7. What is meant by doping?

The addition of impurities to a pure semiconductor is known as doping and added impurity is

called as doping agent or dopant.

S.No Intrinsic semiconductor Extrinsic semiconductors

1 Semiconductor in a extremely pure form is called as intrinsic semiconductors.

Semiconductor which is doped with impurities is called extrinsic semiconductors.

2 Charge carriers are produced only due to thermal agitation.

Charge carriers are produced due to impurities and may also be produced due to thermal agitation.

3 Low electrical conductivity High electrical conductivity

4 Low operating temperature High operating temperature

5 At 0K, the Fermi level exactly lies between conducting band and valence band.

At 0K, the Fermi level lies closer to conducting band in “n” type and lies near valence band in “p” type semiconductor.

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

8. What are n-type and p-type semiconductors?. Give examples.

N-type semiconductor is obtained by doping a pure semiconductor with pentavalent

impurity. The majority carriers in N-type semiconductor are electrons. Ex. Germanium doped

with Phosphorous, arsenic, antimony

P-type semiconductor is obtained by doping a pure semiconductor with trivalent

impurity. The majority carriers in P-type semiconductor is holes. Ex Germanium doped with

Boron, aluminium, etc.

9. Give any four differences between n-type and p-type semiconductors.

S.No n-type semiconductors p-type semiconductors

1 n-type semiconductor is obtained by doping an intrinsic semiconductor with pentavalent impurity.

p-type semiconductor is obtained by doping an intrinsic semiconductor with trivalent impurity.

2 Here electrons are majority charge carriers and holes are minority charge carriers.

Here holes are majority charge carriers and electrons are minority charge carriers.

3 It has donar energy levels very close to conduction band.

It has acceptor energy levels very close to valence band.

4 When temperature is increased these semiconductors can easily donate an electron level to the conduction band.

When temperature is increased these semiconductors can accept an electron from valence band to acceptor energy level.

10. Why compound semiconductors are called direct band gap semiconductors? Give its

application.

During recombination the photons are emitted so they are called as direct band gap

Semiconductors.

11. What is meant by donor and acceptor level?

Donors are pentavalent impurity atoms like P, As, etc., which donates an

electron to the pure semiconductors like Ge or Si. These energy levels are donor energy levels.

Acceptors are trivalent impurity atoms like Ga, In etc., which can easily accept

an electron from the pure semiconductors like Ge or Si. These energy levels are acceptor energy

levels.

12. i) Define mobility.

Mobility is defined as the drift velocity acquired by a charge carrier per unit electric field

Strength. Its unit is m2 V-1 s-1

ii) Find the resistance of an intrinsic Ge rod 1 cm long,1 mm wide, and 0.5mm thick at 300K.

For Ge, ni= 2.5 x 1019/m3, µe= 0.39 m2V-1 S-1 and µh= 0.19 m2V-1 S-1 at 300k.

13. Draw the diagram to show the variation of Fermi level with temperature of a p-type

semiconductor for high and low doping level.

14. With increase of temperature the conductivity of semiconductor increases while that of

metals decreases. Give reasons.

At 0K semiconductors behave as insulators. When the temperature is raised or when

impurities are added, their conductivity increases. In the case of conductors when the temperature

increases, the resistivity increases and hence the conductivity decreases.

15. Define Fermi level in the case of semiconductors. Mention its position in intrinsic and

extrinsic semiconductor at 0 K.

Fermi level is located half way between the valence band and conduction band in an intrinsic

Semiconductors at 0 K. EF =

In n-type semiconductor at 0 K, Fermi level lies exactly at the middle of the donor level Ed and

the bottom of the conduction band Ec EF =

In p-type semiconductor at 0 K, Fermi level lies exactly at the middle of the acceptor level Ea and

the top of the valence band Ev EF =

16. For an intrinsic semiconductor with a band gap of 0.7 eV, determine the position of EF at

T=300 k if mh* = 6 me*.

17. Given an extrinsic semiconductor, how will you find whether it is n-type or p-type?

If Rh value is negative then it is n-type semiconductor.

If Rh value is positive then it is p-type semiconductor.

18. What is Hall Voltage?

When a current carrying conductor is placed in a transverse magnetic field, an electric

field is produced inside the conductor in a direction normal to both the current and the magnetic

field. This effect is known as Hall Effect. The generated voltage is called hall voltage.

19. The Hall Co-efficient of a specimen of a doped silicon is found to be 3.66x10-4 m-3/c.

The resistivity of specimen is 8.93x10-3 Ω m. Find the mobility and density of

charge carriers.

Charge carrier density

20. Mention the applications of Hall Effect.

It is used to find the type of semiconductor.

It is used to measure carrier concentration.

It is used to find mobility of charge carrier.

It is used to measure the magnetic flux density using a semiconductor sample of known

hall coefficient.

PART – B

1. Derive an expression for density of electrons in the conduction band, density of holes

in the valence band of an intrinsic semiconductor.

dn = n =

derivation upto n = 2 e (EF – EC )/KT

dp = p =

derivation upto p = 2 e (Ev – EF )/KT

2. Derive the intrinsic carrier concentration for intrinsic semiconductor and also calculate the Fermi level and its variation with temperature.

n = p =ni

derivation upto ni = 2 (me* mh

* ) e-Eg/2KT

Variation of Fermi level with temperature

EF = log ( ) +

EF =

3. i) Derive the Electrical Conductivity of an intrinsic semiconductor.

derivation for I = 2 e (me* mh

* ) e-Eg/2KT (µe µh )

ii) Discuss the variation of electrical conductivity with temperature.

derivation for I = A e-Eg/2KT

log i = log A -

iii) For an intrinsic GaAs, the room temperature of electrical conductivity is 10-6

(Ω m)-1. The electron and hole mobilities are 0.85 and 0.04 m2/V-s, respectively.

Calculate the intrinsic carrier concentration at room temperature.

4. Explain the method of determining the band gap of a semiconductor. Also describe

the experimental determination of Band gap.

I = A e-Eg/2KT

Derivation for log Ri = log C +

Eg = 2 k

5. Obtain an expression for the carrier concentration of electrons in the conduction band

of n-type semiconductor.

Density of electrons per unit volume is n = 2 e (EF – EC )/KT

Derivation upto n = (2Nd)1/2 e-ΔE/2KT

6. Obtain an expression for the carrier concentration of holes in the Valence band of p-

type semiconductor.

p = 2 e (Ev – EF )/KT

Derivation upto p = (2Na)1/2 e-ΔE/2KT

7. i) With neat sketches, explain how Fermi level varies with impurity concentration

and temperature in n-type semiconductor.

Explanation about the following diagram

ii) With neat sketches, explain how Fermi level varies with impurity concentration

and temperature in p-type semiconductor

Explanation about the following diagram

8. i) Explain the variation of carrier concentration with temperature and impurity in a

semiconductor.

Explanation of

i) Intrinsic range

ii) Exhaustion range

iii) Impurity range

ii) Explain the variation of electrical conductivity in extrinsic semiconductor.

Explanation about the following graph

9. ( i) Derive an expression of Hall coefficient and mobility of charge carriers. Describe

an experimental setup for the measurement of Hall coefficient.

Hall Effect statement:

derivation for Hall effect in n-type semiconductor

RH = where RH = -

derivation for Hall effect in p-type semiconductor

RH = where RH =

Mobility of charge carriers

µe = - µh = -

( ii) The Hall co-efficient of certain silicon was found to be -7.35x10-5 m3C-1 from 100 to 400

k. Determine the nature of the semiconductor. If the conductivity was found to be 200 m-1 Ω-1,

calculate the density and mobility of the charge carriers.

(i) Since hall coefficient is negative the given semiconductor is n-type.

(ii) Charge carrier density

(iii) Mobility =

10. (i) For an intrinsic semiconductor with gap width Eg = 0.7 eV, calculate the

concentration of intrinsic charge carriers at 300 k assuming that me*= mh * = m0

(rest mass of electron).

(ii) The resistivity of an intrinsic semiconductor is 4.5 ohm-m at 20°C and 2.0 ohm-

m at 32°C. Find the energy band gap in eV.

1.66 X

(iii) A sample of silicon doped with 1023 phosphorous atoms/m3. Find the Hall

voltage in a sample with thickness =100 µm, current, Ix=1mA and magnetic field

Bz=0.1 Wb/m2.(Assume electron mobility µe=0.07 m2/V.s)

Hall voltage

UNIT -3 - MAGNETIC AND SUPERCONDUCTING MATERIALS

PART – A

1. Classify magnetic materials based on their magnetic moments.

2. What is Bohr Magnetron? Write its value.

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 bohr magnetron. 1 Bohr

magnetron = µB = 9.27 x 10-24 Am2

3. What are paramagnetic materials? Give examples.

Paramagnetic materials can be defined as a type of magnetism in which the material

gets weakly magnetised in the same direction as the applied filed. The paramagnetic susceptibility

is positive and small and relative permeability values are slightly greater than one. Examples:

Platinum, chromium, aluminium, etc.

4. Define intensity of magnetization and flux density.

Intensity of magnetization: The term magnetization means the process of converting a non-

magnetic material into a magnetic material. It is defined as the magnetic moment per unit

volume. I = Unit: Weber /m2.

Flux density: It is defined as the number of magnetic lines of forces passing normally through

unit area of cross section. B = Unit: Weber/m2 or Tesla.

CLASSIFICATION OF MAGNETIC MATERIALS

NOT HAVING PERMANENT MAGNETIC MOMENT

HAVING PERMANENT MAGNETIC MOMENT

DIAMAGNETIC MATERIALS

PARA MAGNETIC MATERIALS

FERRIMAGNETIC MATERIALS

FERROMAGNETIC MATERIALS

5. A magnetic field of 2000 A/m is applied to a material which has a susceptibility of 1000.

Calculate the (i) Intensity of Magnetisation and (ii) Flux density.

Intensity of Magnetization

Flux Density

6. Define magnetic susceptibility and permeability.

Magnetic susceptibility: It is defined as the ratio between intensity of magnetisation and the

magnetic field intensity.

Magnetic permeability: It is defined as the ratio of magnetic flux density (B) inside the

substance to the magnetizing field intensity (H).

7. A magnetic field of 1800 ampere/metre produces a magnetic flux of 3x10-5 Weber in an iron

bar of cross sectional area 0.2 cm2. Calculate permeability.

8. Define retentivity and coercivity.

Retentivity is the amount of magnetic induction retained in the material even after the magnetic

field is removed.

Coercivity is the amount of magnetising field applied in the reverse direction to remove the

residual magnetism completely from the material.

9. What are soft and hard magnetic materials.

S.No Soft Magnetic Materials Hard Magnetic materials

1 They can be easily magnetized and demagnetized.

They cannot be easily magnetized or demagnetized

2 Loop area is less and hence the hysteresis loss is minimum.

The loop area is large and hence the hysteresis loss is maximum.

3 Susceptibility and permeability is high. Susceptibility and permeability is low.

4 Retentivity and Coercitivity are small. Retentivity and Coercitivity are large.

5 They have low eddy current loss. They have high eddy current loss.

6 These materials are free from irregularities like strain or impurities.

These materials have large amount of impurities and lattice defect.

10. Define energy product of a magnetic material.

The product of retentivity (Br) and coercivity (Hc) is called energy product. It represents

the amount of energy stored in the specimen. The value of energy product is very high for

permanent magnets.

Energy product = Br x Hc

11. What is antiferromagnetism? Give examples.

In anti-ferromagnetism, electron spin of neighbouring atoms are aligned antiparallel.

Anti-ferromagnetic susceptibility is small and positive and it depends greatly on temperature.

Eg. MnO, MnS, NiCr, Cr2O3, etc.

12. What is superconductivity?

The phenomenon of sudden disappearance of electrical resistivity in materials when it

is cooled to sufficiently low temperature is called superconductivity. The materials that exhibit

superconductivity and which are in the superconducting state are called superconductors.

13. Mention the condition for the material to behave as a superconductor.

It should be pure i.e., the residual resistivity due to scattering by impurities (ρ0) should be

zero.

It should be cooled below transition temperature i.e., the ideal resistivity due to scattering

by phonons (ρT) should be zero.

Magnetic induction in the material should be zero when it is placed in external magnetic

field.

14. Define critical temperature and critical field.

The temperature at which the normal conductor loses its resistivity and becomes a

superconductor is known as transition temperature or critical temperature (TC). The magnetic field

at which the superconducting property of a material disappears is called as critical magnetic field.

15. What is isotope effect?

In a superconducting material, transition temperature varies with the average isotopic

mass M of its constituents. i.e.,

Where α is called isotope effect coefficient.

16. What is meant by persistent current?

When DC current of large magnitude is induced in a superconducting ring, then due to

the diamagnetic property of the superconductor, the magnetic flux is trapped inside the ring and

hence the current persists in the ring for a longer time. This is called persistent current.

17. What are high Tc superconductors? Give examples.

Any superconductor, if the transition temperature is above 10 K it is called as high TC

superconductor.

Examples:

YBa2Cu3O7 TC = 92 K

La1.85Ba0.05CuO4 TC = 36 K

18. What are cooper pairs?

The attractive interaction between two electrons mediated by means of a phonon

exchange dominates the usual repulsive interaction. Pair of such electrons having opposite spins

which interact interactively in the phonon filed are called a cooper pair.

19. Calculate the critical current for a wire of lead having a diameter of 1mm at 4.2 K. Critical

temperature for lead is 7.18 K and H0 = 6.5x104 A/m.

20. The critical temperature for Hg with isotopic mass 199.5 at 4.184 K. Calculate the critical

temperature when its mass changes to 203.4.

Critical temperature for isotopic mass M1 Tc1 = 4.185 K M1 = 199.5

Critical temperature for isotopic mass M2 Tc2 = 4.133 K M2 = ?

MαTc = constant

Mα

1Tc1 = Mα 2Tc =

=

= x

= 14.124 x 1.01258 = 14.3016

M2 = 204.536

PART – B

1. i) State the origin of magnetic moment.

Three contributions for the magnetic dipole moment of an atom.

i. Orbital magnetic moment: Orbital motion of electrons (i.e., the motion of electrons in the

Closed orbits around the nucleus). Its magnitude is always small.

ii. Spin magnetic moment: Spin motion of the electrons (i.e., due to electron spin angular in

momentum).

iii. Nuclear magnetic moment: Contribution from the nuclear spin (i.e., due to nuclear spin

angular momentum). It is nearly 103 times smaller than that of electron spin; it is not taken

into consideration.

ii) How are magnetic materials classified based on magnetic moments? Compare their

properties. Give also their characteristics and examples.

S.No Properties Diamagnetic Paramagnetic Ferromagnetic

1 Definition No permanent magnetic moment

Permanent magnetic moment

Enormous magnetic moment

2 Spin alignment No spin alignment Random alignment Parallel and orderly alignment

3 Magnetised direction

Opposite to the applied field

Same direction as the applied external magnetic field.

Same direction as the applied external magnetic field.

4 Permeability Very less High Very high

5 Relative permeability

µr < 1 µr > 1 µr >> 1

6 Susceptibility Negative Positive and small Positive and high

2. Explain ferromagnetic domain theory. Briefly explain different types of energy involved in

domain growth.

A ferromagnetic material is divided into a large number of small region is called

domains. (0.1 to 1 of area), each direction is spontaneously magnetized. When the magnetic

field is applied to the Ferromagnetic material, the magnetization is produced by two ways. 1. By the motion of domain walls. 2. By the rotation of domains.

Different energies involved in domain growth is

1. Exchange energy (or) Magnetic field energy.

2. Crystalline energy (or) Anisotropy energy.

3. Domain wall energy (or) Bloch wall energy.

4. Magnetostriction energy

3. i) Draw the B-H curve (Hysteresis) for a ferromagnetic material and explain the same on the

basis of domain theory.

“When a ferromagnetic material is taken through a cycle of magnetization, the variation of B

(magnetic induction) with respect to H (applied field) can be represented by a closed loop

(or) curve (hysteresis loop or curve)” is called hysteresis.

OA - Due to smaller reversible domain wall movement

AB - Due to larger irreversible domain wall movement

BS - Due to smaller irreversible domain wall rotation

ii) Explain how susceptibility varies with temperature for dia, para, and Ferro magnetic

materials in detail.

Material Diamagnetic Paramagnetic ferromagnetic

susceptibility Negative Positive & small Positive & large

A given material has a susceptibility of 900. Determine the relative permeability of the material.

4. i) Explain soft and hard magnetic materials.

Soft magnetic materials

Easily magnetised and demagnetised.

Hysteresis is very small and hysteresis loss is also small.

Coercivity and retentivity is very small.

Materials have very large values of susceptibility and permeability.

Magentostatic energy is very small.

Eddy current loss is very small.

Hard magnetic materials

Cannot be easily magnetised and demagnetised.

Hysteresis is very large and hysteresis loss is also large.

Coercivity and retentivity is very large.

Materials have very small values of susceptibility and permeability.

Magentostatic energy is very large.

Eddy current loss is very large.

ii) Iron has relative permeability of 5000. Calculate its magnetic susceptibility.

iv) Mention the properties of antiferromagnetism.

• The adjacent magnetic dipoles are aligned antiparallel.

• Anti ferromagnetic susceptibility mainly depends on temperature.

• Its susceptibility is very small and it is positive.

• The susceptibility is given by when T > TN.

5. i) What are ferrites? Explain the structure of ferrites, properties and its applications.

These are compounds of iron oxides with oxides of other metals. Ferrites are the magnetic

compounds consisting of two or more different kinds of atoms. Generally ferrites are

expressed by the formula where X is a divalent metal ion.

Structures of ferrites:

Regular spinel

Inverse spinel

Properties

Spin alignment is antiparallel and of different magnitude.

They have high permeability and high resistivity.

Ferromagnetic materials possess net magnetic moment.

They have low eddy current losses and low hysteresis losses.

The susceptibility of a ferromagnetic material is positive and very large.

Applications

They are used to produce ultrasonic waves.

Used in audio and video transformers.

Ferrites rods are used in radio receivers to increase the sensitivity.

ii) Why are Ferrites advantages for use as transformer core?

Ferrites are used as transformer cores for frequencies upto microwaves. This is because

the eddy current problem which prevents the penetration of magnetic flux into the material is

so much less severe in ferrites than in iron.

6. Explain the different properties of superconductors in detail.

a. Zero electrical resistance: The phenomenon of exhibiting zero electrical resistance by

certain material when it is cooled below a certain temperature (critical temperature Tc).

b. Effect of magnetic field: In the superconducting state, a material possesses zero

electrical resistance and behaves as a diamagnetic material. Below Tc if sufficiently

strong magnetic field is applied, the superconducting property of the material is

destroyed and the material reverts back to its normal state. The minimum field required

to destroy the superconducting property is known as the critical magnetic field (Hc) .

c. Effect of electric current: In the superconducting state, a material possesses zero

electrical resistance and behaves as a diamagnetic material. Below Tc if sufficiently

strong magnetic field is applied, the superconducting property of the material is

destroyed and the (material reverts back to its normal state. The minimum field

required to destroy the superconducting property is known as the critical magnetic field

(Hc) .

d. Persistent current: When DC current of large magnitude is induced in a

superconducting ring, then due to the diamagnetic property of the superconductor, the

magnetic flux is trapped inside the ring and hence the current persists in the ring for a

longer time. This is called persistent current.

e. Meissner effect: When a weak magnetic field is applied to a superconducting specimen

at a temperature below transition temperature (Tc), the magnetic flux lines are expelled.

The specimen acts as an ideal diamagnet. This effect is called “Meissner effect”.

f. Isotope effect: In superconducting materials, “the Tc varies with the average isotopic

mass of their constituents”.

Tc α Mα

MαTc = constant

g. Effect of pressure: When pressure is applied to a superconducting material, transition

temperature increases. Thus it is quite possible that the Tc of a material can be made to

approach room temperature by the application of high pressures.

7. i) What is Meissner effect? Prove that all superconductors are perfect dia-magnet in

superconducting state.

Meissner effect: When a weak magnetic field is applied to a superconducting specimen at a

temperature below transition temperature (Tc), the magnetic flux lines are expelled. The

specimen acts as an ideal diamagnet. This effect is called “Meissner effect”.

Superconductors are perfect dia-magnet – Proof

Magnetic induction is given by

For a superconductor, B = 0

Therefore, 0

Since

H + I = 0

I = -H

Thus, when B = 0, magnetic susceptibility of a superconductor is found to be negative

maximum. This is referred to as perfect diamagnetism.

ii) Discuss the important features and the prediction of BCS theory.

Important features of BCS theory

Electrons form pairs which propagate through the lattice.

The cooper pairs are propagated through the lattice without resistance because the

electrons move in resonance with phonons.

Predictions

This could successfully predict the phenomenon of isotope effect.

It explains the existence of an energy gap between the ground state and first excited

state.

It explains the quantization of the magnetic flux through a superconducting ring.

8. i) Differentiate the Type I and Type II superconductors

S.No Type I Superconductors Type II Superconductors

1 Here, the transformation from superconducting to normal state takes place suddenly at the critical magnetic field (HC).

Here, the transformation from superconducting to normal state is gradually from HC1 to HC2.

2 They exhibit complete Meissner effect. They do not exhibit complete Meissner effect.

3 They have only one critical magnetic field

They have two critical magnetic fields.

4 They are known as soft superconductors because the field required to destroy the superconducting state is small.

They are known as hard superconductors because the field required to destroy the superconducting state is high.

5 Examples: Al, Zn, Sn, Pb, etc. Examples: Vanadium, Niobium, etc.

6.

ii) Explain High temperature superconductors in detail.

In a superconductor if the transition temperature is high i.e., greater than 100K, then it

is called high temperature superconductor.

9. i) Discuss the applications of superconductors in detail.

a. Superconductors are used for the production of high magnetic field magnets.

b. By using superconducting materials, it is possible to manufacture electrical generators

and transformers.

c. Superconducting materials are used in the construction of very sensitive electrical

measuring instruments such as galvanometers.

d. Superconducting materials are used for power cables, which will enable transmission of

power over very long distances without any significant power loss or drop in voltage.

e. It is used to study tiny magnetic signals from brain and heart.

ii) Superconducting tin has a critical temperature of 3.7 K at zero magnetic field and a

critical field of 0.0306 tesla at 0 K. Find the critical field at 2 K.

10. Explain the following in detail

i) AC and DC Josephson Effect :

AC Josephson Effect: when a Dc voltage is applied across the Josephson junction through

which super current is flowing, an AC current is noticed.

DC Josephson Effect: the tunnelling of superconducting electron pairs through Josephson

junction leads to the flow of current without the application of voltage.

ii) Cryotron: Cryotron is a magnetically operated switch. The superconducting property

of a material disappears when the applied magnetic field is greater than the critical magnetic

field.

iii) SQUID: SQUID is the acronym for Superconducting Quantum Interference Device. It

is a double junction quantum interferometer. Two Josephson junctions mounted on a

superconducting ring forms this interferometer. Squids are based on the flux quantization in

the superconducting ring. The total magnetic flux passing through the ring is quantized. It is a

weak ultra-sensitive measuring instrument used for detection of very weak magnetic field in

the order of 10-14 tesla.

iv) Magnetic Levitation: The magnetic levitation is based on the diamagnetic property of

a superconductor which is rejection of magnetic flux lines. A superconductor can be

suspended in air against the repulsive force from a permanent magnet. This magnetic

levitation effect can be used for high speed transportation without frictional loss.

UNIT - IV – Dielectric Materials

PART – A

1. What are Dielectrics?

Dielectrics are the insulating materials having electric dipoles permanently or have an

ability to produce enormous induced dipoles by the application of external electric field.

2. Define electric polarization and electric susceptibility.

The process of producing electric dipoles inside the dielectric by an electric field is

called electric polarization in dielectrics.

The polarization vector is proportional to the total electric field intensity and is

in the same direction of E. therefore the polarization vector can be written as,

Where the constant is referred as the electrical

susceptibility and is a characteristic of every dielectric.

=

3. Compare active and passive dielectrics.

S.No Active dielectrics Passive dielectrics

1 Dielectrics which can easily adapt itself to store the electrical energy in it is called active dielectrics.

Dielectrics which restrict the flow of electrical energy in it are called passive dielectrics.

2 They are not pure insulating materials. They are pure insulating materials like glass.

3 Examples : Piezo-electrics, Pyro-electrics, etc.

Examples : Ceramics, Mica, etc.

4 Application : they are used to generate ultrasonic waves

Application : They are used to produce sheets, pies, etc.

4. Define dielectric constant.

It is the ratio between absolute permittivity of the medium (ε) and the permittivity of free space

(ε0)

If the medium has high dielectric constant, it is easily polarized and behaves as a good electric

insulator.

5. What are the types of polarization mechanism involved in dielectric material?

Electronic polarization

ionic polarization

Orientational polarization

Space charge polarization

6. What are the differences between polar and non polar molecules?

S.No Polar molecules Non-polar molecules

1 These molecules have permanent dipole moment even in the absence of the electric field.

These molecules do not have permanent dipole moments.

2 The polarization of polar molecules is highly temperature dependent

The polarization of this kind of molecules is temperature independent

3 They do not have center of symmetry They have centre of symmetry

4 Here the +ve and –ve charges do not coincide with each other

Here +ve and –ve charges are coinciding with each other.

5 absorption and emission is found in infrared region

There is no absorption or emission in the infrared region.

6 Examples : HCl,H2O,CHCL3 etc. Examples :CCl4, CO2,H2 etc.

7. Define space charge polarization.

In some materials containing two or more phases, the application of an electrical field

causes the accumulation of charges at the interfaces between the phases or electrodes. As a result

of this, polarization is produced. This type of polarization is known as space charge polarization.

8. Define dielectric loss and dielectric breakdown.

Dielectric loss: When a dielectric material is subjected to an A.C voltage the electrical energy is

absorbed by the material and is dissipated in the form of heat. This dissipation of energy is called

dielectric loss.

Dielectric breakdown: When the dielectric is subjected to a heavy voltage, beyond the critical

value it loses its insulation property because electrons start jumping from valence band to the

conduction band. Hence a dielectric material loses its resistivity and permits large value of current

to flow through it. This is known as dielectric breakdown.

9. Define dielectric strength.

The magnitude of electric field strength at which dielectric breakdown occurs is called

dielectric strength. It is the breakdown voltage per unit thickness of the material.

10. What are the requirements of good insulating materials?

It should have low dielectric constant

It should possess low dielectric loss.

It must have high resistance.

It must possess high dielectric strength.

It should have adequate chemical stability.

It must have high moisture resistance.

11. Calculate the polarization produced in a dielectric medium of dielectric constant 6 and it is

subjected to an electric field of 100 V/m. Given ε0 = 8.85 x 10-12F/m.

P = 4.425 X

12. A capacitor consists of two conducting plates of area 200 cm2 each separated by a dielectric

constant ε = 3.7 of thickness 1mm. Find the capacitance and the electric flux density when a

potential of 300 V is applied. (ε0 = 8.85 x 10-12F/m).

Capacitance

Electric Flux Density

13. What are the factors affecting dielectric loss.

Dielectric loss may increase due to the following factors.

• high frequency of the applied voltage.

• high value of the applied voltage.

14. The dielectric constant of a He gas at NTP is 1.0000684. Calculate the electronic

polarizability of He atoms if the gas contains 2.7 x 1025 atoms/m3 and hence evaluate the

radius of the He atoms. Given ε0 = 8.85 x 10-12F/m.

Electronic polarizability

15. Write the Clausius – Mosotti relation. What is its significance?

= • It relates the microscopic quantity with macroscopic quantity

• It relates the dielectric constant of the material with polarisability.

16. If a NaCl crystal is subjected to an electric field of 1000V/m and the resulting polarization is

4.3x10-8 C/m2. Calculate the relative permeability of NaCl.

17. State the properties of ferroelectric materials.

The dielectric constant of these materials does not vary with respect to temperature.

The dielectric constant (εr) reaches a maximum value only at a particular temperature

called Curie temperature

The polarization does not vary linearly with respect to electric field and hence these

materials are also called as non-linear dielectrics.

Ferro-electric materials exhibit hysteresis, similar to that of ferro-magnetic materials.

18. Find the capacitance of layer of Al2O3 having 0.5 μm and area 2500 mm2 with εr = 8.

Capacitance

19. What is Piezoelectricity?

When a dielectric material acquires electric polarization due to the external mechanical

pressure, it is called Piezo electricity and these materials are known as Piezo electrics.

20. Mention the applications of dielectrics.

Ferro-electric materials are used to produce ultrasonics.

They are used in the production of Piezo-electric materials and in turn to make microphones.

Ferro-electrics are also used in SONAR, Strain gauges, etc.

They are also used as frequency stabilizers and crystal controlled oscillators.

Electrets are also used to bond the fractured bones in the human body.

Ferro-electric semiconductors are used to make positors, which in turn are used to measure and

control the temperature.

PART – B

1. Define the following. Give the necessary equations relating the above quantities.

a .Dielectric constant.

It is defined as the ratio of absolute permittivity of the medium to the

permittivity of free space.

b. Polarizability .(α)

The average dipole moment of a system is proportional to the applied electric field.

c. Polarization vector.

If is the average dipole moment per molecule and N is the number of

molecules per unit volume , the polarisation vector is defined as dipole moment per unit

volume of the dielectric material.

= N

d. Electric flux density : The number of electric lines passing through unit area of

cross section.

e. Electric Susceptibility.

The polarisation vector is proportional to the applied electric field strength.

= = - 1

2. Discuss electronic and ionic polarizations with examples in detail.

ELECTRONIC POLARISATION

Electronic polarisation is due to the displacement of positively charged nucleus and

negatively charged electrons of an atom in opposite directions on the application of electric

field. Ex : Inert gases

Derivation for αe

Without electric field:

With electric field: When the dielectric is placed in an electric field (E), two phenomenon

occur. (i) Lorentz force due to the electric field tends to separate the nucleus and the electron

cloud from their equilibrium position. FL = -ZeE

(ii) After separation, an attractive coulomb force arises between the nucleus and electron

cloud which tries to maintain the original equilibrium position.

µind = αe E where αe = 4 IONIC POLARISATION Ionic polarisation is due to the displacement of cations and anions in opposite

directions. This occurs in ionic dielectrics by the influence of external electric field. Ex: Ionic

Without electric field With electric field

Derivation for ( =

3. Explain the different types of polarization mechanisms involved in a dielectric material.

Electronic Polarisation: Electronic polarisation is due to the displacement of positively

charged nucleus and negatively charged electrons of an atom in opposite directions on the

application of electric field. Ex : Inert gases

Derivation for αe

Without electric field

With electric field

µind = αe E where αe = 4 Ionic Polarisation.

Ionic polarisation is due to the displacement of cations and anions in opposite

directions. This occurs in ionic dielectrics. by the influence of external electric field. Ex: Ionic

Derivation for ( =

Orientation Polarisation.

Orientation polarisation takes place only in polar dielectrics. Polar dielectrics

have molecules with permanent dipole moments even in the absence of an electric field.

Derivation for

αo =

Space charge Polarisation.

Space charge polarisation occurs due to the accumulation of charges at the electrodes

or at the interfaces of multiphase dielectric materials.

Total polarisation α = αe+ + αo + αs

4. Explain the frequency and temperature dependence of all type of polarization in dielectrics.

If an alternating electric field is applied across the material, polarisation occurs as a

function of time. ie P(t) = P [1 - e-t/tr ]

Frequency range in Hz Type of polarisation ~ 1015 Electronic ~1013 Ionic + Electronic ~106 -1010 orientation +Ionic + Electronic ~102 Space charge + orientation +

Ionic+ Electronic

Temperature Dependence

Electronic Polarisation: Temperature independent.

Ionic Polarisation: Temperature independent.

Orientatinal Polarisation: increase in temperature decrease the polarisation.

Space charge polarisation: increase in temperature increases the polarisation.

5. What is meant by local field in a dielectric? And how it is calculated for a cubic structure?

Deduce the Clausius – Mosotti relation.

When a dielectric is kept in an external electric field (E), two fields are exerted due to (i)

external Electric field (ii) dipole moment created. These long range Coulomb field which is

responsible for polarization of each atom or molecule is known as internal field or local

field and is given by,

Eint = E1 + E2 +E3 + E4

E1 = Electric field due to charges on the plates of the capacitor.

E2 = Electric field due to polarised charges on the plane surface of the dielectric.

E3 = Electric field due to polarised charges induced on the surface of the imaginary

spherical cavity.

E4 = Electric field due to permanent dipoles of atoms inside the spherical cavity considered.

Eint = E + E3

Eint = E +

=

6. i) A solid contains 5x1028 identical atoms /m3, each with a polarizability of 2x10-40 Fm2.

Assuming that internal field is given by the Lorentz relation, calculate the ratio of internal

field to the applied field. (ε0 = 8.85 x 10-12F/m)

v) The dielectric constant of water is 80. Is water a good dielectric? Is it useful for energy

storage in capacitors? Justify your answer.

Though the dielectric constant of water is 80 at 20C or 293K it is not a good dielectric

material because of the following reasons.

1. At normal temperature the electrical conductivity of water is very high

2. At normal temperature current lead the voltage by a very less angle i.e., cos θ is

very less which leads to very high loss angle (δ) and hence tan δ is very high

3. The dissipation factor and hence the power loss is very high. Thus water cannot be used

as dielectric for energy storage in capacitors.

7. What is dielectric loss? Derive the expression for dielectric power loss.

Dielectric loss: When a dielectric material is subjected to an A.C voltage the electrical energy

is absorbed by the material and is dissipated in the form of heat. This dissipation of electrical

energy is called dielectric loss.

Expression for dielectric loss

PL = VI cos θ

PL = VI cos 90

PL = 0

For a practical dielectric material

PL = 2 f C V2 tan δ

tan δ – power factor

8. What are the different types of dielectric break down in dielectric medium? Discuss in detail

the various types of dielectric breakdown.

Dielectric breakdown: When the dielectric is subjected to a heavy voltage, beyond the critical

value it loses its insulation property because electrons start jumping from valence band to the

conduction band. Hence a dielectric material loses its resistivity and permits large value of current

to flow through it. This is known as dielectric breakdown.

Types of dielectric breakdown

• Intrinsic breakdown and avalanche breakdown

• Thermal breakdown

• Chemical and Electrochemical breakdown

• Discharge breakdown

• Defect breakdown

Explanation of each breakdown and its characteristics.

9. What is ferroelectricity? Explain the properties of ferro electric material. Give examples and

mention the applications.

Ferro electric materials

Some materials which exhibit electronic polarisation even in the absence of the applied

electric field and thus they produce electricity. This electricity is known as

Ferroelectricity. Materials are called Ferroelectric materials.

Ex:

1. Barium titanate. (Ba Tio3)

2. Potassium dihydrogen phosphate(KH2PO4)

3. Ammonium dihydrogen phosphate.(NH4H2PO4)

4. Lithium Niobate(LiNbO3)

5. Rochelle salt(NaKC4H4O6, 4H2O)

10. Explain the uses of dielectrics in capacitors and in Transformers.

Insulating materials are used in power and distribution transformers and capacitors.

In Capacitors :

Explanation about the various types of dielectric materials in various capacitors like

a) Power Capacitors.

b) Electrolytic Capacitors.

c) Ceramic Capacitors.

Explanation about the Uses of insulating materials in power and distribution

transformers .

UNIT – V ADVANCED ENGINEERING MATERIALS

PART – A

1. Define Met glasses.

Metallic glasses are new type of engineering materials which shares the properties of

both metals and glasses. They are metal-metal or metal-metalloid alloy glasses having

amorphous nature.

2. What is meant by glass transition temperature?

The temperature at which metals in the molten form transforms into glasses ie liquid to

solid is known as glass transition temperature (Tg).

3. What do you understand by the term quenching?

Quenching is a technique used to form metallic glasses. Quenching means rapid

cooling. Actually atoms of any material move freely in a liquid state. Atoms can be arranged

regularly when a liquid is cooled slowly. Instead, when a liquid is quenched, there will be an

irregular pattern, which results in the formation of metallic glasses.

4. Mention the merits of metallic glasses as transformer core materials.

Metallic glasses are ferromagnetic. They posses low magnetic losses, high

permeability and saturation magnetization with low coercivity. They also have extreme

mechanical hardness and excellent initial permeability.

These properties make them useful as transformer core materials. Moreover power transformers

made of metallic glasses are smaller in size and efficient in their performance.

5. Mention any four properties of met glasses.

Metallic glasses have tetrahedral closely packed (TCP) structure rather than hexagonal

closely packed structure (HCP) structure.

The metallic glasses are very strong in nature

They posses malleability, ductility, etc.

They exhibit very low hysteresis loss and hence transformer core loss is very less.

6. Write any four application of met glasses.

Metallic glasses are very strong and hard and they can be used to make different kinds of

springs.

Metallic glasses have high resistivity and they can be used to make computer memories,

magneto-resistance sensors etc.

Since Metallic glasses can behave as superconductors, they are used in the production of

high magnetic fields.

Metallic glasses are not affected by irradiation and so they can be used in nuclear reactors.

Metallic glasses are high corrosion resistant and they can be used in reactor vessels,

surgical clips, marine cables, etc.

7. What is transformation temperature?

In SMA, the shape recovery process occurs not at a single temperature; rather it

occurs over a range of temperature. Thus the range of temperature at which the SMA switches

from new shape to its original shape is called transformation temperature or memory transfer

temperature.

8. What is meant by shape memory effect?

The ability of shape memory alloys to get severely deformed on cooling and then

regaining its original shape by heating effect is known as shape memory effect.

9. What do you understand by Martensite and Austenite phase.

Martensite and Austenite are the two solid phases which occur in SMA.

Martensite is relatively soft and it is easily deformable phase which exists at low

temperature (Monoclinic).

Austenite is a phase occurs at high temperature having crystal structure and high degree

of symmetry (Cubic).

10. What is pseudo elasticity?

The ability of the SMA to return to its original shape upon unloading after a substantial

deformation is known as super elasticity or pesudoelasticity. This is based on the stress induced

martensite transformations.

11. What is one-way and two-way shape memory alloys?

One-way shape memory alloy (SMA):

A material which exhibits shape memory effect during both heating and cooling.

Two-way shape memory alloy (SMA):

A material which exhibits shape memory effect during both heating and cooling.

12. Mention the advantages and disadvantages of shape memory alloys.

Advantages

They are simple, compact and high safe.

They have good bio-compatibility.

They have diverse applications and offer clean, silent and spark free working condition.

They have good mechanical properties and strong corrosion resistant.

Disadvantages

They have poor fatigue properties.

They are expensive.

They have low energy efficiency

13. What are nanophase materials? Give examples.

Nanophase materials or nanomaterials are newly developed materials

with grain size at the nanometer range (10-9) ie in the order of 1-100 nm.

14. Mention few techniques for synthesis of nano phased materials.

Plasma-arching

Chemical vapor deposition

Sol-gel technique

Electro-deposition

Mechanical crushing- Ball milling

Laser synthesis

Inert gas condensation

15. What is top down and bottom up approach in nano materials?

Top down process: In this process bulk materials are broken into nano sized particles. In top

down process, the buildings of nanostructures starting with small components like atoms and

molecules that are removed from a bulk material so as to obtain desired microstructure.

Examples: Nanolithography, Ball milling, etc.

Bottom up process: This process involves building larger objects from smaller building blocks.

Nanotechnology seeks to use atoms and molecules as those building blocks. This is the opposite

of top-down process. Instead of taking the material away to make structures, the bottom up

process selectively adds atoms to create structures.

Example: pulsed lased deposition, chemical vapor deposition, etc.

16. Give any four non-linear optical properties.

The few non-linear optical phenomena observed are

Higher harmonic generation

Optical mixing

Optical phase conjugation

Solition

17. What is meant by second harmonic generation?

In some non-linear crystals which lack centre of symmetry, when intense radiation is

passed through, the frequency of radiation is doubled, i.e., wavelength is halved. Such process is

known as second harmonic generation.

18. What are bio-materials?

The materials which are used for structural applications in the field of medicine are

known are known as biomaterials. They are used to make device to replace damaged or diseased

body parts in human and animal bodies.

19. What are the types of bio-materials?

They are classified s four types.

Metals and alloys biomaterials

Ceramics biomaterials

Polymer biomaterials

Composite biomaterials

20. What are the applications of biomaterials?

Applications:

Stainless steel is a predominant alloy widely used in implant and orthopedic applications.

Protosal from cast alloy of Co-Cr-Mo is used to make stem and head of implant hip

endoprosthesis.

PART – B

1. What are metallic glasses? Describe the preparation and properties and applications of metallic

glasses.

Metallic Glasses

Metallic glasses are amorphous metallic solids which combines both the properties of

metals and glasses. They have high strength, good metallic and corrosion resistance

properties. Eg: Alloys of Fe, Ni, Al, Mn, Cu, etc.

Preparation Technique

Melt spinning method

Properties of Metallic Glasses

Structural properties

Mechanical properties

Electrical properties

Magnetic properties

Chemical properties

Application of Metallic Glasses

Structural application

Electrical and Electronics

Metallic glasses as transformer core material

Nuclear reactor engineering.

Bio medical application.

2. How are metallic glasses prepared? Explain how the melt spinner device can be used to

produce met glasses.

Preparation of metallic glasses

1. Twin roller

2. Melt spinning

3. Melt extraction

Melt spinning technique – Extreme rapid cooling of the molten metal alloy (rapid quenching).

Due to rapid cooling atoms are arranged irregularly and form metallic glasses.

3. Explain the properties and application of metallic glasses also mentions its types with

examples.

Properties of Metallic Glasses

Structural properties

They do not have any crystal defects.

They have tetrahedral close packing structure.

Mechanical properties

They have extremely high strength.

They have high elasticity.

They have high ductility.

They are not work-harden but they are work-soften.

Electrical properties

Electrical resistivity of metallic glasses is high and it does not vary with temperature.

Eddy current loss is very small.

Temperature coefficient is zero or negative.

Magnetic properties

They have both soft and hard magnetic properties.

They exhibit high saturation magnetisation.

They have less core losses.

Metallic glasses have very narrow hysteresis loop.

Chemical properties

They are highly resistance to corrosion due to random ordering.

They are highly reactive and stable.

They act as a catalyst.

Application of Metallic Glasses

Structural application

Electrical and Electronics

Metallic glasses as transformer core material

Nuclear reactor engineering.

Bio medical application.

4. What are shape memory alloys? Describe the characteristics of SMA and its applications.

Shape Memory Alloys: The ability of metallic alloys to regain its original shape while

heating or cooling is called as shape memory alloys. Ex: Ni-Ti alloys, Cu Zn Al, Cu Al Ni, Au-

Cd, Ni-Mn-Ga and Fe based alloys.

Phases of Shape Memory Alloys

Austenite – High temperature phase having cubic structure.

Martensite – Low temperature phase having monoclinic structure.

Characteristics of Shape Memory Alloys

1. Shape Memory Effect: the change of shape of a material at low temperature by loading

and regaining of original shape by heating it, is known as Shape Memory Effect.

2. Pseudo elasticity

3. Super elasticity

4. Hysteresis: It is defined as the difference between the temperatures at which a material is

50% transformed into austenite while heating and 50% transformed to martensitic while

cooling.

Advantages

1. Good bio-compatibility

2. Good mechanical properties and good corrosion resistance.

3. High power and weight ratio

Disadvantages

1. Poor fatigue properties.

2. Low energy efficiency

Applications

1. Eye glass frames

2. Toys

3. Helicopter blades

4. Coffee valves

5. Mirco- surgical instruments

6. Dental arch wires

5. i) Mention the properties of Ni –Ti alloy.

Ni-Ti alloy has high shape memory alloy.

The density of Ni-Ti is 6450 Kg m-3.

It is more flexible.

It has high melting point.

The transformation temperature varies from -200C to 110C.

It has high thermal stability and corrosion resistance.

The thermal conductivity varies from 8.5 to 18 WK-1M-1.

ii) Explain the applications of SMA

Eye glass frames

Toys

Helicopter blades

Coffee valves

Mirco- surgical instruments

Dental arch wires

iii) Explain the advantages and disadvantages of SMA.

Advantages

Good bio-compatibility

Good mechanical properties and good corrosion resistance.

High power and weight ratio

Disadvantages

Poor fatigue properties.

Low energy efficiency

6. What are nanomaterials? Explain the preparation, properties and applications nanomaterials.

Nanomaterials are materials having grain sizes in the order of 1 to 100nm.

Synthesis

1. Top down approach – process of breaking down bulk material to nano size.

2. Bottom up approach – nanomaterials are made by building atom by atom.

Synthesis of nanomaterial’s

1. Pulsed laser deposition

2. Chemical vapor deposition

Chemical Vapor Deposition (CVD): Precursor gases are delivered into the reaction chamber

at elevated temperature usually under vacuum. As they pass over or come into contact with a

heated substrate, they react or decompose forming a solid phase. They are deposited onto the

substrate. The substrate temperature is critical and can influence what reactions will take

place.

Properties of Nanomaterials

1. Electrical properties

2. Optical properties

3. Chemical properties

4. Mechanical properties

5. Magnetic properties

Applications of Nanomaterials

1. Used to produce color paints

2. Used in thermal protection and current controlling devices.

3. Used in data storage

4. Used in hydrogen storage devices.

5. Used in drug delivery.

6. Used in semiconductor laser and CD’s.

7. Used as sunscreens.

7. Describe the method of producing nano materials using

i) Pulsed laser deposition: The laser pulses of high intensity and energy is used to

evaporate carbon from graphite. These evaporated carbon atoms are condensed to form

nanotubes.

ii) Chemical Vapor Deposition (CVD): Precursor gases are delivered into the reaction

chamber at elevated temperature usually under vacuum. As they pass over or come into

contact with a heated substrate, they react or decompose forming a solid phase. They

are deposited onto the substrate. The substrate temperature is critical and can influence

what reactions will take place.

8. i) Explain Birefringence.

The appearance of double refraction under the influence of an external agent is known

as artificial double refraction or induced birefringence. The property of double refraction

can be induced in an isotropic material by the application of (i) a mechanical strain (ii) an

electric field or (iii) a magnetic field.

ii) Explain optical Kerr effect.

Anisotropy induced in an isotropic medium under the influence of an electric field is

known as Kerr effect.

The change in refractive index is given by

iii) Explain non-linear materials. Give examples.

Non – linear optics: In some materials, the intensity and frequency of the output is not

linearly proportional to the intensity and frequency of the incident light input. This effect is

called nonlinear effect.

Ex: Lithium tantalite, Lithium niobate, Barium sodium niobate, etc.

9. Explain the origin of non-linear optics? How are second harmonic wave generated.

Non – linear optics: In some materials, the intensity and frequency of the output is not

linearly proportional to the intensity and frequency of the incident light input. This effect is

called nonlinear effect.

Ex: Lithium tantalite, Lithium niobate, Barium sodium niobate, etc.

Second harmonic generation: second order generation represents the generation of new

frequencies with the help of crystals such as quartz, LiO3, etc.

In nonlinear medium for higher fields i.e., higher intensities of light the nonlinear effects are

observed.

10. Explain biomaterials, classification and its modern application in field of medicine.

The materials which are used for structural applications in the field of medicine are

known as biomaterials. These materials are used to make devices to replace damaged or

diseased body parts in human and animal bodies.

Classification of biomaterials

Metals and alloys biomaterials

Ceramics biomaterials

Polymer biomaterials

Composite biomaterials

Applications

Protosal from cast alloy of Co-Cr-Mo is used to make stem and used for implant hip

endoprosthesis.

Ni-Ti shape memory alloy is used in dental arch wires, micro surgical instruments,

blood clot filters, guide wires, etc.

Ceramic implants such as Al2O3 with some SiO2 and alkali metals are used to make

femoral head.

Carbon has good biocompatibility with bone and other tissues.