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AppliedPhysics - I

Applied Physics – I

University Examination

Solved Question Papers

2013 to 2018

1

Applied Physics – I (Sem-I) May-2013 Con. 6874-13. (REVISED COURSE) GS-S193 Time : 2 Hours Total Marks : 60 _________________________________________________________________________________ N.B. : (1) Question No.1 is compulsory.

(2) Attempt any three questions from remaining Question Nos. 2 to 6.

(3) Assume suitable data wherever required.

(4) Figures to the right indicate marks.

1. Attempt any five (Each carry equal weightage) :- 15

(a) Draw unit cells showing position of the atoms for -

(i) a monoatomic BCC Crystal

(ii) a monoatomic SC Crystal

(iii) CsCI Crystal.

(b) Fermi level in potassium is 2.1 eV. What are the energies for which the probabilities of

occupancy at 300 K are 0·99 and 0.01.

(c) Draw the energy band diagram of an unbiased p-n junction and mark the barrier

potential and depletion region.

(d) Write the relation between polarization and dielectric susceptibility and the relation

between dielectric susceptibility and dielectric constant.

(e) Why soft magnetic material are used in core of transformers.

(f) Calculate the change in intensity level when the intensity of sound increases 1000 times

its original intensity.

(g) Explain cavitation effect

2. (a) Derive an expression for Fermi level for an intrinsic semiconductor. 8

Draw the energy level diagram only, to show the effect of

(i) temperature

(ii) impurity atom concentration in low range and (iii) impurity atom concentration in

high range.

(b) An elemental crystal has a density of 8570 kg/m3 packing fraction is 0·68. Determine the

mass' of one atom if the nearest neighbour distance is 2.86 Ao. 7

3. (a) Prove that in a ferromagnetic material, power loss per unit volume in a hysteresis cycle

is equal to the area under hysteresis loop. (4 + 4)

An iron ring of mean circumferential length 30 cm and cross-sectional area 1 cm2 is

wound uniformly with 300 turns of a wire. When a current of 0·032 Amp flows in it, the

flux produced in the ring is 2 x 10-6wb. Find the flux density, magnetic field intensity

and permeability of iron.

(b) Derive Bragg's law. Explain why x- rays and not y-ray are used for crystal structure

analysis. What data about the crystal structure can be obtained from the x-ray

diffraction pattern of a crystal. (4 + 2 + 1)

2

4. (a) Find out the critical radius ratio of an ionic crystal in ligancy 6 configuration. What is the

maximum size of cation in ligancy 6 configuration when the radius of anion is 2.02 Å.

5

(b) In an n type semiconductor, the Fermi level lies 0.4 eV below the conduction band. If

the concentration of donor atom is doubled, find the new position of the Fermi level

with respect to the conduction band. 5

(c) Explain the origin of electronic, ionic and orientational polarization and temperature

dependence of respective polarizability. 5

5. (a) Find out the intercepts made by the planes (1 0 1) and (4 1 4) in a cubic unit cell. 5

Draw [T 2 1] and [1 2 4] in a cubic unit cell.

(b) A bar of n type Ge of size 0.010m x 0.001m x 0.001m is mounted in a Magnetic field of

2 x 10-1T. The electron density in the bar is 7 x 1021/m3. If one milli volt is applied across

the long ends of the bar, determine the current through the bar and the voltage between

Hall electrodes placed across the short dimensions of the bar. Assume μe = 0·39 m2/vs. 5

(c) Define reverberation time. Write Sabine's formula explaining every term. What are the

factors which determine the average absorption co-efficient of a material? 5

6. (a) Explain the differences between three different liquid crystal phases with respect to the

order in the arrangement of molecules, with the help of diagram. Which property of the

liquid crystal is used for display? 5

(b) How a p-n junction diode is used to generate a potential difference in a photovoltaic

solar cell? 5

(c) What is piezoelectric effect? Explain the working of a piezoelectric oscillator used to

produce ultrasonic wave. 5

**************************

3

Applied Physics – I (Sem-I) May – 2013 Solutions

Q.1. Attempt any five (Each carry equal weightage): - 15

(a) Draw unit cells showing position of the atoms for -

(i) a monoatomic BCC Crystal

(ii) a monoatomic SC Crystal

(iii) CsCl Crystal.

(b) Fermi level in potassium is 2.1 eV. What are the energies for which the probabilities

of occupancy at 300 K are 0·99 and 0.01.

(c) Draw the energy band diagram of an unbiased p-n junction and mark the barrier

potential and depletion region.

(d) Write the relation between polarization and dielectric susceptibility and the relation

between dielectric susceptibility and dielectric constant.

(e) Why soft magnetic material are used in core of transformers.

(f) Calculate the change in intensity level when the intensity of sound increases 1000

times its original intensity.

(g) Explain cavitation effect

Ans.1 (a) (i) a monoatomic BCC Crystal:

Ans.1 (a) (ii) a monoatomic SC Crystal:

Ans.1 (a) (iii) CsCl Crystal:

4

Sol.1 (b) EF = 2.1 eV, f(E1) = 0.99, f(E2) = 0.01, T = 300 K

K = 1.38 × 10–23 J/K = 23

19

1.38 10

1.6 10

−

−

= 86.25 × 10–6 eV/K

F (EC) = ( )C FE E /KT

1

1 e−

+

F(E1) = ( )1 FE E /KT

1

1 e −+

0.99 = ( )1 FE E /KT

1

1 e −+

( )1 FE E /KT1 e −+ = 1.01

e(E1 – EF)/KT = 0.01

E1 – EF = – 0.1187 eV

E1 = 1.9813 eV

And f(E2) = ( )2 FE E /KT

1

1 e −+

0.01 = ( )2 FE E /KT

1

1 e −+

E2 – EF = 0.1187 eV

E2 = 2.2187 eV

E1 = 1.9813 eV, E2 = 2.2187 eV.

Ans.1 (c)

Ans.1 (d) D = E = 0 r E …..(3)

= 0 (1+ e)E

Since r =1+ e

Where e = electrical susceptibility

Relation between P & E

As D = 0 (1+ e)En & P = 0 e E

5

Therefore D = 0 E + P ….(4)

Equating (3) & (4) 0 r E = 0 E + P

Therefore P/E = 0 ( r -1 ) …..(5)

Electrical Susceptibility:

The polarization vector P is proportional to the applied electric field E, for Field strength

that are too large.

Therefore, P E

P = 0 e E

Where e is a characteristic of every dielectric and it called electrical Susceptibility

Therefore e = P

ϵ0 E

Since P

E = 0 ( r -1 )

e = ϵ0 ( ϵr −1 )

ϵ0

Therefore e = r -1 or r = e + 1

Ans.1 (e) Soft magnetic materials are used in devices that are subjected to alternating magnetic

fields and in which energy losses must be low; it is used in transformer cores the reason

is the relative area within the hysteresis loop must be small; it is characteristically thin

and narrow. Consequently, a soft magnetic material must have a high initial

permeability and a low coercivity. A material possessing these properties may reach its

saturation magnetization with a relatively low applied field (i.e., is easily magnetized

and demagnetized) and still has low hysteresis energy losses.

Sol.1 (f)

3

0

10=I

I

intincrease in ensity level in dB with respect to original intensity

10

0

10log

=

IL

I 3

1010log (10 )=

30=L dB – This is the change in intensity level.

Ans.1 (g) When ultrasonic waves of very high frequency pass through the liquid, formation of

small bubbles, called micro bubbles takes place. This is because excessive stress on the

liquid breaks it apart and bubbles are formed. The bubbles are highly unstable and they

soon collapse producing a high vacuum within. Due to this action, implosion takes

place. The area surrounding the bubble has a tremendous pressure. The particles in the

vicinity of these bubbles are strongly pulled towards the centers of the bubbles. This

process of creating cavity is called cavitation.

Q.2. (a) Derive an expression for Fermi level for an intrinsic semiconductor. 8

Draw the energy level diagram only, to show the effect of

(i) temperature

(ii) impurity atom concentration in low range and (iii) impurity atom concentration in

high range.

6

(b) An elemental crystal has a density of 8570 kg/m3 packing fraction is 0·68. Determine

the mass' of one atom if the nearest neighbour distance is 2·86 Ao. 7

Ans.2 (a) For intrinsic semiconductors, EF lies mid-way between conduction and valence band.

At any temperature T > 00 K,

ne = Number of electrons in conduction band

nv = Number of holes in valence band

We have

ne = Nce –(EC-EF)/KT …..(1)

where Nc = Effective density of states in conduction band

and nv = Nve –(EF – EV) / KT …..(2)

Where Nv = effective density of states in valence band

For best approximation Nc = Nv …..(3)

For intrinsic semiconductor

nC = nv

NC . e –(EC – EF) / KT = Nv . e –(EF – EV) / KT

( )

( )

C F

F V

E EKT

V

E EKT C

Ne

Ne

− −

−=

e –(EC – EF – EF + EV) / KT = cN

vN

e –(EC – EV – 2EF ) / KT = cN

vN

as Nv = Nc = 1

e –(EC – EV – 2EF ) / KT = 1

Taking In on both sides

KT

EEE Fvc )2( −+− = 0

(EC + EV) = 2 EF

EF = 2

vE+CE … (4)

Thus, Fermi level in an intrinsic semiconductor lies at the center of forbidden energy

gap.

N-type P-type

7

Sol.2 (b) = 8570 kg/m2

A.PF = 0.68

2r = 2.86 × 10–10 m

For A.P.F. = 0.68

It is BCC hence n = 2

2r = 3a

2

a = ( )102 2.86 102d

3 3

− = 103.30 10 m−=

x3 = A

n M

.N

Mass of an atom = 3M a

N n

=

( ) ( )33 103.3 10 8570

2

− = 30153990 10 kg−=

25

A

M1.5399 10 kg

N−=

Q.3 (a) Prove that in a ferromagnetic material, power loss per unit volume in a hysteresis

cycle is equal to the area under hysteresis loop. (4 + 4)

An iron ring of mean circumferential length 30 cm and cross-sectional area 1 cm2 is

wound uniformly with 300 turns of a wire. When a current of 0·032 Amp flows in it,

the flux produced in the ring is 2 x 10-6wb. Find the flux density, magnetic field

intensity and permeability of iron.

(b) Derive Bragg's law. Explain why x- rays and not y-ray are used for crystal structure

analysis. What data about the crystal structure can be obtained from the x-ray

diffraction pattern of a crystal. (4 + 2 + 1)

Sol.3 (a) l = 30 cm, A = 1 cm2 = 10–4 m2, N = 300, I = 0.032 A, = 2 × 10–6 wb

B = 6

2 2

4

2 102 10 wb/m

A 10

−−

−

= =

H = 2

2

NI 300 0.032A0.32 10

L 30 10 m−

−

= =

= 32 A/m

Ans.3 (b) Derivation of Bragg’s Law – O

P

E

G H

F

N

M

D

A

B

CP

2d sin

Fig.: X-ray Diffraction

Fig. shows a particular set of atomic planes in a crystal, d being interplanar spacing.

Suppose an X-ray beams in incident at a glancing angle θ. It is scattered by the atom like

A and B in random direction.

8

Constructive interference takes place only between those scattered waves which are

reflected and have a path difference of nλ, where λ is the X-ray wavelength and n is an

integer.

The path difference for the wave reflected from adjacent given by plane is

CB + BD = d sinθ + d sinθ = 2d sinθ

For constructive interference,

Path difference = nλ

n = 1, 2, 3 ....

According to Bragg’s equation 2d sin θ = nλ

For 1st and 3rd order the values of n are 1, 2, and 3 respectively and values of θ are θ1, θ2,

θ3 respectively.

Sin θ1 : Sin θ2 : Sin θ3 = 1: 2: 3

From the graph, the glancing angles θ1, θ2, θ3 can be measured and it can be seen that

Sin θ1 : Sin θ2 : Sin θ3 = 1: 2: 3

From the observed values of θ and known values of d and n, the wavelength of X-ray

can be calculated.

Suppose the wavelength of incident X-ray is known then the ratio of the interplanar

spacing can be determined.

Suppose, for a particular crystal used on a Bragg’s spectrometer, strong reflections from

thesets of planes (100), (110) and (111) are obtained for angles θ1, θ2, θ3 respectively in the

first order. then from Bragg’s equation we have

2d100 sinθ1 = λ

2d110 sinθ2 = λ

2d111 sinθ3 = λ

Hence, d100 : d110 : d111 = 1/sinθ1 : 1/sinθ2 : 1/sinθ3

Bragg found that for certain crystal: θ1 = 5o23’, θ2 = 7o36’ and θ3 = 9o25’

Hence for crystal

d100 : d110 : d111 = ''' 259sin

1:

367sin

1:

235sin

1ooo

= 744.1

1:

414.1

1:1

= 3

1:

2

1:1 for SC

2d sinθ = nλ

9

Theoretically, this ratio is found to hold a simple cubic lattice structure. Hence, it is

concluded that crystal has a simple cubic structure.

when the first order reflection from the three planes (100), (110) and (111) of NaCl are

compared, the ratios between the interplanar spacing are found as

For FCC d100 : d110 : d111 = 1 2

1 : :2 3

Which agree with the theoretical values for a face centered cubic structure. He

concluded that NaCl has fcc structure.

For BCC d100 : d110 : d111 = 2 1

1 : :2 3

Q.4 (a) Find out the critical radius ratio of an ionic crystal in ligancy 6 configuration. What is

the maximum size of cation in ligancy 6 configuration when the radius of anion is

2.02 A. 5

(b) In an n type semiconductor, the Fermi level lies 0.4 ev below the conduction band. If

the concentration of donor atom is doubled, find the new position of the Fermi level

W.r.t. the conduction band. 5

(c) Explain the origin of electronic, ionic and orientational polarization and temperature

dependence of respective polarizability. 5

Sol.4 (a) Critical radius ratio for ligancy 6:

A

B C

Fig1.26: Ligancy 6

In the above fig. Cation is in between 4 anions in a plane. The fifth anion lies on the upper layer and sixth anion in the bottom layer. Join A& B the cation-anion centers

respectively giving a complete picture of a .In , AC = BC

AB = &

BC =

Also

= = 0.7071

= 1/0.7071

For ligancy (6) rC/rA = 0.4142

ACB ACB 090=ACB

00 45&45 == BACABC

AC rr +

Ar

045coscos =ABC

AC

A

rr

r

+

A

AC

r

rr +

4142.11 =+A

C

r

r

10

Given- ra = 2.02

c

a

r

r = 0.414

rc = ra × 0.414 = 2.02 × 0.414

Sol.4 (b) Given:

EC – EF = 0.4 eV,

nd' = 2 nd,

T = 300 K

Formula:

nd = NC e(EF – EC)/KT

Calculation:

nd’ = NC e(EF’ – EC)/KT

d

d

n '

n =

( ) ( )F C F CE ' E E E /KTe

− − −

EC – EF’ = (EC – EF) – KT ln 2

= 0.4 – 0.018

EC – EF’ = 0.382 eV

The few fermi level will be 0.382 eV below the conduction band. Hence, the fermi level

will be shifted towards the conduction band by an amount (0.4 – 0.382) eV = 0.018 eV.

Ans.4 (c) (1) Electronic polarization: a displacement of the electronic cloud w.r.t the nucleus.

(2) Ionic polarization: separation of +ve and -ve ions in the crystal. (3) Orientational polarization: alignment of permanent dipoles (molecules). (4) Space-charge polarization: free electrons are present but are prevented from

moving by barriers such as grain boundaries - the electrons "pile up".

Q.5 (a) Find out the intercepts made by the planes. (1 0 1) and (4 1 4) in a cubic unit cell. 5

Draw [T 2 1] and [1 2 4] in a cubic unit cell.

(b) A bar of n type Ge of size 0.010m x 0.001m x 0.001m is mounted in a Magnetic field of

2 x 10-1T. The electron density in the bar is 7 x 1021/m3. If one milli volt is applied

across the long ends of the bar, determine the current through the bar and the voltage

between Hall electrodes placed across the short dimensions of the bar. Assume

μe = 0·39 m2/vs. 5

(c) Define reverberation time. Write Sabine's formula explaining every term. What are

the factors which determine the average absorption co-efficient of a material? 5

11

Ans.5 (a)

Sol.5 (b) H 21 19

1 1R

ne 7 10 1.6 10− −= =

28.92 10−=

dV E= H BH

R I; V

t= 3 30.39

10 39 100.01

− −= =

dI neAV= ( )221 19 37 10 1.6 10 0.001 39 10− −= 64.36 10 A−=

2 6 1

H

8.92 10 4.36 10 2 10V

0.001

− − − = 677.7824 10 V−=

Ans.5 (c) • Reverberation is the persistence of sound in a particular space after the original

sound is removed.

• A reverberation, or reverb, is created when a sound is produced in an enclosed

space causing a large number of echoes to build up and then slowly decay as the

sound is absorbed by the walls and air.

• This is most noticeable when the sound source stops but the reflections continue,

decreasing in amplitude, until they can no longer be heard. In a more reflective

room, it will take longer for the sound to die away and the room is said to be 'live'.

In a very absorbent room, the sound will die away quickly and the room will be

described as acoustically 'dead'.

• But the time for reverberation to completely die away will depend upon how loud

the sound was to begin with and will also depend upon the acuity of the hearing

of the observer. To quantitatively characterize the reverberation, the parameter

called the reverberation time is used.

It is defined as the time for the sound to die away to a level 60 decibels below its

original level.

Reverberation time = time required to drop original sound by 60dB below original level

http://en.wikipedia.org/wiki/Sound

http://en.wikipedia.org/wiki/Echo_%28phenomenon%29

http://en.wikipedia.org/wiki/Reflection_%28physics%29

http://en.wikipedia.org/wiki/Amplitude

http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/revtim.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/sound/db.html#c1

12

Prof. Wallace C. Sabine (1868 - 1919) of Harvard University investigated architectural

acoustics scientifically, particularly with reference to reverberation time. He deduced

experimentally, that the reverberation time is:

• directly proportional to the volume of the hall (V).

• inversely proportional to the effective absorbing surface area of the walls and the

materials inside the hall (A)

Revolumeof thehall

verberationtime Tabsorption

Reverberation time T = constantvolumeof thehall

absorption

=kV

TA

……(1)

Where k is proportionality constant having a value of 0.161 when the dimensions are

metric units.

0.161VT

A= ..…(2) It is rewritten as

1

1 1 2 2 3 3

0.161

0.161......(3)

........

=

=+ + + +

n

n n

n n

VT

a S

VT

a S a S a S a S

1 1 2 2 3 3

1

........n

n n n na S a S a S a S a S= + + + +

• Increasing the effective area of complete absorption like, changing the wall

materials or adding more furniture may decrease an excessive reverberation time

for a hall.

• But this also decreases the intensity of a steady tone.

• Also, too much absorption will make the reverberation time too short and cause

the room to sound acoustically 'dead'.

• Hence, the optimum reverberation time is a compromise between clarity of sound

and its intensity.

Determination of absorption coefficient: If T1 is reverberation time of an empty room,

then

Where 1

n

n nA a S= denotes the absorption due to the walls, flooring and ceiling of the

empty room.

Then a certain amount of absorbing material of area S and absorption coefficient a’ is

added in the room and again the reverberation time is measured.

1

1

0.161 0.161= =

n

n n

V VT

Aa S

13

Let it be T2

2

1 2

1 2

0.161

'

1 1 '

0.161

0.161 1 1'

VT

A a S

a S

T T V

Va

S T T

=+

− =

= −

Knowing the quantities on the RHS, of the above equation, a’ of the material under test

can be calculated.

Q.6 (a) Explain the differences between three different liquid crystal phases with respect to

the order in the arrangement of molecules, with the help of diagram. Which property

of the liquid crystal is used for display. 5

(b) How a p-n junction diode is used to generate a potential difference in a photovoltaic

solar cell. 5

(c) What is piezoelectric effect. Explain the working of a piezoelectric oscillator used to

produce ultrasonic wave. 5

Ans.6 (a) Liquid Crystal Phase: Depending upon the magnitude of forces between the molecules,

there are three types of mesophase of liquid crystal. They are

1. Smectic 2. Nematic 3. Cholesteric

(1) Smectic Mesophase

• These phases are found at lower temperature than nematic phase.

• Smectic crystals have a layer structure that can slide over another like soap.

Molecules align themselves in parallel layers gliding one relative to other, thereby

promoting fluidity.

• The long axes of molecules line up either perpendicular to the planes of layers

(smetic A phase) or form a certain angle with the normal to these planes

(smetic c phase).

• The value of this angle depends on the temperature of liquid crystal and can be

altered strongly by an externally applied field.

• There is a very large number of different smetic phase, all characterized by

different types and degrees of positional and orientational order.

…………..(4)

14

(2) Nematic Mesophase

• In this phase the molecules have no positional order, but they do have long range

orientational order.

• In this phase the molecules also have their axes parallel to each other, but the

location of their centers of gravity is as irregular as in conventional liquids (Fig.

20a).

• The entire nematic liquid crystal consists of small regions, each having its

molecules aligned parallel to a unique axes, but this generally varries in direction

in different regions of the crystal. Therefore nematic liquid crystal have high

optical inhomogennity due to which it appears opaque in the transmitted and

reflected light.

• The ordering of the molecules is a function of temperature.

• The molecular orientation (and hence the material’s optical properties) can be

controlled with applied electric fields.

• An externally applied field orients the molecules of all regions in the same

direction and the liquid crystal become transparent.

(3) Cholesteric Mesophase

• This phase is first observed for cholesterol derivatives so this name is Given.

• Only chiral molecules (those that lack inversion symmetry)

can given rise to such a phase.

• This phase exhibits a twisting of the molecules along the

director, with molecular axis perpendicular to the director.

• In going from one plane to the other, the vector L, called the

director, turns by a certain angle and it describes a helix with

a pitch of about 0.2 to 20 m

• The chiral pitch refer to the distance (along the director) over

which the mesogens undergo a full 360O twist (but note that

the structure repeats itself every half pitch, since the positive

and negative direction along the director are equivalent)

• The periodicity of the structure along helical axis results in

Bragg’s reflection of a light at wavelength equal to the pitch

devided by refractive index of the liquid crystal.

15

Liquid Crystal Displays (Based on 2, module 1, Ist edition, page 19)

• A liquid crystal display (LCD) is a thin, flat display device made up of any number of

colour or monochrome pixels arranged in front of a light source or reflector.

• It is suitable for use in battery- powered electronic devices because it uses very small

amount of electronic power.

• LCD is not a light emitting phenomenon (like LED), but rather requires a separate light

source and controls the reflection or transmission of that light.

• Liquid crystal display are passive devices that modify light scattering.

• The operation of liquid crystal display may be based on one of the two processes.

(i) Twisted Nematic field (ii) Dynamic scattering

• The most common liquid crystal device is twisted nematic device (See Fig.)

1. Vertical filter film to polarize the light it enters.

2. Glass substance with ITO electrodes. The shape of these electrodes will determine the

dark shapes that will appear when the LCD is turned on or off vertical ridges etched on

the surface are smooth.

3. Twisted nematic liquid crystals.

4. Glass substrate with common electrode film with horizontal ridges to line up with the

horizontal filter.

5. Horizontal filter film to block/allow through light.

6. Reflective surface to send light back to viewer.

Ans.6 (b) A solar cell (also called a photovoltaic cell) is an electrical device that converts the

energy of light directly into electricity by the photovoltaic effect. It is a form of

photoelectric cell (in that its electrical characteristics-- e.g. current, voltage, or

resistance-- vary when light is incident upon it) which, when exposed to light, can

generate and support an electric current without being attached to any external voltage

source.

The solar cell works in three steps:

1. Photons in sunlight hit the solar panel and are absorbed by semiconducting materials,

such as silicon.

2. Electrons (negatively charged) are knocked loose from their atoms, causing an electric

potential difference. Current starts flowing through the material to cancel the potential

and this electricity is captured. Due to the special composition of solar cells, the

electrons are only allowed to move in a single direction.

3. An array of solar cells converts solar energy into a usable amount of direct current (DC)

electricity.

http://en.wikipedia.org/wiki/Light

http://en.wikipedia.org/wiki/Electricity

http://en.wikipedia.org/wiki/Photovoltaic_effect

http://en.wikipedia.org/wiki/Photon

http://en.wikipedia.org/wiki/Sunlight

http://en.wikipedia.org/wiki/Electrons

http://en.wikipedia.org/wiki/Direct_current

16

Applications

• Used to power space satellites and smaller items like calculators and watches.

• Today, thousands of people power their homes and businesses with individual solar PV

systems.

• Utility companies are also using PV technology for large power stations.

Solar panels used to power homes and businesses are typically made from solar cells

combined into modules that hold about 40 cells. A typical home will use about 10 to 20

solar panels to power the home. The panels

Ans.6 (c) When mechanical compression is applied to the opposite faces of crystal-like quartz, a

voltage proportional to the applied pressure appears across the crystal.

Also, if the faces are subjected to tension instead of compression, the change is still

developed but of the opposite sign.

This phenomenon is known as piezo electric effect.

Conversely, when a voltage is applied to crystal surfaces the crystal is distorted (i.e.

compression or tension) is developed. This is identified as inverse piezoelectric effect.

Quartz SliceQuartz Slice

Fig (2) Piezo-electric effect

Piezoelectric Oscillator

Principle: Resonance is obtained between natural frequency of appropriately cut

piezoelectric crystal and a suitable frequency superimposed on it which is generated by

oscillator.

Circuit diagram

Fig (3) Valve based piezoelectric oscillator

About circuit: In the diagram we have a thin plate of piezoelectric crystal cut on such

axis so that we get inverse piezoelectric effect i.e. on application of high frequency

electric field, it can vibrate in resonance.

17

Which represent order of harmonic. In practice we take k = 1 that is fundamental

frequency. It is important to know that natural frequency n is inversely proportional to

thickness t. Hence for higher frequency, t has to be reduced.

How it works?

• It is a triode valve oscillator.

• The plate coil L2 is inductively coupled to grid coil L1.

• When the circuit is switched on, the valve starts functioning as oscillator producing

oscillations at a frequency given by

Where,

L2 is the inductor

C is the capacitor

The frequency of oscillations can be controlled by varying the capacitor C. By

transformer action, an emf is induced in the coil L which is parallel to the crystal. The

capacitor is varied till the frequency of oscillator matches with the natural frequency of

the piezoelectric crystal.

f = n

where,

f is frequency of oscillator

n is the natural frequency of the piezoelectric crystal

Under this condition crystal will generate oscillations with highest amplitude. The

crystal subjected to ac voltage produces ultrasonic waves in the surrounding air.

Advantages

• Higher frequency range.

• Small size and economical.

• Better waveform

Disadvantages

➢ Low power handling capacity.

18

Applied Physics – I (Sem-I) May-2014

(Revised Course)

QP Code: NP-17709

Time : 2 Hours Total Marks : 60

N.B.:- (1) Question no. 1 is compulsory.

(2) Attempt any three questions from Q.2 to 6.

(3) Use suitable data wherever required.

(4) Figures to the right indicate full marks.

1. Solve any five from the following: - 15

(a) What is x-ray? Why the x-rays are preferred to study crystalline solid.

(b) Represent the following in a cubic unit cell (021), (123), [121].

(c) Find the Miller Indices of a set of parallel planes which makes intercepts in the ratio

3a : 4b on the x and y axes and parallel to Z-axis.

(d) What is Fermi level and Fermi energy? Write Fermi-Dirac distribution function.

(e) Explain the concept of hole in a semiconductor.

(f) Draw the structure of quartz crystal and explain its various axes.

(g) State and explain ohm’s law in magnetic circuit?

2. (a) Describe the formation of energy band in solid? Explain how it helps to classify the

solids in to conductors, insulators and semiconductors with proper diagram. 8

(b) Explain Dimond crystal structure with proper diagram and determine its APF?

7

3. (a) Derive Bragg’s law and describe the powder method to determine crystal structure

of powdered specimen. 8

(b) The magnetic field strength of copper is 106 ampere/ metre and magnetic

susceptibility is –0.8 × 10–3. Calculate magnetic flux density and magnetization in

copper. 7

19

4. (a) What is liquid crystal state of matter? Draw the diagram to describe molecular

arrangement in their different phases? 5

(b) Mention different types of polarizability in a dielectric? Explain electronic

polarizability? 5

(c) Calculate electron and hole concentration in intrinsic silicon at room temperature if

its electrical conductivity is 4 × 10–4 mho/m. (mobility of electron = 0.14 m2/v-s &

mobility of hole = 0.040 m2/v-s) 5

5. (a) Explain with neat diagram construction and working of solar cell.

(b) State the acoustic requirements of good auditorium. Explain how these

requirements can be achieved. 5

(c) If the x-rays of wavelength 1.549 A° will be reflected from crystal having spacing of

4.225 A°, calculate the smallest glancing angle and highest order of reflection that

can be observed. 5

6. (a) Explain with neat diagram Hysterisis effect in ferromagnetic material. 5

(b) Explain piezoelectric oscillator to produce USW? 5

(c) Explain the formation of barrier potential in P-N Junction. 5

******************

20


Solutions

Q.1 Solve any five from the following: - 15

(a) What is x-ray? Why the x-rays are preferred to study crystalline solid.

(b) Represent the following in a cubic unit cell (021), (123), [121].

(c) Find the, miller indices of a set of parallel panes which makes intercepts in the

ratio 3a : 4b on the x and y axes and parallel to Z-axis.

(d) What is Fermi level and Fermi energy? Write Fermi-Dirac distribution function.

(e) Explain the concept of hole in a semiconductor.

(f) Draw the structure of quartz crystal and explain its various axes.

(g) State and explain ohm’s law in magnetic circuit?

Ans.1 (a) X-rays are electromagnetic radiation with wavelengths between about 0.02 A and

100 A (1A = 10-10 meters). The wavelength of X-rays is of the order of inter atomic

spacing and is much smaller than that of visible light (3000 to 8000 A). Since X-rays

have a smaller wavelength than visible light, they have higher energy and are more

penetrative. Its ability to penetrate matter, however, is dependent on density of the

matter. Therefore, X-rays are useful in exploring structures of atoms.

Ans.1 (b)

Sol.1 (c) Plan Miller to z-axis

Intercepent of plane 3a : 4b : ……(1)

1

3 ,

1

4 , 0 …….(1)

Miller indices are (4 3 0) ……..(1)

Ans.1 (d) Fermi – Energy or Fermi Level

• When electrons are filled in the energy levels, the universal rule is that the

lowest energy level gets filled first. However, there will be many more

allowed energy levels left vacant as shown in Fig. below,

The Fermi-Dirac distribution function, also called Fermi function, provides the

probability of occupancy of energy levels by Fermions (electrons are known as

21

Fermions). Fermions are half-integer spin particles, which obey the Pauli exclusion

principle. The Pauli exclusion principle states that only one Fermion can occupy a

single quantum state. Therefore, the Fermi dirac function provides the probability

that an energy level at energy, E, in thermal equilibrium with a large system, is

occupied by an electron. The system is characterized by its temperature, T, and its

Fermi energy, EF. The Fermi dirac function is given.

(2.5.1)

Fermi energy or Fermi level:

The energy of the highest occupied level at absolute zero-degree temperature is

called the Fermi energy and the level is referred to as the Fermi level EF.

• All the energy levels above the Fermi level at T = 00 K are empty and those

lying below are completely filled. EF may or may not be an allowed state.

• Provides a reference with which other energy levels can be compared.

Ans.1 (e) Semiconductors distinguish themselves from metals and insulators by the fact that

they contain an "almost-empty" conduction band and an "almost-full" valence band.

This also means that we will have to deal with the transport of carriers in both

bands.

Holes are missing electrons. They behave as particles with the same properties as

the electrons but opposite charge would have occupying the same states except that

they carry a positive charge.

Ans.1 (f) Quartz crystal belongs to trigonal system.

In its natural form it is a hexagonal prism with pyramids at each end.

The axis joining two end points is called optic axis (ZZ’)

The section of crystal at right angle to optic axis is a hexagon.

22

Fig. Quartz crystal

Various axis on quartz crystal

▪ Optic axis

▪ Electrical axis

▪ Mechanical axis

1) Optic axis

▪ The axis joining two end points of the pyramid is called optic axis.

2) Electrical axis

▪ Refer fig. where a section of quartz crystal, a hexagon is depicted.

▪ The lines passing through the opposite corner of the crystal are called X- axis or

electrical axis.

▪ Three such lines X1 X1 , X2 X2 , X3 X3 are denoted.

The piezoelectric effect is most marked along electrical axis.

3) Mechanical axis

▪ Refer fig. where a section of quartz crystal, a hexagon with three lines passing

through the sides of hexagon are shown.

▪ They are called Y- axis or mechanical axis.

▪ Three such lines Y1 Y1 , Y2 Y2 , and Y3 Y3 are shown. The inverse piezoelectric effect

is marked on mechanical axis.

Applications

Ultrasonic has its applications divided into two main categories:

1. Low intensity applications

2. High intensity applications

Sol.1 (g)

=NI

Hl

. .= =B A H A 0= rHA 0= r

INA

l

23

0

1.

=

r

NI

l

A

⎯⎯→NI MMF

0

⎯⎯→r

lRELUCTANCE

A

Q.2 (a) Describe the formation of energy band in solid? Explain how it helps to classify

the solids in to conductors, insulators and semiconductors with proper diagram.

8

(b) Explain Dimond crystal structure with proper diagram and determine its APF?

7

Ans.2 (a) Consider silicon electronic configuration is 1s2, 2s2, 2p6, 3s2, 3p2

• Point d : 3p and 3s levels along with 1s, 2s and 2p are not at all splitted.

• Point c : 3s and 3p electrons will be affected by presence of neighboring atoms and

at the same time 1s, 2s and 2p electrons will not be affected.

• Point b : splitting of 3s and 3p levels will be such that they start overlapping and a

composite band is developed. 8N states (6N from 3p + 2N from 3s) and 4N

electrons (2N from 3p + 2N from 3s) in 4N levels (3N from 3p + N from 3s). This

band is half filled 4N electrons occupy only 2N levels.

• Point a : Point a is obtained when interatomic spacing is further reduced. It is

noticed that composite band again splits into two (i.e. upper and lower bands). 4N

energy levels are equally divided between two bands i.e. 2N levels in upper and 2N

levels in lower each capable of filling 4N electrons.

• Lower band is called valence band which is full at low temperature.

• Upper band is called conduction band which is vacant

• Gap between valence and conduction bands are said to be forming forbidden

energy band gap.

Fig.: Energy level splitting in Si Crystal

24

Classification of Solids

Important characteristic of solid is its electrical conductivity. Solids are classified

into conductor, insulator and semiconductor, based upon the energy band structure.

Sol.2 (b) The nature of bonding in diamond is partly covalent and partly ionic. It consists of

two inter-penetrating face centered cubic sub lattices placed at (0, 0, 0) and (1/4,

1/4, 1/4) positions. The two sub-lattices, made up of carbon atoms are displaced

from each other along the body diagonal through a distance equal to one quarter of

the body diagonal. Hence, unit cell of diamond consist of eight atoms of carbon.

Thus coordination number is 4.

Assuming one corner atom as origin, the positional coordinates can be written as

follows.

(i) 8 corner atoms (each shared by 8 unit cells) with

(ii) 6 face centre atoms (each shared by 2 unit cells)

(iii) 4 interior atoms

(1) Number of atoms per unit cell = 42

16

8

18 ++

n = 8

Now dralbodydiagon == 24

1 dra == 23

4

1

(2) 8

3ar = is the atomic radius for diamond.

(3) Coordination no. = 4.

(4) Packing fraction = 3

3

3

4

a

rn

= 3

3

8

3

3

48

a

a

= 16

3

= 0.34

25

a = 0.357 nm0

a0

(5) Packing efficiency = 34%

(6) Void = 66%

Uses of diamond: • The primary use of diamond is as a hard object -- the hardest mineral known, which

makes it ideal for cutting tools and precision instruments. Seventy-five percent of diamonds are used this way, industrially. Examples include cutting tools, polishing hard metal, videodisc needles, and bearings for laboratory instruments. Other uses are – cutting and grinding tools such as drill bits and saws – abrasives that cut and polish other materials, including other gemstones – fine engraving tools with detailed precision

• Some blue diamonds are natural semiconductors, in contrast to most diamonds,

which are excellent electrical insulators. Naturally occurring diamonds are formed

over billions of years under intense pressure and heat. They are often brought to the

Earth’s surface by deep volcanic eruptions.

Q.3 (a) Derive the Bragg’s law and describe the powder method to determine crystal

structure of powdered specimen. 8

(b) The magnetic field strength of copper is 106 ampere/ metre and magnetic

susceptibility is –0.8 × 10–3. Calculate magnetic flux density and magnetization in

copper. 7

Ans.3 (a) Bragg’s Law: X-rays have short wavelength of the order of 10-10m. An ordinary

grating which has 6000 lines per cm cannot produce diffraction of X-rays. X-ray

diffraction was first studied by Max Laue by periodic array of atoms. The arranged

atoms in crystal grating correspond to the grating lines and the distance between

two atoms, which is of order 10-8 cm, forms a grating element which can be used for

X-ray diffraction. The difference between crystal grating and optical grating is that

atomic centers of diffraction in the crystal grating are not all in one plane but

distributed in space, while in case of optical grating they are limited to one plane.

26

The crystal is thus a three – dimensional space grating rather than a two –

dimensional plane grating. Bragg has given a very simple interpretation of the

diffraction pattern obtained by Laue. A crystal may be regarded as a stack of

parallel planes of atoms. There are several sets of planes oriented in different

directions. Bragg has shown that any diffracted ray can be regarded as if it were

reflected from one of these system of planes as though it were reflected from a

mirror parallel to planes. The diffracted beams are found only when reflection from

parallel planes of atom interfere constructively. X-rays penetrate the crystal, they are

scattered by the atom in the successive planes.

Derivation of Bragg’s Law – Fig. shows a particular set of atomic planes in a

crystal, d being interplanar spacing. Suppose an X-ray beams in incident at a

glancing angle θ. It is scattered by the atom like A and B in random direction.

Constructive interference takes place only between those scattered waves which are

reflected secularly and have a path difference of nλ, where λ is the X-ray wavelength

and n is an integer.

The path difference for the wave reflected from adjacent from plane is

CB + BD = d sinθ + d sinθ = 2d sinθ



n = 1, 2, 3 .... i.e. 2d sinθ = nλ

This is Bragg’s law

Rotating Crystal Method or a Bragg’s X-ray Spectrometer [Dec. 2003, May 2004,

Dec. 2004, May 2006, May 2010] (Based on 1, module 16, 16th edition, page 498)

Bragg devised an apparatus to investigate the structure of single crystal by using X-

rays. It is used to measure glancing angle θ.

27

He used crystal as reflection grating and not as transmission grating. The

experiment arrangement is shown in the diagram. X-rays from X-ray tube is

collimated using two adjustable slits, S1, and S2 are made to fall on a crystal C with

which in wax on the spectrometer table. The vernier scale V is attached to table on

which the crystal is placed. The vernier scale V is capable to move the circular scale

S and determine the position of the crystal. A strong monochromatic X-ray beam is

made to fall on the crystal face. The reflected beam after passing through the slit S3,

enters an ionisation chamber D mounted on an arm which can be rotated about the

same axis as the crystal. Its position can be read by second veiner V2. The gas in the

chamber is ionisation by the X-rays. The resulting ionisation current, measured by

the electrometer E, is a measure of the intensity of X-rays reflected by the crystal.

The crystal is rotated through small angles (while the arm carrying the isonisation

current is rotated through double the angles) and the ionisation current is measured

each time. The curve of intensity which is measure of ionisation current or

ionisation current against glancing angle θ is plotted. For certain value of glancing

angle θ, The intensity of ionisation current increase abruptly (Fig).

According to Bragg’s equation 2d sin θ = nλ

For 1st and 3rd order the values of n are 1, 2, and 3 respectively and values of θ are θ1,

θ2, θ3 respectively.

Sin θ1 : Sin θ2 : Sin θ3 = 1: 2: 3

From the graph, the glancing angles θ1, θ2, θ3 can be measured and it can be seen that

Sin θ1 : Sin θ2 : Sin θ3 = 1: 2: 3

From the observed values of θ and known values of d and n, the wavelength of X-

ray can be calculated.

Suppose the wavelength of incident X-ray is known then the ratio of the

interplanar spacing can be determined.

Suppose, for a particular crystal used on a Bragg’s spectrometer, strong reflections

from the sets of planes (100), (110) and (111) are obtained for angles θ1, θ2, θ3

respectively in the first order. then from Bragg’s equation we have

2d100 sinθ1 = λ

2d110 sinθ2 = λ

28

2d111 sinθ3 = λ

Hence, d100: d110: d111 = 1/sinθ1: 1/sinθ2: 1/sinθ3


Hence for crystal

d100 : d110 : d111 = ''' 259sin

1:

367sin

1:

235sin

1ooo =

744.1

1:

414.1

1:1

= 3

1:

2

1:1

for SC


concluded that crystal has a simple cubic structure.

When the first order reflection from the three planes (100), (110) and (111) of NaCl

are compared, the ratios between the interplanar spacing are found as

For FCC d100 : d110 : d111 = 1 2

1 : :2 3



For BCC d100 : d110 : d111 = 2 1

1 : :2 3

Sol.3 (b) H = 106 A/m, z = –0.3 × 10–3

Magnetization M = xH = –0.3 × 10–3 × 106 = –0.8 × 103 A/m

Magnetic field strength B = (M + H)

B = 4ST × 10–7 (–0.8 × 103 + 106) = 1.2556 wb/m2

Q.4 (a) What is liquid crystal state of matter? Draw the diagram to describe molecular

arrangement in their different phases? 5

(b) Mention different types of polarizability in a dielectric? Explain electronic

polarizability? 5

(c) Calculate electron and hole concentration in intrinsic silicon at room temperature

if its electrical conductivity is 4 × 10–4 mho/m. (mobility of electron = 0.14 m2/v-s &

mobility of hole = 0.040 m2/v-s) 5

Ans.4 (a) 1. Some crystalline solids, when heated pass through an intermediate phase

while going from solid to liquid state.

2. In these transitional states the substance retains the anisotropy of properties of

solid such as dielectric, optical, magnetic and others inherent in a crystal, and

they also simultaneously acquire the properties specific to liquid – fluidity to

form droplets viscocity etc.

3. The substances exhibiting such an intermediate, mesomorphic phase are

known as liquid crystal.

Liquid crystal are the substance that exhibit a phase of matter that has properties

between those of a conventional liquid, and those of a solid crystal.

29

Temperature of transition (Ts) : The temperature at which crystal solid converts

from solid phase to mesophase (liquid crystal phase) is called the temperature of

transition.

Transparency temperature (Tt): The temperature at which the liquid crystal

converts to the isotropic transparent liquid is called the transparency temperature

(Tt).

The temperature interval st TTT −= vary within wide limits for different liquid

crystal.

(Some crystalline solids, when heated pass through as intermediate phase while

going grow solid to liquid state)

Mesogens: The molecules that exhibit liquid crystal phase are called mesogens.

They are rigid and anisotropic (longer in one direction than another). Mesogen are

of two types.

1. Calamitic mesogens – They are like rigid rod which orient based on their long axis.

2. Discotic mesogens – They are disc like and these orients in the direction of their

short axis.

Weak van der Waal’s forces provide the ordered arrangement of molecules in the

liquid crystalline state. In addition to molecules, polymers and colloidal suspensions

can also form LC phase.

Liquid Crystal Phase

Depending upon the magnitude of forces between the molecules, there are three

types of mesophase of liquid crystal. They are

1. Smectic

2. Nematic

3. Cholesteric

Smectic Mesophase

• These phases are found at lower temperature than nematic phase.

• Smectic crystals have a layer structure that can slide over another like soap.

Molecules align themselves in parallel layers gliding one relative to other, thereby

promoting fluidity.

• The long axes of molecules line up either perpendicular to the planes of layers

(smetic A phase) or form a certain angle with the normal to these planes (smetic

c phase).

• The value of this angle depends on the temperature of liquid crystal and can be

altered strongly by an externally applied field.

• There is a very large number of different smetic phase, all characterized by different

types and degrees of positional and orientational order.

30

Nematic Mesophase

• In this phase the molecules have no positional order, but they do

have long range orientational order.

• In this phase the molecules also have their axes parallel to each

other, but the location of their centres of gravity is as irregular as in

conventional liquids (Fig. 20a).

• The entire nematic liquid crystal consists of small regions, each

having its molecules aligned parallel to a unique axe, but this

generally varries in direction in different regions of the crystal.

Therefore, nematic liquid crystal have high optical inhomogennity

due to which it appears opaque in the transmitted and reflected

light.

• The ordering of the molecules is a function of temperature.

• The molecular orientation (and hence the material’s optical

properties) can be controlled with applied electric fields.

• An externally applied field orients the molecules of all regions in the

same direction and the liquid crystal become transparent.

31

• Placing a thin layer of nematic liquid between the two electrodes enables aligning

the molecules both parallel and perpendicular (i.e. normal) to the planes of

conducting plates.

• The first type of alignment where molecules are parallel to conducting plates is said

to be planar and the second type of alignment where molecules are perpendicular to

the conducting planes is said to be hemotropic.

• The hemotropic structure unlike planar, has no effect on the polarization of light

transmitted through the cell at right angle to the layer of the liquid crystal.

• Special treatment of the surface of plate electrodes or requisite chemical aligning

agents introduced in the liquid give the desired orientation of molecules.

• An electric field procedure between the plates and various other factors can change

this orientation and this alter the optical properties of the liquid crystal.

Cholesteric Mesophase

• This phase is first observed for cholesterol derivatives so this name is Given.

• Only chiral molecules (those that lack inversion symmetry) can given rise to such a

phase.

• This phase exhibits a twisting of the molecules along the director, with molecular

axis perpendicular to the director.

• In going from one plane to the other, the vector L, called the director, turns by a

certain angle and it describes a helix with a pitch of about 0.2 to 20 m

• The chiral pitch refers to the distance (along the director) over which the mesogens

undergo a full 360O twist (but note that the structure repeats itself every half pitch,

since the positive and negative direction along the director are equivalent)

• The periodicity of the structure along helical axis results in Bragg’s reflection of a

light at wavelength equal to the pitch divided by refractive index of the liquid

crystal.

Ans.4 (b) (Beyond Syllabus) The type of polarization may be additionally subdivided into

the following categories:

1) Electronic polarization: a displacement of the electronic cloud w.r.t the nucleus.

2) Ionic polarization: separation of +ve and -ve ions in the crystal.

3) Orientational polarization: alignment of permanent dipoles (molecules).

4) Space-charge polarization: free electrons are present but are prevented from

moving by barriers such as grain boundaries - the electrons "pile up".

1) Electrical Polarization

When an electric field is applied to the dielectrics, the field exerts a force on each

positive charge in its own direction whereas negative charges are pushed in the

opposite direction. This results in creation of electric doublet or dipole in all the

atoms inside the dielectrics. The field pulls electrons more than it repels the nucleus

because electrons are far lighter than the protons and neutrons. The timescale for

32

polarization of a atom due to a field is 10 -15 seconds. This is about as fast as one can

do things.

3

-Zechargedensityof the chargedsphere = or

4 r

3

3

-3Zechargedensityof the chargedsphere =

4 r

lorentzforce=coulombforce

lorentzforce=charge×field =- ZeE

coulombforce=charge×field =- Ze×E

20

Qcoulombforce=Ze×

4 ε x

20

total - ve charge(Q)enclosedin sphereof raadiusxcoulombforce= charge×

4 ε x

heretotal - vecharge(Q)enclosedinsphereof raadiusx =

chargedensityof electrons×volumeof thesphere

3

3

-3Ze 4πxtotal - ve charge(Q)enclosedin sphereof raadiusx = ×

4 r 3

3

3

-ZexQ =

R

3

2 23

2 30 0

-Zex-Z e xRcoulombforce= Ze× =

4 ε x 4 ε R

ATEQUILIBRIUM

lorentzforce=coulombforce

2 2

30

-Z e x-ZeE =

4 ε R

30

-Z e xE =

4 ε R

33

304 ε R E

x =Ze

=Dipolemoment Ze x

3

0e

4=

R EZe

Ze

3

e 04= R E

e E

e = eE

3

04=e R iscalled electronic polarization

Sol.4 (c) 6i = 4 × 10–4 mho/m

e = 0.14 m2/vs, n = 0.040 m2/vs

6i = nie (e + n)

ni = ( )

i6

e e n + =

( )

–4

–19

4 10

1.6 10 0.14 0.040

+ = 1.388 × 1016 /m3

Q.5 (a) Explain with neat diagram construction and working of solar cell.

(b) State the acoustic requirements of good auditorium. Explain how these

requirements can be achieved. 5

(c) If the x-rays of wavelength 1.549 A° will be reflected from crystal having spacing

of 4.225 A°, calculate the smallest glancing angle and highest order of reflection

that can be observed. 5

Ans.5 (a) A solar cell (also called a photovoltaic cell) is an electrical device that converts the




generate and support an electric current without being attached to any external

voltage source.


1. Photons in sunlight hit the solar panel and are absorbed by semiconducting

materials, such as silicon.

2. Electrons (negatively charged) are knocked loose from their atoms, causing an

electric potential difference. Current starts flowing through the material to cancel

the potential and this electricity is captured. Due to the special composition of solar

cells, the electrons are only allowed to move in a single direction.

3. An array of solar cells converts solar energy into a usable amount of direct current

(DC) electricity.








34

Ans.5 (b) Following are the acoustic requirements of good auditorium:

Site Selection:

• Plan the auditorium on quiet exposure, far away from highways, flight path, and

railway stations.

• Location within the building, use corridors and quiet buffer spaces to isolate the

auditorium.

• Treat corridors and lobbies with sound absorbing materials. All doors should be

solid and bracketed around their entire perimeter.

Purpose :

Depending on the purpose a good frequency response sound system will be

required.

Volume:

• The auditorium is so shaped that the audience is as close to the sound source as

possible.

• A fan shaped auditorium with a balcony is desirable to ensure a free flow of direct

sound waves to listeners.

• In an auditorium with cushioned seats and a sound absorbing rear wall for echo

control, the average ceiling height H is usually H=20T where T is the mid frequency

reverberation time in seconds.

• Seating geometry is arranged to give all the audience good sight lines and at the

shortest distance from the stage.

Reverberation time:

• Sabine’s formula for reverberation time is used for suitable acoustic treatment.

• Where ‘V’ is the volume of the hall, ‘S’ is the surface area and ‘a’ is the absorbing

coefficient.

Ceiling:

Central area of the ceiling should be sound reflecting. The perimeter and rear to be

provided with sound absorbing materials like acoustic tiles.

Sidewalls:

35

Sidewalls should be sound reflecting and diffusing with as many irregularities as

possible. For example, making doorway wider at one side of the wall keeping

windows etc., and the back wall is treated with deep sound absorbing finish.

Floor :

All aisles are carpeted except in front of the stage to make full noise control. Fabric

upholstered seats are used. Absorptive and cushioned seats will give stable

reverberation.

Balconies :

Use balconies to increase seating capacity and to reduce the distance to the farthest

row of seats.

Sound reinforcement system:

In large halls a sound amplification system to reinforce the sound to a weak source

in a large room is required. In addition there should be adequate loudness in every

part of the auditorium uniform distribution (diffusion) of sound energy in the room.

The hall should be free from echoes, long delayed reflections, flatter echoes, sound

concentrations, distortions, and sound shadow and room resonance.

Seats:

To make the hearing conditions satisfactory when the room is full or partly full,

upholstered seats with absorbing material at the bottom are used, so that the

absence or presence of audience does not affect the reverberation time.

Sol.5 (c) = 1.549 × 10–10 m

d = 4.255 × 10–10 m

For smallest glancing angle , n = 1

Bragg’s law 2dsin =

= 1sin2d

−

= 1.4875

For highest order of reflection sinmax = 1

2d = nmax

Nmax = 2d

= 5.4913

5

Q.6 (a) Explain with neat diagram Hysterisis effect in ferromagnetic material. 5

(b) Explain piezoelectric oscillator to produce USW? 5

(c) Explain the formation of barrier potential in P-N Junction. 5

Ans.6 (a) (Beyond Syllabus) A great deal of information can be learned about the magnetic

properties of a material by studying its hysteresis loop. A hysteresis loop shows the

relationship between the induced magnetic flux density (B) and the magnetizing

force (H). It is often referred to as the B-H loop. An example hysteresis loop is

shown below.

36

FIG: Hysterisis Curve

The loop is generated by measuring the magnetic flux of a ferromagnetic material

while the magnetizing force is changed. A ferromagnetic material that has never

been previously magnetized or has been thoroughly demagnetized will follow the

dashed line as H is increased. As the line demonstrates, the greater the amount of

current applied (H+), the stronger the magnetic field in the component (B+). At

point "a" almost all of the magnetic domains are aligned and an additional increase

in the magnetizing force will produce very little increase in magnetic flux. The

material has reached the point of magnetic saturation. When H is reduced to zero,

the curve will move from point "a" to point "b." At this point, it can be seen that

some magnetic flux remains in the material even though the magnetizing force is

zero. This is referred to as the point of retentivity on the graph and indicates the

remanence or level of residual magnetism in the material. (Some of the magnetic

domains remain aligned but some have lost their alignment.) As the magnetizing

force is reversed, the curve moves to point "c", where the flux has been reduced to

zero. This is called the point of coercivity on the curve. (The reversed magnetizing

force has flipped enough of the domains so that the net flux within the material is

zero.) The force required to remove the residual magnetism from the material is

called the coercive force or coercivity of the material.

As the magnetizing force is increased in the negative direction, the material will

again become magnetically saturated but in the opposite direction (point "d").

Reducing H to zero brings the curve to point "e." It will have a level of residual

magnetism equal to that achieved in the other direction. Increasing H back in the

positive direction will return B to zero. Notice that the curve did not return to the

origin of the graph because some force is required to remove the residual

magnetism. The curve will take a different path from point "f" back to the saturation

point where it with complete the loop. From the hysteresis loop, a number of

primary magnetic properties of a material can be determined.

1. Retentivity - A measure of the residual flux density corresponding to the saturation

induction of a magnetic material. In other words, it is a material's ability to retain a

37

certain amount of residual magnetic field when the magnetizing force is removed

after achieving saturation. (The value of B at point b on the hysteresis curve.)

2. Residual Magnetism or Residual Flux - the magnetic flux density that remains in a

material when the magnetizing force is zero. Note that residual magnetism and

retentivity are the same when the material has been magnetized to the saturation

point. However, the level of residual magnetism may be lower than the retentivity

value when the magnetizing force did not reach the saturation level.

3. Coercive Force - The amount of reverse magnetic field which must be applied to a

magnetic material to make the magnetic flux return to zero. (The value of H at point

c on the hysteresis curve.)

4. Permeability, - A property of a material that describes the ease with which a

magnetic flux is established in the component.

5. Reluctance - Is the opposition that a ferromagnetic material shows to the

establishment of a magnetic field. Reluctance is analogous to the resistance in an

electrical circuit.

Ans.6 (b) Principle: Resonance is obtained between natural frequency of appropriately cut

piezoelectric crystal and a suitable frequency superimposed on it which is generated

by oscillator.

Circuit diagram

Fig: Valve based piezoelectric oscillator

About circuit: In the diagram we have a thin plate of piezoelectric crystal cut on

such axis so that we get inverse piezoelectric effect i.e. on application of high

frequency electric field, it can vibrate in resonance.

Which represent order of harmonic. In practice we take k = 1 that is fundamental

frequency. It is important to know that natural frequency n is inversely proportional

to thickness t. Hence for higher frequency, t has to be reduced.

38

How it works?

• It is a triode valve oscillator.

• The plate coil L2 is inductively coupled to grid coil L1.

• When the circuit is switched on, the valve starts functioning as oscillator producing

oscillations at a frequency given by

Where,

L2 is the inductor

C is the capacitor

The frequency of oscillations can be controlled by varying the capacitor C . By

transformer action, an emf is induced in the coil L which is parallel to the crystal.

The capacitor is varied till the frequency of oscillator matches with the natural

frequency of the piezoelectric crystal.

f = n

where,

f is frequency of oscillator

n is the natural frequency of the piezoelectric crystal

Under this condition crystal will generate oscillations with highest amplitude. The

crystal subjected to ac voltage produces ultrasonic waves in the surrounding air.

Advantages

• Higher frequency range.


• Better waveform

Disadvantages


Ans.6 (c) The negative terminal of the external source causes an increase in electron energy

and hence an upward shift of all energy levels on the n-side. Similarly, the positive

terminal connected to the p-side causes an increase in hole energy and hence a

lowering of all levels on p-side . As the displacements of the energy levels occur in

opposite directions, the Fermi levels EFn and EFp get separated by a value eV. And

also the bending of the bands near the depletion region is reduced. In other words,

the heights of the conduction hill and valence hill are reduced by an amount of

energy eV. Due to reduction in barrier heights the movements of the majority

carriers is promoted. As a result, the components Jhp and Jen increase.

39

Fig.: Enegery band structure of a forward bised pn junction

Reverse biased pn- junction:

If the semiconductor diode is now connected to the DC power supply. The holes in

the p-type material and the electrons in the n-type material will be attracted by the

negative and positive terminals of the supply and will therefore, move away from

the junction. The flow of charge across the junction will therefore be zero. The

region around the junction in fact loses its charge and becomes an insulator. Under

these conditions very little electrical current can pass through the p-n junction. The

diode has a very high electric resistance and is said to be reverse biased.

Fig. Reverse biased pn-junction

Energy band structure of reverse biased pn junction:

The positive terminal of the external source connected to the n-side reduces the

energy of the electrons in general. Therefore, the energy levels on n-side are pulled

down. The negative terminal connected to p-side reduces the energy of holes in p-

region and hence the energy levels are pushed up on p-side. Due to such

displacement of energy levels on both sides, the Fermi levels EFp and EFn get

separated by an amount eV.

Fig. Energy band structure of feverse biased pn-junction

40

Applications of P-N junction diode

LED

• LEDs are semiconductor diodes, electronic devices that permit current to flow in

only one direction.

• The diode is formed by bringing two slightly different materials together to form a

PN junction (Figure ).

• In a PN junction, the P side contains excess positive charge ("holes," indicating the

absence of electrons) while the N side contains excess negative charge (electrons).

When a forward voltage is applied to the semiconducting element forming the PN

junction (heretofore referred to as the junction), electrons move from the N area

towards the P area and holes move towards the N area. Near the junction, the

electrons and holes combine. As this occurs, energy is released in the form of light

that is emitted by the LED.

Applications

• Used extensively in optical fiber and Free Space Optics communications.

• Used in remote controls, such as for TVs, VCRs, and LED Computers, where

infrared LEDs are often used.

• Opto-isolators use an LED combined with a photodiode or phototransistor to

provide a signal path with electrical isolation between two circuits.

• used as indicator lights on electronic equipment

• Car with LEDs used in its high-mounted brake light

• LEDs used on a train for both overhead lighting and destination signage.

• LED digital display that can display digits and points

http://www.lrc.rpi.edu/programs/nlpip/lightingAnswers/led/whatIsAnLED.asp

http://www.lrc.rpi.edu/programs/nlpip/lightingAnswers/led/whatIsAnLED.asp

http://en.wikipedia.org/wiki/Optical_fiber

http://en.wikipedia.org/wiki/Free_Space_Optics

http://en.wikipedia.org/wiki/Remote_control

http://en.wikipedia.org/wiki/Opto-isolator

http://en.wikipedia.org/wiki/Photodiode

http://en.wikipedia.org/wiki/Phototransistor

http://en.wikipedia.org/wiki/File:PnJunction-LED-E.svg

41


(Revised Course) QP Code: 1027






1. Solve any five from the following: - 15

(a) Draw the following in a cubic unit cell

(0 1 2), (1 2 3) , [1 2 1]

(b) Define the term space lattice, unit cell and lattice parameter.

(c) Determine the lattice constant for FCC lead crystal of radius 1.746 A° and also

find the spacing of (2 2 0) plane.

(d) Define : drift current, diffusion current and mobility of charge carriers.

(e) What is the probability of an electron being thermally promoted to conduction

band in diamond at 27°C, if bandgap is 5.6 eV wide.

(f) Why soft magnetic materials are used in core of transformers?

(g) Calculate the electronic polarizability of Ar. Given number of Ar atoms at NTP =

2.7 × 1025 /m3 and dielectric constant of Ar = 1.0024.

2. (a) Show that for intrinsic semiconductors the Fermi level lies midway between the

conduction band and the valence band. Draw the energy level diagram as a

function of temperature for n-type of semi-conductor. 8

(b) Cu has FCC structure. If the interplanar spacing d is 2.08 A° for the set of (111)

planes. Find the density and diameter of C atom. Given atomic weight of Cu is

63.54. 7

3. (a) What is hysteresis? Draw a hysteresis loop for ferromagnetic material and

explain the various important points on it. For a transformer which kind of

material will you prefer-the one with small hysteresis area of the big one? 8

42

(b) Derive Bragg’s law. X-rays of unknown wavelength give first order Bragg’s

reflection at glancing angle of 20° with (2 1 2) planes of copper having FCC

structure. Find the wavelength of X-rays, if the lattice constant for copper is 3.615

A°. 7

4. (a) Discuss Diamond structure with neat diagram and also determine the effective

number of atoms/unit cell, co-ordination number and atomic radius in terms of

lattice constant. 5

(b) Classify solids on the basis of energy band diagram. 5

(c) Explain orientational polarization with suitable diagram and write the

mathematical expression of orientational polarizability. 5

5. (a) Calculate the number of atoms per unit cell of a metal having the lattice

parameter 2.9 A° and density 7.87 gm/cm3. Atomic weight of metal 55.85.

Avogadro number is 6.023 × 1023/gm mole. 5

(b) What is Hall effect? Mention its significance. How mobility can be determined

by using Hall effect? 5

(c) The reverberation time is found to be 1.5 second for an empty Hall and it is

found to be 1.0 second when a curtain cloth of 20m3 is suspended at the centre of

the Hall. If the dimensions of the hall are 10 × 8 × 6 m3, calculate the coefficient

of absorption of curtain cloth. 5

6. (a) Describe principle, construction and working of magnetostriction oscillator to

produce ultrasonic waves. 5

(b) Explain various point defects in crystals. 5

(c) Explain how a voltage difference is generated in a p-n junction when it is used in

a photovoltaic solar cell. 5

*********************

43


Solution

Q.1 Solve any five from the following: - 15

(a) Draw the following in a cubic unit cell

(0 1 2), (1 2 3) , [1 2 1]

(b) Define the term space lattice, unit cell and lattice parameter.

(c) Determine the lattice constant for FCC lead crystal of radius 1.746 A° and also

find the spacing of (2 2 0) plane.

(d) Define: drift current, diffusion current and mobility of charge carriers.

(e) What is the probability of an electron being thermally promoted to conduction

band in diamond at 27°C, if bandgap is 5.6 eV wide.

(f) Why soft magnetic materials are used in core of transformers?

(g) Calculate the electronic polarizability of Ar. Given number of Ar atoms at

NTP = 2.7 × 1025 /m3 and dielectric constant of Ar = 1.0024.

Ans.1 (a)

Ans.1 (b) Space Lattice: The three-dimensional network of regularly arranged point is

known as space lattice or a lattice.

Unit Cell: The smallest three-dimensional portion which repeats itself in

different directions to generate the complete space lattice. This is called a unit

cell. It is the smallest volume that carries a full description of the entire lattice.

Lattice Parameters: Three primitive vectors a, b and c define the length of the

three edges of the unit cell and represent the crystallographic axes. Let α, β, γ

represent the angles between the edges a and b, b and c, c and a respectively.

The axial length a, b, c and the three inter axial angles α, β, and γ are known as

lattice parameters.

44

Sol. 1 (c) Given: r = 1.746 A°, Plane (2 2 0)

a = 2 2r = 2 × 1.44 × 1.746 = 4.938 A°

d = 2 2 2

a

h k l+ +

( )2 20 2 2

4.938d 1.746A

2 2 0= =

+ +

Ans.1 (d) Net displacement in the electron's position per unit time caused by the

application of an electric field becomes a constant at the steady state. Velocity of

the electrons in the steady state in an applied electric field is called the drift

velocity.

Mobility:

The mobility of electron is defined as the magnitude of the drift velocity

acquired by the electrons in a unit field.

E - applied electric field

d -drift velocity

-mobility

= E

d ..….(6)

J = E

= E

J and J =

A

I

= AE

I

Current = I = ned A*

= AE

Adνne = ne

= ne

σ ….(7)

mobility of electron = 1350 cm2/ v-s

mobility of hole is = 480 cm2/v-s

The ease with which electrons could drift in material under the influence of an

electric field called as mobility.

Mobility of electron is > mobility of holes

Current (I):

I = nedA

A- area of cross section, I-current, v - velocity of electrons,

Sol.1 (e) Given: T = 27° C, Eg = 5.6 eV

( )1

F ECEg

1 exp2KT

=

+

( )1

F EC5.6eV

1 exp2 0.026

=

+

= 1.7 × 10–47

45

Ans.1 (f) In order to lower the magnetizing current drawn by the transformer.

Even if both windings are tightly coupled (wound very close together), so that

there is approximately no leakage between them, a magnetomotive force is

necessary to magnetize the air shared by both windings.

If you compare both inductances (air and soft-iron), the inductance of air is

much lower. Therefore, for a given flux linkage (time integral of voltage), the

(magnetizing) current will be much higher for an air-core transformer.

Sol.1 (g) (Beyond the syllabus)

Given: N = 2.7 × 1025/m3, Er = 1.0024

P = E0 (Er – 1) E’ …….(1)

P = de NE

Thus de = P

NE

( )0E Er 1

N

−=

( )12

25

8.85 10 1.0024 1

2.7 10

− −=

40 27.9 10 F m−= −

Q.2 (a) Show that for intrinsic semiconductors the Fermi level lies midway between

the conduction band and the valence band. Draw the energy level diagram as

a function of temperature for n-type of semi-conductor. 8

(b) Cu has FCC structure. If the interplanar spacing d is 2.08 A° for the set of (111)

planes. Find the density and diameter of C atom. Given atomic weight of Cu is

63.54. 7

Ans.2 (a) EF lies mid-way between conduction and valence band.

At any temperature T > 00 K,

ne = Number of electrons in conduction band


We have

ne = Nce –(EC-EF)/KT …..(1)

Nc = Effective density of states in conduction band

and nv = Nve –(EF – EV) / KT …..(2)

Nv = effective density of states in valence band



nC = nv

( ) ( )C F F VE E E E

KT KTC VN e N e

− −− −

=

( )

( )

C F

F V

E EKT

V

E EKT C

Ne

Ne

−−

−−

=

C F F VE E E E

VKT

C

Ne

N

− + + −

=

( )F C V2E E E

VKT

C

Ne

N

− +

=

as V

C

N

N= 1 for intrinsic semiconductor & taking ‘ln’ on both side

46

( )F C V2E E E

0KT

− +=

(EC + EV) = 2 EF

EF = 2

vE+CE … (4)

Therefore, Fermi level in an intrinsic semiconductor lies at the center of

forbidden energy gap.

Sol.2 (b) Given: n = 4, d = 2.08 A° = 708 × 10–10 m (h k l) = (1 1 1)

2 2 2

ad

h k l=

+ + a 2.08 3 = = 3.60 A°

3 Aa n

N =

3

nA 1.

N a =

( )326 10

4 63.54 1

6.023 10 3.60 10−

=

39038.8 kg/m =

For FCC, a 2

r4

= 3.6 2

4

= = 1.27 A°

Diameter (d) = 2r = 2.54 A°

Q.3 (a) What is hysteresis? Draw a hysteresis loop for ferromagnetic material and

explain the various important points on it. For a transformer which kind of

material will you prefer-the one with small hysteresis area of the big one? 8

(b) Derive Bragg’s law. X-rays of unknown wavelength give first order Bragg’s

reflection at glancing angle of 20° with (2 1 2) planes of copper having FCC

structure. Find the wavelength of X-rays, if the lattice constant for copper is

3.615 A°. 7

Ans.3 (a) (Beyond the syllabus)

A great deal of information can be learned about the magnetic properties of a

material by studying its hysteresis loop. A hysteresis loop shows the relationship

between the induced magnetic flux density (B) and the magnetizing force (H). It

is often referred to as the B-H loop. An example hysteresis loop is shown below.

47


The loop is generated by measuring the magnetic flux of a ferromagnetic

material while the magnetizing force is changed. A ferromagnetic material that

has never been previously magnetized or has been thoroughly demagnetized

will follow the dashed line as H is increased. As the line demonstrates, the

greater the amount of current applied (H+), the stronger the magnetic field in the

component (B+). At point "a" almost all of the magnetic domains are aligned and

an additional increase in the magnetizing force will produce very little increase

in magnetic flux. The material has reached the point of magnetic saturation.

When H is reduced to zero, the curve will move from point "a" to point "b." At

this point, it can be seen that some magnetic flux remains in the material even

though the magnetizing force is zero. This is referred to as the point of

retentivity on the graph and indicates the remanence or level of residual

magnetism in the material. (Some of the magnetic domains remain aligned but

some have lost their alignment.) As the magnetizing force is reversed, the curve

moves to point "c", where the flux has been reduced to zero. This is called the

point of coercivity on the curve. (The reversed magnetizing force has flipped

enough of the domains so that the net flux within the material is zero.) The force

required to remove the residual magnetism from the material is called the

coercive force or coercivity of the material.





positive direction will return B to zero. Notice that the curve did not return to

the origin of the graph because some force is required to remove the residual

magnetism. The curve will take a different path from point "f" back to the

saturation point where it with complete the loop.

From the hysteresis loop, a number of primary magnetic properties of a material

can be determined.

1. Retentivity - A measure of the residual flux density corresponding to the

saturation induction of a magnetic material. In other words, it is a material's

ability to retain a certain amount of residual magnetic field when the

48

magnetizing force is removed after achieving saturation. (The value of B at point

b on the hysteresis curve.)

2. Residual Magnetism or Residual Flux - the magnetic flux density that remains

in a material when the magnetizing force is zero. Note that residual magnetism

and retentivity are the same when the material has been magnetized to the

saturation point. However, the level of residual magnetism may be lower than

the retentivity value when the magnetizing force did not reach the saturation

level.

3. Coercive Force - The amount of reverse magnetic field which must be applied to

a magnetic material to make the magnetic flux return to zero. (The value of H at

point c on the hysteresis curve.)





electrical circuit.

Sol.3 (b) Derivation of Bragg’s Law –Fig.shows a particular set of atomic planes in a

crystal, d being interplanar spacing. Suppose an X-ray beams in incident at a

glancing angle θ. It is scattered by the atom like A and B in random direction.

Constructive interference takes place only between those scattered waves which

are reflected secularly and have a path difference of nλ, where λ is the X-ray

wavelength and n is an integer.

The path difference for the wave reflected from adjacent from plane is

CB + BD = dsinθ + d sinθ

= 2dsinθ

O

P

E

G H

F

N

M

D

A

B

CP

2d sin

Fig 1.18 X rays



Path difference = CB + BD

= dsin + dsin

= 2dsin

i.e. n = 1, 2, 3 ....

This is Bragg’s law

2dsinθ = nλ

49

Given: = 20°, a = 3.615 A°, (h k l) = (2 1 2)

2 2 2

ad

h k l=

+ + =

2 2 2

3.615

2 1 2+ += 1.205 A°

Bragg’s law is, n 2dsin =

2dsin

n

=

2 1.205 sin 20

1

=

0.824A =

Q.4 (a) Discuss Diamond structure with neat diagram and also determine the

effective number of atoms/unit cell, co-ordination number and atomic radius

in terms of lattice constant. 5

(b) Classify solids on the basis of energy band diagram. 5

(c) Explain orientational polarization with suitable diagram and write the

mathematical expression of orientational polarizability. 5

Ans.4 (a) Structure of diamond is similar to the structure of Zinc sulphide. When the Zinc

and sulphur ions are replaced by identical carbon atoms, the diamond structure

results. Silicon and germanium also have diamond structure.

The nature of bonding in diamond is partly covalent and partly ionic. It consists

of two inter-penetrating face centred cubic sub lattices. The two sub-lattices,

made up of carbon atoms are displaced from each other along the body diagonal

through a distance equal to one quarter of the body diagonal. Hence, unit cell of

diamond consists of eight atoms of carbon.

Thus, coordination number is 4.


follows.

(i) 8 corner atoms (each shared by 8 unit cells) with positional coordinates

(000) (l00) (010) (001)

(110) (101) (011) (111)

(ii) 6 face centre atoms (each shared by 2 unit cells) with positional coordinates

(½ ½ 0) (½ 0 ½) (0 ½ ½)

(½ ½ 1) (½ 1 ½) (1 ½ ½)

(iii) 4 interior atoms with positional coordinates

(¼ ¼ ¼) (¾ ¾ ¼) (¾ ¼ ¾) (¼ ¾ ¾)

Thus the structure of diamond contains eight atoms per unit cell.

50

(1) Number of atoms per unit cell = 42

16

8

18 ++

n = 8


1

dra == 234

1

(2) 8


(3) Coordination no. = 4.

(4) Packing fraction = 3

3

3

4

a

rn

= 3

3

8

3

3

48

a

a

= 16

3 = 0.34

(5) Packing efficiency = 34%

(6) Void = 66%

Diamond is a giant covalent network structure, having each Carbon atom

sharing electrons with four other Carbon atoms, therefore having four single

covalent bonds formed. These Carbon covalent bonds are extremely strong and

account for two of diamond’s most prominent physical properties among all

elements, hardness and a high melting point.

Ans.4 (b) "Energy band structure determines whether solid is a conductor, an insulator or

a semiconductor".

• Electronic configuration of Si is 1s2, 2s2, 2p6, 3s2, 3p2

• Point d : 3p and 3s levels along with 1s, 2s and 2p are not at all splitted.

• Point c : 3s and 3p electrons will be affected by presence of neighbouring atoms .

• Point b : 3s and 3p levels start overlapping and a composite band is developed.

• Point a : we get point a when interatomic spacing is further reduced.



• Gap between valence and conduction bands called as forbidden energy band

gap.

Fig. (2) Energy level splitting in Si Crystal

51

2.9.3 Classification of Solids (May-2010, Dec-2008)

(Based on 1, Chapter 17, 7th edition page 17.9)

Solids are classified into conductor, insulator and semiconductor, based upon

the energy band structure.

Sr.No. Conductor Insulator Semiconductor

1. V.B and C.B are

overlapping.

V.B is full & C.B is

completely empty.

V.B is full and C.B has

few electrons.

2. Large number of

electrons.

electrical conduction in

such solids is

insignificant.

electrons can move

under the influence of

even a small applied field

and cause current flow.

3. Current is due to flow of

electrons.

width of for-bidden gap

is large .

width of forbidden gap

is small

4. Eg: Li, Be Eg: Diamond

(Eg = 5.47 eV )

Eg: Si (Eg = 1.12 eV),

Ge (Eg = 0.72 eV)

Ans.4 (c) (Beyond Syllabus) Orientation polarization: it is even slower than ionic

polarization. The relaxation time for orientation polarization in liquid is less than

that in solids. For example; the relaxation time for orientation polarization is 10-10

sec in liquid propyl alcohol while it is 3x10-6 sec in solid ice. The orientation

polarization occurs, when the frequency of the applied voltage is in the audio

range (i.e 20 Hz to 20000 Hz).

Summary of polarizations

52

Space charge polarization: it is the slowest process, as it involves the diffusion

of ions over several inter atomic distances. The relaxation time for this process is

related to the frequency of successful jumps of ions under influence of applied

field, a typical value being 102 Hz. Correspondingly, space charge polarization

occurs at power frequencies (50-60 Hz)

Q.5 (a) Calculate the number of atoms per unit cell of a metal having the lattice

parameter 2.9 A° and density 7.87 gm/cm3. Atomic weight of metal 55.85.

Avogadro number is 6.023 × 1023/gm mole. 5

(b) What is Hall effect? Mention its significance. How mobility can be

determined by using Hall effect? 5

(c) The reverberation time is found to be 1.5 second for an empty Hall and it is

found to be 1.0 second when a curtain cloth of 20m3 is suspended at the centre

of the Hall. If the dimensions of the hall are 10 × 8 × 6 m3, calculate the

coefficient of absorption of curtain cloth. 5

Sol.5 (a) a = 2.9 A° = 2.9 × 10–8 cm

N = 6.023 × 1023 / 9m – mole

3 Ma n

N =

3Na

nM

=

( )323 86.023 10 2.9 10 7.87

55.85

− = 2=

Hence it is BCC.

Ans.5 (b) Based on 1, Chapter 18 , 7th edition page 18.24)

• If a metal or semiconductor, carrying a current I is placed in a transverse

magnetic field B, an electric field E is induced in the direction perpendicular to

both I and B. This phenomenon is known as Hall effect and the electric field or

voltage induced is called Hall voltage (VH).

Fig (14) : Hall effect

Experimental determination

• In equilibrium condition,

q EH = B q v

EH = VH / d

J = n e v = I / w d

VH = EH d = B v d = new

BI

ne

BJd= ….(1)

53

As EH = ne

BJBv

d

VH == …..(2)

now J = nev = A

I …..(3)

also, J = E

= E

J =

A

I. E

1 …..(4)

EH = E

E.

A

I.

ne

B=

A

I.

ne

B=

ne

BJ

)E

1.

A

I(.

ne

B=

E

HE

Using Equation (4) ρne

B=σ.

ne

B=

E

HE ……(5)

where = σ

1

ρne

B=

E

HE …..(6)

• Hall coefficient

RH = BI

WHV=ne

1 .….(7)

• conductivity and mobility are

= ne ….(8)

• mobility

= RH ……(9)

• Hall coefficient also be

RH = ne8

π3

= (π3

σ8) RH ….(10)

Determination of type of majority carrier:

As VH = new

BI =

ne

BJd=

Ane

dBI=

newd

dBI

Hall field per unit current density per unit magnetic induction is called Hall

coefficient RH.

RH = JB

d/HV=JB

HE

Also, VH = ne

dBJ

RH = nedJBne

dBJ

JB

d

ne

BJd

1)(

==

54

The sign of the Hall voltage is positive for p-type & for n-type semiconductor,

the Hall voltage will be negative

Applications:

1. VH is proportional to magnetic field B, for the given current I, hence Hall effect is

used in magnetic field meter.

2. Charge carrier concentration can be determined.

3. Mobility of charge carriers can be determined.

4. Nature of semiconductor (P-type or N-type) can be determined.

Sol.5 (c) T1 = 1.5 sec, T2 = 1 sec

Scustain = 20 m2, V = 10 × 8 × 6 m3 = 480 m3, S1 = 2 × 20m2 = 40m2

1

1 2 1

0.161 V 1 1a

S T T

= −

(As the cloth is suspended at the centre of the hall, both its surfaces will absorb

sound)

1

0.161 480 0.667a

40

=

1a 0.644=

Q.6 (a) Describe principle, construction and working of magnetostriction oscillator to

produce ultrasonic waves. 5

(b) Explain various point defects in crystals. 5

(c) Explain how a voltage difference is generated in a p-n junction when it is used

in a photovoltaic solar cell. 5

Ans.6 (a) Magnetostriction effect

Method is used to produce waves in the frequency range of 20kHz to 100 kHz

▪ G.W. Pierce was the first to design an ultrasonic oscillator basing on the

phenomenon of magnetostriction.

▪ The Pierce oscillator is a triode valve oscillator and is schematically shown in fig.

Principle

▪ Resonance is obtained by superimposing a frequency which obtained from

oscillator and maintained equal to the natural frequency of the rod. Natural

frequency is selected as the order of ultrasound which we want.

Fig 1 Circuit diagram for magnetostriction oscillator

55

About circuit

In the circuit diagram we have a rod of ferromagnetic material of length l, which

is cut in such a way that its length provides its natural frequency which we want

as ultrasonic output.

▪ dc supply maintains the rod permanently in magnetized form.

▪ Coil L2 and capacitor C forms a tuned circuit, so that by varying C we can

control the oscillating frequency of oscillator.

▪ Coil L, which is in grid cathode circuit provides necessary feedback.

Working Principle

▪ When the current through coil L2, changes, it causes a corresponding change in

the magnetization of the rod, due to magnetostriction effect a small change in the

length of rod will be noticed

▪ This change in length will give rise to change of flux linked with the coil L1 and

it will induce an voltage.

▪ This induce voltage will change grid voltage.

▪ The changed grid voltage will be amplified and come out at plate circuit, and

cycle will continue.

▪ The frequency of this oscillator is

By adjusting value of capacitor C, we obtain that value of f which equals natural

frequency n of the rod f = n

At this point rod vibrates under resonance with frequency f = n

Vibrations are noticed at both ends of rod as shown in fig.

Importance of resonance

At resonance amplitude of vibration is considerably increased without any extra

energy used.

Advantages

• Low cost and easy maintenance.

• Large power output.

Disadvantages

• Useful for frequencies below 100 kHz.

• Frequency gets affected by temperature.

56

Ans.6 (b) All the atoms in a solid posses vibrational energy and at all temperature above

absolute zero, there will be a finite number of atoms which have sufficient

energy to break the bonds which hold them in their equilibrium position.

• Once the atoms become free from their lattice sites, they give rise to point

defects.

• Also due to presence of impurity atoms point defects are likely to come in the

crystal.

Types of point defects:

(a) Vacancy

(b) Interstitial

(c) Substitutional impurities

(d) Interstitial impurities

(e) Shottky defects

(f) Frankel defect

(a) Vacancy:

Vacancy is produced due to the removal of an atom from its regular position in

the lattice.

The removal atom does not vanish. It travels to the surface of the material. For

low concentration of vacancies, a relation is

n = N exp (-EV / kT)

Where n = Number of vacancies

N = Total number of atoms

T = Temperature in (oK)

EV= Average energy required to create a vacancy

(b) Interstitial:

An extra atom of the same type is fitted into the void between the regularly

occupied sites.

Since in general the size of atom is larger than the void into which it is fitted, so

the energy required for interstitial formation is higher than that of vacancy

formation.

Fig

(c) Substitutional impurities:

In this, a foreign atom is found occupying a regular site in a crystal lattice,

57

Fig

(d) Interstitial impurities:

Here a foreign atom is found at a non-regular sute.

Fig

(e) Schottky Defect : [ May 2008]

The point imperfect in ionic crystal occurs when a negative ion vacancy is

associated with a positive ion vacancy. It is therefore a localised vacancy pair of

positive and negative ions. This type of defect maintains the crystal electrically

neutral, it is called shottkay defect.

Fig

For ionic crystal number of pair iron production is

n = Ne-Ep / 2kT

Where N = Number of lattice site

k = Boltzman constatn

Ep = Energy required to create a pair of ion vacancy

inside crystal lattice

T = Temperature in ok

(f) Frankel Defect : [ May 2008, May 2008, 7 marks]

• For ionic crstal when a negative ion vacancy is associated with a interstitial

negative ion or a positive ion vacancy is associated with an interstitial positive

ion then it is called Franckel defect.

• That is when an ion (generally cation) shifts from its position to interstitial

position in the crystal lattice, then a vacancy created. The defect is known as

Frankel defects.

58

• Since cations are smaller in size compare to anion and a cation may occupy the

void between the anions i.e. a cation occupies the interstitial position between

the anions.

• This defect can occur in ionic crystal when,

(i) The anion is much larger than cation.

(ii) The ion has low – co-ordination number

Fig.

Number of Frankel defects creation is

n = (NN’) ½ e – E / 2kT

Where

N = Number of lattice site

N’ = Number of interstitial site

E = Energy required to remove an atom from lattice site to an interstitial.

Ans.6 (c) A solar cell (also called a photovoltaic cell) is an electrical device that converts

the energy of light directly into electricity by the photovoltaic effect. It is a form

of photoelectric cell (in that its electrical characteristics-- e.g. current, voltage, or

resistance-- vary when light is incident upon it) which, when exposed to light,

can generate and support an electric current without being attached to any

external voltage source.






the potential and this electricity is captured. Due to the special composition of

solar cells, the electrons are only allowed to move in a single direction.

3. An array of solar cells converts solar energy into a usable amount of direct

current (DC) electricity.









59

Figure 1 – I-V Curve of PV Cell and Associated Electrical Diagram

Theory of I-V Characterization

PV cells can be modeled as a current source in parallel with a diode. When there

is no light present to generate any current, the PV cell behaves like a diode. As

the intensity of incident light increases, current is generated by the PV cell, as

illustrated in Figure 1.

In an ideal cell, the total current I is equal to the current Iℓ generated by the

photoelectric effect minus the diode current ID, according to the equation:

where I0 is the saturation current of the diode, q is the elementary charge 1.6x10-

19 Coulombs, k is a constant of value 1.38x10-23J/K, T is the cell temperature in

Kelvin, and V is the measured cell voltage that is either produced (power

quadrant) or applied (voltage bias).

Applications


• Today, thousands of people power their homes and businesses with individual

solar PV systems.


For large electric utility or industrial applications, hundreds of solar arrays are

interconnected to form a large utility-scale PV system.

60


(Revised Course) QP Code: 28603






1. Attempt any five from the following: - 15

(a) Draw (a) (112) (b) (0 4 0) (c) [0 4 0] with reference to a cubic unit cell.

(b) What is the probability of an electron being thermally promoted to the conduction

band in diamond at 27°C, if the bandgap is 5.6 eV wide?

(c) Define drift current, diffusion current and mobility of charge carries.

(d) What is dielectric polarization and dielectric susceptibility? Write the relation

between them.

(e) State and explain Ohm’s law in magnetic circuit.

(f) Write Sabine’s formula and explain the terms used in it.

(g) Calculate the length of an iron rod which can be used to produce ultrasonic waves

of 20kHz Given – Y = 11.6 × 1010 N/m2, = 7.23 × 103 kg/m3.

2. (a) Explain formation of energy bands in solids and explain classification on the basis

of energy band theory. 8

(b) Zn has hcp structure. The nearest neighbor distance is 0.27nm. The atomic weight

of Zn is 65.37. Calculate the volume of unit cell, density and atomic packing

fraction of Zn. 7

3. (a) What is hysteresis? Draw a hysteresis loop for ferromagnetic material and explain

various important parameters.

A magnetic material has a magnetization of 2300 A/m and produces a flux density

of 0.00314 wb/m2, Calculate magnetizing force and relative permeability of the

material. 8

(b) Explain the statement “crystal act as three dimensional grating with X-rays”.

61

Monochromatic X-ray beam of wavelength = 5.8189 A° is reflected strongly for

a glancing angle of = 75.86° in first order by certain planes of cubic of lattice

constant 3A°. Determine Miller indices of the possible reflecting planes. 7

4. (a) Define Ligancy. Find the value of critical radius ratio for ligancy4. 5

(b) An impurity of 0.01 ppm is added to Si. The semiconductor has a resistivity of

0.25 m at 300K. Calculate the hole concentration and its mobility. Atomic weight

of Si is 28.1, density of Si = 2.4 × 103kg/m3. 5

(c) Explain the origin of electronic, ionic and orientational polarization and

temperature dependence of respective polarizability. 5

5. (a) The density of copper is 8980kg/m3 and unit cell dimensions is 3.61 A°. Atomic

weight of copper is 63.54. Determine type of crystal structure. Calculate atomic

radius and interplanar spacing of (1 1 0) plane. 5

(b) What is Hall effect? Derive expression for Hall voltage with neat labelled diagram.

5

(c) Explain how the reverberation time is affected by (i) size (ii) nature of wall surface

(iii) audience in an auditorium. 5

6. (a) Estimate the ratio of vacancies at (i) –119°C (ii) 80°C where average required to

create vacancy is 1.8eV. 5

(b) How a p-n junction diode is used to generate a potential difference in a

photovoltaic solar cell? 5

(c) Explain with neat labelled diagram the construction and working of piezoelectric

oscillator. 5

******************

62


Solution

Q.1 Attempt any five from the following: - 15

(a) Draw (a) (112) (b) (0 4 0) (c) [0 4 0] with reference to a cubic unit cell.

(b) What is the probability of an electron being thermally promoted to the

conduction band in diamond at 27°C, if the bandgap is 5.6 eV wide?

(c) Define drift current, diffusion current and mobility of charge carries.

(d) What is dielectric polarization and dielectric susceptibility? Write the relation

between them.

(e) State and explain Ohm’s law in magnetic circuit.

(f) Write Sabine’s formula and explain the terms used in it.

(g) Calculate the length of an iron rod which can be used to produce ultrasonic

waves of 20kHz Given – Y = 11.6 × 1010 N/m2, = 7.23 × 103 kg/m3.

Ans.1 (a)

Sol.1 (b) T = 27°C = 300°K

Eg = 5.6 eV

K = 1.38 × 10–23 J/K = 86.25 × 10–6 eV

also,

Eg = C VE E

2

− = 5.6 eV

F(E) = C V

1

E E

21 exp

KT

− +

for insulator =

6

1

5.61 exp

2 86.25 10 300−

+

F(E) = 1.7 × 10–47

Ans.1 (c) Net displacement in the electron's position per unit time caused by the application

of an electric field becomes a constant at the steady state. Velocity of the electrons

in the steady state in an applied electric field is called the drift velocity.

63

Mobility:

The mobility of electron is defined as the magnitude of the drift velocity

acquired by the electrons in a unit field.

E - applied electric field

d -drift velocity

-mobility

= E

d ..….(6)

J = E

= E

J and J =

A

I

= AE

I

Current = I = ned A*

= AE

Adνne = ne

= ne

σ ….(7)

mobility of electron = 1350 cm2/ v-s

mobility of hole is = 480 cm2/v-s

The ease with which electrons could drift in material under the influence of an

electric field called as mobility.

Mobility of electron is > mobility of holes

Current (I):

I = nedA

A- area of cross section, I-current, v - velocity of electrons,

Ans.1 (d) Dielectric Polarization: When an electric field is applied across a dielectric

material then the dielectric material becomes polarized. This mechanism is called

dielectric polarization

Electric susceptibility: In electromagnetism, the electric susceptibility a

dimensionless proportionality constant that indicates the degree of polarization of

a dielectric material in response to an applied electric field

Electrical Susceptibility:

The polarization vector P is proportional to the applied electric field E, for Field

strength that are too large.

Therefore P α E

P = 0 e E

Where e is a characteristics of every dielectric material and is called as electrical

Susceptibility.

Therefore e = P

0 E

64

Since P

E = 0 ( r -1 )

e =

Therefore e = ( r -1 ) or r = e +1

The dielectric constant determines the share of the electric stress which is absorbed

by the material. It is the ratio between the absolute permittivity and the

permittivity of free space 0 and is given by,

r= / 0

Ans.1 (e) Recall the equation

CauseEffect

Opposition=

For magnetic circuits, the effect desired is the flux . The cause is the

magnetomotive force (mmf) , which is the external force (or “pressure”) required

to set up the magnetic flux lines within the magnetic material.

The opposition to the setting up of the flux is the reluctance.

Substituting, we have

F

R =

The magnetomotive force is proportional to the product of the number of turns

around the core (in which the flux is to be established) and the current through

the turns of wire. In equation form,

F = NI (ampere-turns, At)

This equation clearly indicates that an increase in the number of turns or the

current through the wire will result in an increased “pressure” on the system to

establish flux lines through the core.

Although there is a great deal of similarity between electric and magnetic circuits,

one must continue to realize that the flux is not a “flow” variable such as current

in an electric circuit. Magnetic flux is established in the core through the alteration

of the atomic structure of the core due to external pressure and is not a measure

of the flow of some charged particles through the core.

Sol.1 (f) Sabine’s formulae:

T= 0.161V/A

Where T= Reverberation Time

V= volume

A= Total absorption of sound

65

Sol.1 (g) 20kHz=

3 37.23 10 kg/m =

y = 11.6 × 1010 N/m2

y1

2l =

y1

l2

=

10

3 3

1 11.6 10

2 20 10 7.23 10

=

= 0.1 m

length of iron rod = 0.1 m.

Q.2 (a) Explain formation of energy bands in solids and explain classification on the

basis of energy band theory. 8

(b) Zn has hcp structure. The nearest neighbor distance is 0.27nm. The atomic

weight of Zn is 65.37. Calculate the volume of unit cell, density and atomic

packing fraction of Zn. 7

Ans.2 (a) "Energy band structure determines whether solid is a conductor, an insulator or a

semiconductor".

When large number of atoms are brought closer to form solids, there is significant

interaction between the outermost electrons. Hence the energy levels of the outer

shell electrons, which are share by more than one atom in the crystal are changed

considerably.

Si (silicon) 14 1s2 2s2 2p2 3s2 3p2

When the interatomic distance is sufficient large so that there is negligible

interaction between electrons of different atoms, all atoms will have the same

energy is the outermost ‘s’ and ‘p’ subshells. There will be 2N electrons in the 2N

possible quantum states for the ‘s’ subshell energy level and 2N electrons in the

6N possible quantum states of the ‘p’ subshell energy level and 2N electrons in the

6N poss quantum states of the ‘p’ subshells level.

Band Structure



• Gap between valence and conduction bands calledas forbidden energy band gap.

Fig. 3.1Energy level splitting in Si Crystal

66

Classification of Solids

Solids are classified into conductor, insulator and semiconductor, based upon

the energy band structure.

Fig.3.2 Energy band diagram of Insulator, Semiconductor and Conductor

Difference between Conductor, semiconductor and Insulator

Sr.No. Conductor Insulator Semiconductor

1. V.B and C.B are

overlapping.

V.B is full &C.Bis

completely empty.

V.B is full and C.B has

few electrons.

2. Large number of

electrons.

electrical conduction in

such solids is

insignificant.

electrons can move

under the influence of

even a small applied

field and cause current

flow.

3. Current is due to flow of

electrons.

width of for-bidden gap

is large .

width of forbidden gap

is small

4. Eg: Li, Be Eg: Diamond (Eg = 5.47

eV )

Eg: Si (Eg = 1.12 eV),

Ge (Eg = 0.72 eV)

5.

Sol.2 (b) a = 0.27 nm, M = 65.37, density = Massof uniteV

V

V = Volume of unit cell of HCP = 23 3a C

2

For HCP unit cell C 8

a 3=

67

23 3 8V a . a

2 3= 33 2a= ( )

3 273 2 0.27 10−=

27 3V 0.083507 10 m−=

Mass of unit cell = 65.37

6No

26

65.376

6.023 10=

2965.12 10 kg−=

density = 29

27

65.12 10

0.083507 10

−

−

37.798kg/m=

Q.3 (a) What is hysteresis? Draw a hysteresis loop for ferromagnetic material and

explain various important parameters.

A magnetic material has a magnetization of 2300 A/m and produces a flux

density of 0.00314 wb/m2, Calculate magnetizing force and relative permeability

of the material. 8

(b) Explain the statement “crystal act as three dimensional grating with X-rays”.

Monochromatic X-ray beam of wavelength = 5.8189 A° is reflected strongly for

a glancing angle of = 75.86° in first order by certain planes of cubic of lattice

constant 3A°. Determine Miller indices of the possible reflecting planes. 7

Sol.3 (a) (Beyond the Syllabus)

A great deal of information can be learned about the magnetic properties of a

material by studying its hysteresis loop. A hysteresis loop shows the relationship

between the induced magnetic flux density (B) and the magnetizing force (H). It

is often referred to as the B-H loop. An example hysteresis loop is shown below.


The loop is generated by measuring the magnetic flux of a ferromagnetic material

while the magnetizing force is changed. A ferromagnetic material that has never

been previously magnetized or has been thoroughly demagnetized will follow the

dashed line as H is increased. As the line demonstrates, the greater the amount of

current applied (H+), the stronger the magnetic field in the component (B+). At

point "a" almost all of the magnetic domains are aligned and an additional increase

in the magnetizing force will produce very little increase in magnetic flux. The

material has reached the point of magnetic saturation. When H is reduced to zero,

the curve will move from point "a" to point "b." At this point, it can be seen that

68

some magnetic flux remains in the material even though the magnetizing force is

zero. This is referred to as the point of retentivity on the graph and indicates the

remanence or level of residual magnetism in the material. (Some of the magnetic

domains remain aligned but some have lost their alignment.) As the magnetizing

force is reversed, the curve moves to point "c", where the flux has been reduced to

zero. This is called the point of coercivity on the curve. (The reversed magnetizing

force has flipped enough of the domains so that the net flux within the material is

zero.) The force required to remove the residual magnetism from the material is

called the coercive force or coercivity of the material.





positive direction will return B to zero. Notice that the curve did not return to the

origin of the graph because some force is required to remove the residual

magnetism. The curve will take a different path from point "f" back to the

saturation point where it with complete the loop.

From the hysteresis loop, a number of primary magnetic properties of a material

can be determined.

1. Retentivity - A measure of the residual flux density corresponding to the

saturation induction of a magnetic material. In other words, it is a material's ability

to retain a certain amount of residual magnetic field when the magnetizing force

is removed after achieving saturation. (The value of B at point b on the hysteresis

curve.)

2. Residual Magnetism or Residual Flux - the magnetic flux density that remains in

a material when the magnetizing force is zero. Note that residual magnetism and

retentivity are the same when the material has been magnetized to the saturation

point. However, the level of residual magnetism may be lower than the retentivity

value when the magnetizing force did not reach the saturation level.

3. Coercive Force - The amount of reverse magnetic field which must be applied to

a magnetic material to make the magnetic flux return to zero. (The value of H at

point c on the hysteresis curve.)





electrical circuit.

M = 2300 A/M, B = 0.00314 wb/m2

B = ( )0 M H +

M

H = 0 1= −

0

BH M

1= − − 7

0.003142300

4 10−= −

198.7326A/m=

69

M

r 1H

= +2300

1198.7326

= + 12.57=

Sol.3 (b) X-rays have short wavelength of the order of 10-10m. An ordinary grating which

have 6000 lines per cm cannot produce diffraction of X-rays. This fact was first

studied by Max Laue and Paul Petr Ewald in Munich by studying periodic array

of atoms. The atoms arranged as crystal grating correspond to the grating lines

and the distance between two atoms, which is of order 10-8 cm, forms a grating

element which can be used for X-ray diffraction. The difference between crystal

grating and optical grating is that atomic centers of diffraction in the crystal

grating are not all in one plane but distributed in space, while in case of optical

grating they are limited to one plane.

The crystal is thus a three – dimensional space grating rather than a two –

dimensional plane grating.

5.8189A = , 75.86 = , n = 1, a = 3A°

n 2dsin =

( )2 2 2

ad

h k l=

+ +

n

2sin

=

( )

1 5.8189

2 sin 75.86

=

3A=

d = a

i.e. 2 2 2h k l 1+ + =

The miller indices of the possible reflecting planes are

(1 0 0), (0 1 0), (0 0 1), ( )100 , ( )0 10 & ( )00 1

Q.4 (a) Define Ligancy. Find the value of critical radius ratio for ligancy4. 5

(b) An impurity of 0.01 ppm is added to Si. The semiconductor has a resistivity of

0.25 m at 300K. Calculate the hole concentration and its mobility. Atomic

weight of Si is 28.1, density of Si = 2.4 × 103kg/m3. 5

(c) Explain the origin of electronic, ionic and orientational polarization and

temperature dependence of respective polarizability. 5

Ans.4 (a) Ligancy: The number of anions surrounding a central cation is called Ligancy. In

short it is a coordination number in ionic solids.

Critical radius ratio for ligancy – 4 (Tetrahedral configuration)

Consider four anions one each at the vertices of the regular tetrahedron. A cation

fits in the mid formed by these four anions touching each other (three on a close

packed plane and a fourth on tither top or bottom plane) as shown in Fig. All the

four anions touch each other as well as the central cation. The critical radius ration

can be determined using figure as follows:

70

H

D

A B

E

G

a

aa

a

aa2

a2

C

a3

r + r

a

c

a2

ra

A

D G

EJ

Fig1.25: Ligancy 4

From above figure we can write,

Sol.4 (b) Impurity level = 0.01 ppm

= 0.25 ohm-m

T = 300° K

M = 28.1

density of Si = 2.4 × 103 kg/m3

= n b

1

e

Number of Si atoms/unit volume = Avogadro'snumber Density

Atomicweight

= 26 36.023 10 2.4 10

28.1

= 5.144 × 1028 atoms/m3

Here,

PPM = Particle per million

1 PPM = 1 impurity atom/ 106 Si atoms

0.01 PPM = 1 impurity atom/ 108 Si atoms

Number of impurity atom = 28

8

5.144 10

10

20 35.144 10 atoms/m=

71

Each impurity contributes one hole.

5.144 × 1020 impurity atoms/m3 introduces the hole concentration of

h = 5.144 × 1020 holes/m3

Mobility h

h

1

e =

19 20

1

0.25 1.6 10 5.144 10−=

h = 0.0486 m2/V-sec

Ans.4 (c) (Beyond the Syllabus) :

Q.5 (a) The density of copper is 8980kg/m3 and unit cell dimensions is 3.61 A°. Atomic

weight of copper is 63.54. Determine type of crystal structure. Calculate atomic

radius and interplanar spacing of (1 1 0) plane. 5

(b) What is Hall effect? Derive expression for Hall voltage with neat labelled

diagram. 5

(c) Explain how the reverberation time is affected by (i) size (ii) nature of wall

surface (iii) audience in an auditorium. 5

Sol.5 (a) (i) The effective number of atoms/unit cell (n) = 3Na

M

( )

326 108980 6.024 10 3.61 10n

63.54

− =

254.32

63.54= = 4 atoms/unit cell

n = 4 atoms/unit cell

Copper exhibits FCC structure.

(ii) For FCC structure,

Electronic polarization Ionic Polarization Orientation Polarisation

1. Exhibited by the displacement

of the electronic cloud with

respect to the nucleus.

1.Exhibited by alternately

placed Positive and

negative ions.

1. Exhibited by polar

dielectrics

2. This is fastest 2. It is faster 2. It is even slower than

ionic polarization

3. The timescale for electronic

polarization of an atom due to

electric field is 10 -15 seconds.

3. The frequency with

which ions that are

displaced is of the same

order as lattice vibration

frequency(approx 1013 Hz)

3. The relaxation time for

orientation polarization in

liquid is less than that in

solids

4.

4.

4.

72

a

r2 2

= 3.61

2 1.4142=

1.276A=

r = 1.276 A°

(iii) The interplanar spacing is given by,

( )2 2 2

ad

h k l=

+ + ( )

3.61

1 1 0=

+ +2.553A=

Ans.5 (b) If a metal or semiconductor, carrying a current I is placed in a transverse magnetic

field B, an electric field E is induced in the direction perpendicular to both I and

B. This phenomenon is known as Hall effect and the electric field or voltage

induced is called Hall voltage (VH).

Fig.: Hall effect


• In equilibrium condition,

Force due to Electric Field = Force due to Magnetic Field

q EH = Bqv

EH = B v

Also EH = VH / d

J = nev = I / wd = I/A ( since A = w × d)

v = J/ne

VH = EH d = Bvd = new

BI

ne

BJd= ….(1)

As EH = ne

BJBv

d

VH == ....(2)

Now J = nev = A

I ....(3)

also J = E

= E

J =

A

I. E

1 …..(4)

EH =E

E.

A

I.

ne

B=

A

I.

ne

B=

ne

BJ

)E

1.

A

I(.

ne

B=

E

HE

73

Using Equation (4)

ρne

B=σ.

ne

B=

E

HE .…(5)

Where = σ

1

ρne

B=

E

HE ….(6)

• Hall coefficient

RH = BI

WHV=ne

1 ….(7)

• conductivity and mobility are

= ne ….(8)

• mobility

= RH .…(9)

• Hall coefficient also be

RH = ne8

π3

= (π3

σ8) RH ….(10)

Determination of type of majority carrier:

As VH = new

BI

= ne

BJd=

Ane

dBI=

newd

dBI

Ans.5 (c) Reverberation time:

• Sabine’s formula for reverberation time is used for suitable acoustic treatment.

• Where ‘V’ is the volume of the hall, ‘S’ is the surface area and ‘a’ is the absorbing

coefficient.

Ceiling:

Central area of the ceiling should be sound reflecting.

The perimeter and rear to be provided with sound absorbing materials like

acoustic tiles.

Sidewalls:

Sidewalls should be sound reflecting and diffusing with as many irregularities as

possible. For example, making doorway wider at one side of the wall keeping

windows etc., and the back wall is treated with deep sound absorbing finish

74

Floor :

All aisles are carpeted except in front of the stage to make full noise control.

Fabric upholstered seats are used.

Absorptive and cushioned seats will give stable reverberation.

Balconies :

Use balconies to increase seating capacity and to reduce the distance to the farthest

row of seats.

Sound reinforcement system:

In large halls a sound amplification system to reinforce the sound to a weak source

in a large room is required.

In addition there should be adequate loudness in every part of the auditorium

uniform distribution (diffusion) of sound energy in the room.

The hall should be free from echoes, long delayed reflections, flatter echoes, sound

concentrations, distortions, and sound shadow and room resonance.

Audience in Auditorium:

To make the hearing conditions satisfactory when the room is full or partly full,

upholstered seats with absorbing material at the bottom are used, so that the

absence or presence of audience does not affect the reverberation time.

Learning outcome: Students will be able to plan Design and development of an

acoustic auditorium

Q.6 (a) Estimate the ratio of vacancies at (i) –119°C (ii) 80°C where average required to

create vacancy is 1.8eV. 5

(b) How a p-n junction diode is used to generate a potential difference in a

photovoltaic solar cell? 5

(c) Explain with neat labelled diagram the construction and working of

piezoelectric oscillator. 5

Sol.6 (a) K = 1.38 × 10–23 J/K

EV = 1.8 eV = 1.8 × 1.6 × 10–19 J

t1 = –119° C

t2 = 80° C

n = Ev

N expKT

−

Ev/KTne

N−=

For t1 = –119°C = –119 + 273 = 154° K

( )19 231.8 1.6 10 /1.38 10 353n

eN

− −− =

26n2.11 10

N−=

Ans.6 (b) A solar cell (also called a photovoltaic cell) is an electrical device that converts

the energy of light directly into electricity by the photovoltaic effect.









75



the potential and this electricity is captured. Due to the special composition of

solar cells, the electrons are only allowed to move in a single direction.


(DC) electricity.

Fig. (21)Energy band structure of reverse biased pn-junction

Applications


• Today, thousands of people power their homes and businesses with individual

solar PV systems.


For large electric utility or industrial applications, hundreds of solar arrays are

interconnected to form a large utility-scale PV system.

Ans.6 (c) Principle

Piezoelectric oscillator is based on the principle of resonance between the natural

frequency of appropriately cut piezoelectric crystal and a suitable frequency

generated by LC oscillator.

Fig (6) Valve based piezoelectric oscillator



76

Circuit analysis

The quartz crystal is placed between two metal plates A and B.

• The plates are connected to the primary (L1) of a transformer which is inductively

coupled to the electronics oscillator.

• The electronic oscillator circuit is a base tuned oscillator circuit.

• The coils L of oscillator circuit are taken from the secondary of a transformer T.

• The Plate coil L1 is inductively coupled to Grid coil L2.

• The coil L2 and variable capacitor C form the tank circuit of the oscillator.

Working

• When H.T. battery is switched on, the oscillator produces high frequency

alternating voltages with a frequency f.

• Due to the transformer action, an oscillatory e.m.f. is induced in the coil L. This

high frequency alternating voltages are fed back to the crystal plates A and B.

• Inverse Piezo-electric effect takes place and the crystal contracts and expands

alternatively. The crystal is set into mechanical vibrations.

• The frequency of the vibration is given by

Natural frequency of crystal slab cut is given by

Here t = Thickness of crystal slab

k Yn

2t =

Y = Young's modulus

= Density

k = 1, 2, 3, ..... (integer)

K Which represent order of harmonic. In practice we take k = 1 that is fundamental

frequency. It is important to know that natural frequency n is inversely

proportional to thickness t. Hence for higher frequency, thickness t has to be

reduced.

In the above diagram we have a thin plate of piezoelectric crystal cut on such axis

so that we get inverse piezoelectric effect i.e. on application of high frequency

electric field, it can vibrate in resonance with LC oscillator.

Advantages

• Ultrasonic frequencies as high as 5 x 108Hz or 500 MHz can be obtained with this

arrangement.

• The output of this oscillator is very high.

• It is not affected by temperature and humidity.


• Better waveform

Disadvantages


➢ The cost of piezoelectric quartz is very high

➢ The cutting and shaping of quartz crystal are very complex.

***********

77

Applied Physics – I (Sem-I) May 2017 F.E. SEM-1 (CBCGS)

Q.P. CODE: -18533 TIME- 2 HOURS TOTAL MARKS-60 -------------------------------------------------------------------------------------------------------------------------------

N.B: -

1) Question number 1 is compulsory.

2) Attempt any three from Q.No.2 to Q.No. 6

3) Assume any data whenever required.

4) Figures to the right indicates full marks.

Q.1 Solve any five of the following: 15

(a) Draw the unit cell of HCP structure & work out the no. of atoms per unit cell.

(b) The mobility of holes is 0.025 m2/V-sec. What would be the resistivity of n-type Si if

the Hall coefficient of the sample is 2.25 × 10-5 m3/C.

(c) What is the principle of Solar cell? Write it’s advantages & disadvantages.

(d) An electron is confined in a box of dimension 1A0. Calculate minimum uncertainty

in it’s velocity.

(e) Explain the factors on which reverberation time depends.

(f) Explain Cavitation effect.

(g) What is Maglev? How it can have very high speed?

Q.2 (a) Draw the following: ( )1 1 3 , ( )2 0 0 , 001

An electron is accelerated through 1200 volts & is reflected from a crystal. The

second order reflection occurs when glancing angle is 600. Calculate the inter-planer

spacing of the crystal. 8

(b) Explain the concept of Fermi level. Prove that the Fermi level exactly at the centre of

the Forbidden energy gap in intrinsic semiconductor. 7

Q.3 (a) Find the following parameters for DC (Diamond Cubic) structure: 8

(i) No. of atoms per unit cell.

(ii) Co-ordination No.

(iii) Nearest atomic distance

78

(iv) Atomic radius

(v) APF

(b) Define drift current, diffusion current & P-N junction. The electrical conductivity of

a pure silicon at room temperature is 4 ×10-4mho/m. If the mobility of electron is

0.14 m2/V-S & that of hole is 0.04 m2/V-S. Calculate the intrinsic carrier density.

7

Q.4 (a) Distinguish between Type I & Type II superconductors. 5

(b) A classroom has dimensions 10×8×6 m3. The reverberation time is 3 sec. Calculate

the total absorption of surface & average absorption. 5

(c) Explain the principle, construction & working of a Magnetostriction Oscillator.

5

Q.5 (a) Write Fermi-Dirac distribution function. With the help of diagram, explain the

variation of Fermi level with temperature in n-type semiconductor. 5

(b) Derive Schrodinger’s time dependent wave equation for matter waves. 5

(c) Find the depth of sea water from a ship on the sea surface if the time interval of 2 sec

is required to receive the signal back. Given that: -temperature of sea water is 200,

salinity of sea water is 10 gm/lit. 5

Q.6 (a) Define the term critical temperature. Show that in the superconducting state the

material is perfectly diamagnetic. 5

(b) In a solid the energy level is lying 0.012eV below Fermi level. What is the probability

of this level not being occupied by an electron.? 5

(c) What is the wavelength of a beam of neutron having: 5

(i) an energy of 0.025eV?

(ii) an electron & photon each have wavelength of 2 A0. What are their

momentum & energy?

Mn=1.676×10-27Kg, h=6.625×10-34 J-sec.

***********************

79


Solution

Q.1 Solve any five of the following: 15

(a) Draw the unit cell of HCP structure & work out the no. of atoms per unit cell.

(b) The mobility of holes is 0.025 m2/V-sec. What would be the resistivity of n-type Si

if the Hall coefficient of the sample is 2.25 × 10-5 m3/C.

(c) What is the principle of Solar cell? Write it’s advantages & disadvantages.

(d) An electron is confined in a box of dimension 1A0. Calculate minimum

uncertainty in it’s velocity.

(e) Explain the factors on which reverberation time depends.

(f) Explain Cavitation effect.

(g) What is Maglev? How it can have very high speed?

Ans.1 a)

1. Each corner atom in the top & bottom layer is shared by six unit cells . Hence the

fraction of each corner atom belonging to a unit cell is 1

6 .As there are 12 corner

atoms in all, the number of corner atoms belonging to unit cell =1

6 × 12=2

2. Each central atom in the top & bottom layers is shared by two unit cells. Hence

number of central atoms belonging to unit cell= 1

2 ×2=1

3. The three atoms within the body of the cell are not shared with other cells. Hence

number of these cells belonging to a unit cell =3

4. So, the number of atoms per unit cell = 2+1+3=6.

Sol.1 (b) Given things:- µh= 0.025m2/V-sec

n=?

RH = 2.25×10-5 m3/C

Formula: = RH

µ .

= 9×10-4 ohm-m

80

Ans.1 (c) A solar cell is an electrical device that converts the energy of light directly into

electricity by the photovoltaic effect. It is a form of photoelectric cell. Its electrical

characteristics-e.g. current, voltage, or resistance- vary when light is incident upon it

which, when exposed to light, can generate and support an electric current without

being attached to any external voltage source.

It consist of a silicon PN junction diode with a glass window on top surface layer of

p material which is made extremely thin so that incident light photons may easily

reach the PN junction. When these photons collides with valence electrons they

support them sufficient energy as to leave their parent atoms. In this way free

electrons & holes are generated on both sides of the junction. Due to these holes &

electrons, current is produced. This current is directly proportional to illumination

& also depends on size of the surface area being illuminated.

Advantages:

1) It is clean source of energy .

2) Pollution free

3) Suitable for low power applications like calculators

4) Maintenance free which makes them cost efficient in long run

5) Useful in remote areas & satellite where no other source of energy can be frequently

transported.

Disadvantages:

1) Affected by availability of sunlight which has day-night , season-to-season & place-

to-place variations

2) Requires large space for high power applications

3) High cost

4) The output is d.c. which can not be transported through large distance without

significant losses.

Sol.1 (d) Δx= 1 A0 = 10-10 m

Δv = ?

By HUP, (Δx) (Δp) ≥ ħ

(Δx)maximum (Δp)minimum ≥ ħ

(Δx)maximum (mΔv)minimum ≥ ħ

(Δv)minimum ≥ħ

𝑚 𝛥𝑥

(Δv)minimum ≥ 1.05 × 10−34

9.1 × 10−31 × 10−10

(Δv)minimum ≥ 11.5 × 105 m/s.




81

Ans.1 (e) If reverberation time is too low then sound disappears quickly and become

inaudible. If reverberation time is too high then sounds exist for a long period of

time-an overlapping of successive sounds cannot hear the information clearly. For

the good audibility, the reverberation time should be kept at an optimum value.

Reverberation time can be reduced by installing sound absorbing materials like

windows and openings, arranging full capacity of audience, completely covering

the floor with carpets, heavy curtains with folds and decorating the walls with

drawing boards, picture boards

Ans.1 (f) Cavitation Effect: - When an ultrasonic transducer is placed in a liquid , bubbles

are formed due to the ultrasonic waves. The bubbles have a very short life time after

which they collapse which is called ‘implosion’. During implosion of the bubble, the

pressure of the shock wave created near the bubble is very high. The local pressure

may increase to a few thousand atmospheres & the temperature. The bubbles also

hinder the propagation of ultrasonic waves. This phenomenon is called”

Cavitation”.

Cavitation is used in ultrasonic cleaning of electronic components, precious

ornaments & medical & optical instruments. The specimen to be cleaned is kept in a

cleaning solution & ultrasonic waves are passed through it. The tiny bubbles which

form near the surface of specimen exert a strong pull on the surface due to which

the dirt particles are pulled off.

Ans.1 (g) MAGLEV means Magnetic Levitation. When a magnet is brought near a

superconducting coil, current is induced in the superconductor which in turn

produces a magnetic field which repels the magnet. The magnet can thus be made to

float in air (i.e. levitation) on the superconducting coil. This concept has been used

in the development of MAGLEV trains which can attain speeds up to 500km/h as

there is no friction between the rails & the wheels.

Fig: Mechanism of MAGLEV Train

Q.2 (a) Draw the following: ( )1 1 3 , ( )2 0 0 , 001

An electron is accelerated through 1200 volts & is reflected from a crystal. The

second order reflection occurs when glancing angle is 600. Calculate the inter-

planer spacing of the crystal. 8

(b) Explain the concept of Fermi level. Prove that the Fermi level exactly at the centre

of the Forbidden energy gap in intrinsic semiconductor. 7

82

Sol.2 (a)

Given things:-

V=1200 Volts , n=2 ,θ=600 ,d=?

By Bragg’s law 2dsinθ = 𝑛𝜆

Also λ = h

2mqV =

34

31 19

6.67 10

2 9.1 10 1.6 10 1200

−

− −

= 3.22A0

d=𝑛𝜆

2𝑠𝑖𝑛𝜃 =

2×3.22

2×sin 60 = 3.72A0 .

Ans.2 (b) The Fermi level is the total chemical potential for electrons (or electrochemical

potential for electrons) and is usually denoted by µ or EF. The Fermi level of a body

is a thermodynamic quantity, and its significance is the thermodynamic work

required to add one electron to the body (not counting the work required to remove

the electron from wherever it came from) Pure semiconductors are called intrinsic

semiconductors.

nc = Number of electrons in conduction band


We have

nC = Nce–(EC-EF)/KT …..(1)

Nc = Effective density of states in conduction band

and nv = Nve–(EF – EV) / KT …..(2)

Nv = effective density of states in valence band



nC = nv

NC . e –(EC – EF) / KT = Nv . e –(EF – EV) / KT

( )

( )

C F

F V

E E /KT

V

E E /KTC

Ne

Ne

− −

− −=

e –(EC – EF – EF + EV) / KT =cN

vN

https://en.wikipedia.org/wiki/Chemical_potential

https://en.wikipedia.org/wiki/Electrochemical_potential

https://en.wikipedia.org/wiki/Electrochemical_potential

https://en.wikipedia.org/wiki/Thermodynamic

https://en.wikipedia.org/wiki/Thermodynamic_work

83

e –(EC + EV – 2EF ) / KT = cN

vN

as Nv = Nc = 1

e –(EC + EV – 2EF ) / KT = 1

Taking ln on both sides

KT

EEE Fvc )2( −+− = 0

(EC + EV) = 2 EF

EF = c vE E

2

+ … (4)

Therefore Fermi level in an intrinsic semiconductor lies at the center of forbidden

energy gap.

Q.3 (a) Find the following parameters for DC (Diamond Cubic) structure: 8

(i) No. of atoms per unit cell.

(ii) Co-ordination No.

(iii) Nearest atomic distance

(iv) Atomic radius

(v) APF

(b) Define drift current, diffusion current & P-N junction. The electrical conductivity

of a pure silicon at room temperature is 4 ×10-4mho/m. If the mobility of electron

is 0.14 m2/V-S & that of hole is 0.04 m2/V-S. Calculate the intrinsic carrier density.

7

Ans.3 (a) Structure of diamond is similar to the structure of Zinc sulphide. When the Zinc and

sulphur ions are replaced by identical carbon atoms, the diamond structure results.

Silicon and germanium also have diamond structure.

The nature of bonding in diamond is partly covalent and partly ionic. It consists of

two inter-penetrating face centred cubic sub lattices. The two sub-lattices, made up

of carbon atoms are displaced from each other along the body diagonal through a

distance equal to one quarter of the body diagonal. Hence, unit cell of diamond

consist of eight atoms of carbon.

Thus coordination number is 4.


follows.

(i) 8 corner atoms (each shared by 8 unit cells)

(ii) 6 face centre atoms (each shared by 2 unit cells)

(iii) 4 interior atoms

Number of atoms per unit cell = 42

16

8

18 ++

n = 8

Thus , the structure of diamond contains eight atoms per unit cell.

84


1

dra == 234

1

8


Coordination no. = 4.

Atomic Packing Fraction = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑐𝑐𝑢𝑝𝑖𝑒𝑑 𝑏𝑦 𝑎𝑡𝑜𝑚 𝑖𝑛 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙

𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙

If “a” is the lattice constant for cubic unit cell , the total volume of the unit cell is a3

Atomic Packing fraction = 3

3

3

4

a

rn

= 3

3

8

3

3

48

a

a

= 16

3 = 0.34

APF = 34%

Sol.3 (b) Given Things: - µe= 0.14 m2/Volt-Sec.

µh= 0.04 m2/Volt-Sec.

ni=?

e =1.6×10-19 C

σ = 4×10-4mho/m .

Formula σi = ni e (µe+µh)

4×10-4 = ni ×1.6×10-19 (0.14+ 0.04)

ni = 1.39×1016 per m3 .

Q.4 (a) Distinguish between Type I & Type II superconductors. 5

(b) A classroom has dimensions 10×8×6 m3. The reverberation time is 3 sec. Calculate

the total absorption of surface & average absorption. 5

(c) Explain the principle, construction & working of a Magnetostriction Oscillator.

5

85

Ans.4 (a)

Sr.No. Type I superconductors Type II superconductors

1. Very pure samples of lead,

mercury, and tin are examples of

this kind

Niobium, Vanadium, Technetium, Diamond and

Siliconniobium-titanium, niobium-tin are the T-II SC

2. Type I superconductors have Hctoo

low

Type II superconductors have much larger Hc2 values

to be very useful.

3. Below Hc1 the superconductor

excludes all magnetic field lines. no

mixed state

At field strengths between Hc1 and Hc2 the field

begins to intrude into the material called the mixed

state

4. Exhibit complete Meissner Effect Do not show complete Meissner Effect

5. Only One Tc Two Tc ; Tc1 &Tc2

6. Know as soft SC Known as HARD SC

7.

Sol.4 (b) T = 0.167 𝑉

𝛴𝑎𝑆

Total absorption = aS =0.167 𝑉

𝑇

T=3sec , V =10×8×6 m3= 480 m3 .

aS=0.167×480

3 = 26.72 Sabin-m2 .

Average Absorption Coefficient a = 𝛴𝑎𝑆

𝛴𝑆

S = 2 × (10 × 8 + 8 × 6 + 6 × 10) = 376 m2

a=26.72

376 = 0.07106 O.W.U.

Ans.4 (c) G.W. Pierce was the first to design an ultrasonic oscillator basing on the

phenomenon of magnetostriction.

The Pierce oscillator is a triode valve oscillator and is schematically shown in fig.

Principle

Resonance is obtained by superimposing a frequency which obtained from oscillator

and maintained equal to the natural frequency of the rod. Natural frequency is

selected as the order of ultrasound which we want.

http://niobium.co.tv/

http://vanadium.co.tv/

http://technetium.co.tv/

http://covalentsuperconductors-diamond.co.tv/

http://covalentsuperconductors-silicon.co.tv/

http://niobium-titanium.co.tv/

http://niobium-tin.co.tv/

86

In the circuit diagram we have a rod of ferromagnetic material of length l ,Which is

cut in such a way that its length provides its natural frequency which we want as

ultrasonic output.

Natural frequency is given by

n = k Y

2l

where n = Natural frequency

Y = Young's modulus

= Density

l = Length of rod

k = Order of harmonic in practice we take fundamental harmonic

i.e. k =

• DC supply maintains the rod permanently in magnetized form.

• Coil L2 and capacitor C forms a tuned circuit, so that by varying C we can control

the oscillating frequency of oscillator.

• Coil L, which is in grid cathode circuit provides necessary feedback.

Working: -

1. When the current through coil L2 , changes, it causes a corresponding change

in the magnetization of the rod, due to magnetostriction effect a small change

in the length of rod will be noticed

2. This change in length will give rise to change of flux linked with the coil L1

and it will induce an voltage.

3. This induce voltage will change grid voltage.

4. The changed grid voltage will be amplified and come out at plate circuit, and

cycle will continue.

5. The frequency of this oscillator is

2

1 1f

2 L C= .....(2)

By adjusting value of capacitor C, we obtain that value of f which equals natural

frequency n of the rod

f = n

At this point rod vibrates under resonance with frequency f = n

87

Vibrations are noticed at both ends of rod as shown in fig.

Importance of resonance

At resonance amplitude of vibration is considerably increased without any extra

energy used.

Advantages

1. The design of this oscillator is very simple.

2. At low ultrasonic frequencies, the large power output can be produced without the

risk of damage of the oscillatory circuit.

3. Low cost and easy maintenance.

Disadvantages

1. It has low upper frequency limit and cannot generate ultrasonic frequency above

3000 kHz (ie. 3MHz).

2. The frequency of oscillations depends on temperature.

3. There will be losses of energy due to hysteresis and eddy current.

Q.5 (a) Write Fermi-Dirac distribution function. With the help of diagram, explain the

variation of Fermi level with temperature in n-type semiconductor. 5

(b) Derive Schrodinger’s time dependent wave equation for matter waves. 5

(c) Find the depth of sea water from a ship on the sea surface if the time interval of 2

sec is required to receive the signal back. Given that: -temperature of sea water is

200, salinity of sea water is 10 gm/lit. 5

Ans.5 (a) Fermi distribution function

F(E) = P(E) = 1

1+𝑒^(𝐸−𝐸𝑓

𝑘𝑇) =

F

1

E E1 exp

KT

− +

Where, P(E) = Probability of an electron occupying the energy state E = F(E)

Ef = Fermi Energy

K = Boltzmann constant

T = Absolute temperature

At 00K the donor levels are filled.

As the temperature increases, the donor atoms get ionized and donor electrons go

into the conduction band. This is temperature range is called ionization region.

Once all electrons from donor levels are excited into conduction band any further

temperature increase does not create additional electrons. This is called depletion

region.

At high temperatures the number of electron transitions become so large that the

intrinsic electron concentration exceeds the electron concentration due to donors.

This is called intrinsic region.

88

For n type semiconductor at 00K Fermi level is located below EC and above the ED

and is denoted as EFn = 2

DE+CE

With increase in temperature, donor levels get depleted and therefore Fermi level

shifts downward. At the temperature of depletion Td, the Fermi level coincides with

the donor level Ed, (i.e. EFn = ED).

As the temperature grows further above Td, the Fermi level shifts down further, till

intrinsic region starts. At this temperature the Fermi level approaches the intrinsic

value.

EFn = 2

CE+FE=2

gE

Further increase in temperature will transform extrinsic semiconductor into intrinsic

semiconductor and Fermi level will become independent of temperature.

Fig. given below shows the dependence of Fermi level on temperature in n-type

semiconductor.

Ans.5 (b) The general differential equation of a wave travelling in x-direction with velocity ‘u’

having wave function ψ is given by,

2 2

2 2 2

1

x v t

=

----------(1)

89

The general solution of this equation is of the form ψ=ψ0 e i(kx-ωt) ---------(2), where ψ0

is a constant . k=2𝜋

𝜆

By de-Broglie hypothesis , λ=ℎ

𝑝

k= 2𝜋𝑝

ℎ =

𝑝

ђ ----------(3)

E=h =ℎ

2𝜋×2π

E=ђω

ω = 𝐸

ђ ----------(4)

substituting equations 3& 4in equation 2

ψ = ψ0 e 𝑖(𝑝𝑥−𝐸𝑡)

ђ -------(5)

Differentiating equation 5 patially w.r.t. ‘t’

𝜕𝜓

𝜕𝑡 = (

−𝑖𝐸

ђ) ψ0 e

𝑖(𝑝𝑥−𝐸𝑡)

ђ = (

−𝑖𝐸

ђ) ψ

Eψ=(−ђ

𝑖)

𝜕𝜓

𝜕𝑡 =iђ

𝜕𝜓

𝜕𝑡 --------(6)

Differentiating equation (5) partially w.r.t. ‘x’ twice ,

𝜕𝜓

𝜕𝑥= (

𝑖𝑝

ђ)ψ0e


ђ

2

2x

= (

𝑖𝑝

ђ)^2 ψ0 e


ђ = -

𝑝2

ђ2 ψ

p2ψ = - ђ2 2

2x

-----------(7)

The total energy ‘E’ is a sum of kinetic energy 1

2 mv2 & potential energy V

E= 1

2 mv2 + V =

2 2m v

2m

+V

Eψ = 𝑝2

2𝑚ψ +Vψ -----------(8)

Substituting 6,7 in equation 8 we will get ,

2 2

2i V

t 2m x

− = +

This equation is known as shrodinger’s time dependent wave equation in one

dimension.

Sol.5 (c) V = Velocity

T = Temperature in °C

S = Salinity

V= V0 + 1.14 S +4.21 T – 0.037 2T

V0= 1510m/s , T= 200C , S= 10 gm/lit

V= 1510+1.14×10+ 4.21×20 – 0.037×(20)2 = 1590.8m/s

Time=2s

Depth of sea =𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑢𝑙𝑡𝑟𝑎𝑠𝑜𝑛𝑖𝑐 𝑤𝑎𝑣𝑒𝑠 𝑖𝑛 𝑠𝑒𝑎 ×𝑇𝑖𝑚𝑒

2 =

1590.8×2

2 = 1590.8 m

90

Q.6 (a) Define the term critical temperature. Show that in the superconducting state the

material is perfectly diamagnetic. 5

(b) In a solid the energy level is lying 0.012eV below Fermi level. What is the

probability of this level not being occupied by an electron.? 5

(c) What is the wavelength of a beam of neutron having: 5

(i) an energy of 0.025eV?

(ii) an electron & photon each have wavelength of 2 A0. What are their

momentum & energy? Mn=1.676×10-27Kg, h=6.625×10-34 J-sec.

Ans.6 (a) Critical temperature (Tc):- The highest temperature at which superconductivity

occurs in a material. Below this transition temperature Tc the resistivity of the

material is equal to zero.

Superconductors exhibits Meissner effect. i.e. When a specimen is placed in weak

magnetic field & cooled below critical temperature, the magnetic flux originally

present in the specimen is ejected from the specimen. The magnetic induction inside

the specimen is given by

B=µo(H+M) ………..(1)

B=µo(1+ χ) H ………..(2)

At T is less than TC

B=0

µo(H+M) =0 ………..(3)

it follows that

M= -H

χ=M/H= -1………………………………..(2)

Susceptibility is –ve for Diamagnetic materials, hence from the above results gives

the theoretical proof the superconductors are perfectly Diamagnetic materials.

Sol.6 (b) P(E) = 1

1+𝑒^(𝐸−𝐸𝑓

𝑘𝑇) =

F

1

E E1 e P

KT

− +

= F(E)

E-Ef= -0.012eV = 0.012×1.6×10-19 J = 1.92 × 10-21J, K = 1.38 × 10-23 J/K, T = 300 K

By putting all these values in above formula, we get P(E) = 0.386 = F(E)

91

P(E) is the probability of occupancy. Probability of electron not occupying the

energy is = 1 – P(E) = 1 – F(E) = 1 – 0.386 = 0.614

Sol.6 (c) λ =?

(i) E= 0.025 eV = 0.025 ×1.6×10-19 J, h= 6.63×10-34 J-S, m=1.676×10-27kg

λ=ℎ

√2𝑚𝐸 =1.81 A0 =

34

27 19

6.63 10

2 1.676 0.025 1.6 10

−

− −

(ii) λ =ℎ

𝑝

λ=2A0=2×10-10 m

p = 34

10

6.63 10

2 10

−

−

= 3.31×10-24 kg-m/s

E = 2

2

h

2m=

( )

( )

234

227 10

6.63 10

2 1.67 10 2 10

−

− −

= 0.329×10-20 J

E = 20

19

0.329 10

1.6 10

−

−

eV = 0.02056 eV

92

Applied Physics – I (Sem-I) May 2018 F.E. SEM-1 (CBCGS)

TIME- 2 HOURS TOTAL MARKS-60

__________________________________________________________________ N.B.: - 1) Question no. 1 is compulsory

2) Attempt any three questions from Q.2 to Q.6

3) Assume suitable data wherever required.

4) Figures to right indicate marks.

Q.1 Attempt any five.

a) Why X-rays are used to study the crystal structure?

b) Calculate the frequency and wavelength of photons whose energy is 75 eV.

c) Draw the energy band diagram of p-n junction diode in forward and reverse bias

condition.

d) “Superconductor is a perfect diamagnetic “, Explain.

e) What is reverberation time? How is it important? Write the factor affecting

reverberation time.

f) A quartz crystal of thickness 1.5 mm vibrating with resonance. Calculate it’s

fundamental frequency if the Young’s modulus of quartz crystal is 7.9 × 1010 N/m2

and density is 2650 kg/m3.

g) Mobility’s of electron and hole in a sample of Ge at room temperature are 0.36

m2/V-sec and 0.17 m2/V-sec respectively. If electron and hole densities are equal

and it is 2.5 × 1013 /cm3, calculate its conductivity.

Q.2 a) Arrive at Heisenberg’s uncertainty principle with single slit electron diffraction. An

electron has a speed of 300 m/sec with uncertainty of 0.01%. Find the accuracy in its

position.

b) Write the Fermi Dirac distribution function and terms in it. What is the probability

of an electron being thermally excited to the conduction band in Si at 300 C. The

band gap energy is 1.12 eV.

Q.3 a) With neat diagram of unit cell, explain the structure of NaCl crystal and calculate

the no. of ions per unit cell, coordination no. and lattice constant. Calculate the

packing factor of NaCl crystal assuming the radius of Na+ is 0.98 A0 and the radius

of Cl- is 1.81 A0.

b) State the Hall effect. Derive the expression for Hall Voltage and hall coefficient with

neat diagram.

Q.4 a) What is working principle of SQUID? Explain how it is used to detect the magnetic

field?

93

b) A hall of dimensions 25 × 18 × 12 m3 has an average absorption coefficient 0.2 . Find

the reverberation time. If a curtain cloth of area 150 m2 is suspended at the centre of

hall with coefficient of absorption 0.75, What will be the reverberation time?

c) State the piezoelectric effect. With neat circuit diagram explain the principle and

working of piezoelectric oscillator.

Q.5 a) With energy band diagram, explain the variation of Fermi energy level with

impurity concentration in extrinsic semiconductor.

b) Explain with example how to determine crystal structure by Bragg’s X-ray

spectrometer.

c) Obtain one dimensional time independent Schrodinger equation.

Q.6 a) Define ligancy and critical radius ratio. Calculate critical radius ratio for ligancy 8.

b) What is the significance of wave function? Derive the expression for energy eigen

values for free particle in one dimensional potential well.

c) What is photovoltaic effect? Explain the principle and working of Solar cell.

*********************

94

Applied Physics – I (Sem-I) May 2018

Solution Q.1 Attempt any five.

a) Why X-rays are used to study the crystal structure?

b) Calculate the frequency and wavelength of photons whose energy is 75 eV.

c) Draw the energy band diagram of p-n junction diode in forward and reverse bias

condition.

d) “Superconductor is a perfect diamagnetic “, Explain.

e) What is reverberation time? How is it important? Write the factor affecting

reverberation time.

f) A quartz crystal of thickness 1.5 mm vibrating with resonance. Calculate it’s

fundamental frequency if the Young’s modulus of quartz crystal is 7.9 × 1010 N/m2

and density is 2650 kg/m3.

g) Mobility’s of electron and hole in a sample of Ge at room temperature are 0.36

m2/V-sec and 0.17 m2/V-sec respectively. If electron and hole densities are equal

and it is 2.5 × 1013 /cm3, calculate its conductivity.

Ans.1 (a) (i) For diffraction pattern to be studied incident wavelength should be

comparable with obstacle dimensions.

(ii) X-rays can penetrate solids since these are very high energetic radiations of

very short wavelength of 1 A0. In a crystalline solid the atoms are very closely

distributed in crystal planes. The crystal planes, thus form a three-dimensional

slit system with a spacing 1 A0. Due to this fact X-rays get strongly diffracted

from various crystal planes.

Sol.1 (b) Given Things:-E= 75eV = 75 × 1.6 × 10-19 = 1.2 × 10-17 J

Formula: - E = hѵ =ℎ𝑐

𝜆

Solution:-ѵ = 𝐸

ℎ =

1.2 ×10−17

6.63 ×10−34 = 18.13 × 1510 Hz.

λ = ℎ𝑐

𝐸 =

6.63 ×10−34×3×108

1.2 ×10−17 =165.575 A0 .

95

Ans.1 (c)

Ans.1 (d) When a specimen is placed in weak magnetic field & cooled below critical

temperature, the magnetic flux originally present in the specimen is expelled out

from the specimen. This effect, called the Meissner effect.

In normal state the magnetic flux penetrates through the material which is shown in

1st fig. & when it is cooled below TC it expels out as shown in 2nd one. This effect is

reversible & specimen can return to it’s normal state when temperature is raised

above it’s critical temperature. Thus the superconductor behaves like perfect

diamagnets (as it weakly repels magnetic field.

(i) T > TC (ii) T < TC

Fig: - External magnetic field applied to a super conductor

All Magnetic flux is expelled below Tc .

Mathematical proof: The magnetic induction inside the specimen is given by

B = µo(H+M) ……. (1)

96

B = 0

MH 1

H

+

……. (2)

B = µo(1+ χ) H ……. (3)

Where, B=Magnetic Induction inside specimen

H = Applied magnetic field

M= Magnetization produced

χ = Susceptibility

0 =Permeability of free space

At T ≤ TC flux expels out that means magnetic induction inside the specimen

becomes zero i.e. B=0

So, µo (1 + χ) H = 0 ……. (4)

As 0 & H can’t be zero

(1+ χ) = 0 ……. (5)

it follows that

χ = -1 ……. (6)

χ = M/H= -1 ……. (7)

Susceptibility is –ve for Diamagnetic materials, hence from the above results it gives

the theoretical proof that the superconductors are perfectly Diamagnetic materials.

Ans.1 (e) Reverberation Time: - The time taken by the sound to fall from it’s average

intensity to inaudibility is called the “Reverberation time”.

Formula for reverberation time is T= 0.161 × 𝑉

𝐴 .

Where, T is Reverberation time.

V= Volume of the hall

A= Total absorption in hall

a) Too much absorption will make the reverberation time too short and cause the room

to sound acoustically 'dead'.

b) Increasing the effective area of complete absorption may decrease an excessive

reverberation time for a hall but this also decreases the intensity of sound.

Hence, the optimum reverberation time is a compromise between clarity of sound

and its intensity.

Sol.1 (f) Given Things: - t = 1.5 mm = 1.5 × 10-3m , Y = 7.9 × 1010 N/m2 , = 2650 kg/m3 .

Formula: - f = 1

2𝑡

Y

= 1

2 × 1.5 ×10−3 [107.9 10

2650

]

= 1.82 MHz.

Sol.1 (g) Given Things: - µe = 0.36 m2 / V-sec, µh = 0.17 m2 /V-sec, ni = 2.5 × 1013 / cm3.

Formula: - = ni (µe + µh) e

= 2.5 × 1013 (0.36 + 0.17) × 1.6 × 10-19 = 2.12 mho/meter

97

Q.2 a) Arrive at Heisenberg’s uncertainty principle with single slit electron diffraction.

An electron has a speed of 300 m/sec with uncertainty of 0.01%. Find the accuracy

in its position.

b) Write the Fermi Dirac distribution function and terms in it. What is the

probability of an electron being thermally excited to the conduction band in Si at

300 C. The band gap energy is 1.12 eV.

Sol.2 (a) Single slit diffraction

Fig2.5 . Diffraction at single slit

An electron diffraction experiment. The graph at the right shows the degree of

exposure of the film, which in any region is proportional to the number of electrons

striking that region. The components of momentum of an electron striking the outer

edge of the central maximum, at angle θ1, are shown.

➢ Condition for first order minimum is

dsin = nλ

for n = 1

sin = d

….(1)

➢ When passes through slit, its position is known to within an uncertainty which is

equal to the slit width.

y = d …..(2)

➢ After passing through the slit, photon has its value uncertain as it makes an angle

with horizontal.

uncertainty in its y component of momentum is at least as large as p sin .

yp

sinp

= …..(3)

Using Equation (18)

py> p sin …..(4)

98

py> pd

…..(5)

Now, p =

h (de – Broglie’s hypothesis) …..(6)

py>d

h

.

>d

h

>y

h

d y=

y. py> h …..(7)

good agreement with uncertainty principle.

Given Things :- v= 300m/s , ∆𝑣

𝑣 = 0.01 % v 0.01%, =

0.01v 300

100=

Formula:- ∆x.∆p≥ ђ

Calculation:- ∆x. (m. ∆v) ≥ ђ

∆v = 300×0.01

100 = 0.03

∆x ≥ђ

𝑚∆𝑣≥

6.63×10−34

2×3.14×9.1×10−31×0.03 = 3.86 × 10-3 m.

Sol.2 (b) The carrier occupancy of the energy state is represented by a continuous

distribution function known as the Fermi-Dirac distribution function.

Fermi Distribution Function

f(E) =( )rE E /KT

1

1 e −+

This indicates the probability that a particular quantum state at

the energy level E is occupied by an electron.

K= Boltzmann’s constant

T=Absolute temperature.

Ef = Fermi energy.

Given Things:- T= 300 C = 303 0K , Eg = 1.12 eV .

K = 1.38 × 10-23 J/K = 1.38×10−23

1.6 ×10−19 = 86.25 × 10-6 eV/K

Solution:- Si is intrinsic semiconductor. Hence EC – EF = 𝐸𝑔

2 = 0.56 eV

f(E) =1

1+ ( )/C FE E KT

e−

= 1

1+6

0.56( )86.25 10 303e

−

= 4.94 × 10-10.

Q.3 a) With neat diagram of unit cell, explain the structure of NaCl crystal and calculate

the no. of ions per unit cell, coordination no. and lattice constant. Calculate the

packing factor of NaCl crystal assuming the radius of Na+ is 0.98 A0 and the

radius of Cl- is 1.81 A0.

b) State the Hall effect. Derive the expression for Hall Voltage and hall coefficient

with neat diagram.

99

Sol.3 (a) Given Things: - rC = 0.98 A0 , rA = 1.81 A0

Formula: - APF = 2𝜋

3 × [

3 3

3( )

C A

C A

r r

r r

+

+ ]

Solution: - APF = 2𝜋

3 × [

(0.98)3+ (1.81)3

(0.98+1.81)3 ]= 0.66

In sodium chloride, sodium atom loses its outer electron and so acquire an excess of

positive charge while the chlorine atom gains the electron lost by sodium atom and

so acquires an excess of negative charge, Two ions will attract one another because

of the electrostatic force of attraction

The NaCl structure which is common to LiCl, KBr, RbI, MgO, CaO and AgCl. This

structure can be viewed two different ways: face-centered cubic in chloride ions

with sodium ions in every octahedral hole, or as two interpenetrating face-centered

cubic structures.

The lattice for NaCl is Face centred cube, the basic consist of the Na atom and one Cl

atom separated by one half the length of cube i.e. NaCl has FCC diatomic lattice.

Therefore, it can be considered a two face centered cube-sub lattice, one of Na+ ions

having origin at the point 0, 0, 0 the other of Cl ions having its origin midway along

a cube edge, at point 2

a, 0, 0. There are four units of NaCl in each unit cube with

atoms in positions.

Na ions : 8 corner ions (each shared by 8 unit cells) with positional coordinates

Cl–

Na+

a

Fig: NaCl structure

(000) (100) (010) (001) 6 face centre ions (each shared by 2 unit cells) with positional

coordinates (½ ½ 0) (½ 0 ½) (0 ½ ½)

Cl ions : 12 edge centre ions (each shared by 4 unit Cells) with positional coordinates

(½ 0 0 ) (0 ½ 0 ) (0 0 ½ ) one body centre ion at (½ ½ ½ ).

Thus, the structure of NaCl contains Each atom has as nearest neighbours six atoms

of opposite kind. Many chlorides and oxides such as LiF, MgO, FeO, BaO, exhibit

this structure.

1. No. of atoms/unit cell = 4 Chlorine ions + 4 Sodium this structure = 4 NaCl

Molecule

2. Atomic radius of Na+ ions = r +

Atomic radius of Cl- ions = r-

From Fig., a = 2(r+ + r-) = d+ + d-

100

3. Coordination no. = 6

4. P.F. = Volume of unit cell

VolumeoccupiedbyNa ions Volume occupied by Cl inos+ −+

=

3 3

3

4 4

3 3

a

A Cn r n r +

=

3

33

3

33

33

33

)(

)()(3

2

)(

)()(3

2

)(2

)()(3

44

−+

−+

−+

−+

−+

−+

+

+

=+

+

=+

+

dd

dd

rr

rr

rr

rr

Ans.3 (b) Hall Effect: -

If a metal or semiconductor, carrying a current I is placed in a transverse magnetic

field B, an electric field E is induced in the direction perpendicular to both I and B.

This phenomenon is known as Hall effect and the electric field or voltage induced is

called Hall voltage (VH).


In equilibrium condition,

Force due to Electric Field = Force due to Magnetic Field

q EH = Bqv ….(1)

EH = B v ….(2)

Also EH = VH / d ….(3)

J = nev = I / wd = I/A …(4) ( since A = w × d)

v = J/ne ….(5)

From (3) VH = EH d = Bvd = new

BI

ne

BJd= ….(6)

As EH = ne

BJBv

d

VH == ….(7)

now J = nev = A

I

101

also J = E

= E

J =

A

I. E

1

EH =E

E.

A

I.

ne

B=

A

I.

ne

B=

ne

BJ

)E

1.

A

I(.

ne

B=

E

HE ….(8)

Using Equation (4)

ρne

B=σ.

ne

B=

E

HE ….(9)

Where = σ

1

ρne

B=

E

HE ….(10)

Hall coefficient

RH = BI

WHV=ne

1 ….(11)

conductivity and mobility are

= ne ….(12)

mobility

= RH .…(13)

Hall coefficient also be

RH = ne8

π3

= (π3

σ8) RH ….(14)

Applications: -

1. VH is proportional to magnetic field B, for the given current I, hence Hall effect is

used in magnetic field meter.

2. Charge carrier concentration can be determined.

3. Mobility of charge carriers can be determined.

4. Nature of semiconductor (P-type or N-type) can be determined.

Q.4 a) What is working principle of SQUID? Explain how it is used to detect the

magnetic field?

b) A hall of dimensions 25 × 18 × 12 m3 has an average absorption coefficient 0.2.

Find the reverberation time. If a curtain cloth of area 150 m2 is suspended at the

centre of hall with coefficient of absorption 0.75, What will be the reverberation

time?

102

c) State the piezoelectric effect. With neat circuit diagram explain the principle and

working of piezoelectric oscillator.

Ans.4 (a) (i) SQUID is acronym for Superconducting Quantum Interference Device. A

SQUID can measure an extremely small magnetic field, voltage and current. It

is very sensitive magnetometer in which a superconducting loop is used with

one or more Josephson junctions.

(ii) The current I enters and splits into the two paths, each with currents Ia and Ib.

The thin barriers on each path are Josephson junctions, which together

separate the two superconducting regions. represents the magnetic flux

threading the DC SQUID loop.

Fig 5.10 SQUID Circuit diagram

(iii) The current Ia and Ib undergoes a phase shift while crossing the Josephson

junctions and interfere at the end. By measuring the phase shift at other end

we can calculate the value of the magnetic field. In absence of the magnetic

field phase shift & phase difference are zero.

(iv) In practice, instead of the current, the voltage V across the SQUID is

measured. Thus the SQUID is flux to voltage transducer which converts a

small change in magnetic flux into voltage

Sol.4 (b) Given Things: - V = 25 × 18 × 12 m3, avg= 0.2, curtain = 0.75, Scurtain = 100m2

Formula: - T1 = 0.161 × [.av

V

S]

T2 = 0.161 ×['

.av curtain

V

S S +]

Calculations: - S = 2[ (25×18) + (18×12) + (12×25)] = 1932 m2 .

V = 25 × 18 × 12 m3 = 5,400 m3 .

T1 = 0.161 × 5400

0.2 ×1932= 2.25 sec.

Absorption takes place by both the surfaces of the curtain

S’ = 2 × 150 m2 = 300 m2 .

T2 = 0.161 × 5400

(0.2 ×1932)+ (0.75 ×300) = 1.42 sec.

So the change in reverberation time is = 2.25- 1.42 = 0.83 sec.

Ans.4 (c ) Piezoelectric effect: When Piezoelectric crystals like quartz or tourmaline are

stressed along any pair of opposite faces, electric charges of opposite polarity are

103

induced in the opposite faces perpendicular to the stress. This is known as

Piezoelectric effect.

Principle

Piezoelectric oscillator is based on the principle of resonance between the natural

frequency of appropriately cut piezoelectric crystal and a suitable frequency

generated by LC oscillator.

Fig :- piezoelectric oscillator

Circuit analysis

The quartz crystal is placed between two metal plates A and B.

• The plates are connected to the primary (L1) of a transformer which is inductively

coupled to the electronics oscillator.

• The electronic oscillator circuit is a base tuned oscillator circuit.

• The coils L of oscillator circuit are taken from the secondary of a transformer T.

• The Plate coil L1 is inductively coupled to Grid coil L2.

• The coil L2 and variable capacitor C form the tank circuit of the oscillator.

Working

• When H.T. battery is switched on, the oscillator produces high frequency alternating

voltages with a frequency f.

• Due to the transformer action, an oscillatory e.m.f. is induced in the coil L. This high

frequency alternating voltages are fed back to the crystal plates A and B.

• Inverse Piezo-electric effect takes place and the crystal contracts and expands

alternatively. The crystal is set into mechanical vibrations.

• The frequency of the vibration is given by

K Which represent order of harmonic. In practice we take k = 1 that is fundamental

frequency. It is important to know that natural frequency n is inversely proportional

to thickness t. Hence for higher frequency, thickness t has to be reduced.

104

In the above diagram we have a thin plate of piezoelectric crystal cut on such axis so

that we get inverse piezoelectric effect i.e. on application of high frequency electric

field, it can vibrate in resonance with LC oscillator.

Advantages

• Ultrasonic frequencies as high as 5 x 108Hz or 500 MHz can be obtained with this

arrangement.

• The output of this oscillator is very high.

• It is not affected by temperature and humidity.


• Better waveform

Disadvantages


➢ The cost of piezoelectric quartz is very high

➢ The cutting and shaping of quartz crystal are very complex.

Q.5 a) With energy band diagram, explain the variation of Fermi energy level with

impurity concentration in extrinsic semiconductor.

b) Explain with example how to determine crystal structure by Bragg’s X-ray

spectrometer.

c) Obtain one dimensional time independent Schrodinger equation.

Ans.5 (a) Extrinsic Semiconductor

Extrinsic semiconductors are of two types depending upon the impurity element

introduced.

n-Type Semiconductors (Based on 1, Chapter 18, 7th edition page 18.13, 18.14)

• Pentavalent impurity is added to a pure semiconductor it becomes n-type extrinsic

semiconductor. Impurity is called donor impurity.

105

• Antimony (Sb) added as an impurity has five valence electrons.

• Each Sb atom forms covalent bonds with the surrounding four Si atoms with the

help of four of its five electrons and the fifth valence electron remains loosely bound

to the parent impurity atom which becomes available as current carrier.

Fig. : n-type semiconductor

• Energy required to remove fifth electron is 0.0 5eV. This energy is very small in

comparison to 1.12 eV which is then energy gap for Si and also it represents the

energy to break a covalent bond.

• Electrons in conduction band come from two different ways by donor atom & by

intrinsic process.

• Majority current carriers in n-type semiconductors are electrons & minority current

carriers are holes.

• Addition of an impurity adds an allowed energy level ED at a very small distance

below the conduction band

Fig. Donor level in n-type semiconductor

conductivity is

e = n .e .e ….(1)

106

Boltzman factor a

n α e –(EC – ED)/KT .…(2)

( )C DE E /KT

e 0e− −

= ….(3)

In n-type semiconductors, as there are many free electrons in conduction band, the

Fermi level gets shifted towards the conduction band.

At 0 0K it is between the bottom of conduction band and the level ED.

p-Type Semiconductors

• If a trivalent impurity (Group III) is added to a pure semi-conductor, it becomes p-

type extrinsic semiconductor. The impurity added is called as acceptor impurity.

• As shown in Fig. Boron (B) has been added as impurity which has three electrons.

Each B atom tries to form covalent bonds with surroundings four Si atoms and falls

short of one electron for completing four covalent bonds.

Fig. : P-type semiconductor

• As a result a vacancy is left in the bonding. This vacancy is not a hole. Originally,

the environment in the crystal lattice is electrically neutral.

• The introduction of impurity atom does not disturb the environment and the

vacancy arising due to the non-formation of bond is not a hole.

• Electron from a neighboring bond acquires energy and jumps into this vacancy, it

leaves behind a positively charged environment in the broken bond. Therefore a

hole is generated there.

• The addition of an impurity adds an allowed level EA at a very small distance above

the top of the valence band as shown in Fig.

• At T = 00 K all acceptor levels at EA are vacant as shown in Fig. and valence band is

full & the conduction band is empty.

107

• • As temperature increases, electrons from the valence band jump into acceptor level

and leave behind holes. At moderate temperature all acceptor levels are filled

• p-type semiconductor is having holes as majority carriers and electrons as minority

carriers which results because of intrinsic process.

conductivity of a p-type semiconductor

p = p. e. h ...(1)

hole concentration by Boltzman law as

( )V AE E /KT

p e−

= ...(2)

conductivity of a p-type

= ( )V AE E /KT

0e−

….(3)

Fig. : Acceptor level in p-type semiconductor

In n-type semiconductor, as there are many free holes in valence band, the Fermi

level gets shifted towards the valence band. At 00K it is between top of valence

band and the level EA.

108

Ans.5 (b) Bragg devised an apparatus to investigate the structure of single crystal by using X-

rays. It is used to measure glancing angle θ.

S1 S2

V

C

A

S3

HT

D

Fig:- Bragg’s Spectrometer

He used crystal as reflection grating and not as transmission grating.

The experiment arrangement is shown in the diagram.

X-rays from X-ray tube is collimated using two adjustable slits, S1, and S2 are made

to fall on a crystal C with which in wax on the spectrometer table. The vernier scale

V is attached to table on which the crystal is placed. The vernier scale V is capable to

move the circular scale S and determine the position of the crystal. A strong

monochromatic X-ray beam is made to fall on the crystal face. The reflected beam

after passing through the slit S3, enters an ionization chamber D mounted on an arm

which can be rotated about the same axis as the crystal. Its position can be read by

second veiner V2.

The gas in the chamber is ionization by the X-rays. The resulting ionization current,

measured by the electrometer E, is a measure of the intensity of X-rays reflected by

the crystal. The crystal is rotated through small angles (while the arm carrying the

ionization current is rotated through double the angles) and the ionization current is

measured each time. The curve of intensity which is measure of ionization current

or ionization current against glancing angle θ is plotted. For certain value of

glancing angle θ, The intensity of ionization current increase abruptly (Fig ).

According to Bragg’s equation 2d sin θ = nλ From the graph, the glancing angles θ1, θ2,

θ3can be measured and it can be seen that Sin θ1 : Sin θ2 : Sin θ3 = 1: 2: 3

A1

A2

A3

1 2 3

Glancing angle

Ion

isa

tion

Curr

en

t

Fig: - Profile peaks using X ray diffraction

109

From the observed values of θ and known values of d and n, the wavelength of X-

ray can be calculated. Suppose the wavelength of incident X-ray is known then the

ratio of the interplanar spacing can be determined. Suppose, for a particular crystal

used on a Bragg’s spectrometer, strong reflections from thesets of planes (100), (110)

and (111) are obtained for angles θ1, θ2, θ3respectively in the first order. then from

Bragg’s equation we have

2d100 sinθ1 = λ

2d110 sinθ2 = λ

2d111 sinθ3 = λ

Hence, d100 : d110 : d111 = 1/sinθ1 : 1/sinθ2 : 1/sinθ3


Hence for crystal

d100 : d110 : d111 = ''' 259sin

1:

367sin

1:

235sin

1ooo

= 744.1

1:

414.1

1:1 =

3

1:

2

1:1

for SC


concluded that crystal has a simple cubic structure. When the first order reflection

from the three plane (100), (110) and (111) of NaCl are compared, the ratios between

the interplanar spacing are found as

For FCC d100 : d110 : d111 = 1 2

1 : :2 3



For BCC d100 : d110 : d111 = 2 1

1 : :2 3

Ans.5 (c) For systems in a stationary state (i.e., where the Hamiltonian is not explicitly

dependent on time), the time-independent Schrödinger equation is sufficient.

Approximate solutions to the time-independent Schrödinger equation are

commonly used to calculate the energy levels and other properties of atoms and

molecules.

Schrodinger wave equation as

(x, t) = (x) (t) ….(1)

The TDSE is given by

2 2

22

− + =

V i

m x t

....(2)

Equation (2) can be modified as

2 2

2

( ) ( )( ) ( ) ( ) ( )

2

− + =

x tt V x x t i

m x t

http://en.wikipedia.org/wiki/Stationary_state

http://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics)

http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation#Time-independent_equation

http://en.wikipedia.org/wiki/Energy_level

110

Divide both sides by (x) (t)

2 2

2

1 ( ) 1 ( )( )

2 ( ) ( )

− + =

x tV x i

m x x t t

The above equation can also be written as

2 2

2

1 1( )

2

− + =

V x i

m x t

….(3)

Equation (3) shows that we have separated the Schrödinger equation such that on

LHS is function of x only and RHS is function oft only. Since Equation (3) is valid

for any x and t, both the sides must be equal to a constant say energy E.

2 2

2

1. . ( )

2

− + =

V x E

m x

or 2 2

2( )

2

− + =

V E

m x

….(4)

Eq (4) represent time independent Schrodinger equation(TISE).

Q.6 a) Define ligancy and critical radius ratio. Calculate critical radius ratio for ligancy 8.

b) What is the significance of wave function? Derive the expression for energy eigen

values for free particle in one dimensional potential well.

c) What is photovoltaic effect? Explain the principle and working of Solar cell.

Ans.6 (a) Ligancy:- In a given crystal, the number of anions surrounding a cation is called the

Ligancy.

Critical radius ratio:-The cation anion radius ratio is called the critical radius ratio

C

A

r

r.

Critical radius for ligancy 8:

A

A A

A

a

Fig 1.27 EgCsCl, SC diatomic crystal.

8 anions located at the corners touch along the cube edges. Cation located at body

centred position touches the corner anions.

The cube edge a = 2 Ar and the body diagonal is

( )A C2 r r 3a+ =

111

Putting a = A2r we get

2( ( )A Cr r+ =

3C A

A

r r

r

+=

Ans.6 (b) Matter waves associated with electron or any particle. If we consider wave function

associated with a system of electrons then | |2 d is regarded as a measure of

density of electrons. If is a volume inside which an electron is known to be present,

but where exactly the electron is situated inside is not known.

as wave function then ||2d provides the probability of finding the electron in

certain volume d of . Means | |2 is called the probability function.

Since electron must be somewhere inside the volume .

| |2 d = 1

Wave function has no direct physical significance but | |2 has.

State of a system is completely characterized by a wave function.

Wave functions are usually complex.

i.e. = A + iB

Where A and B are real functions

Integral of the wave function over entire space in the box must be equal to unity

because, there is only one particle and at time it is present somewhere inside the box

only. Therefore,

−

= 1|| 2 dV

A wave function which obeys above equation is said to be normalized.

Outside the box V = and particle cannot have infinite energy therefore it cannot

exist outside the box.

Schrodinger's equation is written as

0)(8

2

2

2

2

=−+

Eh

m

dx

d

Inside the box, V = 0

Schrodinger’s equation is written as

08

2

2

2

2

=+

Eh

m

dx

d

Ar2.3

732.1131 =−=+A

C

r

r

732.0=A

C

r

r

112

Fig. One dimensional potential well of infinite height

Equation (39) may be simplified as

02

2

2

=+

Kdx

d

where K2 = 2

28

h

mE

or K2 =2

2

mE

Solution of Equation (40) is written as,

= A cosKx + B sin Kx

When x = 0 at = 0 we get

0 = A cos 0 + B sin 0

A = 0 (Since cos 0 = 1)

When x = a, = 0

0 = A cosKa + B sin Ka

But A = 0

B sin Ka = 0

Here B need not be zero

sin Ka = 0 only when

Ka =

mE2a = n (where n = 0, 1, 2, 3 ….)

where n = Quantum number

n = B sin xa

n

n = 0 is not acceptable because for n = 0, = 0, means electron is not present inside

the box which is not true.

as K2 = 2

28

h

m and K =

a

n

2

2)(

a

n =

2

28

h

mE

113

En =

2 2

28

n h

ma (n = 1, 2, 3, ….)

Energy of the particle can have only certain values which are Eigen values.

Ans.6 (c) Photovoltaic Effect:-In photoelectric effect when radiation is incident on a metal

surface electron are ejected. In photovoltaic effect, certain materials being exposed to

radiation generates electron-hole pairs available for conduction. As a result a

voltage is developed across the material.

A solar cell (also called a photovoltaic cell) is an electrical device that converts the




generate and support an electric current without being attached to any external

voltage source.






the potential and this electricity is captured. Due to the special composition of solar

cells, the electrons are only allowed to move in a single direction.


(DC) electricity.








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