and answers sem-i/ii

Engineering Physics

( 17PHY12/22)

MODULE WISE

QUESTIONS AND

ANSWERS

SEM-I/II

VTU ENGINEERING PHYSICS

C.REDDAPPA, M.Sc; B.Ed; Prof. & HOD,CBIT,KOLAR.

IMPORTANT IA TEST & VTU EXAMINATION QUESTIONS

MODULE-1

Modern Physics and Quantum Mechanics

1. State Planck’s quantum hypothesis(postulates) and explain black body radiation spectra

based on planck’s radiation law.

2. State Planck’s law of radiation and deduce Wein’s law and Rayleigh-Jeans law from Planck’s

law of radiation.

3. Explain Compton effect experimentally.

4. What are matter(de-Broglie) waves ? Mention the properties of matter waves.

5. Derive the relation between Phase velocity and Group velocity.

6. Define Phase velocity and Group velocity and show that Group velocity is equal to Particle

velocity.

7. State and explain Heisenberg’s uncertainty Principle and Show that electrons non exist in the

nucleus using Heisenberg’s uncertainty Principle .

8. Set up Schrodinger 1D( one dimensional) time independent wave equation.

9. Apply Schrodinger time independent wave equation to a particle in an infinite depth

potential well/Box to find eigen energies and eigen functions.

10. Mention the properties of wave function.

MODULE-2

Electrical properties of Materials.

1. Explain the terms : Drift velocity, Mean collision time, Mean free path, Relaxation time.

2. Explain the failures of CFET(Classical free electron theory)

3. State the assumptions of QFET(Quantum free electron theory) and explain the merits of

QFET.

4. Discuss the dependence of Fermi factor on temperature and energies.

5. Derive the expression for electrical conductivity based on QFET.

6. State law of mass action and Derive an expression for concentration of electrons and holes

( carrier density) in intrinsic semiconductors.

7. Write a note on Meissner’s effect and Maglev vehicles.

8. Distinguish between Type-I and Type-II superconductors.

9. Explain BCS theory qualitatively.

10. State and explain Matheissen’s rule ( Dependence of resistance on impurities and

temperature)

11. Write a note on Fermi-Dirac statistics.

Continued on inner side of Back page

REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM

VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 1

ENGINEERING PHYSICS (15PHY12/22)

SEMESTER - I / II

[ As per CBCS Scheme with effect from the academic year 2015-2016 ]

MODULE-1 : MODERN PHYSICS AND QUANTUM MECHANICS

MODERN PHYSICS:

Q:What is meant by Black Body radiation spectra ? explain it briefly.

1. If we plot intensity of the thermal radiations emitted by a Black Body against the

corresponding wave lengths at different temperatures, we get Black body radiation

(BBR)spectra as shown in the graph.

2. From the graph it is clear that the wavelength (𝜆𝑚) corresponding to the maximum

intensity 𝑬𝒎varies inversely as the absolute temperature(T ) of the black body. This is

called Wein’s displacement law , 𝒊𝒆: 𝝀𝒎 ∝𝟏

𝑻 or 𝝀𝒎 𝑻 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 According to this law

′𝜆𝑚′ shifts towards shorter wavelength side with the increase of ‘ T ’ of the body.

3. Wein’s law states that the energy density in the wave length interval λ & λ +dλ

is 𝑈𝜆 dλ = 𝐶1𝜆−5𝑒−𝐶2/𝜆𝑇dλ where 𝐶1 & 𝐶2 are constants. Wein’s law explained only

the shorter wavelength region of BBR spectra below 𝜆𝑚 and it failed to

explain the longer wavelength region of the BBR spectra beyond 𝜆𝑚.

4. Rayleigh-Jean’s law states that the energy density in the wave length interval λ &

λ +dλ is 𝑈𝜆 dλ = 8𝜋𝑘𝑇

𝜆4 dλ ,where k = Boltsmann’s constant. Rayleigh-Jean’s law

explained the longer wavelength region of BBR spectra beyond 𝜆𝑚 and it failed

explain the shorter wavelength region of BBR spectra below 𝜆𝑚.

E

Rayleigh-Jean’s law

𝑬𝒎 Experimental B.B.R spectra

Wein’s law

𝟎 𝝀𝒎 𝝺



5. Planck’s law states that the energy density in the wavelength interval λ & λ +dλ is

𝑈𝜆 dλ = 8𝜋ℎ𝑐

𝜆5

1

𝑒(

ℎ𝑐𝜆𝑘𝑇

) −1

dλ .

Planck’s law explained the BBR spectra completely ie: both shorter and longer

wavelength sides of the spectra.

Q: State Planck’s quantum postulates and explain Planck’s explanation of BBR specta.

The quantum postulates are :

1. The black body is made up of a large number of simple Harmonic oscillating particles,

which can oscillate in all possible frequencies.

2. An oscillating particle can have a set of energies ,which are the integral multiples of a

lowest finite quanta of energy(h𝜈) i.e: 𝐸𝑛 = n.hν ,where n = Quantum number.

3. The oscillating particles absorb or emit energy in discrete units of hν, only when they

undergo transitions between any two allowed energy states.

Planck based on quantum postulates, derived an expression for energy density

in the wavelength range λ and λ+dλ called Planck’s law given by

𝑈𝜆 dλ =

8𝜋ℎ𝑐

𝜆5

1

𝑒(

ℎ𝑐𝜆𝑘𝑇

) −1

dλ.

Based on this, Planck explained the BBR spectrum completely ie: both shorter and longer

wavelength side of the spectrum.

E

Rayleigh-Jean’s law

𝑬𝒎 Experimental B.B.R spectra

Wein’s law

𝟎 𝝀𝒎 𝝺



Q:Deduce/derive Wein’s law & Rayleigh-Jeans law from planck’s law of radiation or

Reduce Planck’s law to Wein’s law and Rayleigh-Jeans law

Planck’s law of radiation is given by 𝑈𝜆dλ = 8𝜋ℎ𝑐

𝜆5

1

𝑒(

ℎ𝑐𝜆𝑘𝑇

) −1

dλ ……………(1)

a) Wein’s law:

For shorter wave lengths, ‘λ ‘is very small ,hence 𝑈𝜆 & 𝑒(ℎ𝑐

𝜆𝑘𝑇) are

Large i.e: 𝑒(ℎ𝑐

𝜆𝑘𝑇) ≫ 1 so that 𝑒(

ℎ𝑐

𝜆𝑘𝑇) − 1 = 𝑒(

ℎ𝑐

𝜆𝑘𝑇) ……….(2)

∴ from eqn 1 &2,we get 𝑈𝜆dλ = 8𝜋ℎ𝑐

𝜆5

1

𝑒(

ℎ𝑐𝜆𝑘𝑇

)

dλ

∴ 𝑼𝝀 dλ = 𝑪𝟏𝝀−𝟓𝒆−𝑪𝟐𝝀𝑻 dλ 𝑇ℎ𝑖𝑠 𝑖𝑠 𝑤𝑒𝑖𝑛’𝑠 𝑙𝑎𝑤 .where 𝐶1 = 8𝜋ℎ𝑐 & 𝐶2 =

ℎ𝑐

𝑘

b) Rayleigh-Jeans law:

For longer wave lengths, ‘λ ‘is very large,

hence 𝑒(ℎ𝑐

𝜆𝑘𝑇) 𝑖𝑠 𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙 ,Expanding 𝑒(

ℎ𝑐

𝜆𝑘𝑇) as power series ,

we get , 𝑒(ℎ𝑐

𝜆𝑘𝑇) = 1+ (

ℎ𝑐

𝜆𝑘𝑇) + (

ℎ𝑐

𝜆𝑘𝑇)2 +……….

= 1 + (ℎ𝑐

𝜆𝑘𝑇) , 𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑖𝑛𝑔 ℎ𝑖𝑔ℎ𝑒𝑟 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 (

ℎ𝑐

𝜆𝑘𝑇)

𝑖𝑒: 𝑒(

ℎ𝑐

𝜆𝑘𝑇)

− 1 = ℎ𝑐

𝜆𝑘𝑇 ………(3)

∴ from eqn 1 &3,we get 𝑈𝜆dλ = 8𝜋ℎ𝑐

𝜆5

1ℎ𝑐

𝜆𝑘𝑇 dλ

𝒊𝒆: 𝑼𝝀 dλ = 𝟖𝝅𝒌𝑻

𝝀𝟒 dλ , 𝑇ℎ𝑖𝑠 is Rayleigh-Jeans law

Q: What is meant by “ultraviolet catastrophe” or failures of Rayleigh-Jeans law ?

1. As per Rayleigh-Jeans law, 𝑈𝜆 → ∞ 𝑎𝑠 𝜆 → 0

2. 𝐵𝑢𝑡 experimentally observed that 𝑈𝜆 → 0 𝑎𝑠 𝜆 → 0.

3. This failure of R-J’s law beyond ultra-violet region is called “ultraviolet catastrophe”.



Q:Explain Compton effect experimentally & mention it’s significance.

1. Compton experimental arrangement is as shown in the diagram.

2. X-ray photons of wave length λ from X-ray tube are collimated by passing through two

slits S1 &S2.

3. The collimated X-rays are made to fall on graphite target.

4. X-rays collide with the electrons of the target at rest and transfer part of their energy

resulting in the recoil of electrons in a direction making an angle ∅ 𝑎𝑛𝑑 𝑆𝑐𝑎𝑡𝑡𝑒𝑟𝑒𝑑 𝑋 −

𝑟𝑎𝑦𝑠 𝑚𝑎𝑘𝑒𝑠 𝑎𝑛 𝑎𝑛𝑔𝑙𝑒 𝜃 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 𝑤𝑖𝑡ℎ the incident X-rays direction.

5. The intensity of the scattered X-rays in different directions are measured using Bragg’s

spectrometer.

6. Compton calculated the wavelengths of scattered X-rays at different scattering angles and

found that scattered X-rays consists of two components namely ‘unmodified’ component

having the same wavelength(λ) as incident X-rays & ’modified’ component having slightly

higher wavelength 𝜆′

7. This scattering of X-rays due to recoil of electrons is called “Compton effect”.

8. The change in wavelength (𝜆′ − 𝜆)is called ‘Compton wavelength’ ,which depends only

𝜃 𝑎𝑛𝑑 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝞴.

9. Based on conservation of energy & momentum laws, considering the collision between X-

rays & electrons as ‘particle-particle’ collision , Compton wavelength is given by

(𝜆′ − 𝜆) = ∆𝜆 = ℎ

𝑚0𝐶 (1−𝐶𝑜𝑠𝜃)

where ℎ

𝑚0𝐶= 𝜆0 called “Compton wavelength of electron”= 0.02426 Å(constant)

10. ∆𝝀 𝑣𝑎𝑟ies from ‘0’ for 𝜃 = 0° 𝑡𝑜 2𝜆0 for 𝜃 = 𝟏𝟖𝟎°

11. 𝐶𝑜𝑚𝑝𝑡𝑜𝑛 𝑒𝑓𝑓𝑒𝑐𝑡 𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑒𝑠 𝑡ℎ𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑛𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑋 − 𝑟𝑎𝑦𝑠(𝑙𝑖𝑔ℎ𝑡)

Q: What is meant by Dual nature of matter ?

Matter is made up of particles like electrons, protons, neutrons, atoms etc ,also

waves are associated with these matter particles under suitable conditions.

Thus matter exhibiting both particle & wave nature is called dual nature of matter.

Bragg’s spectrometer

Slits

X-rays Collimated X-rays 𝜃

𝑆1 𝑆2 ∅

Graphite target Recoil electron



Q: What are de-Broglie (matter) waves ?

de-Broglie(matter) waves are the waves associated with material particles in

motion.

Q: Derive an expression for de-broglie wavelength(λ).

According to Einstein’s mass-energy relation E=m𝐶2……(1)

where m=mass, C=speed of light.

also, from Planck’s quantum theory, E =ℎ𝐶

𝜆 …………(2) where h=planck’s contant.

ν=frequency. From eqns 1&2, m𝐶2 = ℎ𝐶

𝜆

∴ 𝝀 =𝒉

𝒎𝑪

Similarly, for a particle of mass ‘m’ moving with a velocity ‘𝒗 ′,

De-broglie wavelength, λ = 𝒉

𝒎𝓥 or λ =

𝒉

𝑷

Q: De-broglie’s wavelength of an electron accelerated in a potential difference ’V’ volt/

Show that λ = 𝟏𝟐.𝟐𝟖

√𝑽 Å for an electron accelerated in a potential difference ’V’ volt

The K.E of an electron of mass ‘m’,charge ‘e’ accelerated in a potential ‘V’ volt is given

by 1

2𝑚𝑣2 = 𝑒𝑉 , multiplying both Nr. and Dr. of LHS by ‘m’

we get, 𝑚2𝑣2

2𝑚= 𝑒𝑉

𝑖𝑒; 𝑚2𝑣2 = 2𝑚𝑒𝑉

𝑃2 = 2𝑚𝑒𝑉 ∵ 𝑚𝑣 = 𝑃

𝑷 = √𝟐𝒎𝒆𝑽

But λ = 𝒉

𝑷 ∴ 𝝀 =

𝒉

√𝟐𝒎𝒆𝑽

= 𝟔.𝟔𝟐𝟓𝒙10−34

√𝟐𝒙 𝟗.𝟏𝒙𝟏𝟎−𝟑𝟏𝒙𝟏.𝟔𝒙𝟏𝟎−𝟏𝟗𝑽

= 𝟏.𝟐𝟐𝟖𝒙𝟏𝟎−𝟗

√𝑽 m

= 1.228

√𝑽𝑛𝑚 =

𝟏𝟐.𝟐𝟖

√𝑽 Å

Q:Mention the characteristics of matter waves or Write a note of matter waves.

1. MW are non mechanical waves.

2. MW are associated with moving particles .

3. MW are non electromagnetic waves.

4. MW of microscopic particles can be measured .

5. MW are charge independent .



6. MW of macroscopic particles cannot be measured .

7. Different MW have different phase velocity.

8. Phase velocity of MW is greater than that of light.

9. Velocity of MW depends on the velocity of material particles.

Q:Define Phase velocity/What is meant by Phase velocity ?

“Phase Velocity ” is defined as the velocity with which the uni-phase particles on

the wave travels and is given by 𝑣𝑃 =𝜔

𝑘

where 𝜔 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑛𝑑 𝑘 = 𝑤𝑎𝑣𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

Q: Define Group velocity/What is meant by Group velocity ?

“Group Velocity ” is defined as the velocity with which the resultant wave packet of

the group of waves travels and is given by 𝑣𝑔 =𝑑𝜔

𝑑𝑘

where d𝜔 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑛𝑑 𝑑𝑘 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑤𝑎𝑣𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

Q: Derive the relation between group velocity (𝒗𝒈) & phase velocity(𝒗𝑷)/

Show that 𝒗𝒈 = 𝒗𝑷 − 𝝀𝒅𝑉𝑃

𝒅𝝀

Consider a particle of mass ‘m’ having phase velocity ‘ 𝒗𝑷 ‘ and group velocity ‘′𝒗𝒈′.

Let the wave have the wavelength ’λ’ , frequency ‘ν, wave number k & angular velocity ω

We know that 𝑣𝑃 =𝜔

𝑘 …(1) and 𝑣𝑔 =

𝑑𝜔

𝑑𝑘 ….(2) 𝑉𝑃

From eqns 1 & 2 , we get 𝑣𝑔 =𝑑(𝑉𝑃.𝑘)

𝑑𝑘

= 𝑣𝑃 + 𝑘𝑑𝑉𝑃

𝑑𝑘

= 𝑣𝑃 + 𝑘.𝑑𝑉𝑃

𝑑𝜆

𝑑𝜆

𝑑𝑘 ……(3)

Also, λ = 2𝜋

𝑘 ∴

𝑑𝜆

𝑑𝑘 =

𝑑

𝑑𝑘(

2𝜋

𝑘)

= −2𝜋

𝑘2 ……..(4)



From eqns 3 & 4 ,we get 𝑣𝑔 = 𝑣𝑃 + 𝑘.𝑑𝑉𝑃

𝑑𝜆(

−2𝜋

𝑘2 )

= 𝑣𝑃 − 2𝜋

𝑘 𝑋

𝑑𝑉𝑃

𝑑𝜆

ie: 𝒗𝒈 = 𝒗𝑷 − 𝝀𝒅𝑉𝑃

𝒅𝝀 ∵ λ =

2𝜋

𝑘

Note: 𝒗𝒈 < 𝒗𝑷 for a dispersive medium

and 𝑣𝑔 = 𝑣𝑃 = 𝐶 𝑓𝑜𝑟 𝑎 𝑛𝑜𝑛 𝑑𝑖𝑠𝑝𝑒𝑟𝑠𝑖𝑣𝑒 𝑣𝑎𝑐𝑐𝑢𝑚 𝑚𝑒𝑑𝑖𝑢𝑚

Q: Deduce the relation between group velocity(𝒗𝒈) & particle velocity(𝒗𝑷𝒂𝒓𝒕𝒊𝒄𝒍𝒆)/

Show that 𝒗𝒈 = 𝒗𝑷𝒂𝒓𝒕𝒊𝒄𝒍𝒆 or 𝒗𝒈 = 𝒗

Consider a particle of mass ‘m’ having particle velocity ‘𝒗𝑷𝒂𝒓𝒕𝒊𝒄𝒍𝒆′ & 𝑔𝑟𝑜𝑢𝑝 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ′𝒗𝒈′.

We know that, 𝑣𝑔 =𝑑𝜔

𝑑𝑘 … … . . (1)

but 𝜔 = 2𝜋𝜈 & 𝐸 = ℎ𝜈 ,we get 𝜔 = 2𝜋𝐸

ℎ

∴ d𝜔 = (2𝜋

ℎ) 𝑑𝐸 … (2)

Also from k = 2𝜋

𝜆 & 𝜆 =

ℎ

𝑃 we get k = (

2𝜋

ℎ)P

∴ dk = (2𝜋

ℎ)dP ……(3)

From eqns 1,2 &3,we get 𝑣𝑔 =𝑑𝜔

𝑑𝑘=

(2𝜋

ℎ)𝑑𝐸

(2𝜋

ℎ)𝑑𝑃

=𝑑𝐸

𝑑𝑃 …….(4)

Also , from E = ½m𝑣2 =𝑚2𝑣2

2𝑚 & P = m𝓥 ,

𝑤𝑒 𝑔𝑒𝑡 𝐸 = 𝑃2

2𝑚

∴ dE = 2𝑃

2𝑚𝑑𝑃

𝑖𝑒; 𝑑𝐸

𝑑𝑃=

𝑃

𝑚 =

𝑚 𝑣𝑃𝑎𝑟𝑡𝑖𝑐𝑙𝑒

𝑚

= 𝑣𝑃𝑎𝑟𝑡𝑖𝑐𝑙𝑒 ….(5)

From eqns 4 & 5,we get 𝒗𝒈 = 𝒗𝑷𝒂𝒓𝒕𝒊𝒄𝒍𝒆 Thus group velocity =Particle velocity



QUANTUM MECHANICS

Q: State & explain Heisenberg’s Uncertainty Principle (HUP). Mention its significance.

HUP states that “the product of the uncertainty ‘Δ𝑥’ in the position and the uncertainty

‘ΔP𝑥’ in the momentum of a particle at any instant is equal to or greater than ( h/4𝛑) “

i.e.: Δ𝑥 ΔP𝑥 ≥ ℎ

4𝜋 ,Where h is Planck’s constant.

The significance of HUP is that, it is impossible to determine simultaneously both the position and

momentum of the particle accurately at the same instant.

NOTE : Other HUP relations are ∆𝐸.∆𝑡 ≥ ℎ

4𝜋 , 𝑤ℎ𝑒𝑟𝑒 ∆𝐸 = 𝑒𝑛𝑒𝑟𝑔𝑦, ∆𝑡= time,

& ∆𝐿.∆𝜃 ≥ ℎ

4𝜋

where , ∆𝐿 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 & ∆𝜃 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡.

Q: Show that electrons do not present in the nucleus using Heisenberg’s Uncertainty Principle.

Electron to be present in the nucleus, maximum uncertainty in position Δ𝑥=10-14 m (diameter)

According to HUP,

The minimum uncertainty in momentum ΔP𝑥 ≥ ℎ

4𝜋 𝛥𝑥

≥ 6.625 𝑥 10−34

4 𝑥 3.14 𝑥 10−14

ΔP𝑥 ≥ 5.275 x 10-21 kg m/s = P(say)

Using E = m𝐶2 , 𝑃 = 𝑚𝑉 𝑎𝑛𝑑 𝑚 = 𝑚0

√1−𝑉2

𝐶2

, 𝑤𝑒 𝑐𝑎𝑛 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 the minimum energy of the

electron in the nucleus is given by 𝐸2 = P2c2 +𝑚𝑜2 𝑐4, but the rest mass energy 𝑚𝑜

2 𝑐4 is

very very small compared to P2c2,neglecting 𝑚𝑜2𝑐4 ,

we have E ≈ PC = 5.275 x 10−21 x 3 x 108 J

= 5.275 𝑥 10−21 𝑥 3 𝑥 108

1.6 𝑥 10−13 MeV

= 9.89 MeV

= 10 MeV

But the maximum energy of the electrons(𝛃-particle) emitted

from the nucleus does not exceed 4MeV,hence electrons do not

present in the nucleus.

NOTE:

𝐸2 = 𝑚2𝐶4 = 𝑚𝑜

2𝐶4

(1−𝑣2

𝐶2) =

𝑚𝑜2𝐶6

(𝐶2−𝑣2)

𝑃2𝐶2 = 𝑚2𝑣2𝐶2 = 𝑚𝑜

2𝑣2𝐶2

(1−𝑣2

𝐶2) =

𝑚𝑜2𝐶4𝑣2

(𝐶2−𝑣2)

𝐸2 − 𝑃2𝐶2 = 𝑚𝑜

2𝐶4(𝐶2−𝑣2)

(𝐶2−𝑣2)

∴ 𝑬𝟐 = 𝑷𝟐𝑪𝟐 + 𝒎𝒐𝟐𝑪𝟒



Q: Set up 1D time independent Schrodinger wave equation for a free particle.

One dimensional wave function 𝚿 describing the de-broglie wave for a particle moving

freely in the positive direction of 𝑥-direction is given by 𝚿 = A𝑒𝑖(𝑘𝑥−𝜔𝑡)

= A𝑒𝑖𝑘𝑥 𝑒−𝜔𝑡 ,

Where A𝑒𝑖𝑘𝑥 represent the time independent part of the wave function and is

represented by 𝜓 = A𝑒𝑖𝑘𝑥 ……(1) differentiating eqn (1) w.r.t ‘𝑥 ‘ twice we get

𝑑𝜓

𝑑𝑥= 𝐴(𝑖𝑘) 𝑒𝑖𝑘𝑥 and

𝑑2𝜓

𝑑𝑥2 =A(𝑖𝑘) (𝑖𝜔) 𝑒𝑖𝑘𝑥

= 𝑖2 𝑘2 A𝑒𝑖𝑘𝑥 …….(2)

From eqns 1 & 2 we get

𝑑2𝜓

𝑑𝑥2 = −

4𝜋2

𝜆2 𝜓 ……..(3) ∵ 𝒊𝟐 = −1 & 𝒌 =

𝟐𝝅

𝝀

But, de-broglie wave length λ = ℎ

𝑚𝑣

∴ 1

𝜆2=

𝑚2𝑣2

ℎ2

= 2𝑚(½𝑚𝑣2)

ℎ2

= 2𝑚(𝐸𝐾)

ℎ2 ∵ 𝐸𝐾 = ½𝑚𝑣2

Also , the kinetic energy (𝐸𝐾 )in terms of the total energy (E) & the potential energy( V) is

given by 𝐸𝐾=(E−𝑽)

∴ 1

𝜆2=

2𝑚(𝐸−𝑽)

ℎ2 ……(4)

From eqns 3 & 4 ,we get 𝑑2𝜓

𝑑𝑥2= −

4𝜋22𝑚(𝐸−𝑽)

ℎ2 𝜓

𝑑2𝜓

𝑑𝑥2+


ℎ2𝜓 = 0 ……(5)

This is Schrodinger time independent equation for a particle

For a free particle ,V=0

∴ 𝑑2𝜓

𝑑𝑥2+

8𝜋2𝑚𝐸

ℎ2𝜓 = 0

This is Schrodinger time independent equation for free particle.



Q: Obtain normalized wave function for a free particle in a infinite walled potential

Box/Well using Schrodinger 1D time independent equation.

a

∞ ∞

V=∞ V=0

Box/Well

particle

𝑥=0 𝑥=a

Consider a particle of mass ‘m’ moving by reflection at infinitely high walls of a Box/well

of width ‘a’ moving between 𝑥=0 & 𝑥=a Potential V=0 inside the Box/well and V=∞

outside the Box/well .one dimensional Schrodinger equation for particle is given by

𝑑2𝜓

𝑑𝑥2+


ℎ2𝜓 = 0 ……(1).

For a particle inside the Box/well V=0

∴ 𝑑2𝜓

𝑑𝑥2+

8𝜋2𝑚𝐸

ℎ2𝜓 = 0 ….(2)

Putting 8𝜋2𝑚𝐸

ℎ2= 𝐾2 …..(3) in equation (2) ,

we get 𝑑2𝜓

𝑑𝑥2+ 𝐾2𝜓 = 0 ……(4)

The general solution of the quadratic equation (4) is of the form ψ (𝑥)=A sin(K𝑥) +B Cos(K𝑥)

……(5)

where A & B are constants determined from boundary conditions as follows :

ψ (𝑥)=0 at 𝑥=0 from eqn (5), 0=A x 0 +B x 1

∴ B=0

also, ψ (𝑥)=0 at 𝑥=a ∴ from eqn(4) 0=C Sin(Ka)+0xCos(Ka)

0= C Sin(Ka)

Sin(Ka) = 0 as C≠ 0

∴ Sin(Ka) =0= Sin(n𝜋)

ie: Ka=n𝜋 or K=𝑛𝜋

𝑎 …..(6)

From eqns 3 & 6 ,we get 8𝜋2𝑚𝐸

ℎ2=

𝑛2𝜋2

𝑎2

∴ E = 𝑛2ℎ2

8𝑎2𝑚 or

In general En= 𝑛2ℎ2

8𝑎2𝑚 …(7)



where n=1,2,3..called quantum number.

The values of En are called Eigen energy values which satisfy Schrodinger wave equation.

n=1 gives E1 = ℎ2

8𝑎2𝑚 = Eo called end point /ground state energy .

n=2 gives E2 = 4ℎ2

8𝑎2𝑚 = 4 Eo called 1st excited state energy

n=3 gives E3 = 9ℎ2

8𝑎2𝑚 =9 Eo ,called 2nd excited state energy & so on

Substituting the values of B=0 & K= 𝑛𝜋

𝑎 in eqn 5, we get

𝜓𝑛(𝑥) = A Sin( 𝑛𝜋𝑥

𝑎) …….(8) this represents the permitted solutions

To find ’A’ by normalization:

Applying the normalization condition ∫ 𝐼𝛹𝐼2𝑑𝑥 = 1+∞

−∞ to eqn 8 for x=0 & x=a, we get

∫ 𝐴2 𝑆𝑖𝑛2 ( 𝑛𝜋𝑥

𝑎)𝑑𝑥 = 1

𝑎

0

𝐴2 ∫ ½[1 − 𝐶𝑜𝑠 ( 2 𝑛𝜋𝑥

𝑎)]𝑑𝑥 = 1

𝑎

0 ∵ 𝑆𝑖𝑛2 (𝜃)=½[1-Cos(2𝜃)]

½𝐴2 [∫ 1𝑑𝑥 − ∫ 𝐶𝑜𝑠( 2 𝑛𝜋𝑥

𝑎)𝑑𝑥

𝑎

0 ] = 1

𝑎

0

½𝐴2 [𝑥]0𝑎 − [

𝑎

2𝑛𝜋 Sin(

2𝑛𝜋𝑥

𝑎)]0

𝑎 = 1 ∵ ∫ 𝐶𝑜𝑠(𝑚𝑥) 𝑑𝑥 = 𝑆𝑖𝑛(𝑚𝑥)

𝑚

½𝐴2 [𝑎 − 0] − [ 𝑎

2𝑛𝜋 Sin(2n𝜋) −

𝑎

2𝑛𝜋 Sin(0)] =1

½𝐴2 [𝑎 − 0] − [0 − 0] =1

𝐴2 𝑎 = 2 𝐴2 = 2/𝑎 or A =√𝟐/𝒂 …(9)

From eqns 8 &9, the normalized Eigen functions are 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 𝝍𝒏(𝑥)= √𝟐/𝒂 Sin( 𝒏𝝅𝒙

𝒂) …..(10)

𝑛 = 3 𝝍3 𝐼𝝍3𝐼2

.

𝑛 = 2 𝝍2 𝐼𝝍2𝐼2

𝑛 = 1 𝝍1 𝐼𝝍1𝐼2

𝑥=0 𝑥=a 𝑥=0 𝑥=a

𝐸𝑖𝑔𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝝍1 ,

𝝍2 ,𝝍3 , … . 𝑎𝑛𝑑 𝑡ℎ𝑒𝑖𝑟 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑖𝑒𝑠 𝐼𝝍1𝐼2, 𝐼𝝍2𝐼2, 𝐼𝝍3𝐼2 …

are represented as shown in the diagrams.The probability of finding the particle at the

anti-nodes is maximum and at nodes is zero.(The particle never found at nodes)

NOTE: Eigen Functions are the acceptable wave functions [𝜓𝑛(𝑥)]

Eigen Energy values are energy values for which Schrodinger equation

can be solved.

This can be omitted



Q: What is a wave function and mention it’s properties/limitations

The variable quantity that characterizes the de-broglie wave of the particle is called a

‘Wave function’ denoted mathematically by the symbol ‘𝚿’(Psi)

Properties of the wave functions are:-

1. Wave function (𝚿) is single valued everywhere ,

2. 𝚿 is finite everywhere

3. 𝚿 is continuous every where

4. First derivatives of 𝚿 are continuous every where

5. 𝐼𝛹𝐼2 or 𝛹𝛹 * is called probability density.

6.𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑓𝑖𝑛𝑑𝑖𝑛𝑔 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑖𝑛 𝑠𝑝𝑎𝑐𝑒 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 ∫ 𝐼𝛹𝐼2𝑑𝑉 = 1+∞

−∞

PROBLEMS SECTION

MODULE-1 : MODERN PHYSICS AND QUANTUM MECHANICS

Formulae needed : λ = ℎ

𝑃 where P = mv = √2𝑚𝐸 =√2𝑚𝑒𝑉 ,

E = 1

2𝑚𝑉2 =

𝑃2

2𝑚 ;Photon energy E’ = hν =

ℎ𝐶

𝜆 ; ∆𝑥. ∆𝑝 ≥

ℎ

4𝜋 and

∆𝑝 = 𝑚. ∆𝑣 ; ∆𝐸. ∆𝑡 ≥ℎ

4𝜋 𝑎𝑛𝑑 ∆𝐸 =

ℎ𝐶∆𝜆

𝜆2 ; also ∆𝐸 = ℎ ∆𝜈 ;

P = ∫ 𝐼𝛹𝐼2𝑥2

𝑥1dx ; 𝐸𝑛 =

𝑛2ℎ2

8𝑚𝑎2

1. Compare the energy of a photon with that of an electron when both are associated with

wavelength 0.2 nm.(Dec 2014/Jan2015)

Given: 𝜆𝑃 = 𝜆𝑒 = 0.2 𝑛𝑚 = 0.2𝑥10−9 𝑚 , h=6.63x10−34 Js, m=9.1x10−31kg ; C=3x108m/s,

𝐸𝑃

𝐸𝑒= ?

Using 𝐸𝑃 =ℎ𝐶

𝜆𝑃 and 𝐸𝑒 =

ℎ2

2𝑚𝜆𝑒2

We get , 𝐸𝑃

𝐸𝑒 =

𝐶2𝑚𝜆𝑒2

ℎ𝜆𝑃

= 3𝑥108𝑥2𝑥9.1𝑥10−31( 0.2𝑥10−9)2

6.63𝑥10−34 𝑥 0.2𝑥10−9

= 1.647x 102



2. Calculate the kinetic energy of an electron of wavelength 18 nm.(Jun/Jul 14)

Given: 𝜆 = 18 𝑛𝑚 = 18𝑥10−9 𝑚 , h=6.63x10−34 Js, m=9.1x10−31kg ; E= ?

Using λ = ℎ

√2𝑚𝐸

We get ,𝐸 =ℎ2

2𝑚𝜆2

= ( 6.63𝑥10−34 )2

2𝑥9.1𝑥10−31𝑥( 18𝑥10−9)2

= 7.454x10−22 J

=7.454𝑥10−22

1.6𝑥10−19

= 4.66x10−3 eV

3. An excited atom has an average life time of 10−8 seconds. During this period. it emits

photon and returns to the ground state. What is the minimum uncertainty in the frequency

of this photon ? (Jun/Jul 14)

Given: ∆𝑡 = 10−8 S, ℎ = 6.63𝑥10−34 𝐽𝑠 ; ∆𝜈 = ?

Using ∆𝐸 = ℎ. ∆𝜈 and ∆𝐸. ∆𝑡 ≥ℎ

4𝜋

We get , ∆𝜈 ≥1

4𝜋.∆𝑡

≥1

4𝑥𝜋𝑥10−8

= 7.96 x 106 Hz

4. Calculate the wavelength associated with electrons whose speed is 0.01 part

of the speed of light. (Dec 2013/Jan2014)

Given: v =0.01 C = 0.01x 3x108m/s ; h=6.63x10−34 Js; 𝑚 = 9.1𝑥10−31𝑘𝑔 ; 𝜆 = ?

Using λ = ℎ

𝑚𝑣

= 6.63𝑥10−34

9.1𝑥10−31𝑥0.01𝑥 3𝑥108

= 2.43 x10−10 m



5. An electron is bound in one dimensional infinite well of width 0.12nm.Find the

energy value and de-Broglie wavelength in the first excited state. (Dec 2013/Jan2014)

Given: a = 0.12nm = 0.12x10−9 m ; n = 2 ; h=6.63x10−34 Js; 𝑚 = 9.1𝑥10−31𝑘𝑔 ;

𝐸2 = ? & λ =?

Using 𝐸𝑛 = 𝑛2ℎ2

8𝑚𝑎2

𝐸2 = 22(6.63𝑥10−34 )2

8𝑥9.1𝑥10−31𝑥(0.12𝑥10−9)2

= 1.68x10−17 J

Also , λ = ℎ

√2𝑚𝐸2

= 6.63𝑥10−34

2𝑥9.1𝑥10−31𝑥1.68𝑥10−17

= 1.19 x10−10 m

6. Calculate the de-Broglie wavelength associated with an electron of energy 1.5eV. (Jun13)

Given: E=1.5 eV=1.5x1.6x10−19 𝐽 , h=6.63x10−34 Js; m=9.1x10−31kg; e=1.6x10−19 C; λ =

Using λ = ℎ

√2𝑚𝐸

= 6.63𝑥10−34

√(2𝑥9.1𝑥10−31𝑥1.5𝑥1.6𝑥10−19 )

= 1.003 x 10−9 m

7. A spectral line of wavelength 5461Å has a width of 10−4Å. Evaluate the minimum time

spent by the electrons in the upper energy state. (Jun/Jul 2013)

Given:, ℎ = 6.63𝑥10−34 𝐽𝑠 ; 𝜆 = 5461Å = 5461𝑥10−10 𝑚 ; ∆𝑡 = ?

Using ∆𝐸 =ℎ𝐶

𝜆 and ∆𝐸. ∆𝑡 ≥

ℎ

4𝜋

We get , ∆𝑡 ≥𝜆

4𝜋.𝐶

≥5461𝑥10−10

4𝑥𝜋𝑥3𝑥108

= 1.45 x 10−16 S



8. Find the de-Broglie wavelength of an electron accelerated through a potential difference

of 182 volts and object of mass 1 kg moving with a speed of 1 m/s. Compare the results

and comment. (Jun 2012)

Given:V=182V ; 𝑚0 =1 kg, 𝑉0 = 1 m/s , h=6.63x10−34 Js; m=9.1x10−31kg; e=1.6x10−19 C ,

𝜆𝑒 = ? 𝜆0 = ?

Using 𝜆𝑒 =ℎ

√2𝑚𝑒𝑉

=6.63𝑥10−34

√2𝑥9.1𝑥10−31𝑥1.6𝑥10−19 𝑥182

= 9.11 x10−11 m

Also,using 𝜆0 = ℎ

𝑚0𝑉0

= 6.63𝑥10−34

1𝑥1

= 6.63x10−34 𝑚

Comment : Waverlength is inversely proportional to the mass ,as 𝜆𝑒 ≫ 𝜆0

9. A quantum particle confined to one-dimensional box of width ‘a’ is in its first excited

state. What is the probability of finding the particle over an interval of ( 𝑎

2) marked

symmetrically at the centre of box. (𝐽𝑢𝑛

𝐽𝑢𝑙2011)

Given: 1st excited state , n=2 ;the interval ( 𝑎

4 ,

3𝑎

4 )

Using P = ∫ 𝐼𝛹𝐼2𝑥2

𝑥1dx = ∫ (√

2

𝑎𝑆𝑖𝑛 ( 2𝜋𝑥

𝑎))2𝑑𝑥

3𝑎/4

𝑎/4𝑎/2

= ∫𝟐

𝒂𝑆𝑖𝑛2 ( 2𝜋𝑥

𝑎)𝑑𝑥

3𝑎

4𝑎

4

0 a/4 a/2 3a/4 a

= 𝟐

𝒂 ∫ ½[1 − 𝐶𝑜𝑠 ( 2 𝜋𝑥

𝑎)]𝑑𝑥

3𝑎

4𝑎

4

∵ 𝑆𝑖𝑛2 (𝜃)=½[1-Cos(2𝜃)]

= ½𝟐

𝒂 ∫ 1𝑑𝑥 − ∫ 𝐶𝑜𝑠( 2 𝜋𝑥

𝑎)𝑑𝑥

3𝑎

4𝑎

4

3𝑎

4𝑎

4

= 𝑎 [𝑥]𝑎

4

3𝑎

4 − [ 𝑎

2𝜋 Sin(

2𝜋𝑥

𝑎)]𝑎

4

3𝑎

4

= 𝑎 [3𝑎4

− 𝑎4

] − [ 𝑎 2𝜋

Sin( 2𝜋

3𝑎

4

𝑎)−

𝑎

2𝜋 Sin(

2𝜋𝑎

4

𝑎)]

= 𝑎 [𝑎

2] − [0 − 0 = ½ = 0.5 or 50%



10. A particle moving in one dimension box is described by the wave function 𝛹 = 𝑥[√3] 𝑓𝑜𝑟

0<𝑥<1 and Ψ =0 elsewhere. Find the probability of finding the particle within the

interval (0,½). (𝐽𝑢𝑛 2012)

Given : 𝛹 = 𝑥[√3] ; (𝑥1, 𝑥2)( 0,1

2 ) ; P = ?

Using Probability, P = ∫ 𝐼𝛹𝐼2𝑥2

𝑥1dx

= ∫ 3𝑥21

2

0dx = [3.

𝑥3

3]0

1

2

= ( 1

2 )3 – (0)3 =

1

8 = 0.125 or 12.5 %

11. Calculate the energy of electron that produces Bragg’s diffraction of first order at glancing

angle of 22° ,when incident on crystal with inter-planar spacing of 1.8Å. (𝐽𝑢𝑛 2012)

Given : n =1 ; 𝜃 = 22° ; 𝑑 = 1.8 Å = 1.8𝑥 10−10 m ; h=6.63x10−34 Js; m=9.1x10−31kg;

λ = ? & E = ?

Using 1 ) λ = 2𝑑 𝑠𝑖𝑛 𝜃

𝑛

= 2𝑥1.8𝑥 10−10𝑥 𝑠𝑖𝑛 22

1

=1.348x 10−10 m

Also from, λ = ℎ

√2𝑚𝐸

We get, E = ℎ2

2𝑚𝜆2 J OR E =

ℎ2

2𝑚𝜆2 𝑒 eV

= (6.63𝑥10−34)2

2𝑥9.1𝑥10−31𝑥(1.348𝑥 10−10 )2 =

(6.63𝑥10−34)2

2𝑥9.1𝑥10−31𝑥(1.348𝑥 10−10 )2𝑥1.6 𝑥 10−19

= 1.329 x 10−17 J = 83.07 eV = 83.07 eV

12. Find the energy of the neutron in eV whose de-Broglie wavelength is 1Å. (𝐷𝑒𝑐 2011)

Given; λ = 1Å = 1x 10−10 𝑚 ;h=6.63x10−34 Js, m=1.678x10−27kg ; e=1.6x10−19 C ;E = ?

Using , λ = ℎ

√2𝑚𝐸

We get , E = ℎ2

2𝑚𝜆2 joules

= ℎ2

2𝑚𝑒𝜆2 eV

= (6.63𝑥10−34 )2

2𝑥1.678𝑥10−27𝑥1.6𝑥10−19 𝑥(1𝑥 10−10 )2

= 0.082 eV



13. An electron is confined to a box of length 10−9 m, calculate the minimum uncertainty in it’s

velocity. (𝐷𝑒𝑐 2011)

Given: ∆𝑥 = 10−9 m ; h=6.63x10−34 Js, m=9.1x10−31kg ;∆𝑣 = ?

Using ∆𝑥. ∆𝑝 ≥ℎ

4𝜋 and ∆𝑝 = 𝑚. ∆𝑣

We get , ∆𝑣 ≥ℎ

4𝜋𝑚∆𝑥

≥ 6.63𝑥10−34

4𝜋 𝑥9.1𝑥10−31𝑥10−9

= 58 x103m/S

14. Compute the de-Broglie wavelength for a neutron moving with one tenth part of the

velocity of light. (𝐽𝑢𝑛

𝐽𝑢𝑙2011)

Given; V = 𝐶

10 =

3 𝑋 108

10= 0.3 𝑥 108 𝑚/𝑠;h=6.63x10−34 Js, m=1.678x10−27kg ;

C = 3 X 108 m/s ; λ = ?

Using λ = ℎ

𝑚𝑉

= 6.63𝑥10−34

1.678𝑥10−27𝑥0.3 𝑥 108 = 1.317x 10−14 m.

15. An electron has a de-Broglie wavelength 3 nm and rest mass 511keV. Determine its group

velocity and kinetic energy.

Given: λ = 3nm = 3x 10−9 m ; h=6.63x10−34 Js ; 𝑉𝑔 = V = ? and 𝐸𝐾 = ?

m =E/ 𝐶2 =511x103x1.6x 10−19/(3𝑥108)2 kg = 9.08x10−31 kg

Using λ = ℎ

𝑚𝑣 ,we get 𝑉𝑔 = V =

ℎ

𝑚𝜆

= 6.63𝑥10−34

9.08𝑥10−31𝑥3𝑥 10−9

= 2.43x 106 m/s

𝑎𝑙𝑠𝑜, 𝐸𝐾 =1

2𝑚𝑉2 =

1

2𝑥 9.08𝑥10−31 𝑥 (2.43𝑥 106 )2

= 2.681x 10−18 J



16. A spectral line of wavelength 546.1 nm has a width 10−5 nm. Estimate the minimum time

spent by electrons in the excited state during transitions. (𝐷𝑒𝑐 2010)

Given:, ℎ = 6.63𝑥10−34 𝐽𝑠 ; 𝜆 = 546.1 𝑛𝑚 = 546.1𝑥10−9 𝑚 ; ∆𝑡 = ?

Using ∆𝐸 =ℎ𝐶

𝜆 and ∆𝐸. ∆𝑡 ≥

ℎ

4𝜋

𝑤𝑒 𝑔𝑒𝑡 , ∆𝑡 ≥𝜆

4𝜋.𝐶

≥546.1𝑥10−9

4𝜋𝑥3𝑥108 = 1.45x 10−16 S

17. Calculate the momentum of the particle and de-Broglie wavelength associated with an

electron with a kinetic energy of 1.5KeV. (𝐽𝑢𝑛 2010)

Given: E =1.5 Kev =1.5 x 103 x 1.6x10−19 J ; h = 6.625𝑥10−34 𝐽𝑆 ;e = 1.6x10−19 J ;

m = 9.1𝑥10−31 𝑘𝑔 P =? & λ = ?

Ans: Using P = √2𝑚𝐸

= √2𝑥9.1𝑥10−31𝑥1.5 𝑥 103 𝑥 1.6𝑥10−19 = 2.09 x 10−23 kg.m/s

Also, λ = ℎ

𝑃

= 6.625𝑥10−34

2.09 𝑥 10−23 = 3.17x10−11 𝑚

18. An electron is bound in one dimensional potential well of width 0.18 nm. Find the energy

value in eV of the second excited state. (𝐽𝑢𝑛 2010 & 𝐶𝐵𝐶𝑆 − 𝐽𝑢𝑛/𝐽𝑢𝑙16)

Given: n=3(For 2nd excited state), a=0.18 nm =0.18x10−9m,h=6.63x10−34 Js;

m=9.11x10−31kg & e=1.6x10−19 C; 𝐸2 = ?

Using 𝐸𝑛 =𝑛2ℎ2

8𝑚𝑎2 J OR 𝐸𝑛 =

𝑛2ℎ2

8𝑚𝑎2𝑒 eV

𝐸2 = 32(6.63𝑥10−34 )2

8𝑥9.11𝑥10−31𝑥(0.18 𝑥10−9)2 𝐸2 =

32(6.63𝑥10−34 )2

8𝑥9.11𝑥10−31𝑥(0.18 𝑥10−9)2 𝑥1.6𝑥10−19

= 1.675 x10−17 J = 104.7 eV

=1.675 𝑥10−17

1.6𝑥10−19 =104.7 eV



19. Calculate the energy in eV, for the first excited state of an electron in an infinite potential

well of width 2 Å (4 marks)

Given: n=2(For 1st excited state), a=2Å=2x10−10m,h=6.63x10−34 Js; m=9.1x10−31kg &

e=1.6x10−19 C;



𝑛2ℎ2

8𝑚𝑎2𝑒 eV

=22(6.63𝑥10−34 )2

8𝑥9.1𝑥10−31𝑥(2𝑥10−10)2 =

22(6.63𝑥10−34 )2

8𝑥9.1𝑥10−31𝑥(2𝑥10−10)2 𝑥1.6𝑥10−19

= 6.038x10−18 J = 37.74 eV

=6.038𝑥10−18

1.6𝑥10−19 =37.74 eV

20. The wavelength of a fast neutron of mass 1.675x10−27kg is 0.02 nm.Calculate the group

velocity and phase velocity of its de-Broglie waves.( 4 marks)

Given: m=1.675x10−27kg, λ = 0.02 nm=0.02x10−9m, C=3x108m/s, ℎ = 6.63𝑥10−34 𝐽𝑠,

𝑣𝑔 & 𝑣𝑃 =?

𝑈𝑠𝑖𝑛𝑔, 𝑣𝑔 = v =ℎ

𝑚𝜆 =

6.63𝑥10−34

1.675𝑥10−27𝑥0.02𝑥10−9 = 1.979x104 m/s ,

Also 𝑣𝑃 =𝐶2

𝑣𝑔 =

(3𝑥108)2

1.979𝑥104 = 4.55X1012 m/s

21. Calculate the de-Broglie wavelength associated with neutron of mass 1.674 X 10−27 kg

with one tenth part of the velocity of light (4 marks - CBCS Jun/July16)

Given : v = 1

10 C =

1

10 x 3x108m/s = 3x107m/s; h=6.63x10−34 Js; 𝑚 = 9.11𝑥10−31𝑘𝑔 ; 𝜆 = ?

Using λ = ℎ

𝑚𝑣

= 6.63𝑥10−34

9.11𝑥10−31𝑥 3𝑥107

= 2.43 x10−11 m



22. Calculate the deBroglie wavelength of an electron moving with K.E of 50KeV.

Given: E= 50keV= 5x103𝑥 1.6x10−19 𝐽 , h=6.63x10−34 Js; m=9.1x10−31kg;

e=1.6x10−19 C; λ =?

Using λ = ℎ

√2𝑚𝐸

= 6.63𝑥10−34

√(2𝑥9.1𝑥10−31𝑥 50𝑥103𝑥 1.6𝑥10−19 )

= 5.495 x 10−12 m

22. X-rays of wavelength 0.75Å are scattered from a target at an angle of 45°.

Calculate the wavelength of scattered X-rays.

Given: 𝜆 = 0.75Å ; 𝜃 = 45° ; h=6.63x10−34 Js; m=9.1x10−31kg ; C=3x108m/s ;𝝺λ’ = ?

Using , λ’−𝜆 = ℎ

𝑚𝐶 (1−𝑐𝑜𝑠𝜃)

λ’ −0.75Å = 6.63𝑥10−34

9.1𝑥10−31𝑥 3𝑥108 (1−𝑐𝑜𝑠45)

λ’ = 0.75Å + 7.1x10−13m

= 0.75Å + 0.0071Å

= 0.7571Å

223. A particle of mass 940 MeV/𝐶2 has kinetic energy 0.5 KeV. Find its de-

Broglie wavelength , C is velocity of light.

Given: m = 940 MeV/𝐶2 = 940 𝑥 106 𝑒𝑉

(3𝑥108 )2 =

940 𝑥 106 𝑥1.6𝑥10−19 𝐽

(3𝑥108 )2 = 1.671x10−27 𝑘𝑔

E = 0.5 keV = 0.5 𝑥103eV = 0.5 𝑥103𝑥1.6𝑥10−19 𝐽 ; 𝜆 = ?

Using; 𝝺 = ℎ

√(2𝑚𝐸)

= 6.63𝑥10−34

√(2𝑥1.671𝑥10−27𝑥0.5 𝑥103𝑥1.6𝑥10−19)

= 1.282x10−12 m



24. The first excited state energy of an electron in an infinite well is 240 eV. what will be its ground state energy when the width of the potential

Well is doubled.

Given: 𝐸2 = 240 eV ; 𝐸1 = ?


8𝑚𝑎2

𝐸2 =22ℎ2

8𝑚𝑎2

240 =4ℎ2

8𝑚𝑎2

𝑎2 =4ℎ2

8𝑚𝑥240

∴ a = √(ℎ2

480𝑚 )

𝑎𝑙𝑠𝑜 , 𝐸1 =12ℎ2

8𝑚(2√(ℎ2

480𝑚 ) )2

= 480𝑚𝑥ℎ2

8𝑚𝑥4𝑥ℎ2 = 15 eV

𝑂𝑅

Using ; 𝐸𝑛 =𝑛2ℎ2

8𝑚𝑎2 ,we can show that

𝐸2 𝑎22

22 =

𝐸1 𝑎12

12 =

ℎ2

8𝑚

Given; 𝐸2 = 240𝑒𝑉 , 𝑎1 = 2𝑎2

∴ 𝐸1 = 𝐸2 𝑎2

2

𝑎1222

= 240𝑎2

2

4𝑎22𝑥4

= 15 eV

25. Calculate the deBroglie wavelength associated with neutron of mass 1.674 x10−27kg with one tenth part of the velocity of light .

Given; V = 𝐶

10 =

3 𝑋 108

10= 0.3 𝑥 108 𝑚/𝑠;h=6.63x10−34 Js,

m=1.674x10−27kg ; C = 3 X 108 m/s ; λ = ?

Using λ = ℎ

𝑚𝑉 =

6.63𝑥10−34

1.674𝑥10−27𝑥0.3 𝑥 108

= 1.320x 10−14 m.



26. An electron is bound in one dimensional potential well of width

0.18 nm. Find the energy value in eV of the second excited state.

Given: n=3(For 2rd excited state), a=0.18 nm =0.18 x10−9m,h=6.63x10−34 Js;

m=9.1x10−31kg & e=1.6x10−19



𝑛2ℎ2

8𝑚𝑎2𝑒 eV

=32(6.63𝑥10−34 )2

8𝑥9.1𝑥10−31𝑥(0.18 𝑥10−9)2 =

32(6.63𝑥10−34 )2

8𝑥9.1𝑥10−31𝑥1.6𝑥10−19 𝑥(0.18 𝑥10−9)2

= 1.677x10−17 J = 104.83 eV

=6.038𝑥10−18

1.6𝑥10−19

= 104.81 eV

27. Find deBroglie wavelength of a particle of mass 0.58 MeV/𝑪𝟐 has a kinetic

energy 90 eV , where C is the velocity of light. ( 04 Marks) DEC2016

Given: m = 0.58 MeV/𝐶2 = 0.58 𝑥 106 𝑒𝑉

(3𝑥108 )2

=0.58 𝑥 106 𝑥1.6𝑥10−19 𝐽

(3𝑥108 )2 = 1.0324x10−30 𝑘𝑔

E = 90eV = 90𝑥1.6𝑥10−19 𝐽 ; 𝜆 = ?

Using; 𝝺 = ℎ

√(2𝑚𝐸)

= 6.63𝑥10−34

√(2𝑥1.0324𝑥10−30 𝑥 90𝑥1.6𝑥10−19)

= 1.2142𝑥10−10 m

28. The inherent uncertainty in the measurement of time spent by Iridium- 19

nuclei in the excited state is found to be 1.4 x 𝟏𝟎−𝟏𝟎 S. Estimate the

uncertainty that results in its energy in eV in the excited State. ( 04 Marks)

Given:, ℎ = 6.625𝑥10−34 𝐽𝑠 ; ∆𝑡 = 1.4 x 𝟏𝟎−𝟏𝟎 S: ∆𝐸 =?

Using ∆𝐸. ∆𝑡 ≥ℎ

4𝜋 𝑤𝑒 𝑔𝑒𝑡 , ∆𝐸 ≥

ℎ

4𝜋.∆𝑡

≥6.625𝑥10−34

4𝜋𝑥𝟏.𝟒 𝒙 𝟏𝟎−𝟏𝟎 J

≥ 6.625𝑥10−34

4𝜋𝑥𝟏.𝟒 𝒙 𝟏𝟎−𝟏𝟎𝒙𝟏.𝟔𝟎𝟐𝒙𝟏𝟎−𝟏𝟗 eV

≥ 2.351𝑥 10−6 eV



29. A spectral line of wavelength 5896 Å has a width of 𝟏𝟎−𝟓 Å. Evaluate the minimum

time spent by the electrons in the upper energy state between the excitation and

de-excitation processes. ( 04 Marks)

Given: 𝝀=5896 Å = 5896 x𝟏𝟎−𝟏𝟎 m ,∆𝝀 = 𝟏𝟎−𝟓Å = 𝟏𝟎−𝟓 x𝟏𝟎−𝟏𝟎m ,

h = 6.625 x 𝟏𝟎−𝟑𝟒 JS , C = 3 x 𝟏𝟎𝟖𝒎𝒔−𝟏 , ∆𝒕 = ?

Using ∆𝑬. ∆𝒕 ≥ 𝒉

𝟒𝝅 and ∆𝑬 =

𝒉𝒗 ∆𝝀

𝝀𝟐 ∵ E =

𝒉𝒄

𝝀

We get , ∆𝒕 = 𝝀𝟐

𝟒𝝅𝒄∆𝝀

∴ ∆𝒕 = (𝟓𝟖𝟗𝟔 𝒙𝟏𝟎−𝟏𝟎)𝟐

𝟒𝝅𝒙𝟑 𝒙 𝟏𝟎𝟖𝒙𝟏𝟎−𝟓 𝒙𝟏𝟎−𝟏𝟎

= 9.221 x 𝟏𝟎−𝟖 S

29. Compare the energy of a photon with that of a neutron when both are associate

with a wavelength 0.25 nm,mass of neutron is 1.675 X 𝟏𝟎−𝟐𝟕 kg. ( 04 Marks)

Given : 𝝀𝒑 = 𝝀𝒏 = 𝟎. 𝟐𝟓 𝒏𝒎 = 0.25 x 𝟏𝟎−𝟗 m , 𝒎𝒏 = 1.675 X 𝟏𝟎−𝟐𝟕 kg

h = 6.625 x 𝟏𝟎−𝟑𝟒 JS , C = 3 x 𝟏𝟎𝟖𝒎𝒔−𝟏 , 𝝀𝒑

𝝀𝒏 = ?

Using , 𝑬𝒑 =𝒉𝒄

𝝀 =

𝟔.𝟔𝟐𝟓 𝒙 𝟏𝟎−𝟑𝟒𝒙𝟑 𝒙 𝟏𝟎𝟖

𝟎.𝟐𝟓 𝒙 𝟏𝟎−𝟗 = 7.95 x 𝟏𝟎−𝟏𝟔 J

Also , 𝑬𝒏 =𝒉𝟐

𝟐𝒎𝝀𝟐 =

(𝟔.𝟔𝟐𝟓 𝒙 𝟏𝟎−𝟑𝟒)𝟐

𝟐𝒙𝟏.𝟔𝟕𝟓 𝑿 𝟏𝟎−𝟐𝟕𝒙 (𝟎.𝟐𝟓 𝒙 𝟏𝟎−𝟗)𝟐 = 2.096 x𝟏𝟎−𝟐𝟏 J

∴ 𝑬𝒑

𝑬𝒏=

𝟕.𝟗𝟓 𝒙 𝟏𝟎−𝟏𝟔

𝟐.𝟎𝟗𝟔 𝒙𝟏𝟎−𝟐𝟏 = 3.793 x 𝟏𝟎𝟓

**end**


VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 1

MODULE-2:

ELECTRICAL PROPERTIES OF MATERIALS:

Q:Explain the terms: Drift velocity, Mean collision time, Mean free path & relaxation time.

a) Drift velocity(𝒗𝒅) is the average velocity with which the free electrons drift in a direction opposite to that of the applied electric field

b) Mean collision time(𝝉) is the average time elapsed between two consecutive collisions.

c) Mean free path(λ)is the average distance travelled by the electron between two consecutive collisions.

d) Relaxation time(𝝉𝒓) is time during which the drift velocity is reduced to 1

𝑒 of its

velocity when the field is cut off.

Q:Obtain an expression for Drift velocity.

Let an electron of mass ‘m’ ,charge ‘e’ moves with drift velocity ’𝒗𝒅 ‘ when an

electric field ‘E ‘ is applied to it. Then the driving force acting on the electron

F = −𝑒𝐸 ……(1)

Also the retarding force acting on the electron F’= 𝑚𝑣𝑑

𝜏 …..(2)

where 𝜏 = 𝑚𝑒𝑎𝑛 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 𝑡𝑖𝑚𝑒

For a system in steady state F’ = −𝐹

ie: 𝑚𝑣𝑑

𝜏= 𝑒𝐸

∴ 𝒗𝒅 =𝒆𝝉𝑬

𝒎

Q:Derive an expression for electrical conductivity and hence electrical resistivity.

The drift velocity of electron 𝒗𝒅 =𝒆𝝉𝑬

𝒎 ……(1)

Also from Ohm’s law electrical conductivity , 𝜎 =𝒋

𝑬 ..(2)where j=current density &

E=electric field But j = ne𝒗𝒅 … (𝟑)

∴ eqns 1&3 ,we get j = ne. 𝒆𝝉𝑬

𝒎

∴ 𝒋

𝑬 =

𝒏𝒆𝟐𝝉

𝒎 ….(4)

From eqns 2 &4,we get 𝝈 = 𝒏𝒆𝟐𝝉

𝒎 …..(5)

By definition electrical resistivity, 𝜌 =1

𝜎 …(6)

∴ from eqns 5&6, we get 𝝆 = 𝒎

𝒏𝒆𝟐𝝉



Q:Write the assumptions of classical free electron theory(CFET) or Drude-Lorentz theory.

The assumptions of CFET are:-

1. All metals contains a large number of free electrons called conduction electrons .

2. The free electrons are called electron gas which have 3 degrees of freedom as they

are treated as gas molecules.

3. Free electrons obey laws of kinetic theory of gases and hence they have mean free

path, mean collision time, drift velocity.

4. Free electrons move in a constant electric or potential field due to positive

ion cores.

5. Electro-static forces between electron-ions and electron-electrons are negligible.

6. Free electrons move with drift velocity in a direction opposite to the direction

of the applied field.

7. Average kinetic energy of free electrons at temperature T is 3

2𝑘𝑇,

where k = Boltzmann constant.

Q:Explain the failures of classical free electron theory(CFET).

Failures of CFET are:-

1.Specific heat :According to CFET the molar specific heat of electron gas at constant

volume is given by 𝐶𝑣 =3

2𝑅 ,where R=universal gas constant. But experimentally

determined 𝐶𝑣 𝑖𝑠 given by 𝐶𝑣 = 10−4𝑅𝑇 ,where T=absolute temperature. Thus CFET fail to

explain the the dependence of 𝐶𝑣 𝑜𝑛 T and numerical constant.

2.Dependance of electrical conductivity (𝝈) on T:

According to CFET kinetic energy 1

2𝑚𝑣2 =

3

2𝑘𝑇

∴ 𝑣 ∝ √𝑇 …(1)

But mean collision time 𝜏 ∝ 1

𝑣

∴ 𝜏 ∝ 1

√𝑇 … . (2) ∵ 𝑣 ∝ √𝑇

Also from 𝜎 =𝑛𝑒2𝜏

𝑚 ,we get 𝜎 ∝ 𝜏 … . (3)

From eqns 2 & 3 we get 𝝈 ∝ 𝟏

√𝑇 ,but experimentally it has been observed that 𝝈 ∝

𝟏

𝑻 ,

thus CFET fail to explain the dependence of 𝜎 𝑜𝑛 𝑇.

3.Dependance of ′𝝈′𝒐𝒏 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 ‘n’ :

𝐹𝑟𝑜𝑚 𝜎 =𝑛𝑒2𝜏

𝑚 , we get 𝜎 ∝ 𝑛 , according to CFET 𝝈 𝑜𝑓 tri-atomic metals must be more

than that of di-atomic and mono-atomic metals but from experimental results, it has been

found that 𝝈 of mono-atomic metals of low n are more than that of di and tri-atomic metals

of large n .Thus CFET fail to explain dependence of 𝝈 on concentration n.



4. Mean free path’ λ’:

Mean free path of Cu from CFET is given by λ= 𝑣𝜏=2.85nm. But the experimentally

determined value of λ=28.5 nm, which is 10 times more than that λ of CFET.

Q:State/Mention the assumptions of (Sommerfeld’s)quantum free electron theory(QFET)

The assumptions of QFET are:

1.The energy of free electrons are quantized.

2. Free electrons obey Pauli’s exclusion principle.

3. The distribution of free electrons in energy levels is governed by Fermi-Dirac

statistics.

4. Free electrons move in uniform potential field due to ionic cores in a metal .

5. The electrostatic electron-ion attractions and electron-electron repulsions

are negligible.

6. Electrons are considered as wave like particles.

Q:Explain the success of QFET.

QFET explained the following experimental facts which were not explained by CFET.

1.Specific heat: According to QFET the molar specific heat is given by 𝐶𝑣 =2𝑘

𝐸𝐹𝑅𝑇

where k = Boltzmann constant and 𝐸𝐹=Fermi energy. But 2𝑘

𝐸𝐹 = 10−4 ∴ 𝐶𝑣 = 10−4 𝑅𝑇

which agrees with experimental value.

2.Dependance of electrical resistivity ′𝝈′ on ‘T’(temperature):

We know that(WKT) 𝝈 = 𝒏𝒆𝟐𝝉

𝒎∗ and 𝜏 =

𝜆

𝑣𝐹 where 𝑣𝐹=Fermi velocity, λ=wavelength

∴ 𝝈 = 𝒏𝒆𝟐

𝒎∗

𝝀

𝒗𝑭 ∴ 𝝈 ∝ 𝝀 … . . (1)

Also it has been shown that λ ∝1

𝐴∝

1

𝑟2∝

1

𝑇 λ ∝

1

𝑇 …(2)

where A=area, r=radius & T=temperature.

∴ 𝑓𝑟𝑜𝑚 𝑒𝑞𝑛𝑠 1&2,we get 𝝈 ∝𝟏

𝑻 which is the experimentally determined relation.

3.Dependance of ′𝝈′𝒐𝒏′𝒏′:



WKT conductivity 𝝈 = 𝒏𝒆𝟐

𝒎∗

𝝀

𝒗𝑭 , thus 𝝈 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑏𝑜𝑡ℎ 𝑜𝑛 𝒏 𝑎𝑛𝑑

𝝀

𝒗𝑭 .With the decreases of

atomicity, n decreases and 𝝀

𝒗𝑭 increases so that n .

.𝝀

𝒗𝑭 increases ,so that 𝝈 is more for

monoatomic metals than that of di&tri-atomic metals.This explains the conductivity of

mono-atomic metals is more than that of di & tri-atomic metals.

4.Mean free path obtained from QFET is about 28.5 nm which is experimentally

determined value.

Q:Distinguish /(Write the differences )between CFET and QFET. CFET QFET

1.Free electron energy levels are

continuous

1.Free electron energy levels are dis- continuous .

2.Free electrons may possess

same energy

2.No two electrons can possess same energy, as

they obey Pauli’s exclusion principle.

3.Distribution of electrons in

energy levels obeys Maxwell-

Boltzmann statistics

3.Distribution of electrons in energy levels

obeys Fermi-Dirac statistics

Q: Derive an expression for conductivity and hence resistivity based on QFET.

According to de-broglie hypothesis λ = ℎ

𝑚𝑣 ,also λ =

2𝜋

𝑘 where symbols have usual

significance. ∴ 𝑚𝑣 =ℎ𝑘

2𝜋 , where v & k are vectors

differentiating this w.r.t ‘t’,

We get m 𝒅𝒗

𝒅𝒕 =

𝒉

𝟐𝝅

𝒅𝒌

𝒅𝒕 ….(1)

If an electric field E is applied ,the electron experience a force m 𝒅𝒗

𝒅𝒕 = −𝑒𝐸 …(2)

∴ 𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑛𝑠1&2, 𝑤𝑒𝑔𝑒𝑡 , 𝒉

𝟐𝝅

𝒅𝒌

𝒅𝒕 = −𝑒𝐸

∴ 𝑑𝑘 = −2𝜋

ℎ𝑒𝐸 𝑑𝑡

𝑂𝑛 Integration between the limits 0 to t ,we get

∫ 𝑑𝑘𝑡

0 = −

2𝜋

ℎ𝑒𝐸 ∫ 𝑑𝑡

𝑡0

k(𝑡) − 𝑘(0) = −2𝜋

ℎ𝑒𝐸 (𝑡 − 0)

𝜕𝑘 = −2𝜋

ℎ𝑒𝐸 𝑡

The Fermi sphere displace through 𝜕𝑘 in a direction opposite to the direction of

applied electric field E in a time ‘t’ as shown in the diagram.



If t = 𝜏 , 𝑡ℎ𝑒𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛𝑠 𝑡ℎ𝑒 𝑚𝑒𝑎𝑛 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝜕𝑘𝑎𝑣 𝑜𝑓 𝑓𝑒𝑟𝑚𝑖 𝑠𝑝ℎ𝑒𝑟𝑒 is given

by 𝜕𝑘𝑎𝑣 = −2𝜋

ℎ𝑒𝐸 𝜏 ….(3),

Using mv = ℎ

2𝜋𝑘

we get 𝑚 𝜕𝑣𝑎𝑣 = ℎ

2𝜋𝜕𝑘𝑎𝑣

∴ 𝜕𝑣𝑎𝑣 = ℎ

2𝜋𝑚𝜕𝑘𝑎𝑣 ….(4)

From eqns 3&4,we get 𝜕𝑣𝑎𝑣 = −𝑒𝐸

𝑚 𝜏

but 𝜕𝑣𝑎𝑣 = 𝑣𝑑 drift velocity as initial average velocity of electrons is zero.

ie: 𝑣𝑑 = −𝑒𝐸

𝑚 𝜏 … (5)

Also current density J = −𝑛𝑒𝑣𝑑 ….(6)

∴ 𝑓𝑟𝑜𝑚 𝑒𝑞𝑛𝑠 5&6, we get J = 𝑛𝑒2𝜏

𝑚 E

∴ 𝐽

𝐸 =

𝑛𝑒2𝜏

𝑚 ….(7)

But from Ohm’s law conductivity 𝜎 = 𝐽

𝐸 …..(8)

∴ 𝑓𝑟𝑜𝑚 𝑒𝑞𝑛𝑠 7&8 ,we get 𝝈 = 𝒏𝒆𝟐𝝉

𝒎 …(7),

where m is called effective electron mass usually denoted by 𝒎∗

Also resistivity 𝜌 =1

𝜎 …(8)

∴ 𝑓𝑟𝑜𝑚 𝑒𝑞𝑛𝑠 7&8 ,we get 𝝆 =𝒎

𝒏𝒆𝟐𝝉

Q:Explain the terms: Mobility, Fermi velocity, Fermi temperature, Fermi Level &Fermienergy.

a) Mobility(μ) of electrons is defined as the drift velocity (𝒗𝒅)acquired by the electrons per unit

electric field(E). ie: μ = 𝒗𝒅

𝑬 μ =

𝝈

𝒏𝒆 μ =

𝒆𝝉

𝒎

b) Fermi velocity(𝒗𝑭) of an electron is defined as it’s velocity when it’s energy is equal to the

Fermi energy. ie: 1

2𝑚𝑣𝐹

2 = 𝐸𝐹 𝑣𝐹 = (√2𝐸𝐹𝑚

)

c) Fermi temperature(𝑻𝑭) is defined as the temperature at which the average thermal

energy of the electron is equal to the Fermi energy at 0K. ie: 𝑻𝑭 =𝑬𝑭

𝒌 where k=Boltzmann

constant.

d) Fermi level is defined as the highest filled energy level in a metal at 0K.

e) Fermi energy(𝑬𝑭)is defined as the energy of the highest occupied level in a metal at 0K.

E

0

𝜕𝑘



Q:What is Fermi-Dirac statistics ? Explain.

Fermi –Dirac statistics is the statistical rule applied to the distribution of identical,

indistinguishable particles of spin 1

2 called fermions like electrons, which obey Pauli’s

exclusion principle. The probability of occupation of a state by an electron is given

by Fermi factor/Fermi- Dirac distribution function given by

f(E)= 1

1+𝑒(𝐸−𝐸𝐹)/𝑘𝑇

where 𝐸𝐹 = 𝐹𝑒𝑟𝑚𝑖 𝑒𝑛𝑒𝑟𝑔𝑦, E=energy of State, k=Boltzmann constant and

T= temperature.

Case1. At T=0K & E < 𝐸𝐹 ,f(E)= 1

Case2. At T=0k & E > 𝐸𝐹 ,f(E)= 0 and

Case3. At T> 0𝐾 & E = 𝐸𝐹 ,f(E)= 0.5

Variation of f(E) with E is as shown in the graph.

Q:What is meant by Fermi factor ?Explain the variation of Fermi factor with temperature & energy.

Fermi factor is the probability of occupation of a given energy state by an electron in a metal

at thermal equilibrium. It is given by the relation f(E) = 𝟏

𝟏+𝒆(𝑬−𝑬𝑭)/𝒌𝑻 ,where 𝐸𝐹𝐹𝑒𝑟𝑚𝑖 𝑒𝑛𝑒𝑟𝑔𝑦,

E = energy of State, k = Boltzmann constant and T = temperature

The variation of f(E) with temperature and energy is discussed below:

When T = 0K

Case-1: If E < 𝐸𝐹 (𝑬 − 𝑬𝑭)𝑖𝑠 − 𝑣𝑒 ,then 𝒆(𝑬−𝑬𝑭)/𝒌𝑻 = 𝒆−∞ = 0

∴ f(E) = 𝟏

𝟏+𝒆(𝑬−𝑬𝑭)/𝒌𝑻 = 𝟏

𝟏+𝟎 =1,

Thus the probability of occupation up to Fermi level is 100%.

Case-2: If E > 𝐸𝐹 (𝑬 − 𝑬𝑭)𝑖𝑠 + 𝑣𝑒 ,then 𝒆(𝑬−𝑬𝑭)/𝒌𝑻 = 𝒆∞ = ∞

∴ f(E) = 𝟏

𝟏+𝒆(𝑬−𝑬𝑭)/𝒌𝑻 = 𝟏

𝟏+∞=

1

∞ =0,

Thus the probability of occupation above Fermi level is 0%.

Case-3: If E = 𝐸𝐹 (𝑬 − 𝑬𝑭) = 0 ,then 𝒆(𝑬−𝑬𝑭)/𝒌𝑻 = 𝒆𝟎

𝟎 indeterminate

f(E)

1 T = 0K

0.5 T> 𝑜𝐾

0 E

𝐸𝐹



When T > 0K

∴ f(E)= 𝟏

𝟏+𝒆(𝑬−𝑬𝑭)/𝒌𝑻

𝑰𝒇 𝑬 = 𝑬𝑭 (𝑬 − 𝑬𝑭) = 𝟎 𝑡ℎ𝑒𝑛 𝒆(𝑬−𝑬𝑭)/𝒌𝑻 = 𝒆𝟎 = 1

∴ f(E) = 𝟏

𝟏+𝒆(𝑬−𝑬𝑭)/𝒌𝑻 = 𝟏

𝟏+𝟏 =

𝟏

𝟐 Thus the probability

of occupation of Fermi level is 50% above 0K.

The variation of f(E) with temperature(T) and

energy(E)is shown in the graph .

Q:What is meant by Density of states ? Explain.

Density of states g(E) is defined as the number of electronic states present in a unit

energy range. Mathematically g(E) 𝑑𝐸 = 8√2𝜋𝑚3/2

ℎ3 𝐸1/2 𝑑𝐸 is a continuous function and

the product g(E)dE=dN gives the number of states per unit volume in an energy range

(dE)between E and E+dE The number of electrons per unit volume ,n=∫ 𝑔(𝐸)𝑓(𝐸)𝑑𝐸 and

theVariation of g(E) with E is shown in the graph.

Q:What is meant by effective mass? Explain.

When an electric field is applied to a metal, electrons in the K-shell

are not at all accelerated as they are tightly bound to the nucleus.

These electrons possess infinite mass called effective mass denoted by 𝒎∗

Effective mass is equal to the true mass if the electron is in vacuum.

g(E)

0

E

f(E) Note: 𝑇1 = 500𝑘 < 𝑇2=1000k

1 T = 0K

0.5 𝑇1 < 𝑇2

0 E

𝐸𝐹



Q:Explain the dependence of resistivity on temperature and

impurity Or State and explain Matthiessen’s rule.

1. Variation of resistivity 𝝆 of metals with temperature T is given by

Matthiessen’s rule 𝝆 = 𝝆𝒊 + 𝝆𝑷𝒉 (T)

Where 𝜌𝑖 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑖𝑡𝑦 𝑑𝑢𝑒 𝑡𝑜 𝑖𝑚𝑝𝑢𝑟𝑖𝑡𝑖𝑒𝑠 & 𝑖𝑚𝑝𝑒𝑟𝑓𝑒𝑐𝑡𝑖𝑜𝑛𝑠

𝜌𝑃ℎ (T)= 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑖𝑡𝑦 𝑑𝑢𝑒 𝑡𝑜𝑙𝑎𝑡𝑡𝑖𝑐𝑒

𝑝ℎ𝑜𝑛𝑜𝑛𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛𝑠.

2. Variation of resistivity 𝝆 with temperature T is shown graphically.

At 0K, 𝝆𝑷𝒉 (T) = 0 = 𝝆𝒊 ,

3. For pure metals 𝜌𝑖 = 0 ,but 𝑝𝑟𝑎𝑐𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑡𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑒 𝑝𝑢𝑟𝑒 𝑚𝑒𝑡𝑎𝑙𝑠, ∴ 𝜌𝑖 ≠

0

4. and at large T , 𝜌𝑖 is negligible compared to 𝑙𝑎𝑟𝑔𝑒 𝜌𝑃ℎ (T) ∴ 𝜌 = 𝜌𝑃ℎ (T)

Q:Derive an expression for electrical conductivity of a semi-conductor.

In a semiconductor the net current is due to both electrons and holes .

The current due to electrons is 𝐼𝑒 = 𝑛𝑒𝑒𝑎𝑣𝑒 ,where 𝑛𝑒 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑎 = 𝑎𝑟𝑒𝑎 ,

𝑣𝑒 = 𝑑𝑟𝑖𝑓𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 & 𝑒 = 𝑐ℎ𝑎𝑟𝑔𝑒

Also current due to holes is 𝐼ℎ = 𝑛ℎ𝑒𝑎𝑣ℎ

But the total current I= 𝐼𝑒 + 𝐼ℎ

= ea(𝑛𝑒𝑣𝑒 + 𝑛ℎ𝑣ℎ)

𝐼

𝑎 = e (𝑛𝑒𝑣𝑒 + 𝑛ℎ𝑣ℎ)

𝑊𝐾𝑇, 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝐽 =𝐼

𝑎

∴ J= e (𝑛𝑒𝑣𝑒 + 𝑛ℎ𝑣ℎ)

but from Ohm’s law J=𝜎𝐸

ie: 𝜎𝐸 = 𝑒 (𝑛𝑒𝑣𝑒 + 𝑛ℎ𝑣ℎ)

𝜎 = 𝑒 (𝑛𝑒𝑣𝑒

𝐸+ 𝑛ℎ

𝑣ℎ

𝐸)

𝝈 = e(𝒏𝒆𝝁𝒆+𝒏𝒉𝝁𝒉)

where 𝜇𝑒&𝜇ℎ mobilities of electrons and holes respectively.

For an intrinsic semiconductors 𝑛𝑒 = 𝑛ℎ = 𝑛𝑖 called the density of intrinsic charge

carriers.

For n-type semiconductor 𝑛𝑒 ≫ 𝑛ℎ 𝝈 = e𝒏𝒆𝝁𝒆

𝜌

𝜌𝑖

0 T



For p-type semiconductor, 𝑛𝑒 ≪ 𝑛ℎ 𝝈 = e𝒏𝒉𝝁𝒉

Q: State law of mass action. Obtain an expression for the intrinsic carrier density.

Law of mass action states that the product of the concentration of charge carriers

electrons ( 𝑛𝑒)and holes( 𝑛ℎ) in an intrinsic semiconductor is equal to the

square of the intrinsic charge carriers( 𝑛𝑖2) at any temperature.

ie: 𝒏𝒆. 𝒏𝒉 = 𝒏𝒊𝟐

Q: Derive an expression for electron concentration in an intrinsic semiconductor.

or Derive an expression for electron charge carrier density in an intrinsic

semiconductor.

We know that the number of energy levels between the energy interval E and E+dE

is given by g(E)dE = (8√2𝜋(𝑚𝑒

∗)3/2

ℎ3 ) 𝐸1/2𝑑𝐸 ,where 𝑚𝑒∗ = effective mass of

electron.

Also, the probability of occupation of an energy level is given by Fermi factor

f(E) =1

𝑒(𝐸−𝐸𝐹

𝑘𝑇 )

+1

where 𝐸𝐹 = 𝐹𝑒𝑟𝑚𝑖 𝑒𝑛𝑒𝑟𝑔𝑦 & 𝑘 = 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Then ,the concentration of electron charge carriers is given by

𝑛𝑒 = ∫ 𝑓(𝐸)∞

𝐸𝑔 g(E)dE

= ∫1


𝑘𝑇 )

+1

∞

𝐸𝑔 (

8√2𝜋(𝑚𝑒∗ )3/2

ℎ3 ) 𝐸1/2𝑑𝐸

where 𝑚𝑒∗ = effective mass of electron.

On integration we can show that

𝑛𝑒 = 4√2 ( 𝜋𝑚𝑒

∗ 𝑘𝑇

ℎ2)

3/2 𝑒

(𝐸𝐹−𝐸𝑔

𝑘𝑇)

Similarly, the concentration of hole charge carriers is given by

𝑛ℎ = ∫ 𝑓(𝐸)0

−∞ g(E)dE

= ∫1


𝑘𝑇 )

+1

0

−∞ (

8√2𝜋(𝑚ℎ∗ )

3/2

ℎ3 ) 𝐸1/2𝑑𝐸

where 𝑚ℎ∗ = effective mass of electron.

On integration we can show that

𝑛ℎ = 4√2 ( 𝜋𝑚ℎ

∗ 𝑘𝑇

ℎ2 )3/2

𝑒(

−𝐸𝐹𝑘𝑇

)



Q: Based on law of mass action derive an expression for intrinsic charge density

(concentration) in an intrinsic semiconductor.

Law of mass action states that the product of the concentration of charge carriers

electrons ( 𝑛𝑒)and holes( 𝑛ℎ) in an intrinsic semiconductor is equal to the square

of the intrinsic charge carriers( 𝑛𝑖2) at any temperature.

ie: 𝒏𝒊𝟐 = 𝒏𝒆. 𝒏𝒉

But , WKT, 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠, 𝑛𝑒 = 4√2 ( 𝜋𝑚𝑒

∗𝑘𝑇

ℎ2 )3/2

𝑒(

𝐸𝐹−𝐸𝑔𝑘𝑇

) and

Concentration of holes, 𝑛ℎ = 4√2 ( 𝜋𝑚ℎ

∗ 𝑘𝑇

ℎ2)

3/2 𝑒

(−𝐸𝐹𝑘𝑇

)

From law of mass action,

𝑛𝑖2 = 𝑛𝑒 . 𝑛ℎ

∴ 𝑛𝑖2 = 4√2 (

𝜋𝑚𝑒∗𝑘𝑇

ℎ2 )3/2 𝑒(𝐸𝐹−𝐸𝑔

𝑘𝑇) 𝑋 4√2 (

𝜋𝑚ℎ∗𝑘𝑇

ℎ2 )3/2 𝑒(

−𝐸𝐹𝑘𝑇

)

𝑛𝑖 = √4√2 ( 𝜋𝑚𝑒

∗𝑘𝑇

ℎ2 )3/2 𝑒(𝐸𝐹−𝐸𝑔

𝑘𝑇) 𝑋 4√2 (

𝜋𝑚ℎ∗ 𝑘𝑇

ℎ2 )3/2 𝑒(−𝐸𝐹𝑘𝑇

)

on simplification we can we can show that,

𝒏𝒊 = 4√𝟐 ( 𝝅𝒎𝒆

∗𝒌𝑻

𝒉𝟐 )𝟑/𝟒( 𝝅𝒎𝒉

∗ 𝒌𝑻

𝒉𝟐 )𝟑/𝟒 𝒆(

−𝑬𝒈

𝟐𝒌𝑻)

Q:Explain the terms: Fermi velocity & Fermi temperature.

Fermi velocity (𝑣𝐹)is defined as the velocity of the electron having the energy

equal to Fermi energy(𝐸𝐹) ,mathematically ,1

2𝑚𝑣𝐹

2 = 𝐸𝐹 ∴ 𝑣𝐹 = √2𝐸𝐹

𝑚

Fermi temperature (𝑇𝐹)is defined as the ratio of Fermi energy(𝐸𝐹) to

the Boltzmann constant(k). Mathematically,𝑇𝐹 =𝐸𝐹

𝑘



SUPERCONDUCTIVITY:

Q: Explain the terms :-

a. Superconductivity is the phenomenon in which the resistivity/resistance of a substance

becomes zero.

b. Superconducting state is the zero resistivity/resistance state of the substance.

c. Critical/Transition temperature is the minimum temperature at which the substance

changes from normal state to superconducting state.

d. Critical magnetic field is the minimum magnetic field at which the substance changes to

normal state from superconducting state.

e. Flux quantization is the quantization of the magnetic flux through a superconducting ring

and is given by ∅ =𝒏𝒉

𝟐𝒆 ,where n =1,2,3…..h = Planck’s constant and e = electron charge.

f. Isotopic effect states that the transition temperature 𝑇𝐶varies inversely

as the square root of the atomic mass M ie: 𝑇𝐶 ∝1

√𝑀

Q: temperature dependence of resistivity in metals and superconducting materials.

1. The variation of the resistivity of the metals and superconductor is as shown in the graph.

2. When metals are cooled their resistivity decreases and reaches the minimum value

𝜌𝑖 called the residual resistivity at 0K , 𝜌𝑖 is due to impurities and imperfections in

𝑡ℎ𝑒 metals.

3. The resistivity of the metals increases with temperature due to ion vibrations .

4. The resistivity of the metals is given by Mathessian’s rule 𝜌 = 𝜌𝑖 + 𝜌𝑃ℎ(T).

5. But the resistivity of superconductors decreases with the decrease of temperature and

suddenly becomes zero at the temperature 𝑇𝐶 called critical/transition temperature.

6. The transition temperature depends on the impurities present in the metals.

Magnetic materials reduce 𝑇𝐶 to large extent.

𝜌 Normal conductor

𝜌𝑖 Super conductor

0 𝑇𝐶 T



Q:Explain Meissner’s effect.

1) Meissner effect is the phenomenon in which the magnetic flux in the material

Is completely expelled out of the material at temperature called Critical

temperature.

2) Below Critical temperature the material behave as perfect dia-magnet.

3) Consider a superconducting material placed in a magnetic field H, the magnetic

lines penetrate through the conductor.

4) When the temperature of the conductor is cooled below the critical temperature

𝑇𝐶 ,the magnetic flux inside the conductor is completely expelled out of the

conductor.

5) Above 𝑇𝐶 ,B≠ 0 and below 𝑇𝐶 ,B= 0.

6) Meissner effect is the principle of Maglev vehicles, Squid etc.

Q:Write a note on Type-I and Type-II superconductors.

Type-I

Type-I

1) Type-I Superconductors(SC’s) are also called soft SCs.

2) The magnetization of Type-I SC’s increases with the increase of the applied magnetic field

and suddenly drops to zero at the magnetic field 𝐻𝐶 called the critical magnetic field as

shown graphically.

H H

T> 𝑻𝑪 𝑆𝑢𝑝𝑒𝑟 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 T< 𝑻𝑪

−𝑀

Super

Conducting Normal

State State

O 𝐻𝐶 H



3) Below 𝐻𝐶 SC becomes perfectly diamagnetic obeying Meissner effect and beyond 𝐻𝐶 SC

changes to normal state.

4) The Type-I SC’s exists only in two states namely superconducting and normal

Type-II

1) Type-II Superconductors(SC’s) are also called hard SC’s.

2) The magnetization of Type-II SC’s increases with the increase of the applied magnetic

field and begin to decrease at the magnetic field 𝐻𝐶1 called the lower critical field and

finally becomes zero at the upper critical magnetic field 𝐻𝐶2.

3) Up to 𝐻𝐶1 Type-II SC’s exhibit Meissner effect with perfect diamagnetism.

4) Between 𝐻𝐶1 & 𝐻𝐶2 they exhibit meissner effect partially.ie: Magnetic flux begins to

penetrate at 𝐻𝐶1 𝑎𝑛𝑑 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑙𝑦 penetrates at 𝐻𝐶2.

5) Thus Type-II SC’s exists in three states namely SC state up to 𝐻𝐶1, intermediate state

between 𝐻𝐶1 & 𝐻𝐶2 and normal state beyond 𝐻𝐶2.

6) Type-II SC’s exhibit superconductivity electrically even in the intermediate state .They can

carry large currents.

7) These SC’s are used as coils for producing very large magnetic fields.

8) Type-I can be changed in Type-II , Type-I SC Lead can be changed to Type-II by adding

indium as impurity to it.

−𝑀

Super Intermediate Normal

conducting state state

state

O 𝐻𝐶1 𝐻𝐶 𝐻𝐶2 H



Q; Explain the temperature dependence of critical field.

1) Critical magnetic field(𝐻𝐶) is the minimum magnetic field applied to the superconductor

to change from superconducting state to normal state.

2) The variation of the critical field with temperature is given by the relation

𝐻𝐶 = 𝐻0 [1 −𝑇2

𝑇𝐶2],

where 𝐻0 is called the critical field at 0K , T = temperature and 𝑇𝐶 =Critical temperature.

3) Thus critical field decreases with the increase of temperature.

Q:Explain BCS theory of superconductivity qualitatively.

1) BCS(Bardeen-Cooper-Schrieffer) theory successfully explained superconductivity

quantum mechanically.

2) BCS theory is based on electrons interactions through phonons/lattice as

mediators.

3) An electron distorts the lattice due to electrostatic attraction when it approach a

positive ion lattice, Smaller the mass of the ion distortion is more.

4) The oscillatory distorted lattice is quantized in terms of phonons .

5) This distorted lattice attract another electron by reducing its energy.

6) Thus the second electron attracts the first electron through lattice. This results in the

pairing of two electrons called “Cooper Pair”. The attraction between the electrons is

maximum when the two electrons have equal and opposite spin and momentum.

( 𝑘1 − 𝑞 ) ( 𝑘2 +q )

q

𝑘1 𝑘2

e e

H

𝐻𝑂

Superconducting normal

State state

O 𝑇𝐶 T



7) The formation of cooper pair is considered as emission and absorption of the phonon by

the two electrons as shown in the diagram.

8) The cooper pair formation begins at critical temperature 𝑇𝐶 and is completed at 0K.

9) Thus at 0K all the electrons are paired in to ordered cooper pairs, which can pass through

lattice without collisions when a potential difference is applied. This explains the zero

resistivity of superconductors.

10) When the temperature increases the breaking of the cooper pairs begin beyond

0K and completed 𝑎𝑡 𝑇𝐶 as a result the superconductor attains normal state.

11) BCS theory explained the existence of energy gap in superconductors, Meissner

effect ,Coherence length and Flux quantization, etc.

Q: Write a note on High temperature superconductors(HTSC).

1. Generally Superconductors of transition temperature more than 10K are called

High temperature superconductors(HTSCs).

2. First HTSC found was ceramic material (oxide of lanthanium, barium &

copper)whose 𝑇𝐶 =30~35K, by Bednorz and Muller

3. Later many HTSCs are found of 𝑇𝐶 more than 150K using liquid Nitrogen.

In 2015 HTSC found was Hydrogen Sulphide of 𝑇𝐶 = 203K at very high pressure.

4. Also 1-2-3 compound HTSCs are found which are alloys of 1 rare Earth element

atom, 2 barium atoms,3 Copper atoms and 7 Oxygen atoms. Ex:Y𝐵𝑎2𝐶𝑢3𝑂7 etc

5. HTSCs are all Type-II superconductors having complicated structures called

Perovskite , Tetragonal and Orthorhombic.

6. In all the HTSCs 1 to 4 Cu-O layers are placed between other layers.

7. HTSCs are used in Squids, MagLev vehicles ,MRI, NMR, LHC, and other applications

in various fields like Industry, Medicine, Physics, Chemistry, Engineering ,Scientific

research etc

8. HTSCs are not perfectly diamagnetic since they exhibit partial Meissner’s effect.

Q;Explain the Principle,Construction and working of Maglev vehicles.

1) Maglev vehicles works on the principle of Meissner effect & magnetic levitation.

RW RW

Vehicle

Superconducting magnet

Aluminium guide way



2) Floating of a magnet placed over a superconducting magnet is called magnetic levitation.

3) The schematic diagram of Maglev vehicle is as shown in the diagram. Superconducting

magnet is fitted in to the base of the vehicle, which is placed over aluminium guide way.

4) Magnetic field produced due to current in the guide way lifts/levitate the vehicle and also

propel it.

5) The vehicle is provided with retractable wheels which can be pulled out when the vehicle

is brought to rest and can be pulled in when levitated and propelled.

6) Maglev vehicle trains have attained very high speeds of 603 km/hr.

PROBLEMS SECTION

MODULE-2 : ELECTRICAL PROPERTIES OF MATERIALS

Formulae needed : f(E) = 1

1+𝑒𝐸−𝐸𝐹

𝑘𝑇

; 𝜌 = 𝑚

𝑛𝑒2𝜏 ; 𝜎 =

1

𝜌 ; 𝜎 =

𝑛𝑒2𝜏

𝑚 ;

n = 𝑥𝐷𝑁𝐴

𝑀 ; 𝜇 =

1

𝑛𝑒𝜌 ; ½ m𝑉𝐹

2 = 𝐸𝐹 ; 𝑉𝑑 = 𝑒𝐸𝜏

𝑚 ; λ = 𝑉. 𝜏 ; n =

𝜎

𝜇𝑒=

1

𝜌𝜇𝑒 ; 𝐻𝐶 =

𝐻0 [1 −𝑇2

𝑇𝐶2]

1. Find the relaxation time of conduction electrons in a metal of resistivity 1.54 x10−8 Ω-m, if

the metal has 5.8 x1028 Conduction electrons per 𝑚3.( JAN2015)

Given: 𝜌 = 1.54 x10−8 Ω-m ; n = 5.8 x1028 /𝑚3 ; 𝑚 = 9.1𝑥10−31𝑘𝑔 ;

𝑒 = 1.6𝑥10−19𝐶 ; 𝜏𝑟= 𝜏 =?

Using 𝜌 = 𝑚

𝑛𝑒2𝜏 ,we get 𝜏 =

𝑚

𝜌𝑛𝑒2

= 9.1𝑥10−31

1.54 𝑥10−8𝑥5.8 𝑥1028𝑥(1.6𝑥10−19)2

= 3.98 𝑥10 −14 S



2. Calculate the mobility and relaxation time of an electron in copper assuming that each

atom contributes one free electron ,whose density is 8.92 x 103kg/𝑚3 and 𝑁𝐴=6.02 x

1026/kg mole. (Jun/Jul 14) Given: 𝜌 = 1.73 x 10−8 Ω-m ; M = 63.5 ; D = 8.92 x 103kg/𝑚3 ; 𝑁𝐴=6.02 x 1026/kg mole ; 𝑚 =

9.1𝑥10−31𝑘𝑔 ; 𝑒 = 1.6𝑥10−19𝐶 ; 𝑥 = 1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛/𝑎𝑡𝑜𝑚, 𝑛 = ? ; 𝜇 = ? & 𝜏 = ?

𝑈𝑠𝑖𝑛𝑔 n = 𝑥𝐷.𝑁𝐴

𝑀

= 1𝑥8.92 𝑥 103𝑥.6.02 𝑥 1026

63.5

= 8.456x1028 atoms.

𝜇 = 1

𝑛𝑒𝜌

𝜇 = 1

8.456𝑥1028𝑥1.6𝑥10−19𝑥1.73 𝑥 10−8 = 4.27x10−3

also 𝜇 = 𝑒𝜏

𝑚 , we get 𝜏 =

𝜇𝑚

𝑒

= 4.27𝑥10−3𝑥9.1𝑥10−31

1.6𝑥10−19

= 2.43 x10−14 S

3. Find the temperature at which there is 1.0% probability that a state with an energy 0.5eV

above Fermi energy will be occupied. (Dec 2013/Jan2014)

Given: (𝐸 − 𝐸𝐹) = 0.5 𝑒𝑉 = 0.5𝑥1.6𝑥10−19 J , k=1.38x10−23 J/K ; f(E) = 1 % = 1/100 =0.01 ; T= ?

Using f(E)= 1

1+𝑒(𝐸−𝐸𝐹)/𝑘𝑇 ,we get T = (𝐸−𝐸𝐹)

𝑘.𝐼𝑛(1

𝑓(𝐸) –1)

= 0.5𝑥1.6𝑥10−19

1.38𝑥10−23.𝐼𝑛(1

0.01 –1)

= 0.5𝑥1.6𝑥10−19

1.38𝑥10−23.𝐼𝑛(100 –1)

= 1262 K =1.262x103K

4. Calculate the conductivity of sodium given 𝜏𝑚 =2 x 10−14 S. Density of sodium is 971 kg/𝑚3,its

atomic weight is 23 and has one conduction electron/atom. (𝐽𝑢𝑛 2012)

Given: M = 23 ; D = 971 kg/𝑚3 ; 𝑁𝐴=6.02 x 1026/kg mole ; 𝑚 = 9.1𝑥10−31𝑘𝑔 ; 𝜏 = 𝜏𝑚 =2 x 10−14 S

𝑒 = 1.6𝑥10−19𝐶 ; 𝑥 = 1𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛/𝑎𝑡𝑜𝑚, 𝑛 = ? ; 𝜎 = ?

𝑈𝑠𝑖𝑛𝑔 n =𝑥𝐷.𝑁𝐴

𝑀

= 1𝑥971𝑥.6.02 𝑥 1026

23

= 2.54x1028 atoms.



Also, 𝜎 = 𝑛𝑒2𝜏

𝑚

= 2.54𝑥1028𝑥( 1.6𝑥10−19)2𝑥2 𝑥 10−14

9.1𝑥10−31

= 1.42x107 Ωm.

5. Calculate the Fermi velocity and the mean free path for the conduction electrons in silver,

given that its Fermi energy is 5.5 eV and relaxation time for electrons is 3.97

X10−14S. (𝐷𝑒𝑐 2011)

Given: 𝐸𝐹 = 5.5 𝑒𝑉 = 5.5 𝑥 1.6𝑥10−19 𝐽 ; 𝜏 = 𝜏𝑟 = 3.97 X10−14S ; 𝑚 = 9.1𝑥10−31𝑘𝑔 𝑉𝐹 = ? :

λ = ?

Using ½ m𝑉𝐹2 = 𝐸𝐹 ,we get 𝑉𝐹 = √

2𝐸𝐹

𝑚

= √2𝑥5.5 𝑥 1.6𝑥10−19

9.1𝑥10−31

= 1.391x 106 m/s and

Also, λ =𝑉𝐹. 𝜏𝑟 = 1.391x 105 x 3.97 X10−14 = 5.522𝑥 10−9 m

6. The Fermi level of potassium is 2.1eV. What are the energies for which the probabilities of

occupancy at 300 K are 0.99 and 0.5 ? (𝐽𝑢𝑛/𝐽𝑢𝑙2011)

Given: 𝐸𝐹 =2.1 eV = 2.1x1.6x10−19 𝐽 ;T = 300K ; f(𝐸1) =0.99 ; f(𝐸2) =0.5 ;

K =1.38x10−23 J/K ; 𝐸1 =? ; 𝐸2 =?

Using f(E)= 1

1+𝑒(𝐸−𝐸𝐹)/𝑘𝑇 ,

we get E = 𝐸𝐹 + 𝑘𝑇 𝐼𝑛 [ 1

𝑓(𝐸) – 1]

∴ 𝐸1 = 2.1𝑥1.6𝑥10−19 + 1.38x10−23𝑥300 𝐼𝑛 [ 1

0.99 – 1]

= 3.36 𝑥10−19 − 0.1902 𝑥10−19

= 3.17x10−19 𝐽

= 3.17𝑥10−19

1.6𝑥10−19 eV

= 1.98eV

Also, 𝐸2 = 2.1𝑥1.6𝑥10−19 + 1.38x10−23𝑥300 𝐼𝑛 [ 1

0.5 – 1]

= 3.36 𝑥10−19 − 0

= 3.36 𝑥10−19 𝐽

= 3.36 𝑥10−19

1.6 𝑥10−19 eV

= 2.1 eV



7. The Fermi level for a metal is 2.1eV, calculate the energies for which the probability of

occupancy at 300 K are 98% and 50%.(𝐷𝑒𝑐 2010)

Given: 𝐸𝐹 =2.1 eV = 2.1x1.6x10−19 𝐽 ;T = 300K ; f(𝐸1) =98%=0.98 ; f(𝐸2)=50%=0.5

k =1.38x10−23 J/K ; 𝐸1? , 𝐸2 =?

Using f(E)= 1

1+𝑒(𝐸−𝐸𝐹)/𝑘𝑇 ,

we get E = 𝐸𝐹 + 𝑘𝑇 𝐼𝑛 [ 1

𝑓(𝐸) – 1]

∴ 𝐸1 = 2.1𝑥1.6𝑥10−19 + 1.38x10−23𝑥300 𝐼𝑛 [ 1

0.98 – 1]

= 3.36 𝑥10−19 − 0.16 𝑥10−19

= 3.2x10−19 𝐽

= 3.2𝑥10−19

1.6𝑥10−19 eV

=2eV

Also , 𝐸2 = 2.1𝑥1.6𝑥10−19 + 1.38x10−23𝑥300 𝐼𝑛 [ 1

0.5 – 1]

= 2.1𝑥1.6𝑥10−19 − 0

= 2.1𝑥1.6𝑥10−19

1.6𝑥10−19 eV

= 2.1 eV

8. A uniform silver wire has resistivity 1.54 x10−8𝛺𝑚 at room temperature for an electric

field 2 V/m. Calculate relaxation time and drift velocity of the electrons, assuming that

there are 5.8 X 1022 conduction electrons per 𝑐𝑚3 of the material. (𝐽𝑢𝑛 2010)

Given: 𝜌 = 1.54 x 10−8 Ω-m ; E = 2 V/m ; 𝑛 = 5.8 𝑋 1022/𝑐𝑚3 = 5.8 𝑋 1022𝑥10−6/

𝑚3 ; 𝑚 = 9.1𝑥10−31𝑘𝑔 ; 𝑒 = 1.6𝑥10−19𝐶 ; 𝜏 = ? & 𝑉𝑑 = ?

𝑈𝑠𝑖𝑛𝑔: 𝜏 = 𝜌𝑚

𝑛𝑒2

= 1.54 𝑥 10−8 𝑥9.1𝑥10−31

5.8 𝑋 1022𝑥10−6( 1.6𝑥10−19)2

= 9.44 x10−18 S

𝑉𝑑 = 𝑒𝐸𝜏

𝑚

= 1.6𝑥10−19𝑥2𝑥9.44 𝑥10−18

9.1𝑥10−31

= 3.32 x 10−6 m/s



9. Calculate the probability of an electron occupying an energy level 0.02 eV above the Fermi

level at 300K.( 4 marks)

Given: (𝐸 − 𝐸𝐹) = 0.02 𝑒𝑉 = 0.02𝑥1.6𝑥10−19 J , k=1.38x10−23 J/K and T=300K

Using f(E) = 1


= 1

1+𝑒(0.02𝑥1.6𝑥10−19)/1.38𝑥10−23 𝑥 300

= 1

1+𝑒0.773

= 1

1+2.166

=1

3.166

=0.32 or 32 %

10. Calculate the concentration at which the acceptor atoms must be added to a germanium

sample to get a P-type semiconductor with conductivity 0.15 per Ohm-metre. Given the

mobility of holes=0.17 𝑚2/Vs. ( 4 marks)

Given: 𝜎 =0.15 /Ωm ,𝜇 = 0.17 𝑚2/𝑉𝑠 , e=1.6x10−19 C, n= ?

Using, n =𝜎

𝜇𝑒

=0.15

0.17𝑥1.6𝑥10−19

= 5.515 x1018 electrons

13 . A superconducting tin has a critical field of 306 gauss at 0K and 217 gauss at 2K.

Find the critical temperature of superconducting tin.

Given : 𝐻𝑜 = 306 𝑔𝑎𝑢𝑠𝑠, 𝐻𝐶 = 217 𝑔𝑎𝑢𝑠𝑠, 𝑇 = 2𝐾 , 𝑇𝐶 =?

Using: 𝐻𝐶 = 𝐻𝑜 [1 −𝑇2

𝑇𝐶2]

We get , 𝑇𝐶 = √𝑇2

[1−𝐻𝐶𝐻𝑜

]

= √22

[1−217

306]

=3.71K



14. Calculate the mobility of electrons in copper assuming that each atom

Contribute one free electron for conduction. Resistivity of copper=1.7 X 10−8

Ωm, atomic weight =63.54,density =8.96 X 103kg/𝑚3( CBCS-Jun/Jul16)

Given: 𝜌 = 1.7 x 10−8 Ω-m ; M = 63.54 ; D = 8.96 x 103kg/𝑚3 ; 𝑁𝐴=6.025 x 1026/kg mole

; 𝑚 = 9.11𝑥10−31𝑘𝑔 ; 𝑒 = 1.6𝑥10−19𝐶 ; 𝑥 = 1𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛/𝑎𝑡𝑜𝑚, 𝑛 = ? ; 𝜇 = ?

𝑈𝑠𝑖𝑛𝑔 n =𝑥𝐷.𝑁𝐴

𝑀

= 1𝑥8.96 𝑥 103𝑥.6.025 𝑥 1026

63.54

= 8.5x1028 atoms.

𝜇 = 1

𝑛𝑒𝜌

= 1

8.5𝑥1028𝑥1.6𝑥10−19𝑥1.7 𝑥 10−8 = 4.325x10−3 𝑚2/vs

15. For intrinsic gallium arsenide, the room temperature electrical conductivity is

10−6 /Ωm.The electron and hole mobilities are respectively 0.85 𝑚2/𝑉𝑠 and

0.04 𝑚2/𝑉𝑠. 𝐶alculate the intrinsic carrier concentration at room temperature.

Given: 𝜎𝑖 = 10−6 /Ωm ; 𝜇𝑒 = 0.85 𝑚2/𝑉𝑠 𝜇𝑒 = 0.04 𝑚2/𝑉𝑠; 𝑛𝑖 =?

𝑈𝑠𝑖𝑛𝑔, 𝜎𝑖 = 𝑛𝑖e (𝜇𝑒 + 𝜇ℎ),

𝑤𝑒 𝑔𝑒𝑡 , 𝑛𝑖 =𝜎𝑖

𝑒 (𝜇𝑒 +𝜇ℎ)

𝑛𝑖 =10−6

1.6𝑥10−19 (0.85+0.04)

= 7.023𝑥1012/𝑚3

16. The effective mass for the electron in germanium is 0.55 𝑚𝑜,where 𝑚𝑜𝑖𝑠 𝑡ℎ𝑒

free electron mass. Find the electron concentration in Germanium at 300 K,

assuming that the Fermi level lies exactly in the middle of the energy gap,

given that the energy gap for Germaium is 0.66 eV.

Given: 𝑚𝑜 = 9.1 x 10−31kg ; 𝑚𝑒∗ = 0.55x 9.1 x 10−31kg ; T= 300K ; 𝐸𝑔 = 0.66𝑒𝑉;

𝐸𝐹 =𝐸𝑔

2 =

0.66

2 = 0.33 𝑒𝑉

Using ; 𝑛𝑒 = 4√2 ( 𝜋𝑚𝑒 𝑘𝑇

∗

ℎ2 )

3/2

𝑒(

𝐸𝐹−𝐸𝑔𝑘𝑇

)

= 4√2 ( 𝜋𝑥0.55𝑥 9.1 𝑥 10−31𝑥1.38𝑥10−23𝑥300

(6.625𝑥10−34)2)3/2 x 𝑒

(0.33−0.66)𝑥1.6𝑥10−19)

1.38𝑥10−23𝑥300

= 5.657x1.807x1024 x2.892x10−6

𝑛𝑒 = 2.956x1019 /𝑚3



17. For intrinsic Gallium Arsenide , the electric conductivity at room temperature

is 𝟏𝟎−𝟔𝒐𝒉𝒎−𝟏𝒎−𝟏. The electron and hole mobilities are respectively 0.85 𝒎𝟐/𝑽𝑺 𝒂𝒏𝒅

𝟎. 𝟎𝟒 𝒎𝟐/𝑽𝑺. Calculate the intrinsic carrier concentration at room temperature.

(04 Marks)

Given: 𝜎𝑖 = 10−6 /Ωm ; 𝜇𝑒 = 0.85 𝑚2/𝑉𝑠 𝜇𝑒 = 0.04 𝑚2/𝑉𝑠; 𝑛𝑖 =?

𝑈𝑠𝑖𝑛𝑔, 𝜎𝑖 = 𝑛𝑖e (𝜇𝑒 + 𝜇ℎ),

𝑤𝑒 𝑔𝑒𝑡 , 𝑛𝑖 =𝜎𝑖

𝑒 (𝜇𝑒 +𝜇ℎ)

𝑛𝑖 =10−6

1.6𝑥10−19 (0.85+0.04)

= 7.023𝑥1012/𝑚3

18. Calculate the probability of finding an electron at an energy level 0.02 eV above

Fermi level at 200 K . ( 04 Marks)

Given: (𝐸 − 𝐸𝐹) = 0.02 𝑒𝑉 = 0.02𝑥1.602𝑥10−19 J , k=1.38x10−23 J/K and T=200K

Using f(E) = 1


= 1

1+𝑒(0.02𝑥1.602𝑥10−19)/1.38𝑥10−23 𝑥 200

= 1

1+𝑒1.161

= 1

1+3.193

= 1

4.193

= 0.2385 or 23.85 %

19. Gold has one free electron/atom.Its density,atomic weight and resistivity are

19300 kg/𝑚3 ,197 and 2.21 x 10−8 Ω𝑚 . Calculate the free electron concentration

and mobility of conduction electron. ( 04 Marks)

Given : D=19300 kg/𝒎𝟑 , M =197 , 𝝆 = 2.21 x 10−8 Ω𝑚 , 𝒙 = 1 electron/atom,

𝑵𝑨 = 6.02 x 𝟏𝟎𝟐𝟖/Kmol , e=1.6 x 𝟏𝟎−𝟏𝟗 C , n =? , 𝝁 = ?

Using, n =𝒙 𝑫 𝑵𝑨

𝑴

∴ 𝒏 =𝟏𝒙 𝟏𝟗𝟑𝟎𝟎𝒙 𝟔.𝟎𝟐𝒙 𝟏𝟎𝟐𝟖

𝟏𝟗𝟕 = 5.898 x 𝟏𝟎𝟑𝟎 electrons/𝒎𝟑

𝑨𝒍𝒔𝒐 , using 𝝁 =𝟏

𝝆𝒏𝒆

∴ 𝝁 =1

2.21 𝑥 10−8𝑥5.898 𝑥 1030𝑥1.6 𝑥 10−19 = 4.795x10−5 𝑚2/Vs



20. Calculate the drift velocity and thermal velocity of conduction electrons in copper

at a temperature of 300K,when a copper wire of length 2 mand resistance 0.02 Ω carries

a current of 15 A. Given the mobility of free electrons in copper is 4.3x𝟏𝟎−𝟑𝒎𝟐/Vs.

( 04 Marks)

Given: T = 𝟑𝟎𝟎𝑲 ,L = 𝟐 𝒎 , 𝑹 = 0.02 Ω , I = 15 A , 𝝁 = 4.3x𝟏𝟎−𝟑 𝒎𝟐/Vs,

m= 𝒎𝒆 = 9.1 x 𝟏𝟎−𝟑𝟒kg , k = 1.38 x 𝟏𝟎−𝟐𝟑 J/K , 𝒗𝒅 =? , 𝒗𝑭 =?

Using 𝒗𝒅 = 𝝁E but E = 𝑽

𝑳 and V= IR

we get 𝒗𝒅 =𝝁𝑰𝑹

𝑳 =

𝟒.𝟑𝒙𝟏𝟎−𝟑𝒙 𝟏𝟓 𝒙 𝟎.𝟎𝟐

𝟐 = 6.45 x 𝟏𝟎−𝟒 m/s

Also using 𝟑

𝟐𝒌𝑻 =

𝟏

𝟐𝒎𝒗𝑭

𝟐

We get , 𝑽𝑭 = √𝟑𝒌𝑻

𝒎 = √

𝟑𝒙𝟏.𝟑𝟖 𝒙 𝟏𝟎−𝟐𝟑𝒙𝟑𝟎𝟎

𝟗.𝟏 𝒙 𝟏𝟎−𝟑𝟒 =1.168x𝟏𝟎𝟓 m/s

@@ end@@

C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM

VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 1

MODULE-3 : LASERS AND OPTICAL FIBRES

LASERS

Q: Explain Einstein’s explanation of interaction of radiation with matter (or)

Explain Induced absorption , Spontaneous emission &Stimulated emission.

Let N1 & N2 be the population of the energy states E1 & E2 respectively

so that (E2 −E1) = hν & E2 >E1

According to Einstein radiation interacts with matter in 3 ways namely:

1) Induced absorption:

Induced absorption is the phenomenon in which an atom(A) in the lower energy state E1

absorb the incident photon of energy ‘hν’ & excite to the higher energy state E2

if (E2 −E1) = hν .

Mathematically it(induced absorption) is represented as

hν +A → 𝑨∗ or photon +atom→ 𝑎𝑡𝑜𝑚∗

Also, Rate of induced absorption = 𝐵12 𝑁1𝑈𝜈 ,where 𝑈𝜈 = 𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 &

𝐵12 = 𝐸𝑖𝑛𝑠𝑡𝑒𝑖𝑛′𝑠 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑎𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡.

2.Spontaneous emission:

Spontaneous emission is the phenomenon in which an atom (A) in the excited state of

energy E2 de-excite to the lower energy state E1 without any external influence by

emitting a photon of energy hν =(E2 −E1).

Mathematically, it is represented as 𝑨∗ → A + hν

Also, Rate of Spontaneous emission = 𝐴21 𝑁2

𝑤ℎ𝑒𝑟𝑒 𝐴21 = 𝐸𝑖𝑛𝑠𝑡𝑒𝑖𝑛′𝑠 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡.

𝐸2 𝑁2 𝑨∗ •

hν ⇒

𝐸1 • 𝑨 𝑁2

𝐸2 𝑨∗ • 𝑁2

⇒ hν

𝐸1 𝑁1 𝑨 •



3.Stimulated emission:

Stimulated emission is the phenomenon in which an atom (𝑨∗) in the excited state of

energy E2 de-excite to the lower energy state E1 under the influence of an external photon

(hν) by emitting an identical photon of energy hν =(E2 −E1).

Mathematically,𝑖𝑡 𝑖𝑠 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑒𝑑 𝑎𝑠 𝒉𝝂 + 𝑨∗ → A +2 hν

Also, Rate of Stimulated emission= 𝐵21 𝑁2𝑈𝜈 ,where 𝑈𝜈 = 𝑒𝑛𝑒𝑟𝑔𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 &

𝐵21 = 𝐸𝑖𝑛𝑠𝑡𝑒𝑖𝑛′𝑠 𝑆𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡.

Q: Explain the terms : Active medium ; Population Inversion ; Pumping ; Meta stable state

and Laser Cavity( Optical resonator) Or Explain the requisites of a laser system.

The requisites of a laser system are :

1. Active medium is a Solid/Liquid/Gas medium in which stimulated emission and

amplification of the radiations can be achieved.

2. Pumping is the supply of energy to the atoms in the lower states in order to excite them to

higher states. The methods of pumping are Optical pumping, Electrical pumping, Forward

bias pumping ,Chemical pumping, Elastic one-one collisions.

3. Population Inversion is condition of system in which the population of higher energy states

exceed the population of lower states.

4. Meta stable state is an intermediate state in which the average life of the atoms is of the

order of 10−3s ie: their life is 105 times more than that of normal states.

5. Laser Cavity( Optical resonator) is a pair of parallel/con-focal/concentric mirrors between

which active medium is placed so that stimulated emitting photons are used to cause

further Stimulated emissions and to amplify the beam.

One mirror is highly silvered and the other partially silvered. The distance between the

mirrors is given by L = 𝑛𝜆

2 ,where λ = wavelength and n = number of stationary waves

produced.

𝐸2 𝑨∗ • 𝑁2

hν hν + hν

𝐸1 𝑁1 •A

𝐸2 𝑁2 •



𝑸:Derive an expression for energy density in terms of Einstein’s coefficients (or)

Derive the relation between Einstein’s coefficients

.

Consider a system of atoms in thermal equilibrium with radiation of energy density

𝑈𝜈 & frequency ‘ν’. Let N1 & N2 be the population of the energy states E1 & E2

respectively, where (E2 >E1 )

We know that,

Rate of induced absorption = 𝐵12 𝑁1𝑈𝜈 ,

where 𝐵12 = 𝐸𝑖𝑛𝑠𝑡𝑒𝑖𝑛′𝑠 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑎𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑜𝑛 .

Rate of Spontaneous emission = 𝐴21 𝑁2 𝑤ℎ𝑒𝑟𝑒 𝐴21 =

𝐸𝑖𝑛𝑠𝑡𝑒𝑖𝑛′𝑠 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑎𝑛𝑑

Rate of Stimulated emission= 𝐵21 𝑁2𝑈𝜈 ,

where 𝐵21 = 𝐸𝑖𝑛𝑠𝑡𝑒𝑖𝑛′𝑠 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑆𝑡𝑖𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛 .

At thermal equilibrium,

Rate of induced absorption = 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑆𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠

𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛 +

𝑅𝑎𝑡𝑒 𝑜𝑓 𝑆𝑡𝑖𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛

ie: 𝐵12 𝑁1𝑈𝜈 = 𝐴21 𝑁2 + 𝐵21 𝑁2𝑈𝜈

∴ 𝑈𝜈( 𝐵12 𝑁1 − 𝐵21 𝑁2) = 𝐴21 𝑁2

𝑖𝑒: 𝑈𝜈 = 𝐴21 𝑁2

( 𝐵12 𝑁1− 𝐵21 𝑁2) dividing both Nr & Dr by 𝐵21 𝑁2 ,we get

= 𝐴21

𝐵21

1

[ 𝐵12 𝐵21

𝑁1𝑁2

−1] ……….(1)

According to Boltzmann’s law, 𝑁1

𝑁2 = 𝑒(𝐸2 −𝐸1)/𝐾𝑇 = 𝑒 ℎ𝜈/𝐾𝑇 …….(2)

From eqns 1 &2 , we get. 𝑈𝜈 = 𝐴21

𝐵21

1

[ 𝐵12

𝐵21 𝑒ℎ𝜈/𝐾𝑇 −1]

……..(3)

But .the energy density given by Planck’s law is 𝑈𝜈 = 8𝜋ℎ𝜈3

𝐶3

1

( 𝑒ℎ𝜈/𝐾𝑇 −1) ……(4)

Comparing eqns 3 & 4 ,we get 𝑨𝟐𝟏

𝑩𝟐𝟏 =

𝟖𝝅𝒉𝝂𝟑

𝑪𝟑 or

𝐴21

𝐵21 = 𝑈𝜈 ( 𝑒

ℎ𝜈

𝐾𝑇 − 1) and 𝑩𝟏𝟐 = 𝑩𝟐𝟏

Thus coefficient of stimulated absorption = coefficient of stimulated emission.

Thus energy density 𝑼𝝂 = 𝑨𝟐𝟏

𝑩𝟐𝟏

𝟏

[𝒆𝒉𝝂/𝑲𝑻 –𝟏]

Induced absorption Stimulated emission

𝐸2 𝑁2 • •

hν hν hν 2hν

𝐸1 • 𝑁1 • •

Spontaneous emission



Q:Explain the fundamental mode of vibration in CO2 molecule.

In CO2 molecule there are three fundamental modes of vibrations, namely

1. Symmetric mode :

Symmetric mode is the mode in which both the oxygen atoms oscillate simultaneously to

& fro about the stationary carbon atom along the molecular axis.

2. Asymmetric mode:

Asymmetric mode is the mode in which both the oxygen atoms move in one direction and

the carbon atom move in the opposite direction along the molecular axis.

3. Bending mode:

Bending mode is the mode in which both oxygen atoms and carbon atom move in opposite

directions perpendicular to the molecular axis.

The internal vibrations of CO2 molecule are the combination of the above three modes.

Molecular axis

O O C

Molecular axis

O O C

Molecular axis

C

O O



Q:Explain the principal, construction and working of C𝑶𝟐 laser.

Q:Explain the construction & working of CO2 Laser.

Principle : CO2 laser works on the principle of stimulated emission.

Construction:

1) The schematic diagram CO2 Laser is as shown in the diagram invented by CKN Patel an

Indian engineer

2) It consists of a (glass)discharge tube of length 5 m & diameter 2.5 cm filled with a

mixture of gases CO2 ,N2 ,He in the ratio 1:2:3

3) High DC voltage can be applied to the gas between the electrodes A&C .

4) Ends of the tube is fitted with ( NaCl ) Brewster windows to get polarized laser beam

5) Two con-focal silicon mirrors coated with aluminum are provided at the ends of the tube

which act as optical resonators.

6) Cold water is circulated through a tube surrounding the discharge tube

𝑪𝑶𝟐 laser diagram

C𝑂2 𝑁2 𝐻𝑒 gases outlet

Water outlet

𝑀1 -optical cavity(oc) 𝑀2 (oc)

5 m

A electrodes C 2.5 cm Laser

BW BW=Brewster window

Water inlet Ba

Energy level diagram

𝑁2 Resonant transfer of energy C𝑂2

𝜈 = 1 𝐶5 (meta stable state)

𝐶4 10.6 𝜇𝑚 laser

Excitations by collission 𝐶3 9.6 𝜇𝑚 laser

𝐶2

𝜈 = 0 Ground state 𝐶1



Working:

1) CO2 Laser is a four level molecular gas laser which produce continuous or pulsed laser

beam.

2) It works on the principle of stimulated emission between the rotational sublevels of an

upper & lower vibrational levels of CO2 molecules.

3) Ionisation takes place due to electric discharge when high DC voltage is applied between

electrodes producing electrons.

4) The accelerated electrons excite both N2& CO2 atoms to their higher energy levels ‘ 𝓥 =1 &

C5 from their ground states 0 & C1 due to collision as follows:

e+𝑵𝟐 → 𝑵𝟐∗

+e’ and e+𝑪𝑶𝟐 → 𝑪𝑶𝟐∗

+e’

where e& e ’are the energies of electron before and after collision.

5) 𝑵𝟐∗ molecule in excited level collide with CO2 molecules in their ground state C1 &

excite it to metastable state C5 by resonant energy transfer as level C5 of CO2 is same as

level 𝓥=1 of 𝑁2 given by 𝑵𝟐∗ +𝑪𝑶𝟐 → 𝑪𝑶𝟐

∗+𝑵𝟐

6) As this process continues due to electric discharge pumping , population inversion takes

place betweenC5 &C4 and C5 & C3.

7) The transitions/de-excitations takes place as follows:

C5 → C4 producing laser 10.6𝜇𝑚 (IR region)

C5 → C3 producing laser 9.6𝜇𝑚 (IR region)

C4 → C2

C3 → C2 Radiation less transitions

C2 → C1

8) Due to high thermal conductivity of He, it removes heat from mixture and de-populate the

lower states C3 &C2 quickly .

9) Laser beam is amplified by using optical resonators.

10) The laser output is 100kW for continuous mode and 10 kW in pulsed mode.

Q: Explain the construction and working of Semi-conductor laser.

Semiconductor laser Energy level giagram Metalic coated(MC) surface P-type pn-Jn n-type RS 𝐸𝐹𝑒 PS CB Ba p P Laser 𝐸𝐹𝑝 h+e

n-type VB Polished surfac(PS) Laser Roughened surface(RS) MC

P-type

Pn-jn

n-type



Principle: SC laser works on the principle of stimulated emission.

Construction:

1. The schematic diagram of GaAs semi-conductor device is as shown in the diagram.

2. It consists of heavily doped n-region of GaAs doped with tellurium and p-region of GaAs

doped with zinc.

3. The upper and lower surfaces are metalized so that pn-junction is forward biased .

4. Two surfaces perpendicular to the Jn are polished so that they act as optical resonators

and the other two surfaces roughened to prevent lasing in that direction.

Working:

1. Semi-conductor laser are made up of highly de-generate semi-conductors having direct

band gap like Gallium Arsenide (GaAs).

2. When GaAS diode is forward biased with voltage nearly equal to the energy gap voltage,

electrons from n-region & holes from p-region flow across the junction creating

population inversion in the active jn region.

3. As the voltage is gradually increased due to forward biasing population inversion is

achieved between the valence band and conduction band which in turn result in

stimulated emission.

4. Photons produced are amplified between polished optical resonator surfaces producing

laser beam.

5. GaAs laser produce laser beam of wavelength 8870Å in IR region , GaAsP produce laser

beam of 6500Å in visible region etc.

Q:Mention the characteristics of laser beam/light.

The laser beam characteristics are:

1. They are highly monochromatic.

2. They are highly coherent.

3. They are highly directional.

4. They are highly focusable.

5. They are least divergent.

Q:Mention the uses of laser beam.

Laser beam is used in Holography, Laser welding, Laser cutting laser drilling and

measurement of atmospheric pollutants.



Q: What is holography ?

Holography is the process in which the details of an object in 3-dimensions can

be recorded on a 2-dimensional aid based on the principle of interference of light.

Q: Construction/Recording of Hologram.

1. The schematic diagram for the construction of a Hologram is as shown in the diagram.

2. A laser beam of wavelength (λ) is made to fall on beam splitter, which split the beam in to

two beams. One beam which passes through splitter is called ‘reference beam’. The other

reflected beam is made to fall on the object whose hologram is to be produced.

3. The beam reflected by the object is called ‘spherical object beam’.

4. The reference and object beams produce concentric circular rings called “Gaber zones

“where both intensity and phase are recorded due to interference on the photographic

plate.

5. On developing photographic plate we obtain ‘hologram’.

6. At each and every point on the hologram, complete information/details of the object are

recorded.

7. If the hologram is cut in to any number of pieces, each piece produce the complete image

of the object with less resolution.

Q: Re-construction of Hologram.

Object

Reflected beam Object spherical beam

Laser beam (λ) Hologram

Beam splitter reference beam

Hologram Observation direction

Laser beam (λ)

Object Virtual image Object Real image



1. The schematic diagram for the reconstruction of the image from Hologram is as shown in

the diagram.

2. The hologram is illuminated with the same laser beam of wavelength(λ) which used for

construction of hologram.

3. The hologram act as diffraction grating. Due to diffraction and interference two images of

the object are produced.

4. One which is formed on the side of incident laser is virtual image and the other formed on

the other side is the real image of the object.

5. By changing the direction of the observation ,we can see all the details of the object

originally hidden from view.

Q:Explain laser welding

1. The schematic arrangement of laser welding is as shown in the diagram.

2. High intensity and high focussability of lasers is used for laser welding.

3. Laser beam is focused on to the spot to be welded. Due to generation of high heat the

material melts in a short period & the impurities float to the surface.

4. On cooling the joint, it becomes homogeneous stronger joint.

5. As laser welding is contactless welding ,no foreign materials enter in to the joint.

6. Laser welding can be done more precisely by using pre-programmed computer assisted

welding.

7. CO2 laser can be used to weld both metallic and non-metallic substances.

Q:Explain laser cutting:

𝑂2/𝑁2 inlet

Condenser lens

material

Laser beam

Condenser lens

Materials to be weld



1. The schematic arrangement of laser cutting is as shown in the diagram.

2. High intensity and high focussability of lasers is used for laser cutting.

3. Laser cutting involve melting and gas assisted blowing out the material.

4. Laser beam focused on to the surface to be cut ,melts it and the high speed

gas O2 /N2 passed through the nozzle blows away the molten material. O2

gas need low power laser than for N2.

5. This process continues till the material is cut.

6. Laser cutting can be done more precisely by using pre-programmed computer

assisted cutting.

7. CO2 laser can be used to cut both metallic and non-metallic substances

Q:Explain laser drilling.

1..The schematic arrangement of laser cutting is as shown in the diagram.

2.High intensity and high focussability of lasers is used for laser cutting.

3.The pulsed laser beam is focused to the spot where the hole to be drilled.

High heat generated melts the material.

4.Melted material absorb the heat further and vaporizes.

5.with continues heating by the laser vapours ionize into plasma state.

6. The plasma state material further absorb laser and emit black body radiations,

which in turn generate detonation waves. These waves and high power pressure

expel the material out of hole.

7.. Laser drilling can be done more precisely by using pre-programmed computer

assisted drilling.

8. CO2 laser can be used to drill both metallic and non-metallic substances.

Laser beam

Condenser lens

material



Q:Measurement of atmospheric pollutants using LIDAR(Light Detection And Ranging).

1. The schematic experimental arrangement to measure atmospheric pollutants is as shown

in the diagram.

2. LIDAR (LIght Detection And Ranging)method is used to find the concentration of

atmospheric pollutants using laser.

3. Laser beam is sent in to the atmosphere in a particular direction and the scattered laser

beam by the pollutants are detected using detector.

4. The received scattered laser beam is analysed using analyser.

5. The measured intensity gives the concentration of the pollutants in a particular distance

and direction.

6. This method does not give information regarding the nature of the pollutants.

7. In order to know the nature of pollutants we have to adopt Raman spectroscopy method .n

this comparing the spectra of atmospheric pollutants and comparing them with standard

spectra of pollutants, we can identify pollutants.

OPTICAL FIBRES:

Q:What is an Optical Fibre?

Optical Fibre is a transparent di-electric material (like glass/plastic) which guides/

carry) light along it based on the principle of total reflection of light.

Optical fibre consists of a cylindrical transparent di-electric material of high refractive

index called core. It is surrounded by another di-electric transparent material of low

refractive index called cladding Cladding in turn is surrounded by cylindrical insulator

called Sheath, which gives mechanical strength & protect the fibre from absorption,

scattering etc

Atmosphere

• • ⋅ • • • ⋅ •

pollutants

• • ⋅ • • • ⋅ •

laser Detector

Analyser



Q:What is an acceptance angle of an optical fibre ? Find the expression for acceptance

angle or Numerical aperture.

Acceptance angle is the maximum angle submitted by the ray with the axis of the fibre so

that light can be accepted and guided along the fibre.

Let 𝑛1 , 𝑛2 & 𝑛𝑜 be the RI of core, cladding and launch medium respectively. Also OA incident

ray,AB refracted ray,BC totally reflected ray, 𝜃𝑖, 𝜃𝑟 & ∅ be the angles of incidence, refraction at A & angle of

incidence at B respectively.

By snell’s law at A, 𝑛𝑜 𝑠𝑖𝑛 𝜃𝑖 = 𝑛1 𝑠𝑖𝑛 𝜃𝑟 𝑠𝑖𝑛 𝜃𝑖 = 𝑛1

𝑛𝑜𝑠𝑖𝑛 𝜃𝑟 ……(1)

But, from ∆𝑙𝑒 𝐴𝐷𝐵, 𝜃𝑟 =(90 −∅) 𝑠𝑖𝑛 𝜃𝑟 = 𝑠𝑖𝑛(90 − ∅)=cos(∅) … . (2)

when 𝜃𝑖 𝑖𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 = 𝜃𝑜 , 𝑎𝑐𝑐𝑒𝑝𝑡𝑎𝑛𝑐𝑒 𝑎𝑛𝑔𝑙𝑒, 𝑡ℎ𝑒𝑛 ∅ = ∅𝐶 Critical angle ……..(3)

From eqns 1,2 &3 ,we get 𝑠𝑖𝑛 𝜃𝑜 = 𝑛1

𝑛𝑜𝑐𝑜𝑠(∅𝐶) …….(4)

Also, 𝑠𝑖𝑛(∅𝐶) =𝑛2

𝑛1 and 𝑐𝑜𝑠(∅𝐶) = √(1 − 𝑠𝑖𝑛2∅𝐶) =

√𝑛12−𝑛2

2

𝑛1 …(5)

From eqns 4 &5,we get 𝑠𝑖𝑛 𝜃𝑜 =√𝑛1

2−𝑛22

𝑛𝑜 or 𝜃𝑜 = 𝑠𝑖𝑛−1

√𝑛12−𝑛2

2

𝑛𝑜

for air 𝑛𝑜 = 1

Q: What is meant by numerical aperture(NA) and mention the expression for it.

Numerical aperture is the ability of the optical fibre to accept the light and guide

along the fibre and is numerically equal to sine of the acceptance angle.

ie: NA = 𝑠𝑖𝑛 𝜃𝑜 = √𝑛1

2−𝑛22

𝑛𝑜

Launch medium (𝑛𝑜 ) Fibre axis

A

𝜃𝑖

O

𝛷 Core (𝑛1)

𝜃𝑟

D

B Cladding ( 𝒏𝟐 )



Q:What is meant by fractional refractive index change(∆).

The ratio of the difference between refractive indices of core and cladding to that

of core is called fractional refractive index change. ie: ∆ =(𝑛1−𝑛2 )

𝑛1

Q:What is attenuation and explain types of attenuation in optical fibres.

Loss of power of light signal as it is guided along the fibre is called attenuation.

Attenuation is measured in terms of dB/km.

There are three types of attenuations in the fibre namely:

1. Absorption losses are the losses due to impurities & material itself and they are

two types namely

a) Impurity losses are the losses due to the impurities(Cu, Fe, etc) present in the

fibre, which can be minimised by taking care during manufacture of the fibre.

b) Intrinsic losses are the losses due to the material itself, these losses decreases with the

increase of wavelength.

2. Scattering losses are the losses due to imperfections of the fibre called Rayleigh

scattering losses which varies inversely as the 𝜆4.

3. Radiation losses are the losses are two types namely:-

a) Microscopic losses are the losses due to non-linearity of the fibre axis ,which can be

minimized by providing compressible jacket & taking care during manufacture of

the fibre.

a) Macroscopic losses are the losses due to large curvature/bending of the fibre when it is

wound over a spool/bent at corners. These losses increase exponentially up to

threshold radius and there afterwards losses becomes large.

Q: Obtain expression for attenuation coefficient in an optical fiber of length L

Lambert’s law states that the rate of decrease of intensity of light with distance(−𝑑𝑃

𝑑𝐿 )

in a medium is directly proportional to the original intensity(P)

ie: −𝑑𝑃

𝑑𝐿 ∝ 𝑃

𝑑𝑃

𝑑𝐿 = −𝛼𝑃 …(1) ,where 𝛼 is called the attenuation coefficient.

Rewriting(1 ) as 𝑑𝑃

𝑃 = −𝛼. 𝑑𝐿 ,

Integrating this between the limits (𝑃𝑖 , 𝑃𝑜) 𝑓𝑜𝑟 𝑃 𝑎𝑛𝑑 (0, 𝐿)𝑓𝑜𝑟 𝐿,

∫𝑑𝑃

𝑃= −𝛼

,𝑃𝑜

𝑃𝑖∫ 𝑑𝐿

𝐿

0

[𝑙𝑜𝑔10P]𝑃𝑖

,𝑃𝑜 = −𝛼 [L]0𝐿

𝑙𝑜𝑔10 (,𝑃𝑜

𝑃𝑖) = −𝛼L……(2) where 𝑃𝑖 , 𝑃𝑜 are 𝑖𝑛 & out put powers.

Equation (2),can be written as 𝜶 = −𝟏𝟎

𝑳 𝒍𝒐𝒈𝟏𝟎 (

,𝑷𝒐

𝑷𝒊) dB/km.

or 𝑷𝒐

𝑷𝒊= 𝒆

−𝜶𝑳

𝟏𝟎



Q:What is meant by modes of propagation.

The paths along which the light is guided in the fibre are called modes of

propagation and the number of modes of the fibre is given by N = 𝑉2

2

Q: What is V-Parameter/number ?

It is the quantity which represent the number of modes of the fibre given by

V = 𝜋𝑑

𝜆 . NA =

𝜋𝑑

𝜆 .

√𝑛12−𝑛2

2

𝑛0

where d = diameter , λ = wavelength,

𝑛1, 𝑛2 &𝑛0 RI’s of Core, Cladding & medium

Q: Explain the construction and working of Types of optical Fibres.

There are three types of optical fibres namely:-

1) Single mode step index fibre(SMF

1. SMF has a core diameter 8 − 10𝜇𝑚 of uniform RI and cladding diameter

60-70𝜇𝑚 has 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑅𝐼

2. The step index, cross section, modes and pulse profiles of SMF are as shown in the

diagram.

3. SMF guides light in a single mode as shown.

4. The V-number is < 2.4.

5. Numerical aperture is < 0.12 .

6. Attenuation is in the range 0.25-0.5 dB/km.

7. Information carrying capacity is very large. They are long haul carriers.

8. The output and input pulses are almost same.

9. Laser source is used. Connectors are costly

Cladding

Core

Input output

Index profile Cross section profile Modes profile Pulse profile



2) Multimode step index fibre(MMF)

Input output

1) MMF has a core diameter 50 − 200𝜇𝑚 of uniform RI and cladding diameter

100-250𝜇𝑚 has 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑅𝐼.

2) The step index, cross section,modes and pulse profiles are as shown in the diagram.

3) MMF guides light in multi-modes as shown .

4) The V-number is > 2.4.

5) Numerical aperture is 0.2 𝑡𝑜 0.3 .

6) Attenuation is in the range 0.5- 4 dB/km.

7) Information carrying capacity is small to medium and short haul carriers.

8) The output pulse is widened.

9) LED source is used & Connectors are cheap.

3) Graded index multimode fibre(GRIN

1) GRIN has a core diameter 50 − 200𝜇𝑚 of variable RI and cladding diameter 100-250𝜇𝑚

has 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑅𝐼.

2) The graded index, cross section ,modes and pulse profiles are as shown in the diagram.

3) MMF guides light in multi-modes as shown in the diagram.

4) The V-number is > 2.4.

5) Numerical aperture is 0.2 𝑡𝑜 0.3 .

Cladding

Core

Input output


Cladding

Core

Input output




6) Attenuation is in the range 0.5- 4 dB/km.

7) Information carrying capacity is large and efficient and are short haul carriers.

8) The output pulse and input pulse are as shown.

9) Laser/LED source is used. Connectors are cheap.

10) Easy to splice and interconnect but expensive.

Q:Explain point to point communication system using optical fibre.

The schematic block diagram of point to point communication system using optical fibre is

as shown in the diagram.

Note: AVS =audio/video signal. AES=analog electrical signal,

BES=binary electrical signal & OS = optical signal.

1. Information receiver-receives , convert input AVS in to AES & fed to coder.

2. Coder- receives, convert AES in to BES and fed in to optical transmitter after

modulating it with carrier signal.

3. Optical transmitter-receives, convert BES in to OS and fed in to carrier optical fibre.

4. Carrier optical fibre-receive OS and guide it along the fibre. Weakened OS is fed

in to repeater.

4. Re-peater( Receiver cum transmitter)-receives the Weakened OS, restore to original

strength and fed back in to carrier optical fibre again, which in turn guide OS and fed in

to optical receiver.

6. Optical receiver- receive ,convert OS in to BES & fed in to de-coder.

7. De-coder-receive, de-modulate & convert BES in to AES & fed in to information

transmitter.

8. Information transmitter-finally receive, convert AES in to AVS as output

Input

Audio-video

Signal

Output

Audio-video

Signal

Information

receiver

AVS AES

Coder

AES BES

Modulator

Optical

transmitter

BES OS

Optical fibre

carrier REPEATER

[ Receiver

Cum

Transmitter ]

Information

transmitter

AES AVS

De-Coder

BES AES

De- modulator

Optical

receiver

OS BES

Optical fibre

carrier



PROBLEMS SECTION

MODULE-3 : LASERS AND OPTICAL FIBRES

Formulae needed: L = 𝑛𝜆

2 ; 𝛼 = −

10

𝐿 log[

𝑃𝑜

𝑃𝑖 ] ;

𝑁2

𝑁1 = 𝑒−(ℎ𝐶/𝜆𝑘𝑇)

; E= P.t and E =n.ℎ𝐶

𝜆

;

NA = √𝑛1

2−𝑛22

𝑛𝑜 ; Sin𝜃𝑜 = NA ; V =

𝜋𝑑

𝜆 X NA ; N =

𝑉2

2 ; ∆ =

(𝑛1−𝑛2)

𝑛1 ;

1. Find the number of modes of standing waves in the resonator cavity of length 1 m in He-

Ne temperature of laser operating at wavelength 632.8nm.( JAN2015)

Given: L= 1 m ; λ = 632.8 nm = 632.8𝑥10−9 m. n = ?

Using L = 𝑛𝜆

2 ,we get n =

2𝐿

𝜆

= 2𝑥1

632.8𝑥10−9

= 3.161 x 106 𝑚𝑜𝑑𝑒𝑠

2. A fiber with an input power of 9μW has a loss of 1.5 dB/km. If the fiber is 3000 m long,

calculate the output power.(JAN2015)

Given: 𝑃𝑖 = 9𝜇𝑤 = 9𝑥10−6𝑤 ,𝛼 = 1.5 𝑑𝐵/𝑘𝑚 ,L = 3000m = 3 km , 𝑃𝑜 = ?

Using 𝛼=−10

𝐿 log ⟦

𝑃𝑜

𝑃𝑖⟧ we get 𝑃𝑜 = 𝑃𝑖 𝑒

−𝛼𝐿

10

= 9𝑥10−6𝑒−1.5𝑥3

10

= 5.74 𝑥10−6𝑤

3. Find the ratio of populations of two energy levels in a laser if the transition between

them produces light of wavelength 6493Å,assuming the ambient temperature as

27.(Jun/Jul 14)

Given: λ = 6493Å = 6493x10−10 m, t=27 T = 273+27 =300K; h=6.63x10−34 Js;

k=1.38x10−23 J/K ;C=3x108m/s ;𝑁2

𝑁1 = ?

Using 𝑁2

𝑁1 = 𝑒

−(ℎ𝐶

𝜆𝑘𝑇)

= 𝑒−(

6.63𝑥10−34𝑥3𝑥108

6493𝑥10−10𝑥1.38𝑥10−23𝑥300)

= 𝑒−79.99

= 7.33x10−33



4. The numerical aperture of an optical fibre is 0.2 when surrounded by air. Determine the

R.I of its core, given the R.I of the cladding is 1.59.Also find the acceptance angle when the

fibre is in water of R.I 1.33(Jun/Jul 14)

Given: 𝑛𝑜𝑥 NA = 0.2 , 𝑛2 = 1.59 , 𝑛′𝑜 = 1.33 , 𝑛1 =? a𝑛𝑑 𝜃′𝑜 = ?

For air , Using NA = √𝑛1

2−𝑛22

𝑛𝑜 , we get 𝑛1 = √(𝑛𝑜𝑥𝑁𝐴)2 + 𝑛2

2

= √( 0.22 + 1.592 )

= 1.603

Also 𝜃′𝑜 = 𝑆𝑖𝑛−1(

√𝑛12−𝑛2

2

𝑛′𝑜

) = 𝑆𝑖𝑛−1( √1.6032−1,592

1.33 ) = 8.65°

5. A pulse laser has an average power output 1.5mW per pulse and pulse duration is

20ns.The number of photons emitted per pulse is estimated to be 1.047 x108.Find the

wavelength of the emitted laser. (Dec 2013/Jan2014)

Given: P=1.5mW = 1.5x10−3 𝑤 ,t=20 ns =20x10−9 s , C=3x108m/s ; h=6.63x10−34 Js,

n = 1.047 x108 ,λ = ?

Using E= Pxt and E = n.ℎ𝐶

𝜆 , 𝑤𝑒 𝑔𝑒𝑡 , λ =

𝑛ℎ𝐶

𝑃𝑡

= 1.047 𝑥108𝑥6.63𝑥10−34𝑥3𝑥108

1.5𝑥10−3 𝑥20𝑥10−9

= 6.942x10−7 m.

6. The angle of acceptance of an optical fibre is 30° 𝑤ℎ𝑒𝑛 kept in air. Find the angle of

acceptance when it is in a medium of refractive index 1.33. (Dec 2013/Jan2014)&(Jun/Jul

2013) (𝐽𝑢𝑛

𝐽𝑢𝑙2011)

Given: 𝜃𝑜 =30° , 𝑛′𝑜=4/3=1.333 , 𝑛𝑜 = 1 𝑓𝑜𝑟 𝑎𝑖𝑟 , 𝜃′𝑜 =?

Using 𝑛′𝑜𝑆𝑖𝑛 𝜃′𝑜 = 𝑛𝑜Sin𝜃𝑜 = √𝑛12 − 𝑛2

2 = constsnt

We get , 𝜃′𝑜 = 𝑆𝑖𝑛−1 (𝑛𝑜𝑆𝑖𝑛𝜃𝑜 𝑛′𝑜⁄ )

= 𝑆𝑖𝑛−1 (1 𝑆𝑖𝑛30° 1.333⁄ )

=22.03°

7. Calculate the NA,V-number and number of modes in an optical fibre of core diameter

50μm,core and cladding refractive indices 1.41 and 1.4 at wavelength 820 nm. (𝐽𝑢𝑛 2012)

Given: 𝑛1 = 1.41 ; 𝑛2 = 1.4 ; λ = 820 nm= 820 x 10−9 m ; d = 50μm =50x 10−6 m ;

𝑛𝑜 =1 for air NA = ? ; V = ? & N = ?

a. NA = √𝑛1

2−𝑛22

𝑛𝑜 =

√(1.41)2−(1.4)2

1 = 0 .168

b. V = 𝜋𝑑

𝜆 X NA =

𝜋𝑥50𝑥 10−6𝑥0 .168

820 𝑥 10−9 = 32.18

c. N = 𝑉2

2 =

(32.17)2

2= 517.5 = 518



8. The refractive indices of the core and cladding of a step index fibre are 1.45 and 1.40

respectively and its core diameter is 45μm.Calculate its relative refractive index

difference, V- number at wavelength 1000 nm and number of modes. (𝐷𝑒𝑐 2011) Given: 𝑛1 = 1.45 ; 𝑛2 = 1.4 ; λ = 1000 nm= 1000 x 10−9 m ; d = 45μm =45x 10−6 m ; 𝑛𝑜 = 1; ∆ = ? ; V = ?

& N = ?

1) ∆ =(𝑛1−𝑛2)

𝑛1 =

(1.45−1.4)

1.45= 0.0345

2) V = 𝜋𝑑

𝜆 𝑋

√𝒏𝟏𝟐−𝒏𝟐

𝟐

𝒏𝒐=

3.14𝑥45𝑥 10−6𝑥√1.452−1.42

1000 𝑥 10−9 𝑥1= 53.34

3) N = 𝑉2

2 =

(53.34)2

2= 1422.6 =1423

9. A He-Ne gas laser is emitting a laser beam with an average power of 4.5 mW. Find the

number of photons emitted per second by the laser. The wavelength of the emitted

radiation is 6328Å. (𝐷𝑒𝑐 11)

Given: P=4.5mW = 4.5x10−3 𝑤 ,t= 1 S , C=3x108m/s ; h=6.63x10−34 Js,

λ = 6328Å = 6328x10−10 m ; n = ?

Using E= Pxt and E =n.ℎ𝐶

𝜆 𝑛 =

𝜆𝑃𝑡

ℎ𝐶

= 6328𝑥10−10

𝑥4.5𝑥10−3𝑥1

6.63𝑥10−34 𝑥3𝑥108

= 1.432x1016 photons.

10. Find the number of modes of standing waves and their frequency separation in the

resonant cavity of 1 m length of He-Ne operating at wavelength of 632.8 nm. (𝐽𝑢𝑛/

𝐽𝑢𝑙2011)

Given: L= 1 m ; λ = 632.8 nm = 632.8𝑥10−9 m. n = ? and

Using L = 𝑛𝜆

2 , we get , n =

2𝐿

𝜆

= 2𝑥1

632.8𝑥10−9

= 3.161 𝑥 106 𝑚𝑜𝑑𝑒𝑠

11. The ratio of population of two energy states in a laser 1.059 x10−30.If the temperature of

the system is 57°𝐶, 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑠𝑒𝑟 ?

Given: 𝑁2

𝑁1 = 1.059x10−30, t = 57°𝐶, T =273+57=330K; h=6.63x10−34 Js; k=1.38x10−23

J/K ;C=3x108m/s ; λ = ?

Using 𝑁2

𝑁1 = 𝑒−(ℎ𝐶/𝜆𝑘𝑇) ,we get , In(

𝑁2

𝑁1) = −(ℎ𝐶/𝜆𝑘𝑇) 𝑆𝑖𝑛𝑐𝑒: 𝑙𝑜𝑔𝑒 (

𝑁2

𝑁1) = In(

𝑁2

𝑁1)

λ = −(ℎ𝐶/𝐼𝑛 (𝑁2

𝑁1) 𝑘𝑇)

= −6.63𝑥10−34 𝑥3𝑥108

𝐼𝑛(1.059𝑥10−30) 𝑥1.38𝑥10−23𝑥330

= 6.328x10−7 𝑚



12. A signal with input power 200mW loses 10% of its power after travelling a distance of

3000 m. Find the attenuation coefficient of the fiber. (𝐷𝑒𝑐 2010)

Given: 𝑃𝑖 = 200𝑚𝑊 = 200𝑥10−3𝑤 , 𝑃𝑜 = 90% 𝑃𝑖 = 0.9𝑥200𝑥10−3𝑤 ;

L = 3000 m = 3 km ; 𝛼 =? ,

Using 𝛼= −10

𝐿 log[

𝑃𝑜

𝑃𝑖] = −

10

3 log [

0.9𝑥200𝑥10−3

200𝑥10−3] = 0.1525 𝑑𝐵/𝑘𝑚

13. Calculate on the basis of Einstein’s theory the number of photons emitted per second by

He-Ne laser source emitting light of wavelength 6328Å with an optical power

10mW. (𝐽𝑢𝑛 2010) Given: P=10mW = 10x10−3 𝑤 ,t=1 s , C=3x108m/s ; h=6.63x10−34 Js, ,

λ = 6328 Å = 6328 x 10−10 m ; n = ?


𝜆 , 𝑤𝑒 𝑔𝑒𝑡 𝑛 =

𝜆𝑃𝑡

ℎ𝐶

= 6328 𝑥 10−10 𝑥10𝑥10−3 𝑥1

6.63𝑥10−34 𝑥3𝑥108

= 3.18 x1016 /𝑚3

14. An optical fiber has core R.I 1.5 and R.I of cladding is 3% less than the core index. Calculate

the numerical aperture, angle of acceptance and internal critical acceptance

angle. (𝐽𝑢𝑛 2010)

Given: 𝑛1 = 1.5 ; 𝑛2 =97% 𝑛1 = 0.97 x1.5 =1.455 ; 𝑛0 = 1 ; 𝑁𝐴 = ? ; 𝜃𝑜 = ? & 𝜃𝐶 = ?

Using 1) NA = √𝑛1

2−𝑛22

𝑛𝑜

= √1.52−1.4552

1 = 0.365

2) 𝑠𝑖𝑛 𝜃0= 𝑁𝐴 𝜃𝑜 = 𝑆𝑖𝑛−1(𝑁𝐴)

= 𝑆𝑖𝑛−1(0.365) = 21.41°

3) 𝑠𝑖𝑛 𝜃𝐶 =𝑛2

𝑛1 𝜃𝐶 = 𝑆𝑖𝑛−1 (

𝑛2

𝑛1) = 𝑆𝑖𝑛

−1(1.455

1.5)

= 75.93°

A 5W pulsed laser emits light of wavelength 694 nm. If the duration of each pulse is

20 ns, calculate the number of photons emitted per pulse.( 4 marks

Given: P=5W , λ =694 nm=694x10−9 m ,t=20 ns =20x10−9 s , C=3x108m/s ;

h=6.63x10−34 Js, n =?

Using E=P.t and E =n.ℎ𝐶

𝜆 ,

𝑤𝑒 𝑔𝑒𝑡 n = 𝑃𝑡𝜆

ℎ𝐶

= 5𝑥20𝑥10−9 𝑥694𝑥10−9

6.63𝑥10−34 𝑥3𝑥108

=3.489x1011 photons/pulse.



15. The angle of acceptance of an optical fiber kept in air is 30°.Find the angle of acceptance

when the fiber is in a medium of refractive index 4/3.( 4 marks)

Given: 𝜃𝑜 =30° , 𝑛′𝑜=4/3=1.333 , 𝑛𝑜 = 1 𝑓𝑜𝑟 𝑎𝑖𝑟 , 𝜃′𝑜 =?

Using 𝑛′𝑜𝑆𝑖𝑛 𝜃′𝑜 = 𝑛𝑜Sin𝜃𝑜 =constant (√𝒏𝟏𝟐 − 𝒏𝟐

𝟐 ) for a fibre

We get 𝜃′𝑜 = 𝑆𝑖𝑛−1 (

𝑛𝑜𝑆𝑖𝑛𝜃𝑜

𝑛′𝑜

)

= 𝑆𝑖𝑛−1 (1 𝑆𝑖𝑛30°

1.333)

= 22.03°

16. The ratio of population of two energy levels out of which one corresponds to metastable

state is 1.059 x10−30.Find 𝑡ℎ𝑒 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑒𝑚𝑖𝑡𝑡𝑒𝑑 𝑎𝑡 330𝐾.(CBCS-Jun/Jul16)

Given: 𝑁2

𝑁1 = 1.059x10−30, T =330K; h=6.63x10−34 Js; k=1.38x10−23 J/K ;

C=3x108m/s ; λ = ?

Using 𝑁2

𝑁1 = 𝑒−(ℎ𝐶/𝜆𝑘𝑇)

we get , In(𝑁2

𝑁1) = −(ℎ𝐶/𝜆𝑘𝑇) 𝑆𝑖𝑛𝑐𝑒: 𝑙𝑜𝑔𝑒 (

𝑁2

𝑁1) = In(

𝑁2

𝑁1)

λ = −(ℎ𝐶/𝐼𝑛 (𝑁2

𝑁1) 𝑘𝑇)

= −6.63𝑥10−34 𝑥3𝑥108

𝐼𝑛(1.059𝑥10−30) 𝑥1.38𝑥10−23𝑥330

= 6.328x10−7 𝑚

17. The refractive indices of the core and cladding of a step index fibre are 1.45 and 1.40

respectively and its core diameter is 45μm.Calculate the refractive index change and

numerical Aperture.(CBCS-Jun/Jul16)

Given: 𝑛1 = 1.45 ; 𝑛2 = 1.4 ; ; d = 45μm =45x 10−6 m ; 𝑛𝑜 = 1, ∆ = ? ; & NA = ?

1) ∆ = (𝑛1−𝑛2)

𝑛1

= (1.45−1.4)

1.45 = 0.0345

2) NA = √𝒏𝟏

𝟐−𝒏𝟐𝟐

𝒏𝒐

= √1.452−1.42

1 = 0.3775



19.The average power output of a laser beam of wavelength 6500Å is 10 mW.Find

the number of photon emitted per second by the laser source.

Given: P =10mW = 10x10−3 𝑤 ; t=1 s ; C=3x108m/s ; h=6.63x10−34 Js ;

λ = 6500 Å = 6500 x 10−10 m ; n = ?


𝜆

We get, 𝑛 = 𝜆𝑃𝑡

ℎ𝐶

= 6500 𝑥 10−10 𝑥10𝑥10−3 𝑥1

6.63𝑥10−34 𝑥3𝑥108

=3.268 x 1016 𝑝ℎ𝑜𝑡𝑜𝑛𝑠/𝑚3

20. An optical signal propagating in a fiber retains 85% of input power after

travelling a distance of 500 m in the fiber. Calculate the attenuation coefficient.

Given: 𝑃𝑜 = 85% 𝑃𝑖 = 0.85 𝑃𝑖 ; L = 500m = 0.5 km; 𝛼 =? ,

Using 𝛼 = −10

𝐿 log[

𝑃𝑜

𝑃𝑖]

= −10

0.5 log [

0.85 𝑃𝑖

𝑃𝑖] = −

10

0.5 log [0.85]

= 1.412 dB/km

21. The attenuation of light in an optical fiber is 2 dB/km. What fraction of its

intensity remains after (i) 2 km, (ii) 5 km ?

Given: 𝛼 = 2𝑑𝐵

𝑘𝑚, (

𝑃𝑜

𝑃𝑖)

𝐿=2𝑘𝑚=?, (

𝑃𝑜

𝑃𝑖)

𝐿=5𝑘𝑚=?

Using (𝑃𝑜

𝑃𝑖)

𝐿 = 10

−𝛼𝐿

10

(𝑃𝑜

𝑃𝑖)

𝐿=2𝑘𝑚= 10

−𝛼𝐿

10 = 10−2𝑥2

10 = 0.3981

(𝑃𝑜

𝑃𝑖)

𝐿=5𝑘𝑚= 10

−𝛼𝐿

10 = 10−2𝑥5

10 = 0.1

22. The refractive indices of the core and cladding of a step-index optical fibre

are 1.45 and 1.40 respectively and its core diameter is 45μm. Calculate its

fractional refractive index change and numerical aperture.

Given: 𝑛1 = 1.45 ; 𝑛2 = 1.4 ; ; d = 45μm =45x 10−6 m ; ∆ = ? ; & NA = ?

1) ∆ = (𝑛1−𝑛2)

𝑛1 =

(1.45−1.4)

1.45 = 0.0345

2) NA = √𝒏𝟏

𝟐−𝒏𝟐𝟐

𝒏𝒐

= √1.452−1.42

1

= 0.3775



23. Find the ratio of population of two energy levels in a medium at thermal equilibrium ,

if the wavelength of light emitted at 291 K is 6928 Å . ( 04 Marks)

Given: λ = 6928Å = 6928x10−10 m, t=27 T =291K; h=6.625x10−34 Js;

k=1.38x10−23 J/K ;C=3x108m/s ;𝑁2

𝑁1 = ?

Using 𝑁2

𝑁1 = 𝑒

−(ℎ𝐶

𝜆𝑘𝑇)

= 𝑒−(

6.625𝑥10−34𝑥3𝑥108

6928𝑥10−10𝑥1.38𝑥10−23𝑥291)

= 𝑒−71.44

= 9.419x10−32 or = 9.441x10−32

24. Calculate the numerical aperture and angle of acceptance for an optical fibe

having refractive indices 1.563 and 1.498 for core and cladding respectively. ( 04 Marks)

Given: 𝑛1 = 1.563; 𝑛2 = 1.498 ,𝑛𝑜 = 1 ; 𝑁𝐴 =? a𝑛𝑑 𝜃𝑜 = ?

Using NA = √𝑛1

2−𝑛22

𝑛𝑜

= √1.5632−1.4982

1

= 0.4461

Also 𝜃𝑜 = 𝑆𝑖𝑛−1(𝑁𝐴 ) = 𝑆𝑖𝑛−1(0.4461) = 26.49°

25. A pulsed laser emits photons of wavelength 820 nm with 22 nW average

power/pulse. Calculate the number of photons contained in each pulse,if the

pulse duration is 12 ns. ( 04 Marks)

Given: 𝝀 = 820 nm = 820 x 10−9 m , P = 22 mW = 22 x 10−3W ,

t = 12 ns = 12 x 10−9s , h = 6.625 x 10−34 JS , C = 3 x 108𝑚𝑠−1 , n = ?

Using E = Pt and E = 𝑛 𝑥 ℎ𝐶

𝜆

We get , n =𝜆𝑃𝑡

ℎ𝐶 =

820 𝑥 10−9𝑥 22 𝑥 10−3𝑥 12 𝑥 10−9

6.625 𝑥 10−34𝑥 3 𝑥 108 = 1.089 x 109 electrons/𝑚3



26. A glass clad fiber is made with core glass of refractive index 1.5 and cladding is

doped to give a fractional index difference of 0.0005.Determine the cladding index

and numerical aperture. ( 04 Marks)

Given: 𝑛1 = 1.5 ,∆ =0.0005 ,𝑛𝑜 = 1 , 𝑛2 = ? , 𝑁𝐴 =?

Using ∆ =(𝑛1−𝑛2)

𝑛1

We get 𝑛2 = 𝑛1 − ∆ 𝑛1 = 1.5 – 0.0005𝑥 1.5 = 1.49925 (1.499)

Also using NA =√𝑛1

2−𝑛22

𝑛0 =

√1.52−1.499252

1 = 0.0474 (0.054)

@@end@@

REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM

VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 1

MODULE-4

CRYSTAL STRUCTURE :

• • • • • • • • • • • • • • •

Lattice sites/points

Space lattice :The 3 dimensional periodic array(arrangement) of geometrical points in

Space is called space lattice/crystal lattice. Each point is called lattice point/site which has

identical surroundings.

Basis/Primitive/Translational vectors: A co-ordinate system is used to represent

latticepoints . Three basic vectors , & 𝑐 used to represent lattice points along x,y & z-

axis are called Basis/Primitive Vectors.

Translational vectors are position vector of lattice point in space given by

= 𝑛1+ 𝑛2+ 𝑛3𝑐 , where 𝑛1, 𝑛2 & 𝑛3 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠.

Q: What is Bravais lattice & non-Bravais lattice ?

Bravais lattice –is the lattice in which all the lattice points are equivalent ,

each lattice point represent an identical set of one or more atoms/molecules.

Non-Bravais lattice- is the lattice in which all the lattice points are not equivalent .

Superposition/overlapping of two or more bravais lattice forms non-bravais

lattice.

Q: What is Basis /Pattern?

Unit assembly of identical composition of atoms/molecules is called

Basis/pattern.

Q: What is crystal structure?

Crystal structure is obtained if basis is added to each and every lattice point.

• • • •

• • + = • •

• • • •

Lattice + Basis = Crystal structure



Q: What is unit cell ? and explain types of cells.

Unit cell-is the smallest block/geometrical figure from which the crystal can be

built by repetition of it in three dimensions.

Primitive cell-is the unit cell having lattice points only at the

corners /vertices of it or is the unit cell containing only one lattice point in it.

Non-Primitive cell-is the unit cell which have lattice point within it in addition to

lattice points at the corners /vertices.

Note: All primitive cells are unit cells but all unit cells are not primitive cells.

Q: What are lattice parameters ?

The Six quantities that describe the unit cell completely are called lattice parameters.

Three basis vectors are 𝑎 , b & c and three interfacial angles 𝛼, 𝛽& 𝛾 𝑎𝑟𝑒 𝑐𝑎𝑙𝑙𝑒𝑑

𝑡ℎ𝑒 𝑙𝑎𝑡𝑡𝑖𝑐𝑒 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟𝑠.

Q:What are Miller Indices of a plane ?

Miller Indices are the three smallest integers which represent the position & orientation

of the crystal planes having the same ratio as the reciprocals of the intercepts of the plane

on x,y and z axes.

Q:What are Crystallographic direction and Crystallographic plane ?

Crystallographic direction is the line joining origin to a lattice point(s) and

Crystallographic plane is the plane passing through all lattice points in a particular

crystallographic direction.

Z c 𝛽 𝛼 0 b y a 𝛾 x



Q: Explain the seven crystal systems.

Based on six lattice parameters all the crystals are classified in to seven crystal systems as

follows:-

1) Cubic 2) Tetragonal

3 ) Orthorhombic 4) Triogonal/Rhombohedral

5 ) Hexagonal 6) Monoclinic

7) Triclinic

𝑎≠𝑏 ≠𝑐 c 𝛼 = 𝛽 = 𝛾=90 Types:SC/BaC/BCC/F 𝛽 0 𝛼 b Ex:KNO3, MgSO3 a 𝛾

𝑎=𝑏 =𝑐 a 𝛼 = 𝛽 = 𝛾 ≠90 Types:SC 𝛽 𝛼 Ex: As, Sb, 0 a a 𝛾

𝑎=𝑏 ≠𝑐 𝛼 = 𝛽 =90 , 𝛾=120 c Types: SC Ex:Zn,Mg, 𝛽 0 𝛼 a a 𝛾

𝑎≠𝑏 ≠𝑐 𝛼 ≠ 𝛽 ≠ 𝛾 ≠90 c

Types:SC Ex: 5H2O CuSO4 𝛽 0 𝛼

a 𝛾 b

𝑎≠𝑏 ≠𝑐 𝛼 = 𝛾 =90 ≠ 𝛽 c Types:SC/BaC Ex: CaSO4,2H2O 𝛽 𝛼 b a 0 𝛾

𝑎=𝑏 =𝑐 a 𝛼 = 𝛽 = 𝛾=90 Types: SC/BCC/FCC 𝛽 𝛼 a Ex: Copper/Gold a 0 𝛾

𝑎=𝑏 ≠𝑐 c 𝛼 = 𝛽 = 𝛾=90 Types: SC/BCC Ex: SnO2,TiO2 𝛽 𝛼

0 a a 𝛾



Q:Explain the procedure for finding miller indices.

1, Find the intercepts of the plane along X,Y & Z axes in terms of basis vectors 𝑎,𝑏 &𝑐.

Ex: Let X,Y &Z axes intercepts are 3𝑎,2𝑏 &1𝑐 respectively.

2. Find the coefficients of 𝑎,b &𝑐.

ie: 3,2 & 1.

3. Find the reciprocals of the coefficients of basis vectors.

ie:The reciprocals are 𝟏

𝟑 ,

𝟏

𝟐,

𝟏

𝟏

4. Find the LCM of the denominators .

ie: The LCM of 3,2 &1 is 6.

5. Multiply each reciprocal term by LCM & write the results within the brackets

like(h k l),which gives the miller indices of the plane.

ie:The miller indices are 𝟏

𝟑𝒙 𝟔 ,

𝟏

𝟐𝒙 𝟔,

𝟏

𝟏𝒙 𝟔 = 𝟐, 𝟑, 𝟏 = (𝟐 𝟑 𝟏)

CHARACTERISTICS OF THREE CUBIC LATTICES

Note: 1. For an intercept at infinity, the miller indices is zero (0)

2. For negative intercepts, the corresponding miller index is negative. ex: h or 2 (Read as

‘bar’ h or 2

3. The miller indices ( h k l ) do not represent a single plane but set of parallel planes.



Q:Derive an expression for inter-planar distance/spacing in terms of miller indices.

1. Consider the plane ABC of a crystal which intersect the X,Y & Z axes at A,B & C

respectively.

2. Let ( h k l ) be the miller indices of the family of ABC planes.

3. Draw ‘ON ’ ⊥𝑙𝑟 to the plane ABC ,then ON=𝒅 represent inter-planar distance/spacing.

Then OA= 𝑎

ℎ , OB=

𝑏

𝑘 &OC =

𝑐

𝑙 ,where 𝑎,𝑏& 𝑐 are basis vectors

From right 𝑎𝑛𝑔𝑙𝑒 ∆𝑙𝑒 OAN, Cos 𝜃𝑋 =𝑂𝑁

𝑂𝐴 =

𝑑

𝑎/ℎ =

ℎ𝑑

𝑎 …….(1)

From ∆𝑙𝑒 OBN, Cos 𝜃𝑌 =𝑂𝑁

𝑂𝐵 =

𝑑

𝑏/𝑘 =

𝑘𝑑

𝑏 …….(2) and

also, from ∆𝑙𝑒 OCN, Cos 𝜃𝑍 =𝑂𝑁

𝑂𝐶 =

𝑑

𝑐/𝑙 =

𝑙𝑑

𝑐 ……(3)

For orthogonal co-ordinates 𝑐𝑜𝑠2 𝜃𝑋+𝑐𝑜𝑠2 𝜃𝑌 + 𝑐𝑜𝑠2 𝜃𝑍 = 1 ……(4)

From eqns 1,2,3 &4 we get, ℎ2𝑑2

𝑎2 +

𝑘2𝑑2

𝑏2+

𝑙2 𝑑2

𝑐2

= 1

ie: 𝑑2 [ ℎ2

𝑎2 +

𝑘2

𝑏2+

𝑙2

𝑐2 ] = 1

∴ d = 𝟏

√ 𝒉𝟐

𝒂𝟐 + 𝒌𝟐

𝒃𝟐+ 𝒍𝟐

𝒄𝟐

For a cubic crystal 𝑎=𝑏 =𝑐

∴ d= 𝒂

√ 𝒉𝟐 +𝒌𝟐 +𝒍𝟐

Z

C

N B Y

𝜃𝑧 𝜃𝑦

0 𝜃𝑥

A X



Q:Derive an expression for space lattice constant ‘𝑎’ for a cubic lattice.

Let 𝜌 𝑏𝑒 𝑡ℎ𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 & M the molecular weight of a cubic crystal of lattice constant’𝑎’.

Then, volume of the unit cell = 𝑎3 and mass of unit cell =𝜌 𝑎3 …….(1)

If n be the number of molecules per unit cell and 𝑁𝐴 be the Avagadro’s number.

Then, mass of each molecule = (M/𝑁𝐴) and

mass of molecules in the unit cell=n (M/𝑁𝐴)….(2)

From eqns 1& 2,we get 𝜌 𝑎3 = n (M/𝑁𝐴) 𝒂𝟑 =𝒏𝑴

𝝆𝑵𝑨

Q:Find the relation between atomic radius (R) and lattice constant(𝑎) for SC ,BCC & FCC

crystals:-

a) Simple cubic(SC):

If ‘ r ’ be the rdius and ‘ 𝑎 ‘ be the lattice constant/side of the unit cell,

then from the diagram 𝑎=2R 𝑎/2 = R

a) Body centered cubic (BCC) :

In a bcc cube AB=BC=CD=𝑎 and

AC=√𝑎2 + 𝑎2=√2𝑎

Also from ∆𝑙𝑒 ACD , 𝐴𝐷2 = 𝐴𝐶2 + 𝐶𝐷2

ie: (4𝑅)2 = (2𝑎)2 +𝑎2

= 3𝑎2

4R = √3𝑎

𝑂𝑟 𝑅 =√3𝑎

4

a

C

a B

D

4R

A

AA A

a

2 R



c) Face centered cubic (FCC)

In a fcc cube AB=BC=𝑎 and AC=4R

Also from ∆𝑙𝑒 ABC,

𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2

ie: (4𝑅)2 = (𝑎2 +𝑎2

= 2𝑎2

Or 4R =√2𝑎

Or R = √2𝑎

4 =

𝑎

2√2

Q:Find the number of atoms in unit SC.BCC & FCC cell.

SC : In SC cell each atom is shared by 8 neighbour cells & hence no of atoms per unit cell

due to Corner atoms = 1

8 x 8 =1

BCC: no of atoms per unit bcc cell= 1

8 x 8 corner atoms + one atom at the body = 1

FCC: no of atoms per unit fcc cell = 1

8 𝑥 8 𝑐𝑜𝑟𝑛𝑒𝑟 𝑎𝑡𝑜𝑚𝑠 +

1

2 x 6 face atoms=1+3=4

Q: What is meant by Co-ordination number ? & mention the co-ordination number of SC,

BCC & FCC cells.

Co-ordination number of any atom in a crystal is the number of nearest equidistant

neighbouring atoms Surrounding that atom.

SC: Co-ordination number of SC =6

BCC: Co-ordination number of bcc = 8

FCC: Co-ordination number of fcc =12

a

B

a

C

4R

A



SC BCC FCC

Q:What is meant by Atomic Packing Factor(APF) ?

Atomic packing factor is defined as the fraction of the volume of the unit cell occupied

by the atoms present in the unit cell.

ie: Atomic packing factor = (𝑛𝑜 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙)𝑥𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑎𝑡𝑜𝑚

(𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙)

ie: APF = 𝑛

𝑎3 𝑥

4

3 𝜋𝑅3 where a = Lattice constant & R = Radius of the atom

Q: Calculate the atomic packing factors of SC,BCC & FCC cells.

For SC, n =1, R = 𝒂

𝟐

Using, APF = 𝑛

𝑎3 𝑥 4

3 𝜋𝑅3

= 1

𝑎3 𝑥 4

3 𝜋(

𝑎

2 )3

= 1

𝑎3 𝑥 4

3 𝜋

𝑎3

8

= 𝜋

6 = 0.52 or 52%

For BCC , n = 2, R = √𝟑𝒂

𝟒

Using, APF = 𝑛

𝑎3 𝑥 4

3 𝜋𝑅3

= 2

𝑎3 𝑥 4

3 𝜋(

√3𝑎

4 )3

= 2

𝑎3 𝑥 4

3 𝜋

3√3𝑎3

64

= √3𝜋

8= 0.68 or 68%

For FCC, n = 4, R = 𝑎

2√2

Using, APF = 𝑛

𝑎3 𝑥 4

3 𝜋𝑅3

= 4

𝑎3 𝑥

4

3 𝜋(

𝑎

2√2 )3

= 4

𝑎3 𝑥 4

3 𝜋

𝑎3

16√2

= 𝜋

3√2 = 0.74 or 74%

•

•

•



NOTE: Important data related to SC , BCC & FCC cells.

Sl.

No

Particulars Simple Body

Centere

Face

Centered

1 Volume of a unit cell 𝑎3 𝑎3 𝑎3

2 No. of atoms per cell 1 2 4

3 Co-ordination number

(No. of nearest neighbours)

6 8 12

4 Nearest neighbour distance

(2R)

𝑎 (√3/2)𝑎 𝑎/√2

5 Packing Faction 0.52 0.68 0.74

Q:Explain the structure of diamond crystal.

Note: = 8 corner atoms, = 6 face atoms & = 4 body diagonal atoms

OR

1. Diamond is formed by two intervening FCC sub-lattices one of them is moved by 𝑎

4

along the body diagonal as shown in the diagram.

2. The origin of one FCC is ( 0,0,0) and the origin of the other FCC is (a/4,a/4,a/4)

3. There are 8 atoms at the corners, 6 atoms at the faces and 4 atoms along the

diagonals.

4. Each diagonal atom form bond with One nearest corner atom and Three nearest face

atoms.

5. The coordination number of diamond is 4.

6. Total atoms in unit diamond cell = 1

8 x 8+

1

2 x 6+4 =8

7. If R be the radius and a the lattice constant ,then we can show that

O 1/2

O

1/2 O 1/2

O 1/2 O

3/4 1/4

1/4 3/4

B 2R A

𝑎

4



AB=2R=√𝑎2

16 +

𝑎2

16+

𝑎2

16

= √3𝑎2

16

= √3𝑎

4 or

R= √𝟑𝒂

𝟖

For Diamond n = 8 & R = √𝟑𝒂

𝟖

∴ Using APF = 𝑛

𝑎3 𝑥

4

3 𝜋𝑅3

= 8

𝑎3 𝑥 4

3 𝜋(

√3𝑎

8 )3

= 8

𝑎3 𝑥 4

3 𝜋

3√3𝑎3

8𝑥8𝑥8

= 𝜋√3

16

= 0.34 or 34%

Q: What is perovskite ? explain.

Perovskie is the common name for all the oxides of type AB𝑶𝟑,where A & B

are different metals.

Ex: 𝐵𝑎𝑇𝑖𝑂3 ( 𝐵𝑎𝑟𝑖𝑢𝑚 𝑡𝑖𝑡𝑎𝑛𝑎𝑡𝑒) , 𝐶𝑎𝑇𝑖𝑂3( calcium titanate) etc

The common structure of perovskite is as follows:

= 8 Ba/Ca atoms form are at the edges of the cell.,

= 6 𝑂3atoms are at the face of the cell and

= 1 𝑇𝑖 atom at the body centre of the cell.

Perovskites exhibit superconductivity.

Perovskies exhibit both piezo-electric & ferro-electric properties .

Perovskies are used as dielectric in capacitors.

Perovskies are used as piezo-electric in microphone.

Type II superconductors have Perovskite structure.



Q: Explain Allotrophy and Polymorphism.

Allotrophy is the property of the material by the virtue of which it can have more

than one type of crystal structures.

All the structures will have the same chemical properties and different physical properties

Carbon,Sulphur ,Phosphorous exhibit allotrophy.

Ex: 1.Diamond and graphite are the allotrophic forms of carbon.

2. Diamond is hard and good insulator .

3. Graphite is soft and good conductor.

Polymorphism is the ability of the materials to crystallize in several solid phases that

posses different lattice structures at different temperatures.

Ex: Iron ( Alpha Iron)exhibit BCC structure from room temperature to 910.

Iron ( Gamma Iron)exhibit FCC structure from 910 to upto 400

Iron ( Delta Iron)exhibit BCC structure from 1400 to 1540( melting point.

All these phases are reversible.

Q:Derive Bragg’s law of diffraction.

Let the incident parallel x-rays of wavelength λ, incident on two atomic planes of

Separation d at glancing angles 𝜃 and reflect along BC and EF respectively.

Draw BG & BH ⊥𝑙𝑟 to DE & EF respectively.

From the ∆𝑙𝑒 BGE & BHE we have GE=BE.Sin𝜃 & 𝐻𝐸 = 𝐵𝐸. 𝑆𝑖𝑛𝜃

where BE = d ,interplanar distance.

∴ 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒, 𝑝𝑑 = 𝐺𝐸 + 𝐻𝐸

= 𝑑 𝑆𝑖𝑛𝜃 + 𝑑 𝑆𝑖𝑛𝜃

= 2𝑑 𝑆𝑖𝑛𝜃 … . (1)

Also for constructive interference pd= nλ ……(2)

∴ 𝑓𝑟𝑜𝑚 𝑒𝑞𝑛𝑠 1 &2 , 𝑤𝑒 𝑔𝑒𝑡 nλ = 𝟐𝒅 𝑺𝒊𝒏𝜽 ,

Incident rays -AD A C reflected rays-CF

Glancing angle

D 𝜃 𝜃 F

₀ ₀ B ₀ ₀ ₀

G 𝜃 𝜃 H d = Interplanar spacing

₀ ₀ ₀ E ₀ ₀



This equation is called the Bragg’s law

Q:Describe the construction & working of Bragg’s X-ray spectrometer.

The labeled schematic diagram of the Bragg’s X-ray Spectrometer is as shown in the

diagram.

To verify Bragg’s law:

1. Collimate the X-rays from X-ray tube by passing them through slits S1 & S2.

2. Allow the collimated X-rays to incident on the surface of a given crystal mounted

vertically at the centre of a turn table at a glancing angle 𝜃

3. The position of the crystal is noted using the vernier V1 and horizontal circular

scale.

4. Adjust the ionization chamber to receive the scattered X-rays and measure the

ionization current (I)which is a measure of intensity of the X-rays using electrometer E.

5. Repeat the experiment as before by increasing the glancing angle 𝜽 gradually and

note the corresponding ionization current each time.

6. Plot a graph of ‘I‘v/s ‘𝜃′ 𝑎𝑛𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ note down the glancing angles 𝜃1, 𝜃2& 𝜃3

for the 1st,2nd &3rd order diffraction respectively.

7 Then substituting for 𝜃1, 𝜃2& 𝜃3 in the Bragg’s equation 2d Sin𝜃 = 𝑛𝜆 ,

It will be found that ,Sin𝜽𝟏: 𝑺𝒊𝒏𝜽𝟐: 𝑺𝒊𝒏𝜽𝟑 = 𝟏: 𝟐: 𝟑,which verifies Bragg’s law.

To find the nature of the crystal:

1. Collimate the X-rays from X-ray tube by passing them through slits S1 & S2.

2. Allow the collimated X-rays to incident on the surface of a given crystal mounted

vertically at the centre of a turn table at a glancing angle 𝜃. The position of the

Vernier- 𝑉1 Circular scale

Slits glancing angle-𝜃 Crystal target

𝑆1 𝑆2

X-rays Vernier-𝑉2

Collimated X-rays

𝑆3

𝑆4 E -electrometer

Ionisation chamber (D) Ba

Intensity

0 𝜃

𝜃1 𝜃2 𝜃3



crystal is noted using the vernier V1 and horizontal circular scale

3. Adjust the ionization chamber to receive the scattered X-rays and measure the

ionization current (I) produced by the X-rays using electrometer E.

4. Adjust 𝑑100 plane of the crystal to the incident X-rays and find the glancing angle

𝜃1 for 1st order Spectrum.

5. Repeat the experiment as before for the crystal surfaces 𝑑110 & 𝑑111 and find

the corresponding 1st order glancing angles 𝜃2 & 𝜃3 .

6. Then using bragg’s equation, 2d Sin𝜃 = 𝑛𝜆 ,

find 𝒅𝟏𝟎𝟎 ∶ 𝒅𝟏𝟏𝟎 ∶ 𝒅𝟏𝟏𝟏 = 𝟏

𝑺𝒊𝒏𝜽𝟏 :

𝟏

𝑺𝒊𝒏𝜽𝟐:

𝟏

𝑺𝒊𝒏𝜽𝟑

7. The nature of the crystal is identified from the ratios of 1

𝑆𝑖𝑛𝜃1 :

1

𝑆𝑖𝑛𝜃2:

1

𝑆𝑖𝑛𝜃3

as follows:

For SC; 𝟏


𝟏


𝟏

𝑺𝒊𝒏𝜽𝟑 = 1:

𝟏

√𝟐 :

𝟏

√𝟑

For BCC; 𝟏


𝟏


𝟏


𝟐

√𝟐 :

𝟏

√𝟑

For FCC; 𝟏


𝟏


𝟏


𝟏

√𝟐 :

𝟐

√𝟑

PROBLEMS SECTION

MODULE-4 : CRYSTAL STRUCTURE

Formulae needed : nλ = 2dSin𝜃 ; d= 𝑎

√ℎ2 +𝑘2 +𝑙2 ;

For SC ,n=1& a = 2R , BCC, n=2 & a = 4

√3 R , FCC , n=4 & a = 2√2 𝑅

Atomic Packing Factor (APF) = 𝑛

𝑎3 x

4𝜋𝑟3

3

1. Copper has fcc structure with atomic radius 0.127nm.Calculate the inter-planar

spacing for (3 2 1) plane.( JAN2015)

Given: r = 0.127 nm = 0.127x10−9𝑚 , (ℎ 𝑘 𝑙 ) = ( 3 2 1),

𝑓𝑜𝑟 𝑓𝑐𝑐 𝑎 = 2√2 𝑟 = 2√2 x 0.127x10−9𝑚 , 𝑑 = ?

Using d = 𝑎

√ℎ2 +𝑘2 +𝑙2

= 2√2 𝑥 0.127𝑥10−9

√32 +22 +12

= 9.60 x 10−11 𝑚



2. 𝐷𝑟𝑎𝑤 the following planes in a cubic unit cell: i) (2 0 0) ii) (2 1 0)iii)(1 3 2)

3. The minimum order of Bragg’s reflection occurs at an angle of 20° in the plane

(2 1 2 ).Find the wavelength of X-rays if lattice constant is 3.614Å. (Dec 2013/Jan2014)

Given: n = 1 ; 𝜃 = 20° ; ( h k l ) = ( 2 1 2 ) ; a = 3.614Å = 3.614x10−10 m ; λ = ?

Using nλ = 2dSin𝜃 & d= 𝑎

√ℎ2 +𝑘2 +𝑙2

We get , λ = 2𝑎 𝑆𝑖𝑛𝜃

𝑛√ℎ2 +𝑘2 +𝑙2

= 2𝑥3.614𝑥10−10 𝑆𝑖𝑛20°

1𝑥√22 +12+22 = 8.240 x 10−11 m

4. Monochromatic X-rays of wavelength 0.82 Å undergo first order reflection from a crystal

of cubic lattice with lattice constant 3 Å at a glancing angle of 7.855°. Identify the possible

planes which give rise to this reflection in terms of Miller indices. (𝐷𝑒𝑐 11 & 𝐽𝑢𝑛 − 𝐽𝑢𝑙16)

Given: n = 1 ; 𝜃 = 7.855° ; a = 3 Å = 3x10−10 m ; λ = 0.82 Å =0.82 x 10−10 m ; ( h k l ) = ?


√ℎ2 +𝑘2 +𝑙2

We get , √ℎ2 + 𝑘2 + 𝑙2 = 2𝑎 𝑆𝑖𝑛𝜃

𝑛𝜆

= 2𝑥3𝑥10−10 𝑆𝑖𝑛(7.855)

1𝑥0.82 𝑥 10−10 = 1

Thus the possible planes ( h k l ) =( 1 0 0 ) ,(0 1 0 ) ,( 0 0 1)

(2 0 0) 𝐱 =

𝟏

𝟐, 𝐲 = ∞ & 𝑧 = ∞

Z ½ 0 y X

(2 1 0) 𝐱 = −𝟏

𝟐, , 𝐲 = 𝟏 & 𝑧 = ∞

Z ½

0 1 Y

X

(1 3 2 ) 𝐱 = 𝟏, 𝐲 = −𝟏

𝟑 & 𝑧 =

𝟏

𝟐

Z ½

−1

3

0 Y 1 X



5. Calculate the glancing angle of the (1 1 0) plane of a simple cubic crystal (a=2.814 Å )

corresponding to second order diffraction maximum for the x-rays of wavelength

0.710 Å.

Given: n = 2 ; ( h k l ) = (1 1 0 ) ; a = 2.814Å = 2.814x10−10 m ; λ = 0.710

Å =0.710 x 10−10 m ; 𝜃 =?


√ℎ2 +𝑘2 +𝑙2

We get , 𝜃 = 𝑠𝑖𝑛−1( 𝑛𝜆√ℎ2 +𝑘2 +𝑙2

2𝑎 )

= 𝑠𝑖𝑛−1(2𝑥0.710 𝑥 10−10𝑥√12 +12 +0

2𝑥2.814𝑥10−10 ) = 20.91°

6. Draw the following planes in the unit cube: i) ( 1 0 2 ) ii) ( 1 1 2 ). (𝐷𝑒𝑐 2010)

Ans: i) ( 1 0 2 ) 𝑥 = −1, 𝑦 = ∞ & 𝑧 =1

2 & ii) (1 1 2 ) 𝑥 = −1, 𝑦 = 1 & 𝑧 = −

1

2

7. A monochromatic X-ray beam of wavelength 1.5Å undergoes second order Bragg

reflection from the plane (2 1 1) of a cubic crystal, at a glancing angle of 54.38°,calculate

the lattice constant. (𝐽𝑢𝑛 2010)

Given: n = 2 ; ( h k l ) = (2 1 1 ) ; λ = 1.5 Å =1.5 x 10−10 m ; 𝜃 = 54.38° ; 𝑎 = ?


√ℎ2 +𝑘2 +𝑙2

We get , 𝑎 = 𝑛𝜆√ℎ2 +𝑘2 +𝑙2

2𝑆𝑖𝑛𝜃

= 2𝑥1.5 𝑥 10−10 𝑥√22 +12 +12

2𝑥𝑆𝑖𝑛 54.38

= 4.52 x 10−10 𝑚

( 1 0 2 ) 𝒙 = −𝟏, 𝒚 = ∞ & 𝑧 =𝟏

𝟐

Z

½ −1

0 Y

x

(1 1 2 ) 𝒙 = −𝟏, 𝒚 = 𝟏 & 𝑧 = −𝟏

𝟐

Z −1

0 1 Y

X(−1

2)



8. In a calcite crystal, second order Bragg’s reflections occur from the planes with d-spacing

3 Å,at a glancing angle of 24° .Calculate the path difference between X-rays reflected from

the two adjacent planes. Also calculate the wavelength of x-rays.( 4 marks)

Given: d = 3 Å =3x10−10 𝑚 , 𝜃 = 24° , n = 2 ,Pd = ? and λ = ?

Path difference,Pd = 2d Sin𝜃 =2x3x10−10 xSin24° =2.44 x10−10 m

Also, λ = 𝑃𝑑

𝑛

= 2𝑑𝑆𝑖𝑛𝜃

𝑛

= 2.44 𝑥10−10

2

= 1.22 x10−10 m

9. The atomic radius of gold is 0.144nm.Determine the interplanar distance for ( 1 1 0)

planes assuming that gold belongs to FCC system.( 4 marks)

Given: r = 0.144 nm =0.144x10−9 m , (h k l)=(1 1 0),

For FCC a =2√2 𝑟 =2√2 𝑥0.144𝑥10−9 𝑚 , d = ?

Using d= 𝑎

√(ℎ2+𝑘2+𝑙2 )

= 2√2 𝑥0.144𝑥10−9

√(12+12+02 )

= 2.88 x 10−10 𝑚

10. Calculate the glancing angle for incidence of of X-rays of wavelength 0.058 nm on the

plane o (1 3 2)of NaCl which results in 2nd order diffraction maxima taking the lattice

spacing as 3.81 Å.

Given: n = 2 ; ( h k l ) = (1 3 2 ) ; a = 3.81Å = 3.81x10−10 m ; λ = 0.058nm=0.058 x 10−9 m

; 𝜃 =?


√ℎ2 +𝑘2 +𝑙2

We get , 𝜃 = 𝑠𝑖𝑛−1( 𝑛𝜆√ℎ2 +𝑘2 +𝑙2

2𝑎 )

= 𝑠𝑖𝑛−1(2𝑥0.058 𝑥 𝟏𝟎−𝟗𝑥√12 +32 +22

2𝑥3.81𝑥10−10 )

= 34.72° ( also d = 1.018 X 10−10 m)



11. Draw the following planes in a cubic unit cell.

i) (2 0 0) ii) (Ι Ī 0) iii) (Ι 0 2) iV) (Ι Ι 2)

12. An X-ray beam of wavwlength 0.7 Å undergoes first order Bragg’s reflection

from the plane ( 302) of a cubic crystal at glancing angle 35° , 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆 𝒕𝒉𝒆

lattice constant. ( 04 Marks)

Given: n = 1 ; ( h k l ) = ( 302 ) ; λ = 0.7 Å =0.7 x 10−10 m ; 𝜃 = 35° ; 𝑎 = ?


√ℎ2 +𝑘2 +𝑙2

We get , 𝑎 = 𝑛𝜆√ℎ2 +𝑘2 +𝑙2

2𝑆𝑖𝑛𝜃

= 1𝑥0.7 𝑥 10−10 𝑥√32 +02 +22

2𝑥𝑆𝑖𝑛 35

= 2.200 x 10−10 𝑚 [ Note: d = 6.102x10−11m]

(102)⇒X = 1,Y = ∞ & 𝑍 = ½ Z

½

0 Y

X 1

(112)⇒X = 1,Y = 1 & 𝑍 = ½

Z

½

0 1 Y

X 1

(Ι Ī 0)⇒X = 1 ,Y = −1 & 𝑍 = ∞

z

0 y

1

x

−1 0 Y

1

X

(200)⇒X = ½ ,Y = ∞ & 𝑍 = ∞ Z

0 Y 1/2

X



13.Draw the crystal planes ( 102 ) ( 111 ) ( 011 ) and ( 002 ) in a cubic crystal. ( 04 Marks)

14. Find the Miller indices of a set of parallel planes which make intercepts in th

ratio 3a : 4b and parallel to z-axis and also calculate the interplanar distance of the

planes taking thr lattice to be cubic with a = b = c = 2 Å. ( 04 Marks)

Given: Intercepts 3a:4b: ∞𝑐 , a= 𝑏 = 𝑐 =2Å=2 x 10−10m , (h k l ) =? ,d = ?

Coefficients of a,b and c are 3 ,4, ∞

Reciprocals of the coefficients are 1

3 ,

1

4 ,0

LCM 0f the denominators 3 & 4 is 12

Miller indices are 1

3𝑥12 = 4 ,

1

4𝑥12 = 3 , 0 x 12= 0

∴ ( h k l ) = ( 4 3 0 )

𝐴𝑙𝑠𝑜 𝑤ℎ𝑒𝑛 𝑎 = 𝑏 = 𝑐

We have d =𝑎

√ℎ2+𝑘2+𝑙2 ∴ d =

2 𝑥 10−10

√42+32+02 = 4 x 10−11m

(102) X =1/1 =1, Y =1/0 = ∞ and Z =1/2

Z

1/2

0 Y

1

X

(111) X =1/1 =1, Y =1/1 =1, and Z 1/1 =1,

Z 1

1

0 Y

X 1

(002) X =1/0 = ∞, Y =1/0 = ∞ and Z =1/2

Z

½

0 Y

X

(011) X =1/0 = ∞, Y =1/1 =1and Z =1/1 = 1

Z 1

0 1 Y

X



14. Draw the following planes in a cubic unit cell,

i) ( 0 0 I ) ii) ( I ī 0 ) iii) ( I I 2) iV) ( 0 2 0 ) ( 04 Marks)

@@end@@

(112) x =1/1= 1, y=1/1=1 & z =1/2

z

½

0 1 y

X 1

(020) x =1/0= ∞, y =1/2 & z = 1/0 = ∞

Z

0 ½ y

x

( I ī 0 ) x =1/1= 1, y =1/-1=−1 & z =1/0 =∞

z

0 1 y

X 1

(001) x =1/0= ∞, y=1/0=∞ & z =1/1=1

Z 1

0 y

X


VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 1

MODULE-5

SHOCK WAVES AND SCIENCE OF NANO MATERIALS:

SHOCK WAVES

Q:What are shock waves ? Explain.

Shock waves are the waves produced due to the sudden release/dissipation of energy.

Shock waves are the waves in which the pressure, density and temperature changes

are large

Ex: Shock waves are produced during the burst of crackers, Explosion of

dynamite/bombs, Volcanic eruptions, etc.

Q:Mention different types of shock waves.

There are FOUR types of shock waves namely :-

Stationary shock waves,

Moving shock waves,

Normal shock waves and

Oblique shock waves,

Q:Mention methods of producing shock waves.

Shock waves can be produced by the following methods ,namely By detonation

of crackers/explosives, by volcanic eruptions, supersonic objects/waves, by

Reddy shock tube in the laboratory.

Q:What are Acoustic waves? Mention the types of acoustic waves.

Acoustic waves are the longitudinal waves which travel with the speed of sound

in a medium (Solid/liquid/gas)

Acoustic waves are classified in to THREE types namely:

1. Infrasonic waves(Infrasonics) are the Acoustic waves of frequency less than 20 Hz.

2. Audible waves are the Acoustic waves of frequency between 20 Hz and 20 kHz.

3. Ultrasonic waves(Ultrasonics) are the Acoustic waves of frequency greater

than 20 kHz. (Elephants detect Infrasonics, Human ear detect Audible waves &

Bats/Dogs detec Ultrasonics)



Q: Define Mach number, subsonic waves supersonic waves and Mach angle.

1. Mach Number(M) is the ratio of the speed of an object (V) through a fluid to the speed of

sound(a) in the fluid at that point. Mathematically, M = 𝑽

𝒂 , M is dimensionless quantity.

2. Subsonic waves are the mechanical waves whose speed is less than that of sound in the

same medium. mach number of Subsonic waves is less than 1.

Ex: Motor cycle, Bus, Train , aeroplanes etc produce subsonic waves.

3. Supersonic waves are the mechanical waves whose speed is greater than that of sound in

the same medium. for which the mach number of Supersonic waves is greater than 1.

Ex: Fighter planes, Rockets, Missiles,tornedo etc produce supersonic waves

4. Mach angle(𝝁) is the half the angle of cone of sound waves formed and is given by

𝝁 = 𝑺𝒊𝒏−𝟏 (𝟏

𝑴)

Q: Explain the basic conservation laws and Rankine-Hugonite/Normal shock wave

Relations.

There are three basic conservation laws namely Conservation of mass, Conservation

of momentum and Conservation of energy.

1. Conservation of mass states that the total mass of the system always remains constant as

the mass can neither be created nor destroyed.

Mathematically, 𝜌𝑣 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 or

∴ 𝜌1𝑣1 = 𝜌2𝑣2 , Where 𝜌1, 𝜌2densities & 𝑣1 , 𝑣2 velocities.

2. Conservation of momentum states that the sum total momentum of the system always

remain constant.

Mathematically,𝑃 + 𝜌𝑣2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑟 𝑃1 + 𝜌1𝑣12 = 𝑃2 + 𝜌2𝑣2

2

Where 𝑃1, 𝑃2 pressures , 𝑣1, 𝑣2 velocities and 𝜌1, 𝜌2densities.

3. Conservation of energy states that the sum total energy of a system is always remains

constant.

Mathematically, ℎ +𝑣2

2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑟 ℎ1 +

𝑣12

2= ℎ2 +

𝑣22

2

Where ℎ1, ℎ2 enthalpies and 𝑣1, 𝑣2 velocities.



Q: RH equations or Normal shock wave equations.

RH equation /Normal shock wave equations are derived from basic conservation laws

,which relate the temperature ratio and density ratio in terms of pressure ratio for normal

shock waves.

The pressure ratio in terms of upstream Mach number M is given by 𝑷𝟐

𝑷𝟏 =

𝟐𝜸

𝜸+𝟏𝑀2 −

𝜸−𝟏

𝜸+𝟏 …….(1)

But temperature ratio, 𝑻𝟐

𝑻𝟏 = (

𝟏

𝑀2+

𝜸−𝟏

𝟐) (

𝟐𝜸

𝜸−𝟏 𝑀2 − 1) (

𝟐(𝜸−𝟏)

(𝜸+𝟏)𝟐) ……..(2)

Substituting for 𝑀2 𝑓𝑟𝑜𝑚 𝑒𝑞𝑛 1, 𝑖𝑛 𝑒𝑞𝑛 2, 𝑤𝑒 𝑔𝑒𝑡

𝑻𝟐

𝑻𝟏 =

[𝟏+𝜸−𝟏

𝜸+𝟏

𝑷𝟐𝑷𝟏

]𝑷𝟐𝑷𝟏

𝑷𝟐𝑷𝟏

+𝜸−𝟏

𝜸+𝟏

…..(3) ∵ 4ab+(𝑎 − 𝑏)2 = (𝑎 + 𝑏)2

Also , density ratio, 𝝆𝟐

𝝆𝟏 =

𝑷𝟐𝑷𝟏𝑻𝟐𝑻𝟏

…….(4)

From eqns 1,3 &4,we get,

𝝆𝟐

𝝆𝟏

=

𝑷𝟐

𝑷𝟏

[𝟏 +𝜸 − 𝟏𝜸 + 𝟏

𝑷𝟐

𝑷𝟏]

𝑷𝟐

𝑷𝟏𝑷𝟐

𝑷𝟏+

𝜸 − 𝟏𝜸 + 𝟏

= (

𝜸+𝟏

𝜸−𝟏

𝑷𝟐𝑷𝟏

+𝟏

𝜸+𝟏

𝜸−𝟏+

𝑷𝟐𝑷𝟏

) ….(5)

eqns 3 & 5 are called RH relations/NS relations.



Q:What is a (Reddy)shock tube ? Describe the construction and working of simple

Reddy shock tube .

1. Reddy Shock tube is a device used to produce and study shock waves in the laboratory.

2. Schematic labeled diagram of the original Reddy shock tube is as shown in the diagram.

Construction :

1. RST consists of a steel tube of length 100 cm and diameter 2.9 cm.

2. A diaphragm of thickness 0.1cm divides the tube in to two compartments of length 49 cm

fitted with piston called Driver section filled with driver gas. The other compartment of

length 51 cm is called Driven section filled with driven gas.

3. Sensor S fitted to driver section measures the rupture pressure 𝑃2,temparature𝑇2.

4. Two sensors 𝑆1 & 𝑆2 separated by a distance ∆𝑋 fitted to driven section measures the

pressures 𝑃4,𝑃5 and temperatures 𝑇4, 𝑇5 respectively.

Working :

1. Driver section is filled with gas at high pressure and Driven section is filled with gas of

low pressure 𝑃1 & 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑇1 .

2. Diaphram is ruptured to produce shock waves by pushing the piston and the rupture

pressure 𝑃2 & temperature is measured using sensor S.

3. The time’𝒕′ taken by the shock wave to travel the distance ‘x’ is measured using CRO. also

the pressures 𝑃4,𝑃5 and temperatures 𝑇4, 𝑇5 are measured using the sensors 𝑆1 & 𝑆2

respectively.

4. The speed of the shock waves is calculated using V = 𝑥

𝑡 .

5. Then the match number of the shock waves is calculated using M = 𝑉

𝑎 ,where a is the

speed of sound at temperature 𝑇1.

6. Also the mach number M can be calculated using the RH relations 𝑷𝟐

𝑷𝟏 =

𝟐𝜸

𝜸+𝟏𝑀2 −

𝜸−𝟏

𝜸+𝟏 , by finding 𝑃2, 𝑃1 for the gas of known 𝛾.

Piston Diaphragm S 𝑆1( 𝑃4) 𝑠𝑒𝑛𝑠𝑜𝑟𝑠 𝑆2(𝑃5)

𝑃2 𝑃1 x =7 cm

Driver section/gas Driven section/gas 2.9cm

Plunger

49 cm 51 cm



Q: Uses of Shock waves.

1. Shock waves (SW)are used in the treatment of kidney stones.

2. SW are used in the pencil industry for softening of pencil wood and painting.

3. Sw are used in the extraction of sandal wood.

4. Sw sre used to rejunevate/activate dried bore wells.

5. Sw are used for needleless drug delivery.

6. Sw are used to push DNA in to the cell.

7. SW are used for the treatment of orthopedic diseases.

SCIENCE OF NANOMATERIALS:

Q:What are nano-particles?

Nano-particles are the material particles whose size is in the range of 1nm to 100 nm

and their properties are size dependant.

Q:What is mesoscopic state?

Mesoscopic state is the size/state of the matter at which its physical properties

changes and becomes size dependent.

Q:Explain different structures(3D,2D,1D & 0D) based on their energy density

graphically.

3D structure/Bulk metal :

1. The 3D structure is as shown in the diagram, which have all the 3 dimensions.

2. The density of states for 3D structure is given by g(E) dE = 𝟖√𝟐𝝅𝒎𝟑/𝟐𝑬𝟏/𝟐

𝒉𝟑 𝑑𝐸

Where m = mass, E = energy, h = Planck’s constant

3. The density of states g(E) increases with energy E is as shown in the graph

4. The electrons are not confined to any direction.

g(E)

0 E



2D structure/Quantum Film/Well:

.

1. Quantum film/well is obtained when 3D structure reduced to nano scale in one dimension

as shown in the diagram.

2. The density of states for quantum well is given by g(E) 𝑑𝐸 = 4𝜋𝑚

ℎ2 dE where m = mass,

h =Planck’s constant.

3. For each quantum state the density of states is constant.

4. The variation of g(E) with E is as shown in the graph.

5. The electrons are confined to 1 direction.

1D structure/Quantum Wire :

0

1. Quantum wire is obtained when 3D structure reduced to nano Scale in two dimensions as

shown in the diagram.

2. The density of states for quantum well is given by g(E)dE = 2√2𝑚1/2𝐸−1/2

ℎ dE

where m = mass, E = energy, h = Planck const.

3. The variation of g(E) with E is as shown in the graph

4. The electrons are confined to 2 directions.

g(E)

0 𝐸1𝐸2𝐸3 𝐸

g(E)

0 𝐸1 𝐸2 𝐸3 𝐸



0D structure /Quantum Dot :

0

1. Quantum Dot is obtained when 3D structure reduced to nano scale all the 3 dimensions

2. The variation of g(E) with E is as shown in the graph.

3. The electrons are confined to all the 3 directions

4. and g(E) has discrete structure as shown.

Q: Explain the TWO approaches/methods adopted to obtain nanoparticles.

Nanoparticals

TOP DOWN approach

1. In the TOP DOWN approach ,bulk material is crushed in to small pieces.

2. Crushed pieces are further grinded in to smaller and smallers pieces.

3. Crushing and grinding process is continued till nanoparticles are obtained.

BOTTAM UP approach

1. Basic atoms/molecules are grouped to form clusters/globules.

2. The clusters /globules are further grouped in to nanoparticles.

g(E)

0 𝐸1 𝐸2 𝐸4 𝐸

Top Down method

Bulk material

Powder

Nanoparticals

Bottom Up method

Nanoparticals

Clusters /Groups

Atoms/molecules



Q: Explain Ball Mill method of synthesis/producing nanoparticles.

1. Ball mill method is a mechanical method based on “ Top Down ” approach to

synthesis nanoparticles of metals and alloys on large scale.

2. It consists of a steel container filled with the material whose nanoparticles to be produced

along with large number of heavy steel balls.

3. The container is capable of rotation about an axis inclined to the horizontal.

4. The steel balls rotate circularly about the axis and spin about their own axis. Due to these

motions of steel balls ,the material is continuously crushed and powdered.

5. As this process is continued finally we get powdered nanoparticles.

6. Merits : By ball mill method we can produce nanoparticles on large scale economically.

7. Demerits: Nanoparticles produced are irregular in shape and impure.

Q: Explain Sol-Gel method of synthesis/producing nanoparticles.

1. Sol-gel method is a wet chemical process in which bottom-up approach is adopted.

2. The precursor is dissolved in suitable liquid to form Sol.

Steel container Steel Balls

Nanopowder

Stand

Precursor + Solution = Sol + Dehydration = Gel

Spray/dip adding Surfactants Slow heating

Gelled spheres Zero gel

Substrate On Calcination

On calcination On calcination Dense ceramic

Nano film Nano powder



3. Sol is then dehydrated to get Gel. 4. On slow drying of gel zerogel is obtained. 5. On calcinations zerogel change in to nano dense ceramic.

6. If surfactants are added to Sol, gelled spheres are formed,which on calcinations form nano

powder 7. If Sol sprayed on to substrate by spinning/dipping, gel is formed on the substrate, which

on calcinations form thin nano film. 8. By sol-gel method we can obtain pure nanoparticles.

Q: what are Carbon nano tubes? Mention and explain structures of different types of CNT’s..

1. Carbon nanotube(CNT) structure is imagined to be the cylinder formed by rolling a

hexagonal

graphene sheet of carbon atoms and then closing the ends with fullerene hemispheres.

2. There are three types of CNT structures namely Aramchair CNT,Zigzag CNT and Chiral

CNT.

Types of CNTs:

1.Sigle walled nanotubes(SWNT) consists of single graphene sheet.

2.Multiwalled nanotubes(MWNT) consists of nanotube with in

nanotubes.

Arm chair CNT Zig-Zag CNT Chiral CNT

Axis



Q:Expain the production of CNTS by Arc method.

1. The schematic diagram of Arc discharge method is as shown in the diagram.

2. It consists of a chamber in to which two graphite electrodes separated by 1 mm and

diameter 5-20 𝜇𝑚 are sealed.

3. Helium gas is circulated through the chamber at a pressure of

500 torrand a voltage of 20-25V can be applied between the electrodes.

4. When a voltage of 20-25 V is applied between the electrodes, graphite evaporates and is

deposited on the cathode.

5. If the anode is coated with catalytic agents like Iron, cobalt or nickel ,SWCNT’s are

produced.

6. MWCNT’s are produced if the anode is not coated with catalytic agents.

7. Pure CNT’s can be obtained by using pure graphite rods.

Q:Expain the production of CNTS by Pyrolysis method.

𝑵𝟐 𝑪𝟐𝑯𝟐

1. The schematic diagram of Pyrolisis method is as shown in the diagram.

𝐻𝑒 𝑔𝑎𝑠 𝑖𝑛𝑙𝑒𝑡 𝐻𝑒 gas outlet

Chamber

Graphite anode

500 torr pressure 5 Ba

Graphite cathode

Furnace

Chamber

C Catalyst

Pressure gauge

𝑵𝟐 𝑪𝟐𝑯𝟐

Substrate on graphite sheet

⊙ ⊙ ⊙ ⊙ ⊙

⊙

⊙ ⊙ ⊙ ⊙ ⊙

⊙

⊙

⊙



2. It consists of a chamber in an electric furnace through which a mixture of nitrogen and

acetylene is passed and let out through the out let.

3. A substrate is placed on the quartz sheet inside the chamber.

4. The temperature in the chamber is maintained at about 700-800.

5. Due to high temperature acetylene breaks down in to carbon atoms.

6. When these atoms come near substrate they get attracted and deposited as carbon

nanotubes in the presence of catalysts which are MWCNT’s.

7. SWCNT’

s are obtained when acetylene is replaced with methane or carbon monoxide at 1200

Q: Mention the properties of CNTS’.

1. CNT’s are highly elastic.

2. Young’s modulus of CNTs is about 9 times sronger than that of steel.

3. CNT’s exhibit large strength in tension.

4. CNT’s can be bent without breaking.

5. Electrical properties of CNTs ranges from semiconductor to good conductor.

6. CNT’s have low resistivity and low heat dissipation.

7. Electrical Conductivity of CNTs is maximum along the axis and very less along the

perpendicular direction.

8. CNT’s exhibit magneto-resistance ie:their resistance decreses with increasing mag.field.

9. Thermal conductivity of CNTs is maximum along the axis and very less along the

perpendicular direction

10. CNTs have very high strength to weight ratio and have low density.

11. CNTs are chemically more inert compared to other forms of carbon.

Q: Mention the uses of CNTS’.

1. CNTs can store lithium hence they are used in the manufacture of batteries

2. CNTs can store hydrogen hence they are used in fuel cells.

3. CNTs are used as atomic force microscope probe tips.

4. CNTs are used to produce flat panel display of television and computer monitors .

5. CNTs are used as light weight shield for electromagnetic radiation.

6. Semiconducting CNTs are used to produce field effect transistors used in computers

whose processing capacity is 104 faster than present processors.

7. CNTs are used to produce light weight high strength materials for aircrafts, rockets

automobiles etc

8. CNTs are used as chemical sensors to detect gases.



Q: Explain the Principle, Construction and Working of SEM

(Scanning Electron Microscope).

Principle:

1. SEM is a device used to produce very high resolving images based on the principle of

wave nature of electrons.

2. The resolving power of SEM is 105 times more than that of a best optical microscope.

3. The schematic diagram of SEM is as shown in the labeled diagram.

Working:

1. The electrons produced from the electron gun are passed through two magnetic

condensing coils 1&2 in a highly evacuated chamber to condense the beam

2. The condensed beam is passed through scanning coil which will scan the entire specimen.

3. The scanning beam is focused on to the specimen by the magnetic objective coil.

4. When the electron beam incident on the specimen a few electrons are reflected back called

back scattered electrons which are detected using the detector 𝐷1.

5. Secondary electrons produced due to interaction with valence electrons are detected by

the detector 𝐷2

6. and X-rays produced due to deep penetration are detected using the detector 𝐷3

7. High resolution 3D image can be seen on the TV monitor to which 𝐷1, 𝐷2& 𝐷3 connected.

8. Biological specimens must be dehydrated and non conducting samples must be coated

with a thin conducting layer.

Electron gun

To Vacuum pump

Magnetic condenser coil -1

Magnetic condenser coil-2

Scanning coil

Objective magnetic coil

𝐷1 back scattered electron detector

𝐷2 Secondary electron detector

𝐷3 X-rays detector

Evacuated chamber Specimen

Base

TV monitor



Q:Mention the uses of SEM.

1. SEM is used to study reflectivity, roughness of the surfaces .

2. SEM is used to study the composition of a compound and the abundance

of the constituents.

3. SEM is used to study biological specimens like pollen grains, blood cells

tissues, bacteria etc

4. SEM is used to study the structures, corroded layers .

5. SEM is used in forensic science for examining gunshot residues etc

6. SEM is used in textile industry for the evaluation of fabric.

@@@@

PROBLEMS SECTION

MODULE-5 : SHOCK WAVES AND SCIENCE OF NANO MATERIALS

Formulae needed:V = 𝑋

𝑡 ; M =

𝑉

𝑎 ; λ =

ℎ

√2𝑚𝑒𝑉 or

1.228 𝑋10−9

√𝑉 ( For an electron)

1. The distance between the two pressure sensors in a shock tube is 100 mm.The time taken

by a shock wave to travel this distance is 200 microsecond.If the velocity of sound under

the same conditions is 340 m/s. find the Mach number of the shock wave. ( 4 marks)

Given: x =100 mm=100x10−3 m,t =200μS=200𝑥10−6 S,a=340 m/s ,M=?

Using, ,V=𝑋

𝑡 and M =

𝑉

𝑎

We get, M = 𝑋

𝑎𝑡 =

100𝑥10−3

340𝑥200𝑥10−6

=1.471

2. In a scanning electron microscope ,electrons are accelerated by an anode potential

difference of 60 kilo volt. Estimate the wavelength of the electrons in the scanning beam.

( 4 marks)

Given: V=60KV=60x103 V , h=6.63x10−34 Js; m=9.1x10−31kg; e=1.6x10−19 C; λ =?

Using λ = ℎ

√(2𝑚𝑒𝑉)

= 6.63𝑥10−34

√(2𝑥9.1𝑥10−31𝑥1.6𝑥10−19 𝑥60𝑥103) = 5.016 x 10−12 m

or λ = 1.228 𝑋10−9

√𝑉

= 1.228 𝑋10−9

√(60𝑥103) = 5.013 x 10−12 m



3. In a Reddy shock tube,it was found that, the time taken to travel between the two sensors

is 195 μs.If the distance between the two sensors is 100mm,find the Match number.(CBCS-

Jun/Jul16)

Given: x=100 mm=100x10−3 m,t =195μS=195𝑥10−6 S,a=?(Not given) ,M=?

Using, ,𝑉𝑆 = 𝑋

𝑡

= 100𝑥1 0−3

195𝑥10−6

= 512.8 m/s and

Also, M = 𝑉𝑆

𝑎

= 512.8

𝑎

4. Calculate the wavelength of an electron accelerated under a potential

difference of 100 V in scanning electron microscope.

Given: V=100V , h=6.63x10−34 Js; m=9.11x10−31kg; e=1.6x10−19 C; λ =?

Using λ = ℎ

√( 2𝑚𝑒𝑉)

=6.63𝑥10−34

√( 2𝑥9.11𝑥10−31𝑥1.6𝑥10−19 𝑥100)

= 1.228 x 10−10 m

or λ = 1.228 𝑋10−9

√𝑉

= 1.228 𝑋10−9

√(100)

= 1.228 x 10−10 m

5. The distance between two pressure sensors in a shock tube is 150 mm. The

Time taken by a shock wave to travel this distance is 0.3 ms.If the velocity

of sound under the same condition is 340m/s.Find the Mach number of the

shock wave.

Given: x =150 mm=150x10−3 m,t = 0.3mS= 0.3𝑥10−3 S, a=340 m/s ,M=?

Using, ,V = 𝑋

𝑡 and M =

𝑉

𝑎

We get, M = 𝑋

𝑎𝑡

= 150𝑥10−3

340𝑥0.3𝑥10−3 = 1.471



6. Calculate the wavelength of an electron accelerated under a potential

difference of 100 V in scanning electron microscope.

Given: V=100V , h=6.63x10−34 Js; m=9.11x10−31kg; e=1.6x10−19 C; λ =?

Using λ = ℎ

√(2𝑚𝑒𝑉)

= 6.63𝑥10−34

√(2𝑥9.11𝑥10−31𝑥1.6𝑥10−19 𝑥100)

= 1.228 x 10−10 m

@@end@@

OR λ = 1.228 𝑋10−9

√𝑉

= 1.228 𝑋10−9

√(100)

= 1.228 x 10−10 m

IMPORTANT IA TEST & VTU EXAMINATION QUESTIONS

MODULE-3

Lasers and Optical Fibres.

1. Explain : Induced absorption, Spontaneous emission , Stimulated emission and Derive an expression for energy

de sit i ter s of Ei stei ’s coefficie ts. 2. Explain the principle,construction and working of CO2 laser and Semiconductor laser.

3. Explain Recording and Reconstruction of images in Holography.

4. Explain types of Optical fibres with diagrams.

5. Explain point to point communication using optical fibres with block diagram.

6. Define attenuation and explain attenuation mechanisms/losses in optical fibres.

7. Derive an expression for acceptance angle (Theory of optical fibre) and Numerical aperture.

8. Explain the applications of lasers: Measurement of atmospheric pollutants, Welding,

Cutting & Drilling.

9. Explain the Requisites of Laser system.

MODULE-4

Crystal Structure.

1. Explain seven crystal systems with diagrams.

2. What are Miller Indices and explain the steps/method of finding the Miller Indices of a plane.

3. Derive an expression for inter-planar spacing/distance.

4. Define Co-ordination number and (APF)Atomic Packing factor).Also find the APF of SC,BCC&FCC.

5. Write a note on structures of Diamond and Perovskites with diagrams.

6. Deri e Bragg’s la a d e plai ho to erif Bragg’s la a d Ide tif cr stals usi g Bragg’s X-ray

spectrometer(Diffractometer)

7. Explain Polymorphism and Allotrophy.

MODULE-5

Shock Waves and Science of Nano Materials.

1. Explain the terms : Acoustic waves, Infrasonic waves, Audible waves, Ultrasonic waves,

Mach waves, subsonic waves , supersonic waves and Shock waves.

2. State and explain basics of conservation of mass, momentum and energy.

3. Derive/explain Rankine-Hugonite( Normal shock wave) equations.

4. Describe the principle, construction and working of operated Reddy shock tube.

5. Explain density of states with diagrams.

6. Explain the synthesis of nano particles by Ball mill method and Sol-Gel method.

7. Explain the synthesis of CNTs by Arc discharge method and Pyrolysis method.

8. Explain the principle, construction , working and uses of SEM(Scanning Electron Microscope)

9. Methods of producing shock waves and properties and uses of shock waves.

10. Explain different types and structures of CNTs and mention properties & uses of CNTs

HINTS FOR SURE SUCCESS……..

“ SLOW AND STEADY WINS THE RACE “

1. Adopt Easy to Difficult approach. First thorough with the

easy questions of the topic and then attempt difficult questions, which you will feel easy.

2. Practice writing diagrams again and again, because no marks are

awarded for derivations without relevant diagrams wherever needed.

3. Memorize all the formulae so that you can solve problems and

score maximum marks with less effort.

4. Always express the given data in terms of simple SI units ,then write

the relevant formula and substitute the data.

5. While answering in the tests/exams, First choose the questions

which you can answer correctly and completely.

6. Attempt all the questions by writing whatever you know about it even

if you are not sure about the answers.

7. Practice the use of calculator by solving the solved problems again

and again, so that you will not find any difficulty in the tests/exams.

8. Take all the tests without fail before the main examination so that

you will know your strengths and weakness .This will help you to

rectify the mistakes and score more marks.

9. Pay attention to the units of the physical quantities so that you can

improve your marks.

10.Write the diagrams neatly and answers legibally.

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and answers sem-i/ii

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