and answers sem-i/ii
TRANSCRIPT
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 Planckβs quantum hypothesis(postulates) and explain black body radiation spectra
based on planckβs radiation law.
2. State Planckβs law of radiation and deduce Weinβs law and Rayleigh-Jeans law from Planckβs
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 Heisenbergβs uncertainty Principle and Show that electrons non exist in the
nucleus using Heisenbergβs 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 Meissnerβs effect and Maglev vehicles.
8. Distinguish between Type-I and Type-II superconductors.
9. Explain BCS theory qualitatively.
10. State and explain Matheissenβs 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 (ππ) corresponding to the maximum
intensity π¬πvaries inversely as the absolute temperature(T ) of the black body. This is
called Weinβs displacement law , ππ: ππ βπ
π» or ππ π» = ππππππππ According to this law
β²ππβ² shifts towards shorter wavelength side with the increase of β T β of the body.
3. Weinβs law states that the energy density in the wave length interval Ξ» & Ξ» +dΞ»
is ππ dΞ» = πΆ1πβ5πβπΆ2/ππdΞ» where πΆ1 & πΆ2 are constants. Weinβs law explained only
the shorter wavelength region of BBR spectra below ππ and it failed to
explain the longer wavelength region of the BBR spectra beyond ππ.
4. Rayleigh-Jeanβs law states that the energy density in the wave length interval Ξ» &
Ξ» +dΞ» is ππ dΞ» = 8πππ
π4 dΞ» ,where k = Boltsmannβs constant. Rayleigh-Jeanβs law
explained the longer wavelength region of BBR spectra beyond ππ and it failed
explain the shorter wavelength region of BBR spectra below ππ.
E
Rayleigh-Jeanβs law
π¬π Experimental B.B.R spectra
Weinβs law
π ππ πΊ
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 2
5. Planckβs law states that the energy density in the wavelength interval Ξ» & Ξ» +dΞ» is
ππ dΞ» = 8πβπ
π5
1
π(
βππππ
) β1
dΞ» .
Planckβs law explained the BBR spectra completely ie: both shorter and longer
wavelength sides of the spectra.
Q: State Planckβs quantum postulates and explain Planckβs 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(hπ) i.e: πΈπ = n.hΞ½ ,where n = Quantum number.
3. The oscillating particles absorb or emit energy in discrete units of hΞ½, 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 Ξ» and Ξ»+dΞ» called Planckβs law given by
ππ dΞ» =
8πβπ
π5
1
π(
βππππ
) β1
dΞ».
Based on this, Planck explained the BBR spectrum completely ie: both shorter and longer
wavelength side of the spectrum.
E
Rayleigh-Jeanβs law
π¬π Experimental B.B.R spectra
Weinβs law
π ππ πΊ
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 3
Q:Deduce/derive Weinβs law & Rayleigh-Jeans law from planckβs law of radiation or
Reduce Planckβs law to Weinβs law and Rayleigh-Jeans law
Planckβs law of radiation is given by ππdΞ» = 8πβπ
π5
1
π(
βππππ
) β1
dΞ» β¦β¦β¦β¦β¦(1)
a) Weinβs law:
For shorter wave lengths, βΞ» βis very small ,hence ππ & π(βπ
πππ) are
Large i.e: π(βπ
πππ) β« 1 so that π(
βπ
πππ) β 1 = π(
βπ
πππ) β¦β¦β¦.(2)
β΄ from eqn 1 &2,we get ππdΞ» = 8πβπ
π5
1
π(
βππππ
)
dΞ»
β΄ πΌπ dΞ» = πͺππβππβπͺπππ» dΞ» πβππ ππ π€πππβπ πππ€ .where πΆ1 = 8πβπ & πΆ2 =
βπ
π
b) Rayleigh-Jeans law:
For longer wave lengths, βΞ» βis very large,
hence π(βπ
πππ) ππ π£πππ¦ π ππππ ,Expanding π(
βπ
πππ) as power series ,
we get , π(βπ
πππ) = 1+ (
βπ
πππ) + (
βπ
πππ)2 +β¦β¦β¦.
= 1 + (βπ
πππ) , πππππππ‘πππ βππβππ πππ€πππ ππ (
βπ
πππ)
ππ: π(
βπ
πππ)
β 1 = βπ
πππ β¦β¦β¦(3)
β΄ from eqn 1 &3,we get ππdΞ» = 8πβπ
π5
1βπ
πππ dΞ»
ππ: πΌπ dΞ» = ππ ππ»
ππ dΞ» , πβππ is Rayleigh-Jeans law
Q: What is meant by βultraviolet catastropheβ or failures of Rayleigh-Jeans law ?
1. As per Rayleigh-Jeans law, ππ β β ππ π β 0
2. π΅π’π‘ experimentally observed that ππ β 0 ππ π β 0.
3. This failure of R-Jβs law beyond ultra-violet region is called βultraviolet catastropheβ.
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 4
Q:Explain Compton effect experimentally & mention itβs significance.
1. Compton experimental arrangement is as shown in the diagram.
2. X-ray photons of wave length Ξ» 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 β πππ ππππ‘π‘ππππ π β
πππ¦π πππππ ππ πππππ π πππ ππππ‘ππ£πππ¦ π€ππ‘β the incident X-rays direction.
5. The intensity of the scattered X-rays in different directions are measured using Braggβs
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 βunmodifiedβ component
having the same wavelength(Ξ») as incident X-rays & βmodifiedβ component having slightly
higher wavelength πβ²
7. This scattering of X-rays due to recoil of electrons is called βCompton effectβ.
8. The change in wavelength (πβ² β π)is called βCompton wavelengthβ ,which depends only
π πππ πππππππππππ‘ ππ π΄.
9. Based on conservation of energy & momentum laws, considering the collision between X-
rays & electrons as βparticle-particleβ collision , Compton wavelength is given by
(πβ² β π) = βπ = β
π0πΆ (1βπΆππ π)
where β
π0πΆ= π0 called βCompton wavelength of electronβ= 0.02426 Γ (constant)
10. βπ π£ππies from β0β for π = 0Β° π‘π 2π0 for π = πππΒ°
11. πΆππππ‘ππ ππππππ‘ π ππππππππ π‘βπ ππππ‘ππππ πππ‘π’ππ ππ π β πππ¦π (πππβπ‘)
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.
Braggβs spectrometer
Slits
X-rays Collimated X-rays π
π1 π2 β
Graphite target Recoil electron
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 5
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(Ξ»).
According to Einsteinβs mass-energy relation E=mπΆ2β¦β¦(1)
where m=mass, C=speed of light.
also, from Planckβs quantum theory, E =βπΆ
π β¦β¦β¦β¦(2) where h=planckβs contant.
Ξ½=frequency. From eqns 1&2, mπΆ2 = βπΆ
π
β΄ π =π
ππͺ
Similarly, for a particle of mass βmβ moving with a velocity βπ β²,
De-broglie wavelength, Ξ» = π
ππ₯ or Ξ» =
π
π·
Q: De-broglieβs wavelength of an electron accelerated in a potential difference βVβ volt/
Show that Ξ» = ππ.ππ
βπ½ Γ for an electron accelerated in a potential difference βVβ volt
The K.E of an electron of mass βmβ,charge βeβ accelerated in a potential βVβ volt is given
by 1
2ππ£2 = ππ , multiplying both Nr. and Dr. of LHS by βmβ
we get, π2π£2
2π= ππ
ππ; π2π£2 = 2πππ
π2 = 2πππ β΅ ππ£ = π
π· = βππππ½
But Ξ» = π
π· β΄ π =
π
βππππ½
= π.ππππ10β34
βππ π.ππππβππππ.ππππβπππ½
= π.ππππππβπ
βπ½ m
= 1.228
βπ½ππ =
ππ.ππ
βπ½ Γ
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 .
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 6
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 ?
βPhase Velocity β is defined as the velocity with which the uni-phase particles on
the wave travels and is given by π£π =π
π
where π = ππππ’πππ π£ππππππ‘π¦ πππ π = π€ππ£π ππππ π‘πππ‘.
Q: Define Group velocity/What is meant by Group velocity ?
βGroup Velocity β is defined as the velocity with which the resultant wave packet of
the group of waves travels and is given by π£π =ππ
ππ
where dπ = πβππππ ππ ππππ’πππ π£ππππππ‘π¦ πππ ππ = πβππππ ππ π€ππ£π ππππ π‘πππ‘.
Q: Derive the relation between group velocity (ππ) & phase velocity(ππ·)/
Show that ππ = ππ· β ππ ππ
π π
Consider a particle of mass βmβ having phase velocity β ππ· β and group velocity ββ²ππβ².
Let the wave have the wavelength βΞ»β , frequency βΞ½, wave number k & angular velocity Ο
We know that π£π =π
π β¦(1) and π£π =
ππ
ππ β¦.(2) ππ
From eqns 1 & 2 , we get π£π =π(ππ.π)
ππ
= π£π + ππππ
ππ
= π£π + π.πππ
ππ
ππ
ππ β¦β¦(3)
Also, Ξ» = 2π
π β΄
ππ
ππ =
π
ππ(
2π
π)
= β2π
π2 β¦β¦..(4)
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 7
From eqns 3 & 4 ,we get π£π = π£π + π.πππ
ππ(
β2π
π2 )
= π£π β 2π
π π
πππ
ππ
ie: ππ = ππ· β ππ ππ
π π β΅ Ξ» =
2π
π
Note: ππ < ππ· for a dispersive medium
and π£π = π£π = πΆ πππ π πππ πππ ππππ ππ£π π£ππππ’π πππππ’π
Q: Deduce the relation between group velocity(ππ) & particle velocity(ππ·πππππππ)/
Show that ππ = ππ·πππππππ or ππ = π
Consider a particle of mass βmβ having particle velocity βππ·πππππππβ² & ππππ’π π£ππππππ‘π¦ β²ππβ².
We know that, π£π =ππ
ππ β¦ β¦ . . (1)
but π = 2ππ & πΈ = βπ ,we get π = 2ππΈ
β
β΄ dπ = (2π
β) ππΈ β¦ (2)
Also from k = 2π
π & π =
β
π we get k = (
2π
β)P
β΄ dk = (2π
β)dP β¦β¦(3)
From eqns 1,2 &3,we get π£π =ππ
ππ=
(2π
β)ππΈ
(2π
β)ππ
=ππΈ
ππ β¦β¦.(4)
Also , from E = Β½mπ£2 =π2π£2
2π & P = mπ₯ ,
π€π πππ‘ πΈ = π2
2π
β΄ dE = 2π
2πππ
ππ; ππΈ
ππ=
π
π =
π π£ππππ‘ππππ
π
= π£ππππ‘ππππ β¦.(5)
From eqns 4 & 5,we get ππ = ππ·πππππππ Thus group velocity =Particle velocity
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 8
QUANTUM MECHANICS
Q: State & explain Heisenbergβs Uncertainty Principle (HUP). Mention its significance.
HUP states that βthe product of the uncertainty βΞπ₯β in the position and the uncertainty
βΞPπ₯β in the momentum of a particle at any instant is equal to or greater than ( h/4π) β
i.e.: Ξπ₯ ΞPπ₯ β₯ β
4π ,Where h is Planckβs 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 βπΈ.βπ‘ β₯ β
4π , π€βπππ βπΈ = ππππππ¦, βπ‘= time,
& βπΏ.βπ β₯ β
4π
where , βπΏ = ππππ’πππ ππππππ‘π’π & βπ = πππ πππππππππ‘.
Q: Show that electrons do not present in the nucleus using Heisenbergβs Uncertainty Principle.
Electron to be present in the nucleus, maximum uncertainty in position Ξπ₯=10-14 m (diameter)
According to HUP,
The minimum uncertainty in momentum ΞPπ₯ β₯ β
4π π₯π₯
β₯ 6.625 π₯ 10β34
4 π₯ 3.14 π₯ 10β14
ΞPπ₯ β₯ 5.275 x 10-21 kg m/s = P(say)
Using E = mπΆ2 , π = ππ πππ π = π0
β1βπ2
πΆ2
, π€π πππ π βππ€ π‘βππ‘ the minimum energy of the
electron in the nucleus is given by πΈ2 = P2c2 +ππ2 π4, but the rest mass energy ππ
2 π4 is
very very small compared to P2c2,neglecting ππ2π4 ,
we have E β PC = 5.275 x 10β21 x 3 x 108 J
= 5.275 π₯ 10β21 π₯ 3 π₯ 108
1.6 π₯ 10β13 MeV
= 9.89 MeV
= 10 MeV
But the maximum energy of the electrons(π-particle) emitted
from the nucleus does not exceed 4MeV,hence electrons do not
present in the nucleus.
NOTE:
πΈ2 = π2πΆ4 = ππ
2πΆ4
(1βπ£2
πΆ2) =
ππ2πΆ6
(πΆ2βπ£2)
π2πΆ2 = π2π£2πΆ2 = ππ
2π£2πΆ2
(1βπ£2
πΆ2) =
ππ2πΆ4π£2
(πΆ2βπ£2)
πΈ2 β π2πΆ2 = ππ
2πΆ4(πΆ2βπ£2)
(πΆ2βπ£2)
β΄ π¬π = π·ππͺπ + ππππͺπ
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 9
Q: Set up 1D time independent Schrodinger wave equation for a free particle.
One dimensional wave function πΏ describing the de-broglie wave for a particle moving
freely in the positive direction of π₯-direction is given by πΏ = Aππ(ππ₯βππ‘)
= Aππππ₯ πβππ‘ ,
Where Aππππ₯ represent the time independent part of the wave function and is
represented by π = Aππππ₯ β¦β¦(1) differentiating eqn (1) w.r.t βπ₯ β twice we get
ππ
ππ₯= π΄(ππ) ππππ₯ and
π2π
ππ₯2 =A(ππ) (ππ) ππππ₯
= π2 π2 Aππππ₯ β¦β¦.(2)
From eqns 1 & 2 we get
π2π
ππ₯2 = β
4π2
π2 π β¦β¦..(3) β΅ ππ = β1 & π =
ππ
π
But, de-broglie wave length Ξ» = β
ππ£
β΄ 1
π2=
π2π£2
β2
= 2π(Β½ππ£2)
β2
= 2π(πΈπΎ)
β2 β΅ πΈπΎ = Β½ππ£2
Also , the kinetic energy (πΈπΎ )in terms of the total energy (E) & the potential energy( V) is
given by πΈπΎ=(Eβπ½)
β΄ 1
π2=
2π(πΈβπ½)
β2 β¦β¦(4)
From eqns 3 & 4 ,we get π2π
ππ₯2= β
4π22π(πΈβπ½)
β2 π
π2π
ππ₯2+
8π2π(πΈβπ½)
β2π = 0 β¦β¦(5)
This is Schrodinger time independent equation for a particle
For a free particle ,V=0
β΄ π2π
ππ₯2+
8π2ππΈ
β2π = 0
This is Schrodinger time independent equation for free particle.
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 10
Q: Obtain normalized wave function for a free particle in a infinite walled potential
Box/Well using Schrodinger 1D time independent equation.
a
β β
V=β V=0
Box/Well
particle
π₯=0 π₯=a
Consider a particle of mass βmβ moving by reflection at infinitely high walls of a Box/well
of width βaβ moving between π₯=0 & π₯=a Potential V=0 inside the Box/well and V=β
outside the Box/well .one dimensional Schrodinger equation for particle is given by
π2π
ππ₯2+
8π2π(πΈβπ½)
β2π = 0 β¦β¦(1).
For a particle inside the Box/well V=0
β΄ π2π
ππ₯2+
8π2ππΈ
β2π = 0 β¦.(2)
Putting 8π2ππΈ
β2= πΎ2 β¦..(3) in equation (2) ,
we get π2π
ππ₯2+ πΎ2π = 0 β¦β¦(4)
The general solution of the quadratic equation (4) is of the form Ο (π₯)=A sin(Kπ₯) +B Cos(Kπ₯)
β¦β¦(5)
where A & B are constants determined from boundary conditions as follows :
Ο (π₯)=0 at π₯=0 from eqn (5), 0=A x 0 +B x 1
β΄ B=0
also, Ο (π₯)=0 at π₯=a β΄ from eqn(4) 0=C Sin(Ka)+0xCos(Ka)
0= C Sin(Ka)
Sin(Ka) = 0 as Cβ 0
β΄ Sin(Ka) =0= Sin(nπ)
ie: Ka=nπ or K=ππ
π β¦..(6)
From eqns 3 & 6 ,we get 8π2ππΈ
β2=
π2π2
π2
β΄ E = π2β2
8π2π or
In general En= π2β2
8π2π β¦(7)
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 11
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 = β2
8π2π = Eo called end point /ground state energy .
n=2 gives E2 = 4β2
8π2π = 4 Eo called 1st excited state energy
n=3 gives E3 = 9β2
8π2π =9 Eo ,called 2nd excited state energy & so on
Substituting the values of B=0 & K= ππ
π in eqn 5, we get
ππ(π₯) = A Sin( πππ₯
π) β¦β¦.(8) this represents the permitted solutions
To find βAβ by normalization:
Applying the normalization condition β« πΌπΉπΌ2ππ₯ = 1+β
ββ to eqn 8 for x=0 & x=a, we get
β« π΄2 πππ2 ( πππ₯
π)ππ₯ = 1
π
0
π΄2 β« Β½[1 β πΆππ ( 2 πππ₯
π)]ππ₯ = 1
π
0 β΅ πππ2 (π)=Β½[1-Cos(2π)]
Β½π΄2 [β« 1ππ₯ β β« πΆππ ( 2 πππ₯
π)ππ₯
π
0 ] = 1
π
0
Β½π΄2 [π₯]0π β [
π
2ππ Sin(
2πππ₯
π)]0
π = 1 β΅ β« πΆππ (ππ₯) ππ₯ = πππ(ππ₯)
π
Β½π΄2 [π β 0] β [ π
2ππ Sin(2nπ) β
π
2ππ Sin(0)] =1
Β½π΄2 [π β 0] β [0 β 0] =1
π΄2 π = 2 π΄2 = 2/π or A =βπ/π β¦(9)
From eqns 8 &9, the normalized Eigen functions are πππ£ππ ππ¦ ππ(π₯)= βπ/π Sin( ππ π
π) β¦..(10)
π = 3 π3 πΌπ3πΌ2
.
π = 2 π2 πΌπ2πΌ2
π = 1 π1 πΌπ1πΌ2
π₯=0 π₯=a π₯=0 π₯=a
πΈππππ ππ’πππ‘ππππ π1 ,
π2 ,π3 , β¦ . πππ π‘βπππ ππππππππππ‘π¦ ππππ ππ‘πππ πΌπ1πΌ2, πΌπ2πΌ2, πΌπ3πΌ2 β¦
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 [ππ(π₯)]
Eigen Energy values are energy values for which Schrodinger equation
can be solved.
This can be omitted
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 12
Q: What is a wave function and mention itβs properties/limitations
The variable quantity that characterizes the de-broglie wave of the particle is called a
βWave functionβ denoted mathematically by the symbol βπΏβ(Psi)
Properties of the wave functions are:-
1. Wave function (πΏ) is single valued everywhere ,
2. πΏ is finite everywhere
3. πΏ is continuous every where
4. First derivatives of πΏ are continuous every where
5. πΌπΉπΌ2 or πΉπΉ * is called probability density.
6.ππππππππππ‘π¦ ππ πππππππ ππππ‘ππππ ππ π ππππ ππ πππ£ππ ππ¦ β« πΌπΉπΌ2ππ = 1+β
ββ
PROBLEMS SECTION
MODULE-1 : MODERN PHYSICS AND QUANTUM MECHANICS
Formulae needed : Ξ» = β
π where P = mv = β2ππΈ =β2πππ ,
E = 1
2ππ2 =
π2
2π ;Photon energy Eβ = hΞ½ =
βπΆ
π ; βπ₯. βπ β₯
β
4π and
βπ = π. βπ£ ; βπΈ. βπ‘ β₯β
4π πππ βπΈ =
βπΆβπ
π2 ; also βπΈ = β βπ ;
P = β« πΌπΉπΌ2π₯2
π₯1dx ; πΈπ =
π2β2
8ππ2
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: ππ = ππ = 0.2 ππ = 0.2π₯10β9 π , h=6.63x10β34 Js, m=9.1x10β31kg ; C=3x108m/s,
πΈπ
πΈπ= ?
Using πΈπ =βπΆ
ππ and πΈπ =
β2
2πππ2
We get , πΈπ
πΈπ =
πΆ2πππ2
βππ
= 3π₯108π₯2π₯9.1π₯10β31( 0.2π₯10β9)2
6.63π₯10β34 π₯ 0.2π₯10β9
= 1.647x 102
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 13
2. Calculate the kinetic energy of an electron of wavelength 18 nm.(Jun/Jul 14)
Given: π = 18 ππ = 18π₯10β9 π , h=6.63x10β34 Js, m=9.1x10β31kg ; E= ?
Using Ξ» = β
β2ππΈ
We get ,πΈ =β2
2ππ2
= ( 6.63π₯10β34 )2
2π₯9.1π₯10β31π₯( 18π₯10β9)2
= 7.454x10β22 J
=7.454π₯10β22
1.6π₯10β19
= 4.66x10β3 eV
3. An excited atom has an average life time of 10β8 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: βπ‘ = 10β8 S, β = 6.63π₯10β34 π½π ; βπ = ?
Using βπΈ = β. βπ and βπΈ. βπ‘ β₯β
4π
We get , βπ β₯1
4π.βπ‘
β₯1
4π₯ππ₯10β8
= 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=6.63x10β34 Js; π = 9.1π₯10β31ππ ; π = ?
Using Ξ» = β
ππ£
= 6.63π₯10β34
9.1π₯10β31π₯0.01π₯ 3π₯108
= 2.43 x10β10 m
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 14
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 = 0.12x10β9 m ; n = 2 ; h=6.63x10β34 Js; π = 9.1π₯10β31ππ ;
πΈ2 = ? & Ξ» =?
Using πΈπ = π2β2
8ππ2
πΈ2 = 22(6.63π₯10β34 )2
8π₯9.1π₯10β31π₯(0.12π₯10β9)2
= 1.68x10β17 J
Also , Ξ» = β
β2ππΈ2
= 6.63π₯10β34
2π₯9.1π₯10β31π₯1.68π₯10β17
= 1.19 x10β10 m
6. Calculate the de-Broglie wavelength associated with an electron of energy 1.5eV. (Jun13)
Given: E=1.5 eV=1.5x1.6x10β19 π½ , h=6.63x10β34 Js; m=9.1x10β31kg; e=1.6x10β19 C; Ξ» =
Using Ξ» = β
β2ππΈ
= 6.63π₯10β34
β(2π₯9.1π₯10β31π₯1.5π₯1.6π₯10β19 )
= 1.003 x 10β9 m
7. A spectral line of wavelength 5461Γ has a width of 10β4Γ . Evaluate the minimum time
spent by the electrons in the upper energy state. (Jun/Jul 2013)
Given:, β = 6.63π₯10β34 π½π ; π = 5461Γ = 5461π₯10β10 π ; βπ‘ = ?
Using βπΈ =βπΆ
π and βπΈ. βπ‘ β₯
β
4π
We get , βπ‘ β₯π
4π.πΆ
β₯5461π₯10β10
4π₯ππ₯3π₯108
= 1.45 x 10β16 S
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 15
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 ; π0 =1 kg, π0 = 1 m/s , h=6.63x10β34 Js; m=9.1x10β31kg; e=1.6x10β19 C ,
ππ = ? π0 = ?
Using ππ =β
β2πππ
=6.63π₯10β34
β2π₯9.1π₯10β31π₯1.6π₯10β19 π₯182
= 9.11 x10β11 m
Also,using π0 = β
π0π0
= 6.63π₯10β34
1π₯1
= 6.63x10β34 π
Comment : Waverlength is inversely proportional to the mass ,as ππ β« π0
9. A quantum particle confined to one-dimensional box of width βaβ is in its first excited
state. What is the probability of finding the particle over an interval of ( π
2) marked
symmetrically at the centre of box. (π½π’π
π½π’π2011)
Given: 1st excited state , n=2 ;the interval ( π
4 ,
3π
4 )
Using P = β« πΌπΉπΌ2π₯2
π₯1dx = β« (β
2
ππππ ( 2ππ₯
π))2ππ₯
3π/4
π/4π/2
= β«π
ππππ2 ( 2ππ₯
π)ππ₯
3π
4π
4
0 a/4 a/2 3a/4 a
= π
π β« Β½[1 β πΆππ ( 2 ππ₯
π)]ππ₯
3π
4π
4
β΅ πππ2 (π)=Β½[1-Cos(2π)]
= Β½π
π β« 1ππ₯ β β« πΆππ ( 2 ππ₯
π)ππ₯
3π
4π
4
3π
4π
4
= π [π₯]π
4
3π
4 β [ π
2π Sin(
2ππ₯
π)]π
4
3π
4
= π [3π4
β π4
] β [ π 2π
Sin( 2π
3π
4
π)β
π
2π Sin(
2ππ
4
π)]
= π [π
2] β [0 β 0 = Β½ = 0.5 or 50%
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 16
10. A particle moving in one dimension box is described by the wave function πΉ = π₯[β3] πππ
0<π₯<1 and Ξ¨ =0 elsewhere. Find the probability of finding the particle within the
interval (0,Β½). (π½π’π 2012)
Given : πΉ = π₯[β3] ; (π₯1, π₯2)( 0,1
2 ) ; P = ?
Using Probability, P = β« πΌπΉπΌ2π₯2
π₯1dx
= β« 3π₯21
2
0dx = [3.
π₯3
3]0
1
2
= ( 1
2 )3 β (0)3 =
1
8 = 0.125 or 12.5 %
11. Calculate the energy of electron that produces Braggβs diffraction of first order at glancing
angle of 22Β° ,when incident on crystal with inter-planar spacing of 1.8Γ . (π½π’π 2012)
Given : n =1 ; π = 22Β° ; π = 1.8 Γ = 1.8π₯ 10β10 m ; h=6.63x10β34 Js; m=9.1x10β31kg;
Ξ» = ? & E = ?
Using 1 ) Ξ» = 2π π ππ π
π
= 2π₯1.8π₯ 10β10π₯ π ππ 22
1
=1.348x 10β10 m
Also from, Ξ» = β
β2ππΈ
We get, E = β2
2ππ2 J OR E =
β2
2ππ2 π eV
= (6.63π₯10β34)2
2π₯9.1π₯10β31π₯(1.348π₯ 10β10 )2 =
(6.63π₯10β34)2
2π₯9.1π₯10β31π₯(1.348π₯ 10β10 )2π₯1.6 π₯ 10β19
= 1.329 x 10β17 J = 83.07 eV = 83.07 eV
12. Find the energy of the neutron in eV whose de-Broglie wavelength is 1Γ . (π·ππ 2011)
Given; Ξ» = 1Γ = 1x 10β10 π ;h=6.63x10β34 Js, m=1.678x10β27kg ; e=1.6x10β19 C ;E = ?
Using , Ξ» = β
β2ππΈ
We get , E = β2
2ππ2 joules
= β2
2πππ2 eV
= (6.63π₯10β34 )2
2π₯1.678π₯10β27π₯1.6π₯10β19 π₯(1π₯ 10β10 )2
= 0.082 eV
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 17
13. An electron is confined to a box of length 10β9 m, calculate the minimum uncertainty in itβs
velocity. (π·ππ 2011)
Given: βπ₯ = 10β9 m ; h=6.63x10β34 Js, m=9.1x10β31kg ;βπ£ = ?
Using βπ₯. βπ β₯β
4π and βπ = π. βπ£
We get , βπ£ β₯β
4ππβπ₯
β₯ 6.63π₯10β34
4π π₯9.1π₯10β31π₯10β9
= 58 x103m/S
14. Compute the de-Broglie wavelength for a neutron moving with one tenth part of the
velocity of light. (π½π’π
π½π’π2011)
Given; V = πΆ
10 =
3 π 108
10= 0.3 π₯ 108 π/π ;h=6.63x10β34 Js, m=1.678x10β27kg ;
C = 3 X 108 m/s ; Ξ» = ?
Using Ξ» = β
ππ
= 6.63π₯10β34
1.678π₯10β27π₯0.3 π₯ 108 = 1.317x 10β14 m.
15. An electron has a de-Broglie wavelength 3 nm and rest mass 511keV. Determine its group
velocity and kinetic energy.
Given: Ξ» = 3nm = 3x 10β9 m ; h=6.63x10β34 Js ; ππ = V = ? and πΈπΎ = ?
m =E/ πΆ2 =511x103x1.6x 10β19/(3π₯108)2 kg = 9.08x10β31 kg
Using Ξ» = β
ππ£ ,we get ππ = V =
β
ππ
= 6.63π₯10β34
9.08π₯10β31π₯3π₯ 10β9
= 2.43x 106 m/s
πππ π, πΈπΎ =1
2ππ2 =
1
2π₯ 9.08π₯10β31 π₯ (2.43π₯ 106 )2
= 2.681x 10β18 J
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 18
16. A spectral line of wavelength 546.1 nm has a width 10β5 nm. Estimate the minimum time
spent by electrons in the excited state during transitions. (π·ππ 2010)
Given:, β = 6.63π₯10β34 π½π ; π = 546.1 ππ = 546.1π₯10β9 π ; βπ‘ = ?
Using βπΈ =βπΆ
π and βπΈ. βπ‘ β₯
β
4π
π€π πππ‘ , βπ‘ β₯π
4π.πΆ
β₯546.1π₯10β9
4ππ₯3π₯108 = 1.45x 10β16 S
17. Calculate the momentum of the particle and de-Broglie wavelength associated with an
electron with a kinetic energy of 1.5KeV. (π½π’π 2010)
Given: E =1.5 Kev =1.5 x 103 x 1.6x10β19 J ; h = 6.625π₯10β34 π½π ;e = 1.6x10β19 J ;
m = 9.1π₯10β31 ππ P =? & Ξ» = ?
Ans: Using P = β2ππΈ
= β2π₯9.1π₯10β31π₯1.5 π₯ 103 π₯ 1.6π₯10β19 = 2.09 x 10β23 kg.m/s
Also, Ξ» = β
π
= 6.625π₯10β34
2.09 π₯ 10β23 = 3.17x10β11 π
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. (π½π’π 2010 & πΆπ΅πΆπ β π½π’π/π½π’π16)
Given: n=3(For 2nd excited state), a=0.18 nm =0.18x10β9m,h=6.63x10β34 Js;
m=9.11x10β31kg & e=1.6x10β19 C; πΈ2 = ?
Using πΈπ =π2β2
8ππ2 J OR πΈπ =
π2β2
8ππ2π eV
πΈ2 = 32(6.63π₯10β34 )2
8π₯9.11π₯10β31π₯(0.18 π₯10β9)2 πΈ2 =
32(6.63π₯10β34 )2
8π₯9.11π₯10β31π₯(0.18 π₯10β9)2 π₯1.6π₯10β19
= 1.675 x10β17 J = 104.7 eV
=1.675 π₯10β17
1.6π₯10β19 =104.7 eV
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 19
19. Calculate the energy in eV, for the first excited state of an electron in an infinite potential
well of width 2 Γ (4 marks)
Given: n=2(For 1st excited state), a=2Γ =2x10β10m,h=6.63x10β34 Js; m=9.1x10β31kg &
e=1.6x10β19 C;
Using πΈπ =π2β2
8ππ2 J OR πΈπ =
π2β2
8ππ2π eV
=22(6.63π₯10β34 )2
8π₯9.1π₯10β31π₯(2π₯10β10)2 =
22(6.63π₯10β34 )2
8π₯9.1π₯10β31π₯(2π₯10β10)2 π₯1.6π₯10β19
= 6.038x10β18 J = 37.74 eV
=6.038π₯10β18
1.6π₯10β19 =37.74 eV
20. The wavelength of a fast neutron of mass 1.675x10β27kg is 0.02 nm.Calculate the group
velocity and phase velocity of its de-Broglie waves.( 4 marks)
Given: m=1.675x10β27kg, Ξ» = 0.02 nm=0.02x10β9m, C=3x108m/s, β = 6.63π₯10β34 π½π ,
π£π & π£π =?
ππ πππ, π£π = v =β
ππ =
6.63π₯10β34
1.675π₯10β27π₯0.02π₯10β9 = 1.979x104 m/s ,
Also π£π =πΆ2
π£π =
(3π₯108)2
1.979π₯104 = 4.55X1012 m/s
21. Calculate the de-Broglie wavelength associated with neutron of mass 1.674 X 10β27 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=6.63x10β34 Js; π = 9.11π₯10β31ππ ; π = ?
Using Ξ» = β
ππ£
= 6.63π₯10β34
9.11π₯10β31π₯ 3π₯107
= 2.43 x10β11 m
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 20
22. Calculate the deBroglie wavelength of an electron moving with K.E of 50KeV.
Given: E= 50keV= 5x103π₯ 1.6x10β19 π½ , h=6.63x10β34 Js; m=9.1x10β31kg;
e=1.6x10β19 C; Ξ» =?
Using Ξ» = β
β2ππΈ
= 6.63π₯10β34
β(2π₯9.1π₯10β31π₯ 50π₯103π₯ 1.6π₯10β19 )
= 5.495 x 10β12 m
22. X-rays of wavelength 0.75Γ are scattered from a target at an angle of 45Β°.
Calculate the wavelength of scattered X-rays.
Given: π = 0.75Γ ; π = 45Β° ; h=6.63x10β34 Js; m=9.1x10β31kg ; C=3x108m/s ;πΊΞ»β = ?
Using , Ξ»ββπ = β
ππΆ (1βπππ π)
Ξ»β β0.75Γ = 6.63π₯10β34
9.1π₯10β31π₯ 3π₯108 (1βπππ 45)
Ξ»β = 0.75Γ + 7.1x10β13m
= 0.75Γ + 0.0071Γ
= 0.7571Γ
223. A particle of mass 940 MeV/πΆ2 has kinetic energy 0.5 KeV. Find its de-
Broglie wavelength , C is velocity of light.
Given: m = 940 MeV/πΆ2 = 940 π₯ 106 ππ
(3π₯108 )2 =
940 π₯ 106 π₯1.6π₯10β19 π½
(3π₯108 )2 = 1.671x10β27 ππ
E = 0.5 keV = 0.5 π₯103eV = 0.5 π₯103π₯1.6π₯10β19 π½ ; π = ?
Using; πΊ = β
β(2ππΈ)
= 6.63π₯10β34
β(2π₯1.671π₯10β27π₯0.5 π₯103π₯1.6π₯10β19)
= 1.282x10β12 m
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 21
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: πΈ2 = 240 eV ; πΈ1 = ?
Using πΈπ =π2β2
8ππ2
πΈ2 =22β2
8ππ2
240 =4β2
8ππ2
π2 =4β2
8ππ₯240
β΄ a = β(β2
480π )
πππ π , πΈ1 =12β2
8π(2β(β2
480π ) )2
= 480ππ₯β2
8ππ₯4π₯β2 = 15 eV
ππ
Using ; πΈπ =π2β2
8ππ2 ,we can show that
πΈ2 π22
22 =
πΈ1 π12
12 =
β2
8π
Given; πΈ2 = 240ππ , π1 = 2π2
β΄ πΈ1 = πΈ2 π2
2
π1222
= 240π2
2
4π22π₯4
= 15 eV
25. Calculate the deBroglie wavelength associated with neutron of mass 1.674 x10β27kg with one tenth part of the velocity of light .
Given; V = πΆ
10 =
3 π 108
10= 0.3 π₯ 108 π/π ;h=6.63x10β34 Js,
m=1.674x10β27kg ; C = 3 X 108 m/s ; Ξ» = ?
Using Ξ» = β
ππ =
6.63π₯10β34
1.674π₯10β27π₯0.3 π₯ 108
= 1.320x 10β14 m.
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 22
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 x10β9m,h=6.63x10β34 Js;
m=9.1x10β31kg & e=1.6x10β19
Using πΈπ =π2β2
8ππ2 J OR πΈπ =
π2β2
8ππ2π eV
=32(6.63π₯10β34 )2
8π₯9.1π₯10β31π₯(0.18 π₯10β9)2 =
32(6.63π₯10β34 )2
8π₯9.1π₯10β31π₯1.6π₯10β19 π₯(0.18 π₯10β9)2
= 1.677x10β17 J = 104.83 eV
=6.038π₯10β18
1.6π₯10β19
= 104.81 eV
27. Find deBroglie wavelength of a particle of mass 0.58 MeV/πͺπ has a kinetic
energy 90 eV , where C is the velocity of light. ( 04 Marks) DEC2016
Given: m = 0.58 MeV/πΆ2 = 0.58 π₯ 106 ππ
(3π₯108 )2
=0.58 π₯ 106 π₯1.6π₯10β19 π½
(3π₯108 )2 = 1.0324x10β30 ππ
E = 90eV = 90π₯1.6π₯10β19 π½ ; π = ?
Using; πΊ = β
β(2ππΈ)
= 6.63π₯10β34
β(2π₯1.0324π₯10β30 π₯ 90π₯1.6π₯10β19)
= 1.2142π₯10β10 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 ππβππ S. Estimate the
uncertainty that results in its energy in eV in the excited State. ( 04 Marks)
Given:, β = 6.625π₯10β34 π½π ; βπ‘ = 1.4 x ππβππ S: βπΈ =?
Using βπΈ. βπ‘ β₯β
4π π€π πππ‘ , βπΈ β₯
β
4π.βπ‘
β₯6.625π₯10β34
4ππ₯π.π π ππβππ J
β₯ 6.625π₯10β34
4ππ₯π.π π ππβππππ.ππππππβππ eV
β₯ 2.351π₯ 10β6 eV
REDDAPPA C MODULE-1 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22) SEM-I/II Page 23
29. A spectral line of wavelength 5896 Γ has a width of ππβπ Γ . Evaluate the minimum
time spent by the electrons in the upper energy state between the excitation and
de-excitation processes. ( 04 Marks)
Given: π=5896 Γ = 5896 xππβππ m ,βπ = ππβπΓ = ππβπ xππβππm ,
h = 6.625 x ππβππ JS , C = 3 x πππππβπ , βπ = ?
Using βπ¬. βπ β₯ π
ππ and βπ¬ =
ππ βπ
ππ β΅ E =
ππ
π
We get , βπ = ππ
ππ πβπ
β΄ βπ = (ππππ πππβππ)π
ππ ππ π ππππππβπ πππβππ
= 9.221 x ππβπ 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 ππβππ kg. ( 04 Marks)
Given : ππ = ππ = π. ππ ππ = 0.25 x ππβπ m , ππ = 1.675 X ππβππ kg
h = 6.625 x ππβππ JS , C = 3 x πππππβπ , ππ
ππ = ?
Using , π¬π =ππ
π =
π.πππ π ππβππππ π πππ
π.ππ π ππβπ = 7.95 x ππβππ J
Also , π¬π =ππ
ππππ =
(π.πππ π ππβππ)π
πππ.πππ πΏ ππβπππ (π.ππ π ππβπ)π = 2.096 xππβππ J
β΄ π¬π
π¬π=
π.ππ π ππβππ
π.πππ πππβππ = 3.793 x πππ
**end**
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
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(ππ ) 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(π) is the average time elapsed between two consecutive collisions.
c) Mean free path(Ξ»)is the average distance travelled by the electron between two consecutive collisions.
d) Relaxation time(ππ) is time during which the drift velocity is reduced to 1
π of its
velocity when the field is cut off.
Q:Obtain an expression for Drift velocity.
Let an electron of mass βmβ ,charge βeβ moves with drift velocity βππ β when an
electric field βE β is applied to it. Then the driving force acting on the electron
F = βππΈ β¦β¦(1)
Also the retarding force acting on the electron Fβ= ππ£π
π β¦..(2)
where π = ππππ ππππππ πππ π‘πππ
For a system in steady state Fβ = βπΉ
ie: ππ£π
π= ππΈ
β΄ ππ =πππ¬
π
Q:Derive an expression for electrical conductivity and hence electrical resistivity.
The drift velocity of electron ππ =πππ¬
π β¦β¦(1)
Also from Ohmβs law electrical conductivity , π =π
π¬ ..(2)where j=current density &
E=electric field But j = neππ β¦ (π)
β΄ eqns 1&3 ,we get j = ne. πππ¬
π
β΄ π
π¬ =
ππππ
π β¦.(4)
From eqns 2 &4,we get π = ππππ
π β¦..(5)
By definition electrical resistivity, π =1
π β¦(6)
β΄ from eqns 5&6, we get π = π
ππππ
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 2
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
2ππ,
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 πΆπ£ =3
2π ,where R=universal gas constant. But experimentally
determined πΆπ£ ππ given by πΆπ£ = 10β4π π ,where T=absolute temperature. Thus CFET fail to
explain the the dependence of πΆπ£ ππ T and numerical constant.
2.Dependance of electrical conductivity (π) on T:
According to CFET kinetic energy 1
2ππ£2 =
3
2ππ
β΄ π£ β βπ β¦(1)
But mean collision time π β 1
π£
β΄ π β 1
βπ β¦ . (2) β΅ π£ β βπ
Also from π =ππ2π
π ,we get π β π β¦ . (3)
From eqns 2 & 3 we get π β π
βπ ,but experimentally it has been observed that π β
π
π» ,
thus CFET fail to explain the dependence of π ππ π.
3.Dependance of β²πβ²ππ ππππππππ πππππππππππππ βnβ :
πΉπππ π =ππ2π
π , we get π β π , according to CFET π ππ tri-atomic metals must be more
than that of di-atomic and mono-atomic metals but from experimental results, it has been
found that π 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 π on concentration n.
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 3
4. Mean free pathβ Ξ»β:
Mean free path of Cu from CFET is given by Ξ»= π£π=2.85nm. But the experimentally
determined value of Ξ»=28.5 nm, which is 10 times more than that Ξ» of CFET.
Q:State/Mention the assumptions of (Sommerfeldβs)quantum free electron theory(QFET)
The assumptions of QFET are:
1.The energy of free electrons are quantized.
2. Free electrons obey Pauliβs 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 πΆπ£ =2π
πΈπΉπ π
where k = Boltzmann constant and πΈπΉ=Fermi energy. But 2π
πΈπΉ = 10β4 β΄ πΆπ£ = 10β4 π π
which agrees with experimental value.
2.Dependance of electrical resistivity β²πβ² on βTβ(temperature):
We know that(WKT) π = ππππ
πβ and π =
π
π£πΉ where π£πΉ=Fermi velocity, Ξ»=wavelength
β΄ π = πππ
πβ
π
ππ β΄ π β π β¦ . . (1)
Also it has been shown that Ξ» β1
π΄β
1
π2β
1
π Ξ» β
1
π β¦(2)
where A=area, r=radius & T=temperature.
β΄ ππππ ππππ 1&2,we get π βπ
π» which is the experimentally determined relation.
3.Dependance of β²πβ²ππβ²πβ²:
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 4
WKT conductivity π = πππ
πβ
π
ππ , thus π πππππππ πππ‘β ππ π πππ
π
ππ .With the decreases of
atomicity, n decreases and π
ππ increases so that n .
.π
ππ increases ,so that π 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 Pauliβs 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 Ξ» = β
ππ£ ,also Ξ» =
2π
π where symbols have usual
significance. β΄ ππ£ =βπ
2π , where v & k are vectors
differentiating this w.r.t βtβ,
We get m π π
π π =
π
ππ
π π
π π β¦.(1)
If an electric field E is applied ,the electron experience a force m π π
π π = βππΈ β¦(2)
β΄ ππππ πππ’ππ 1&2, π€ππππ‘ , π
ππ
π π
π π = βππΈ
β΄ ππ = β2π
βππΈ ππ‘
ππ Integration between the limits 0 to t ,we get
β« πππ‘
0 = β
2π
βππΈ β« ππ‘
π‘0
k(π‘) β π(0) = β2π
βππΈ (π‘ β 0)
ππ = β2π
βππΈ π‘
The Fermi sphere displace through ππ in a direction opposite to the direction of
applied electric field E in a time βtβ as shown in the diagram.
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 5
If t = π , π‘βππ ππ’π π‘π ππππππ ππππ π‘βπ ππππ πππ πππππππππ‘ ππππ£ ππ πππππ π πβπππ is given
by ππππ£ = β2π
βππΈ π β¦.(3),
Using mv = β
2ππ
we get π ππ£ππ£ = β
2πππππ£
β΄ ππ£ππ£ = β
2ππππππ£ β¦.(4)
From eqns 3&4,we get ππ£ππ£ = βππΈ
π π
but ππ£ππ£ = π£π drift velocity as initial average velocity of electrons is zero.
ie: π£π = βππΈ
π π β¦ (5)
Also current density J = βπππ£π β¦.(6)
β΄ ππππ ππππ 5&6, we get J = ππ2π
π E
β΄ π½
πΈ =
ππ2π
π β¦.(7)
But from Ohmβs law conductivity π = π½
πΈ β¦..(8)
β΄ ππππ ππππ 7&8 ,we get π = ππππ
π β¦(7),
where m is called effective electron mass usually denoted by πβ
Also resistivity π =1
π β¦(8)
β΄ ππππ ππππ 7&8 ,we get π =π
ππππ
Q:Explain the terms: Mobility, Fermi velocity, Fermi temperature, Fermi Level &Fermienergy.
a) Mobility(ΞΌ) of electrons is defined as the drift velocity (ππ )acquired by the electrons per unit
electric field(E). ie: ΞΌ = ππ
π¬ ΞΌ =
π
ππ ΞΌ =
ππ
π
b) Fermi velocity(ππ) of an electron is defined as itβs velocity when itβs energy is equal to the
Fermi energy. ie: 1
2ππ£πΉ
2 = πΈπΉ π£πΉ = (β2πΈπΉπ
)
c) Fermi temperature(π»π) is defined as the temperature at which the average thermal
energy of the electron is equal to the Fermi energy at 0K. ie: π»π =π¬π
π where k=Boltzmann
constant.
d) Fermi level is defined as the highest filled energy level in a metal at 0K.
e) Fermi energy(π¬π)is defined as the energy of the highest occupied level in a metal at 0K.
E
0
ππ
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 6
Q:What is Fermi-Dirac statistics ? Explain.
Fermi βDirac statistics is the statistical rule applied to the distribution of identical,
indistinguishable particles of spin 1
2 called fermions like electrons, which obey Pauliβs
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+π(πΈβπΈπΉ)/ππ
where πΈπΉ = πΉππππ ππππππ¦, E=energy of State, k=Boltzmann constant and
T= temperature.
Case1. At T=0K & E < πΈπΉ ,f(E)= 1
Case2. At T=0k & E > πΈπΉ ,f(E)= 0 and
Case3. At T> 0πΎ & E = πΈπΉ ,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) = π
π+π(π¬βπ¬π)/ππ» ,where πΈπΉπΉππππ ππππππ¦,
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 < πΈπΉ (π¬ β π¬π)ππ β π£π ,then π(π¬βπ¬π)/ππ» = πββ = 0
β΄ f(E) = π
π+π(π¬βπ¬π)/ππ» = π
π+π =1,
Thus the probability of occupation up to Fermi level is 100%.
Case-2: If E > πΈπΉ (π¬ β π¬π)ππ + π£π ,then π(π¬βπ¬π)/ππ» = πβ = β
β΄ f(E) = π
π+π(π¬βπ¬π)/ππ» = π
π+β=
1
β =0,
Thus the probability of occupation above Fermi level is 0%.
Case-3: If E = πΈπΉ (π¬ β π¬π) = 0 ,then π(π¬βπ¬π)/ππ» = ππ
π indeterminate
f(E)
1 T = 0K
0.5 T> ππΎ
0 E
πΈπΉ
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 7
When T > 0K
β΄ f(E)= π
π+π(π¬βπ¬π)/ππ»
π°π π¬ = π¬π (π¬ β π¬π) = π π‘βππ π(π¬βπ¬π)/ππ» = ππ = 1
β΄ f(E) = π
π+π(π¬βπ¬π)/ππ» = π
π+π =
π
π 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) ππΈ = 8β2ππ3/2
β3 πΈ1/2 ππΈ 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=β« π(πΈ)π(πΈ)ππΈ 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 πβ
Effective mass is equal to the true mass if the electron is in vacuum.
g(E)
0
E
f(E) Note: π1 = 500π < π2=1000k
1 T = 0K
0.5 π1 < π2
0 E
πΈπΉ
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 8
Q:Explain the dependence of resistivity on temperature and
impurity Or State and explain Matthiessenβs rule.
1. Variation of resistivity π of metals with temperature T is given by
Matthiessenβs rule π = ππ + ππ·π (T)
Where ππ = π‘πππππππ‘π’ππ πππππππππππ‘ πππ ππ π‘ππ£ππ‘π¦ ππ’π π‘π ππππ’πππ‘πππ & πππππππππ‘ππππ
ππβ (T)= π‘πππππππ‘π’ππ πππππππππ‘ πππ ππ π‘ππ£ππ‘π¦ ππ’π π‘ππππ‘π‘πππ
πβπππππ£πππππ‘ππππ .
2. Variation of resistivity π with temperature T is shown graphically.
At 0K, ππ·π (T) = 0 = ππ ,
3. For pure metals ππ = 0 ,but πππππ‘ππππππ¦ πππ‘ πππ π ππππ π‘π πππππ’ππ ππ’ππ πππ‘πππ , β΄ ππ β
0
4. and at large T , ππ is negligible compared to πππππ ππβ (T) β΄ π = ππβ (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 πΌπ = πππππ£π ,where ππ = ππ’ππππ ππππ ππ‘π¦, π = ππππ ,
π£π = πππππ‘ π£ππππππ‘π¦ & π = πβππππ
Also current due to holes is πΌβ = πβπππ£β
But the total current I= πΌπ + πΌβ
= ea(πππ£π + πβπ£β)
πΌ
π = e (πππ£π + πβπ£β)
ππΎπ, ππ’πππππ‘ ππππ ππ‘π¦ π½ =πΌ
π
β΄ J= e (πππ£π + πβπ£β)
but from Ohmβs law J=ππΈ
ie: ππΈ = π (πππ£π + πβπ£β)
π = π (πππ£π
πΈ+ πβ
π£β
πΈ)
π = e(ππππ+ππππ)
where ππ&πβ mobilities of electrons and holes respectively.
For an intrinsic semiconductors ππ = πβ = ππ called the density of intrinsic charge
carriers.
For n-type semiconductor ππ β« πβ π = eππππ
π
ππ
0 T
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 9
For p-type semiconductor, ππ βͺ πβ π = eππππ
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 ( ππ)and holes( πβ) in an intrinsic semiconductor is equal to the
square of the intrinsic charge carriers( ππ2) at any temperature.
ie: ππ. ππ = πππ
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 = (8β2π(ππ
β)3/2
β3 ) πΈ1/2ππΈ ,where ππβ = effective mass of
electron.
Also, the probability of occupation of an energy level is given by Fermi factor
f(E) =1
π(πΈβπΈπΉ
ππ )
+1
where πΈπΉ = πΉππππ ππππππ¦ & π = π΅πππ‘π§ππππ ππππ π‘πππ‘
Then ,the concentration of electron charge carriers is given by
ππ = β« π(πΈ)β
πΈπ g(E)dE
= β«1
π(πΈβπΈπΉ
ππ )
+1
β
πΈπ (
8β2π(ππβ )3/2
β3 ) πΈ1/2ππΈ
where ππβ = effective mass of electron.
On integration we can show that
ππ = 4β2 ( πππ
β ππ
β2)
3/2 π
(πΈπΉβπΈπ
ππ)
Similarly, the concentration of hole charge carriers is given by
πβ = β« π(πΈ)0
ββ g(E)dE
= β«1
π(πΈβπΈπΉ
ππ )
+1
0
ββ (
8β2π(πββ )
3/2
β3 ) πΈ1/2ππΈ
where πββ = effective mass of electron.
On integration we can show that
πβ = 4β2 ( ππβ
β ππ
β2 )3/2
π(
βπΈπΉππ
)
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 10
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 ( ππ)and holes( πβ) in an intrinsic semiconductor is equal to the square
of the intrinsic charge carriers( ππ2) at any temperature.
ie: πππ = ππ. ππ
But , WKT, πππππππ‘πππ‘πππ ππ πππππ‘ππππ , ππ = 4β2 ( πππ
βππ
β2 )3/2
π(
πΈπΉβπΈπππ
) and
Concentration of holes, πβ = 4β2 ( ππβ
β ππ
β2)
3/2 π
(βπΈπΉππ
)
From law of mass action,
ππ2 = ππ . πβ
β΄ ππ2 = 4β2 (
πππβππ
β2 )3/2 π(πΈπΉβπΈπ
ππ) π 4β2 (
ππββππ
β2 )3/2 π(
βπΈπΉππ
)
ππ = β4β2 ( πππ
βππ
β2 )3/2 π(πΈπΉβπΈπ
ππ) π 4β2 (
ππββ ππ
β2 )3/2 π(βπΈπΉππ
)
on simplification we can we can show that,
ππ = 4βπ ( π ππ
βππ»
ππ )π/π( π ππ
β ππ»
ππ )π/π π(
βπ¬π
πππ»)
Q:Explain the terms: Fermi velocity & Fermi temperature.
Fermi velocity (π£πΉ)is defined as the velocity of the electron having the energy
equal to Fermi energy(πΈπΉ) ,mathematically ,1
2ππ£πΉ
2 = πΈπΉ β΄ π£πΉ = β2πΈπΉ
π
Fermi temperature (ππΉ)is defined as the ratio of Fermi energy(πΈπΉ) to
the Boltzmann constant(k). Mathematically,ππΉ =πΈπΉ
π
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 11
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 β =ππ
ππ ,where n =1,2,3β¦..h = Planckβs constant and e = electron charge.
f. Isotopic effect states that the transition temperature ππΆvaries inversely
as the square root of the atomic mass M ie: ππΆ β1
βπ
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
ππ called the residual resistivity at 0K , ππ is due to impurities and imperfections in
π‘βπ metals.
3. The resistivity of the metals increases with temperature due to ion vibrations .
4. The resistivity of the metals is given by Mathessianβs rule π = ππ + ππβ(T).
5. But the resistivity of superconductors decreases with the decrease of temperature and
suddenly becomes zero at the temperature ππΆ called critical/transition temperature.
6. The transition temperature depends on the impurities present in the metals.
Magnetic materials reduce ππΆ to large extent.
π Normal conductor
ππ Super conductor
0 ππΆ T
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 12
Q:Explain Meissnerβs 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
ππΆ ,the magnetic flux inside the conductor is completely expelled out of the
conductor.
5) Above ππΆ ,Bβ 0 and below ππΆ ,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(SCβs) are also called soft SCs.
2) The magnetization of Type-I SCβs increases with the increase of the applied magnetic field
and suddenly drops to zero at the magnetic field π»πΆ called the critical magnetic field as
shown graphically.
H H
T> π»πͺ ππ’πππ πππππ’ππ‘ππ T< π»πͺ
βπ
Super
Conducting Normal
State State
O π»πΆ H
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 13
3) Below π»πΆ SC becomes perfectly diamagnetic obeying Meissner effect and beyond π»πΆ SC
changes to normal state.
4) The Type-I SCβs exists only in two states namely superconducting and normal
Type-II
1) Type-II Superconductors(SCβs) are also called hard SCβs.
2) The magnetization of Type-II SCβs increases with the increase of the applied magnetic
field and begin to decrease at the magnetic field π»πΆ1 called the lower critical field and
finally becomes zero at the upper critical magnetic field π»πΆ2.
3) Up to π»πΆ1 Type-II SCβs exhibit Meissner effect with perfect diamagnetism.
4) Between π»πΆ1 & π»πΆ2 they exhibit meissner effect partially.ie: Magnetic flux begins to
penetrate at π»πΆ1 πππ πππππππ‘ππ¦ penetrates at π»πΆ2.
5) Thus Type-II SCβs exists in three states namely SC state up to π»πΆ1, intermediate state
between π»πΆ1 & π»πΆ2 and normal state beyond π»πΆ2.
6) Type-II SCβs exhibit superconductivity electrically even in the intermediate state .They can
carry large currents.
7) These SCβs 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.
βπ
Super Intermediate Normal
conducting state state
state
O π»πΆ1 π»πΆ π»πΆ2 H
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 14
Q; Explain the temperature dependence of critical field.
1) Critical magnetic field(π»πΆ) 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
π»πΆ = π»0 [1 βπ2
ππΆ2],
where π»0 is called the critical field at 0K , T = temperature and ππΆ =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 βCooper Pairβ. The attraction between the electrons is
maximum when the two electrons have equal and opposite spin and momentum.
( π1 β π ) ( π2 +q )
q
π1 π2
e e
H
π»π
Superconducting normal
State state
O ππΆ T
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 15
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 ππΆ 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 ππ‘ ππΆ 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 ππΆ =30~35K, by Bednorz and Muller
3. Later many HTSCs are found of ππΆ more than 150K using liquid Nitrogen.
In 2015 HTSC found was Hydrogen Sulphide of ππΆ = 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:Yπ΅π2πΆπ’3π7 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 Meissnerβs 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
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 16
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+ππΈβπΈπΉ
ππ
; π = π
ππ2π ; π =
1
π ; π =
ππ2π
π ;
n = π₯π·ππ΄
π ; π =
1
πππ ; Β½ mππΉ
2 = πΈπΉ ; ππ = ππΈπ
π ; Ξ» = π. π ; n =
π
ππ=
1
πππ ; π»πΆ =
π»0 [1 βπ2
ππΆ2]
1. Find the relaxation time of conduction electrons in a metal of resistivity 1.54 x10β8 Ξ©-m, if
the metal has 5.8 x1028 Conduction electrons per π3.( JAN2015)
Given: π = 1.54 x10β8 Ξ©-m ; n = 5.8 x1028 /π3 ; π = 9.1π₯10β31ππ ;
π = 1.6π₯10β19πΆ ; ππ= π =?
Using π = π
ππ2π ,we get π =
π
πππ2
= 9.1π₯10β31
1.54 π₯10β8π₯5.8 π₯1028π₯(1.6π₯10β19)2
= 3.98 π₯10 β14 S
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 17
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/π3 and ππ΄=6.02 x
1026/kg mole. (Jun/Jul 14) Given: π = 1.73 x 10β8 Ξ©-m ; M = 63.5 ; D = 8.92 x 103kg/π3 ; ππ΄=6.02 x 1026/kg mole ; π =
9.1π₯10β31ππ ; π = 1.6π₯10β19πΆ ; π₯ = 1 πππππ‘πππ/ππ‘ππ, π = ? ; π = ? & π = ?
ππ πππ n = π₯π·.ππ΄
π
= 1π₯8.92 π₯ 103π₯.6.02 π₯ 1026
63.5
= 8.456x1028 atoms.
π = 1
πππ
π = 1
8.456π₯1028π₯1.6π₯10β19π₯1.73 π₯ 10β8 = 4.27x10β3
also π = ππ
π , we get π =
ππ
π
= 4.27π₯10β3π₯9.1π₯10β31
1.6π₯10β19
= 2.43 x10β14 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: (πΈ β πΈπΉ) = 0.5 ππ = 0.5π₯1.6π₯10β19 J , k=1.38x10β23 J/K ; f(E) = 1 % = 1/100 =0.01 ; T= ?
Using f(E)= 1
1+π(πΈβπΈπΉ)/ππ ,we get T = (πΈβπΈπΉ)
π.πΌπ(1
π(πΈ) β1)
= 0.5π₯1.6π₯10β19
1.38π₯10β23.πΌπ(1
0.01 β1)
= 0.5π₯1.6π₯10β19
1.38π₯10β23.πΌπ(100 β1)
= 1262 K =1.262x103K
4. Calculate the conductivity of sodium given ππ =2 x 10β14 S. Density of sodium is 971 kg/π3,its
atomic weight is 23 and has one conduction electron/atom. (π½π’π 2012)
Given: M = 23 ; D = 971 kg/π3 ; ππ΄=6.02 x 1026/kg mole ; π = 9.1π₯10β31ππ ; π = ππ =2 x 10β14 S
π = 1.6π₯10β19πΆ ; π₯ = 1πππππ‘πππ/ππ‘ππ, π = ? ; π = ?
ππ πππ n =π₯π·.ππ΄
π
= 1π₯971π₯.6.02 π₯ 1026
23
= 2.54x1028 atoms.
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 18
Also, π = ππ2π
π
= 2.54π₯1028π₯( 1.6π₯10β19)2π₯2 π₯ 10β14
9.1π₯10β31
= 1.42x107 Ξ©m.
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
X10β14S. (π·ππ 2011)
Given: πΈπΉ = 5.5 ππ = 5.5 π₯ 1.6π₯10β19 π½ ; π = ππ = 3.97 X10β14S ; π = 9.1π₯10β31ππ ππΉ = ? :
Ξ» = ?
Using Β½ mππΉ2 = πΈπΉ ,we get ππΉ = β
2πΈπΉ
π
= β2π₯5.5 π₯ 1.6π₯10β19
9.1π₯10β31
= 1.391x 106 m/s and
Also, Ξ» =ππΉ. ππ = 1.391x 105 x 3.97 X10β14 = 5.522π₯ 10β9 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 ? (π½π’π/π½π’π2011)
Given: πΈπΉ =2.1 eV = 2.1x1.6x10β19 π½ ;T = 300K ; f(πΈ1) =0.99 ; f(πΈ2) =0.5 ;
K =1.38x10β23 J/K ; πΈ1 =? ; πΈ2 =?
Using f(E)= 1
1+π(πΈβπΈπΉ)/ππ ,
we get E = πΈπΉ + ππ πΌπ [ 1
π(πΈ) β 1]
β΄ πΈ1 = 2.1π₯1.6π₯10β19 + 1.38x10β23π₯300 πΌπ [ 1
0.99 β 1]
= 3.36 π₯10β19 β 0.1902 π₯10β19
= 3.17x10β19 π½
= 3.17π₯10β19
1.6π₯10β19 eV
= 1.98eV
Also, πΈ2 = 2.1π₯1.6π₯10β19 + 1.38x10β23π₯300 πΌπ [ 1
0.5 β 1]
= 3.36 π₯10β19 β 0
= 3.36 π₯10β19 π½
= 3.36 π₯10β19
1.6 π₯10β19 eV
= 2.1 eV
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 19
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%.(π·ππ 2010)
Given: πΈπΉ =2.1 eV = 2.1x1.6x10β19 π½ ;T = 300K ; f(πΈ1) =98%=0.98 ; f(πΈ2)=50%=0.5
k =1.38x10β23 J/K ; πΈ1? , πΈ2 =?
Using f(E)= 1
1+π(πΈβπΈπΉ)/ππ ,
we get E = πΈπΉ + ππ πΌπ [ 1
π(πΈ) β 1]
β΄ πΈ1 = 2.1π₯1.6π₯10β19 + 1.38x10β23π₯300 πΌπ [ 1
0.98 β 1]
= 3.36 π₯10β19 β 0.16 π₯10β19
= 3.2x10β19 π½
= 3.2π₯10β19
1.6π₯10β19 eV
=2eV
Also , πΈ2 = 2.1π₯1.6π₯10β19 + 1.38x10β23π₯300 πΌπ [ 1
0.5 β 1]
= 2.1π₯1.6π₯10β19 β 0
= 2.1π₯1.6π₯10β19
1.6π₯10β19 eV
= 2.1 eV
8. A uniform silver wire has resistivity 1.54 x10β8πΊπ 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 ππ3 of the material. (π½π’π 2010)
Given: π = 1.54 x 10β8 Ξ©-m ; E = 2 V/m ; π = 5.8 π 1022/ππ3 = 5.8 π 1022π₯10β6/
π3 ; π = 9.1π₯10β31ππ ; π = 1.6π₯10β19πΆ ; π = ? & ππ = ?
ππ πππ: π = ππ
ππ2
= 1.54 π₯ 10β8 π₯9.1π₯10β31
5.8 π 1022π₯10β6( 1.6π₯10β19)2
= 9.44 x10β18 S
ππ = ππΈπ
π
= 1.6π₯10β19π₯2π₯9.44 π₯10β18
9.1π₯10β31
= 3.32 x 10β6 m/s
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 20
9. Calculate the probability of an electron occupying an energy level 0.02 eV above the Fermi
level at 300K.( 4 marks)
Given: (πΈ β πΈπΉ) = 0.02 ππ = 0.02π₯1.6π₯10β19 J , k=1.38x10β23 J/K and T=300K
Using f(E) = 1
1+π(πΈβπΈπΉ)/ππ
= 1
1+π(0.02π₯1.6π₯10β19)/1.38π₯10β23 π₯ 300
= 1
1+π0.773
= 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 π2/Vs. ( 4 marks)
Given: π =0.15 /Ξ©m ,π = 0.17 π2/ππ , e=1.6x10β19 C, n= ?
Using, n =π
ππ
=0.15
0.17π₯1.6π₯10β19
= 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 : π»π = 306 πππ’π π , π»πΆ = 217 πππ’π π , π = 2πΎ , ππΆ =?
Using: π»πΆ = π»π [1 βπ2
ππΆ2]
We get , ππΆ = βπ2
[1βπ»πΆπ»π
]
= β22
[1β217
306]
=3.71K
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 21
14. Calculate the mobility of electrons in copper assuming that each atom
Contribute one free electron for conduction. Resistivity of copper=1.7 X 10β8
Ξ©m, atomic weight =63.54,density =8.96 X 103kg/π3( CBCS-Jun/Jul16)
Given: π = 1.7 x 10β8 Ξ©-m ; M = 63.54 ; D = 8.96 x 103kg/π3 ; ππ΄=6.025 x 1026/kg mole
; π = 9.11π₯10β31ππ ; π = 1.6π₯10β19πΆ ; π₯ = 1πππππ‘πππ/ππ‘ππ, π = ? ; π = ?
ππ πππ n =π₯π·.ππ΄
π
= 1π₯8.96 π₯ 103π₯.6.025 π₯ 1026
63.54
= 8.5x1028 atoms.
π = 1
πππ
= 1
8.5π₯1028π₯1.6π₯10β19π₯1.7 π₯ 10β8 = 4.325x10β3 π2/vs
15. For intrinsic gallium arsenide, the room temperature electrical conductivity is
10β6 /Ξ©m.The electron and hole mobilities are respectively 0.85 π2/ππ and
0.04 π2/ππ . πΆalculate the intrinsic carrier concentration at room temperature.
Given: ππ = 10β6 /Ξ©m ; ππ = 0.85 π2/ππ ππ = 0.04 π2/ππ ; ππ =?
ππ πππ, ππ = ππe (ππ + πβ),
π€π πππ‘ , ππ =ππ
π (ππ +πβ)
ππ =10β6
1.6π₯10β19 (0.85+0.04)
= 7.023π₯1012/π3
16. The effective mass for the electron in germanium is 0.55 ππ,where ππππ π‘βπ
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: ππ = 9.1 x 10β31kg ; ππβ = 0.55x 9.1 x 10β31kg ; T= 300K ; πΈπ = 0.66ππ;
πΈπΉ =πΈπ
2 =
0.66
2 = 0.33 ππ
Using ; ππ = 4β2 ( πππ ππ
β
β2 )
3/2
π(
πΈπΉβπΈπππ
)
= 4β2 ( ππ₯0.55π₯ 9.1 π₯ 10β31π₯1.38π₯10β23π₯300
(6.625π₯10β34)2)3/2 x π
(0.33β0.66)π₯1.6π₯10β19)
1.38π₯10β23π₯300
= 5.657x1.807x1024 x2.892x10β6
ππ = 2.956x1019 /π3
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 22
17. For intrinsic Gallium Arsenide , the electric conductivity at room temperature
is ππβππππβππβπ. The electron and hole mobilities are respectively 0.85 ππ/π½πΊ πππ
π. ππ ππ/π½πΊ. Calculate the intrinsic carrier concentration at room temperature.
(04 Marks)
Given: ππ = 10β6 /Ξ©m ; ππ = 0.85 π2/ππ ππ = 0.04 π2/ππ ; ππ =?
ππ πππ, ππ = ππe (ππ + πβ),
π€π πππ‘ , ππ =ππ
π (ππ +πβ)
ππ =10β6
1.6π₯10β19 (0.85+0.04)
= 7.023π₯1012/π3
18. Calculate the probability of finding an electron at an energy level 0.02 eV above
Fermi level at 200 K . ( 04 Marks)
Given: (πΈ β πΈπΉ) = 0.02 ππ = 0.02π₯1.602π₯10β19 J , k=1.38x10β23 J/K and T=200K
Using f(E) = 1
1+π(πΈβπΈπΉ)/ππ
= 1
1+π(0.02π₯1.602π₯10β19)/1.38π₯10β23 π₯ 200
= 1
1+π1.161
= 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/π3 ,197 and 2.21 x 10β8 Ξ©π . Calculate the free electron concentration
and mobility of conduction electron. ( 04 Marks)
Given : D=19300 kg/ππ , M =197 , π = 2.21 x 10β8 Ξ©π , π = 1 electron/atom,
π΅π¨ = 6.02 x ππππ/Kmol , e=1.6 x ππβππ C , n =? , π = ?
Using, n =π π« π΅π¨
π΄
β΄ π =ππ ππππππ π.πππ ππππ
πππ = 5.898 x ππππ electrons/ππ
π¨πππ , using π =π
πππ
β΄ π =1
2.21 π₯ 10β8π₯5.898 π₯ 1030π₯1.6 π₯ 10β19 = 4.795x10β5 π2/Vs
REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS (17 PHY12/22) SEM-I/II Page 23
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 Ξ© carries
a current of 15 A. Given the mobility of free electrons in copper is 4.3xππβπππ/Vs.
( 04 Marks)
Given: T = ππππ² ,L = π π , πΉ = 0.02 Ξ© , I = 15 A , π = 4.3xππβπ ππ/Vs,
m= ππ = 9.1 x ππβππkg , k = 1.38 x ππβππ J/K , ππ =? , ππ =?
Using ππ = πE but E = π½
π³ and V= IR
we get ππ =ππ°πΉ
π³ =
π.ππππβππ ππ π π.ππ
π = 6.45 x ππβπ m/s
Also using π
πππ» =
π
ππππ
π
We get , π½π = βπππ»
π = β
πππ.ππ π ππβππππππ
π.π π ππβππ =1.168xπππ 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 Einsteinβs 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 βE1) = hΞ½ & 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 βhΞ½β & excite to the higher energy state E2
if (E2 βE1) = hΞ½ .
Mathematically it(induced absorption) is represented as
hΞ½ +A β π¨β or photon +atomβ ππ‘ππβ
Also, Rate of induced absorption = π΅12 π1ππ ,where ππ = ππππππ¦ ππππ ππ‘π¦ &
π΅12 = πΈπππ π‘πππβ²π ππππ’πππ πππ ππππ‘πππ πππππππππππ‘.
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 hΞ½ =(E2 βE1).
Mathematically, it is represented as π¨β β A + hΞ½
Also, Rate of Spontaneous emission = π΄21 π2
π€βπππ π΄21 = πΈπππ π‘πππβ²π π ππππ‘πππππ’π ππππ π πππ πππππππππππ‘.
πΈ2 π2 π¨β β’
hΞ½ β
πΈ1 β’ π¨ π2
πΈ2 π¨β β’ π2
β hΞ½
πΈ1 π1 π¨ β’
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 2
3.Stimulated emission:
Stimulated emission is the phenomenon in which an atom (π¨β) in the excited state of
energy E2 de-excite to the lower energy state E1 under the influence of an external photon
(hΞ½) by emitting an identical photon of energy hΞ½ =(E2 βE1).
Mathematically,ππ‘ ππ ππππππ πππ‘ππ ππ ππ + π¨β β A +2 hΞ½
Also, Rate of Stimulated emission= π΅21 π2ππ ,where ππ = ππππππ¦ ππππ ππ‘π¦ &
π΅21 = πΈπππ π‘πππβ²π πππππ‘πππππ’π ππππ π πππ πππππππππππ‘.
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 10β3s 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 = ππ
2 ,where Ξ» = wavelength and n = number of stationary waves
produced.
πΈ2 π¨β β’ π2
hΞ½ hΞ½ + hΞ½
πΈ1 π1 β’A
πΈ2 π2 β’
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 3
πΈ:Derive an expression for energy density in terms of Einsteinβs coefficients (or)
Derive the relation between Einsteinβs coefficients
.
Consider a system of atoms in thermal equilibrium with radiation of energy density
ππ & frequency βΞ½β. Let N1 & N2 be the population of the energy states E1 & E2
respectively, where (E2 >E1 )
We know that,
Rate of induced absorption = π΅12 π1ππ ,
where π΅12 = πΈπππ π‘πππβ²π πππππππππππ‘ ππ ππππ’πππ πππ ππππ‘πππ .
Rate of Spontaneous emission = π΄21 π2 π€βπππ π΄21 =
πΈπππ π‘πππβ²π πππππππππππ‘ ππ π ππππ‘πππππ’π ππππ π πππ πππ
Rate of Stimulated emission= π΅21 π2ππ ,
where π΅21 = πΈπππ π‘πππβ²π πππππππππππ‘ ππ ππ‘πππ’πππ‘ππ ππππ π πππ .
At thermal equilibrium,
Rate of induced absorption = π ππ‘π ππ πππππ‘πππππ’π
ππππ π πππ +
π ππ‘π ππ ππ‘πππ’πππ‘ππ ππππ π πππ
ie: π΅12 π1ππ = π΄21 π2 + π΅21 π2ππ
β΄ ππ( π΅12 π1 β π΅21 π2) = π΄21 π2
ππ: ππ = π΄21 π2
( π΅12 π1β π΅21 π2) dividing both Nr & Dr by π΅21 π2 ,we get
= π΄21
π΅21
1
[ π΅12 π΅21
π1π2
β1] β¦β¦β¦.(1)
According to Boltzmannβs law, π1
π2 = π(πΈ2 βπΈ1)/πΎπ = π βπ/πΎπ β¦β¦.(2)
From eqns 1 &2 , we get. ππ = π΄21
π΅21
1
[ π΅12
π΅21 πβπ/πΎπ β1]
β¦β¦..(3)
But .the energy density given by Planckβs law is ππ = 8πβπ3
πΆ3
1
( πβπ/πΎπ β1) β¦β¦(4)
Comparing eqns 3 & 4 ,we get π¨ππ
π©ππ =
ππ πππ
πͺπ or
π΄21
π΅21 = ππ ( π
βπ
πΎπ β 1) and π©ππ = π©ππ
Thus coefficient of stimulated absorption = coefficient of stimulated emission.
Thus energy density πΌπ = π¨ππ
π©ππ
π
[πππ/π²π» βπ]
Induced absorption Stimulated emission
πΈ2 π2 β’ β’
hΞ½ hΞ½ hΞ½ 2hΞ½
πΈ1 β’ π1 β’ β’
Spontaneous emission
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 4
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
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 5
Q:Explain the principal, construction and working of CπΆπ 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
πͺπΆπ laser diagram
Cπ2 π2 π»π gases outlet
Water outlet
π1 -optical cavity(oc) π2 (oc)
5 m
A electrodes C 2.5 cm Laser
BW BW=Brewster window
Water inlet Ba
Energy level diagram
π2 Resonant transfer of energy Cπ2
π = 1 πΆ5 (meta stable state)
πΆ4 10.6 ππ laser
Excitations by collission πΆ3 9.6 ππ laser
πΆ2
π = 0 Ground state πΆ1
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 6
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 β π₯ =1 &
C5 from their ground states 0 & C1 due to collision as follows:
e+π΅π β π΅πβ
+eβ and e+πͺπΆπ β πͺπΆπβ
+eβ
where e& e βare the energies of electron before and after collision.
5) π΅πβ 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 π₯=1 of π2 given by π΅πβ +πͺπΆπ β πͺπΆπ
β+π΅π
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 β C4 producing laser 10.6ππ (IR region)
C5 β C3 producing laser 9.6ππ (IR region)
C4 β C2
C3 β C2 Radiation less transitions
C2 β 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 πΈπΉπ PS CB Ba p P Laser πΈπΉπ h+e
n-type VB Polished surfac(PS) Laser Roughened surface(RS) MC
P-type
Pn-jn
n-type
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 7
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 8870Γ in IR region , GaAsP produce laser
beam of 6500Γ 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.
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 8
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 (Ξ») is made to fall on beam splitter, which split the beam in to
two beams. One beam which passes through splitter is called βreference beamβ. 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 βspherical object beamβ.
4. The reference and object beams produce concentric circular rings called βGaber zones
βwhere both intensity and phase are recorded due to interference on the photographic
plate.
5. On developing photographic plate we obtain βhologramβ.
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 (Ξ») Hologram
Beam splitter reference beam
Hologram Observation direction
Laser beam (Ξ»)
Object Virtual image Object Real image
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 9
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(Ξ») 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:
π2/π2 inlet
Condenser lens
material
Laser beam
Condenser lens
Materials to be weld
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 10
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
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 11
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
β’ β’ β β’ β’ β’ β β’
pollutants
β’ β’ β β’ β’ β’ β β’
laser Detector
Analyser
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 12
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 π1 , π2 & ππ be the RI of core, cladding and launch medium respectively. Also OA incident
ray,AB refracted ray,BC totally reflected ray, ππ, ππ & β be the angles of incidence, refraction at A & angle of
incidence at B respectively.
By snellβs law at A, ππ π ππ ππ = π1 π ππ ππ π ππ ππ = π1
πππ ππ ππ β¦β¦(1)
But, from βππ π΄π·π΅, ππ =(90 ββ ) π ππ ππ = π ππ(90 β β )=cos(β ) β¦ . (2)
when ππ ππ πππ₯πππ’π = ππ , ππππππ‘ππππ πππππ, π‘βππ β = β πΆ Critical angle β¦β¦..(3)
From eqns 1,2 &3 ,we get π ππ ππ = π1
πππππ (β πΆ) β¦β¦.(4)
Also, π ππ(β πΆ) =π2
π1 and πππ (β πΆ) = β(1 β π ππ2β πΆ) =
βπ12βπ2
2
π1 β¦(5)
From eqns 4 &5,we get π ππ ππ =βπ1
2βπ22
ππ or ππ = π ππβ1
βπ12βπ2
2
ππ
for air ππ = 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 = π ππ ππ = βπ1
2βπ22
ππ
Launch medium (ππ ) Fibre axis
A
ππ
O
π· Core (π1)
ππ
D
B Cladding ( ππ )
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 13
Q:What is meant by fractional refractive index change(β).
The ratio of the difference between refractive indices of core and cladding to that
of core is called fractional refractive index change. ie: β =(π1βπ2 )
π1
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 π4.
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
Lambertβs law states that the rate of decrease of intensity of light with distance(βππ
ππΏ )
in a medium is directly proportional to the original intensity(P)
ie: βππ
ππΏ β π
ππ
ππΏ = βπΌπ β¦(1) ,where πΌ is called the attenuation coefficient.
Rewriting(1 ) as ππ
π = βπΌ. ππΏ ,
Integrating this between the limits (ππ , ππ) πππ π πππ (0, πΏ)πππ πΏ,
β«ππ
π= βπΌ
,ππ
ππβ« ππΏ
πΏ
0
[πππ10P]ππ
,ππ = βπΌ [L]0πΏ
πππ10 (,ππ
ππ) = βπΌLβ¦β¦(2) where ππ , ππ are ππ & out put powers.
Equation (2),can be written as πΆ = βππ
π³ πππππ (
,π·π
π·π) dB/km.
or π·π
π·π= π
βπΆπ³
ππ
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 14
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 = π2
2
Q: What is V-Parameter/number ?
It is the quantity which represent the number of modes of the fibre given by
V = ππ
π . NA =
ππ
π .
βπ12βπ2
2
π0
where d = diameter , Ξ» = wavelength,
π1, π2 &π0 RIβs 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 β 10ππ of uniform RI and cladding diameter
60-70ππ has π’ππππππ π πΌ
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
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 15
2) Multimode step index fibre(MMF)
Input output
1) MMF has a core diameter 50 β 200ππ of uniform RI and cladding diameter
100-250ππ has π’ππππππ π πΌ.
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 π‘π 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 β 200ππ of variable RI and cladding diameter 100-250ππ
has π’ππππππ π πΌ.
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 π‘π 0.3 .
Cladding
Core
Input output
Index profile Cross section profile Modes profile Pulse profile
Cladding
Core
Input output
Index profile Cross section profile Modes profile Pulse profile
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 16
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
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 17
PROBLEMS SECTION
MODULE-3 : LASERS AND OPTICAL FIBRES
Formulae needed: L = ππ
2 ; πΌ = β
10
πΏ log[
ππ
ππ ] ;
π2
π1 = πβ(βπΆ/πππ)
; E= P.t and E =n.βπΆ
π
;
NA = βπ1
2βπ22
ππ ; Sinππ = NA ; V =
ππ
π X NA ; N =
π2
2 ; β =
(π1βπ2)
π1 ;
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 ; Ξ» = 632.8 nm = 632.8π₯10β9 m. n = ?
Using L = ππ
2 ,we get n =
2πΏ
π
= 2π₯1
632.8π₯10β9
= 3.161 x 106 πππππ
2. A fiber with an input power of 9ΞΌW has a loss of 1.5 dB/km. If the fiber is 3000 m long,
calculate the output power.(JAN2015)
Given: ππ = 9ππ€ = 9π₯10β6π€ ,πΌ = 1.5 ππ΅/ππ ,L = 3000m = 3 km , ππ = ?
Using πΌ=β10
πΏ log β¦
ππ
ππβ§ we get ππ = ππ π
βπΌπΏ
10
= 9π₯10β6πβ1.5π₯3
10
= 5.74 π₯10β6π€
3. Find the ratio of populations of two energy levels in a laser if the transition between
them produces light of wavelength 6493Γ ,assuming the ambient temperature as
27.(Jun/Jul 14)
Given: Ξ» = 6493Γ = 6493x10β10 m, t=27 T = 273+27 =300K; h=6.63x10β34 Js;
k=1.38x10β23 J/K ;C=3x108m/s ;π2
π1 = ?
Using π2
π1 = π
β(βπΆ
πππ)
= πβ(
6.63π₯10β34π₯3π₯108
6493π₯10β10π₯1.38π₯10β23π₯300)
= πβ79.99
= 7.33x10β33
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 18
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: πππ₯ NA = 0.2 , π2 = 1.59 , πβ²π = 1.33 , π1 =? aππ πβ²π = ?
For air , Using NA = βπ1
2βπ22
ππ , we get π1 = β(πππ₯ππ΄)2 + π2
2
= β( 0.22 + 1.592 )
= 1.603
Also πβ²π = πππβ1(
βπ12βπ2
2
πβ²π
) = πππβ1( β1.6032β1,592
1.33 ) = 8.65Β°
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 = 1.5x10β3 π€ ,t=20 ns =20x10β9 s , C=3x108m/s ; h=6.63x10β34 Js,
n = 1.047 x108 ,Ξ» = ?
Using E= Pxt and E = n.βπΆ
π , π€π πππ‘ , Ξ» =
πβπΆ
ππ‘
= 1.047 π₯108π₯6.63π₯10β34π₯3π₯108
1.5π₯10β3 π₯20π₯10β9
= 6.942x10β7 m.
6. The angle of acceptance of an optical fibre is 30Β° π€βππ 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) (π½π’π
π½π’π2011)
Given: ππ =30Β° , πβ²π=4/3=1.333 , ππ = 1 πππ πππ , πβ²π =?
Using πβ²ππππ πβ²π = ππSinππ = βπ12 β π2
2 = constsnt
We get , πβ²π = πππβ1 (πππππππ πβ²πβ )
= πππβ1 (1 πππ30Β° 1.333β )
=22.03Β°
7. Calculate the NA,V-number and number of modes in an optical fibre of core diameter
50ΞΌm,core and cladding refractive indices 1.41 and 1.4 at wavelength 820 nm. (π½π’π 2012)
Given: π1 = 1.41 ; π2 = 1.4 ; Ξ» = 820 nm= 820 x 10β9 m ; d = 50ΞΌm =50x 10β6 m ;
ππ =1 for air NA = ? ; V = ? & N = ?
a. NA = βπ1
2βπ22
ππ =
β(1.41)2β(1.4)2
1 = 0 .168
b. V = ππ
π X NA =
ππ₯50π₯ 10β6π₯0 .168
820 π₯ 10β9 = 32.18
c. N = π2
2 =
(32.17)2
2= 517.5 = 518
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 19
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 45ΞΌm.Calculate its relative refractive index
difference, V- number at wavelength 1000 nm and number of modes. (π·ππ 2011) Given: π1 = 1.45 ; π2 = 1.4 ; Ξ» = 1000 nm= 1000 x 10β9 m ; d = 45ΞΌm =45x 10β6 m ; ππ = 1; β = ? ; V = ?
& N = ?
1) β =(π1βπ2)
π1 =
(1.45β1.4)
1.45= 0.0345
2) V = ππ
π π
βπππβππ
π
ππ=
3.14π₯45π₯ 10β6π₯β1.452β1.42
1000 π₯ 10β9 π₯1= 53.34
3) N = π2
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 6328Γ . (π·ππ 11)
Given: P=4.5mW = 4.5x10β3 π€ ,t= 1 S , C=3x108m/s ; h=6.63x10β34 Js,
Ξ» = 6328Γ = 6328x10β10 m ; n = ?
Using E= Pxt and E =n.βπΆ
π π =
πππ‘
βπΆ
= 6328π₯10β10
π₯4.5π₯10β3π₯1
6.63π₯10β34 π₯3π₯108
= 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. (π½π’π/
π½π’π2011)
Given: L= 1 m ; Ξ» = 632.8 nm = 632.8π₯10β9 m. n = ? and
Using L = ππ
2 , we get , n =
2πΏ
π
= 2π₯1
632.8π₯10β9
= 3.161 π₯ 106 πππππ
11. The ratio of population of two energy states in a laser 1.059 x10β30.If the temperature of
the system is 57Β°πΆ, π€βππ‘ ππ π‘βπ π€ππ£ππππππ‘β ππ π‘βπ πππ ππ ?
Given: π2
π1 = 1.059x10β30, t = 57Β°πΆ, T =273+57=330K; h=6.63x10β34 Js; k=1.38x10β23
J/K ;C=3x108m/s ; Ξ» = ?
Using π2
π1 = πβ(βπΆ/πππ) ,we get , In(
π2
π1) = β(βπΆ/πππ) πππππ: ππππ (
π2
π1) = In(
π2
π1)
Ξ» = β(βπΆ/πΌπ (π2
π1) ππ)
= β6.63π₯10β34 π₯3π₯108
πΌπ(1.059π₯10β30) π₯1.38π₯10β23π₯330
= 6.328x10β7 π
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 20
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. (π·ππ 2010)
Given: ππ = 200ππ = 200π₯10β3π€ , ππ = 90% ππ = 0.9π₯200π₯10β3π€ ;
L = 3000 m = 3 km ; πΌ =? ,
Using πΌ= β10
πΏ log[
ππ
ππ] = β
10
3 log [
0.9π₯200π₯10β3
200π₯10β3] = 0.1525 ππ΅/ππ
13. Calculate on the basis of Einsteinβs theory the number of photons emitted per second by
He-Ne laser source emitting light of wavelength 6328Γ with an optical power
10mW. (π½π’π 2010) Given: P=10mW = 10x10β3 π€ ,t=1 s , C=3x108m/s ; h=6.63x10β34 Js, ,
Ξ» = 6328 Γ = 6328 x 10β10 m ; n = ?
Using E= Pxt and E =n.βπΆ
π , π€π πππ‘ π =
πππ‘
βπΆ
= 6328 π₯ 10β10 π₯10π₯10β3 π₯1
6.63π₯10β34 π₯3π₯108
= 3.18 x1016 /π3
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. (π½π’π 2010)
Given: π1 = 1.5 ; π2 =97% π1 = 0.97 x1.5 =1.455 ; π0 = 1 ; ππ΄ = ? ; ππ = ? & ππΆ = ?
Using 1) NA = βπ1
2βπ22
ππ
= β1.52β1.4552
1 = 0.365
2) π ππ π0= ππ΄ ππ = πππβ1(ππ΄)
= πππβ1(0.365) = 21.41Β°
3) π ππ ππΆ =π2
π1 ππΆ = πππβ1 (
π2
π1) = πππ
β1(1.455
1.5)
= 75.93Β°
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 , Ξ» =694 nm=694x10β9 m ,t=20 ns =20x10β9 s , C=3x108m/s ;
h=6.63x10β34 Js, n =?
Using E=P.t and E =n.βπΆ
π ,
π€π πππ‘ n = ππ‘π
βπΆ
= 5π₯20π₯10β9 π₯694π₯10β9
6.63π₯10β34 π₯3π₯108
=3.489x1011 photons/pulse.
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 21
15. The angle of acceptance of an optical fiber kept in air is 30Β°.Find the angle of acceptance
when the fiber is in a medium of refractive index 4/3.( 4 marks)
Given: ππ =30Β° , πβ²π=4/3=1.333 , ππ = 1 πππ πππ , πβ²π =?
Using πβ²ππππ πβ²π = ππSinππ =constant (βπππ β ππ
π ) for a fibre
We get πβ²π = πππβ1 (
πππππππ
πβ²π
)
= πππβ1 (1 πππ30Β°
1.333)
= 22.03Β°
16. The ratio of population of two energy levels out of which one corresponds to metastable
state is 1.059 x10β30.Find π‘βπ π€ππ£ππππππ‘β ππ πππβπ‘ ππππ‘π‘ππ ππ‘ 330πΎ.(CBCS-Jun/Jul16)
Given: π2
π1 = 1.059x10β30, T =330K; h=6.63x10β34 Js; k=1.38x10β23 J/K ;
C=3x108m/s ; Ξ» = ?
Using π2
π1 = πβ(βπΆ/πππ)
we get , In(π2
π1) = β(βπΆ/πππ) πππππ: ππππ (
π2
π1) = In(
π2
π1)
Ξ» = β(βπΆ/πΌπ (π2
π1) ππ)
= β6.63π₯10β34 π₯3π₯108
πΌπ(1.059π₯10β30) π₯1.38π₯10β23π₯330
= 6.328x10β7 π
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 45ΞΌm.Calculate the refractive index change and
numerical Aperture.(CBCS-Jun/Jul16)
Given: π1 = 1.45 ; π2 = 1.4 ; ; d = 45ΞΌm =45x 10β6 m ; ππ = 1, β = ? ; & NA = ?
1) β = (π1βπ2)
π1
= (1.45β1.4)
1.45 = 0.0345
2) NA = βππ
πβπππ
ππ
= β1.452β1.42
1 = 0.3775
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 22
19.The average power output of a laser beam of wavelength 6500Γ is 10 mW.Find
the number of photon emitted per second by the laser source.
Given: P =10mW = 10x10β3 π€ ; t=1 s ; C=3x108m/s ; h=6.63x10β34 Js ;
Ξ» = 6500 Γ = 6500 x 10β10 m ; n = ?
Using E= Pxt and E =n.βπΆ
π
We get, π = πππ‘
βπΆ
= 6500 π₯ 10β10 π₯10π₯10β3 π₯1
6.63π₯10β34 π₯3π₯108
=3.268 x 1016 πβππ‘πππ /π3
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: ππ = 85% ππ = 0.85 ππ ; L = 500m = 0.5 km; πΌ =? ,
Using πΌ = β10
πΏ log[
ππ
ππ]
= β10
0.5 log [
0.85 ππ
ππ] = β
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: πΌ = 2ππ΅
ππ, (
ππ
ππ)
πΏ=2ππ=?, (
ππ
ππ)
πΏ=5ππ=?
Using (ππ
ππ)
πΏ = 10
βπΌπΏ
10
(ππ
ππ)
πΏ=2ππ= 10
βπΌπΏ
10 = 10β2π₯2
10 = 0.3981
(ππ
ππ)
πΏ=5ππ= 10
βπΌπΏ
10 = 10β2π₯5
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 45ΞΌm. Calculate its
fractional refractive index change and numerical aperture.
Given: π1 = 1.45 ; π2 = 1.4 ; ; d = 45ΞΌm =45x 10β6 m ; β = ? ; & NA = ?
1) β = (π1βπ2)
π1 =
(1.45β1.4)
1.45 = 0.0345
2) NA = βππ
πβπππ
ππ
= β1.452β1.42
1
= 0.3775
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 23
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 Γ . ( 04 Marks)
Given: Ξ» = 6928Γ = 6928x10β10 m, t=27 T =291K; h=6.625x10β34 Js;
k=1.38x10β23 J/K ;C=3x108m/s ;π2
π1 = ?
Using π2
π1 = π
β(βπΆ
πππ)
= πβ(
6.625π₯10β34π₯3π₯108
6928π₯10β10π₯1.38π₯10β23π₯291)
= πβ71.44
= 9.419x10β32 or = 9.441x10β32
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: π1 = 1.563; π2 = 1.498 ,ππ = 1 ; ππ΄ =? aππ ππ = ?
Using NA = βπ1
2βπ22
ππ
= β1.5632β1.4982
1
= 0.4461
Also ππ = πππβ1(ππ΄ ) = πππβ1(0.4461) = 26.49Β°
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: π = 820 nm = 820 x 10β9 m , P = 22 mW = 22 x 10β3W ,
t = 12 ns = 12 x 10β9s , h = 6.625 x 10β34 JS , C = 3 x 108ππ β1 , n = ?
Using E = Pt and E = π π₯ βπΆ
π
We get , n =πππ‘
βπΆ =
820 π₯ 10β9π₯ 22 π₯ 10β3π₯ 12 π₯ 10β9
6.625 π₯ 10β34π₯ 3 π₯ 108 = 1.089 x 109 electrons/π3
C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS ( 17 PHY 12/22 ) SEM-I/II Page 24
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: π1 = 1.5 ,β =0.0005 ,ππ = 1 , π2 = ? , ππ΄ =?
Using β =(π1βπ2)
π1
We get π2 = π1 β β π1 = 1.5 β 0.0005π₯ 1.5 = 1.49925 (1.499)
Also using NA =βπ1
2βπ22
π0 =
β1.52β1.499252
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 :
β’ β’ β’ β’ β’ β’ β’ β’ β’ β’ β’ β’ β’ β’ β’
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 , & π 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
= π1+ π2+ π3π , where π1, π2 & π3 πππ πππ‘πππππ .
Q: What is Bravais lattice & non-Bravais lattice ?
Bravais lattice βis 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.
β’ β’ β’ β’
β’ β’ + = β’ β’
β’ β’ β’ β’
Lattice + Basis = Crystal structure
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 2
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 π , b & c and three interfacial angles πΌ, π½& πΎ πππ ππππππ
π‘βπ πππ‘π‘πππ πππππππ‘πππ .
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 π½ πΌ 0 b y a πΎ x
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 3
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
πβ π β π c πΌ = π½ = πΎ=90 Types:SC/BaC/BCC/F π½ 0 πΌ b Ex:KNO3, MgSO3 a πΎ
π=π =π a πΌ = π½ = πΎ β 90 Types:SC π½ πΌ Ex: As, Sb, 0 a a πΎ
π=π β π πΌ = π½ =90 , πΎ=120 c Types: SC Ex:Zn,Mg, π½ 0 πΌ a a πΎ
πβ π β π πΌ β π½ β πΎ β 90 c
Types:SC Ex: 5H2O CuSO4 π½ 0 πΌ
a πΎ b
πβ π β π πΌ = πΎ =90 β π½ c Types:SC/BaC Ex: CaSO4,2H2O π½ πΌ b a 0 πΎ
π=π =π a πΌ = π½ = πΎ=90 Types: SC/BCC/FCC π½ πΌ a Ex: Copper/Gold a 0 πΎ
π=π β π c πΌ = π½ = πΎ=90 Types: SC/BCC Ex: SnO2,TiO2 π½ πΌ
0 a a πΎ
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 4
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 π,π &π.
Ex: Let X,Y &Z axes intercepts are 3π,2π &1π respectively.
2. Find the coefficients of π,b &π.
ie: 3,2 & 1.
3. Find the reciprocals of the coefficients of basis vectors.
ie:The reciprocals are π
π ,
π
π,
π
π
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 π
ππ π ,
π
ππ π,
π
ππ π = π, π, π = (π π π)
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
βbarβ h or 2
3. The miller indices ( h k l ) do not represent a single plane but set of parallel planes.
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 5
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 βON β β₯ππ to the plane ABC ,then ON=π represent inter-planar distance/spacing.
Then OA= π
β , OB=
π
π &OC =
π
π ,where π,π& π are basis vectors
From right πππππ βππ OAN, Cos ππ =ππ
ππ΄ =
π
π/β =
βπ
π β¦β¦.(1)
From βππ OBN, Cos ππ =ππ
ππ΅ =
π
π/π =
ππ
π β¦β¦.(2) and
also, from βππ OCN, Cos ππ =ππ
ππΆ =
π
π/π =
ππ
π β¦β¦(3)
For orthogonal co-ordinates πππ 2 ππ+πππ 2 ππ + πππ 2 ππ = 1 β¦β¦(4)
From eqns 1,2,3 &4 we get, β2π2
π2 +
π2π2
π2+
π2 π2
π2
= 1
ie: π2 [ β2
π2 +
π2
π2+
π2
π2 ] = 1
β΄ d = π
β ππ
ππ + ππ
ππ+ ππ
ππ
For a cubic crystal π=π =π
β΄ d= π
β ππ +ππ +ππ
Z
C
N B Y
ππ§ ππ¦
0 ππ₯
A X
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 6
Q:Derive an expression for space lattice constant βπβ for a cubic lattice.
Let π ππ π‘βπ ππππ ππ‘π¦ & M the molecular weight of a cubic crystal of lattice constantβπβ.
Then, volume of the unit cell = π3 and mass of unit cell =π π3 β¦β¦.(1)
If n be the number of molecules per unit cell and ππ΄ be the Avagadroβs number.
Then, mass of each molecule = (M/ππ΄) and
mass of molecules in the unit cell=n (M/ππ΄)β¦.(2)
From eqns 1& 2,we get π π3 = n (M/ππ΄) ππ =ππ΄
ππ΅π¨
Q:Find the relation between atomic radius (R) and lattice constant(π) for SC ,BCC & FCC
crystals:-
a) Simple cubic(SC):
If β r β be the rdius and β π β be the lattice constant/side of the unit cell,
then from the diagram π=2R π/2 = R
a) Body centered cubic (BCC) :
In a bcc cube AB=BC=CD=π and
AC=βπ2 + π2=β2π
Also from βππ ACD , π΄π·2 = π΄πΆ2 + πΆπ·2
ie: (4π )2 = (2π)2 +π2
= 3π2
4R = β3π
ππ π =β3π
4
a
C
a B
D
4R
A
AA A
a
2 R
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 7
c) Face centered cubic (FCC)
In a fcc cube AB=BC=π and AC=4R
Also from βππ ABC,
π΄πΆ2 = π΄π΅2 + π΅πΆ2
ie: (4π )2 = (π2 +π2
= 2π2
Or 4R =β2π
Or R = β2π
4 =
π
2β2
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 π₯ 8 ππππππ ππ‘πππ +
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
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 8
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 = (ππ ππ ππ‘πππ ππ π‘βπ π’πππ‘ ππππ)π₯ππππ’ππ ππ πππβ ππ‘ππ
(ππππ’ππ ππ π‘βπ π’πππ‘ ππππ)
ie: APF = π
π3 π₯
4
3 ππ 3 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 = π
π
Using, APF = π
π3 π₯ 4
3 ππ 3
= 1
π3 π₯ 4
3 π(
π
2 )3
= 1
π3 π₯ 4
3 π
π3
8
= π
6 = 0.52 or 52%
For BCC , n = 2, R = βππ
π
Using, APF = π
π3 π₯ 4
3 ππ 3
= 2
π3 π₯ 4
3 π(
β3π
4 )3
= 2
π3 π₯ 4
3 π
3β3π3
64
= β3π
8= 0.68 or 68%
For FCC, n = 4, R = π
2β2
Using, APF = π
π3 π₯ 4
3 ππ 3
= 4
π3 π₯
4
3 π(
π
2β2 )3
= 4
π3 π₯ 4
3 π
π3
16β2
= π
3β2 = 0.74 or 74%
β’
β’
β’
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 9
NOTE: Important data related to SC , BCC & FCC cells.
Sl.
No
Particulars Simple Body
Centere
Face
Centered
1 Volume of a unit cell π3 π3 π3
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)
π (β3/2)π π/β2
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 π
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
π
4
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 10
AB=2R=βπ2
16 +
π2
16+
π2
16
= β3π2
16
= β3π
4 or
R= βππ
π
For Diamond n = 8 & R = βππ
π
β΄ Using APF = π
π3 π₯
4
3 ππ 3
= 8
π3 π₯ 4
3 π(
β3π
8 )3
= 8
π3 π₯ 4
3 π
3β3π3
8π₯8π₯8
= πβ3
16
= 0.34 or 34%
Q: What is perovskite ? explain.
Perovskie is the common name for all the oxides of type ABπΆπ,where A & B
are different metals.
Ex: π΅ππππ3 ( π΅ππππ’π π‘ππ‘ππππ‘π) , πΆππππ3( calcium titanate) etc
The common structure of perovskite is as follows:
= 8 Ba/Ca atoms form are at the edges of the cell.,
= 6 π3atoms are at the face of the cell and
= 1 ππ 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.
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 11
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 Braggβs law of diffraction.
Let the incident parallel x-rays of wavelength Ξ», incident on two atomic planes of
Separation d at glancing angles π and reflect along BC and EF respectively.
Draw BG & BH β₯ππ to DE & EF respectively.
From the βππ BGE & BHE we have GE=BE.Sinπ & π»πΈ = π΅πΈ. ππππ
where BE = d ,interplanar distance.
β΄ πππ‘β ππππππππππ, ππ = πΊπΈ + π»πΈ
= π ππππ + π ππππ
= 2π ππππ β¦ . (1)
Also for constructive interference pd= nΞ» β¦β¦(2)
β΄ ππππ ππππ 1 &2 , π€π πππ‘ nΞ» = ππ πΊπππ½ ,
Incident rays -AD A C reflected rays-CF
Glancing angle
D π π F
β β B β β β
G π π H d = Interplanar spacing
β β β E β β
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 12
This equation is called the Braggβs law
Q:Describe the construction & working of Braggβs X-ray spectrometer.
The labeled schematic diagram of the Braggβs X-ray Spectrometer is as shown in the
diagram.
To verify Braggβs 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 π
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 π½ gradually and
note the corresponding ionization current each time.
6. Plot a graph of βIβv/s βπβ² πππ ππππ π‘βπ ππππβ note down the glancing angles π1, π2& π3
for the 1st,2nd &3rd order diffraction respectively.
7 Then substituting for π1, π2& π3 in the Braggβs equation 2d Sinπ = ππ ,
It will be found that ,Sinπ½π: πΊπππ½π: πΊπππ½π = π: π: π,which verifies Braggβs 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 π. The position of the
Vernier- π1 Circular scale
Slits glancing angle-π Crystal target
π1 π2
X-rays Vernier-π2
Collimated X-rays
π3
π4 E -electrometer
Ionisation chamber (D) Ba
Intensity
0 π
π1 π2 π3
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 13
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 π100 plane of the crystal to the incident X-rays and find the glancing angle
π1 for 1st order Spectrum.
5. Repeat the experiment as before for the crystal surfaces π110 & π111 and find
the corresponding 1st order glancing angles π2 & π3 .
6. Then using braggβs equation, 2d Sinπ = ππ ,
find π πππ βΆ π πππ βΆ π πππ = π
πΊπππ½π :
π
πΊπππ½π:
π
πΊπππ½π
7. The nature of the crystal is identified from the ratios of 1
ππππ1 :
1
ππππ2:
1
ππππ3
as follows:
For SC; π
πΊπππ½π :
π
πΊπππ½π:
π
πΊπππ½π = 1:
π
βπ :
π
βπ
For BCC; π
πΊπππ½π :
π
πΊπππ½π:
π
πΊπππ½π = 1:
π
βπ :
π
βπ
For FCC; π
πΊπππ½π :
π
πΊπππ½π:
π
πΊπππ½π = 1:
π
βπ :
π
βπ
PROBLEMS SECTION
MODULE-4 : CRYSTAL STRUCTURE
Formulae needed : nΞ» = 2dSinπ ; d= π
ββ2 +π2 +π2 ;
For SC ,n=1& a = 2R , BCC, n=2 & a = 4
β3 R , FCC , n=4 & a = 2β2 π
Atomic Packing Factor (APF) = π
π3 x
4ππ3
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 = 0.127x10β9π , (β π π ) = ( 3 2 1),
πππ πππ π = 2β2 π = 2β2 x 0.127x10β9π , π = ?
Using d = π
ββ2 +π2 +π2
= 2β2 π₯ 0.127π₯10β9
β32 +22 +12
= 9.60 x 10β11 π
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 14
2. π·πππ€ 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 Braggβs reflection occurs at an angle of 20Β° in the plane
(2 1 2 ).Find the wavelength of X-rays if lattice constant is 3.614Γ . (Dec 2013/Jan2014)
Given: n = 1 ; π = 20Β° ; ( h k l ) = ( 2 1 2 ) ; a = 3.614Γ = 3.614x10β10 m ; Ξ» = ?
Using nΞ» = 2dSinπ & d= π
ββ2 +π2 +π2
We get , Ξ» = 2π ππππ
πββ2 +π2 +π2
= 2π₯3.614π₯10β10 πππ20Β°
1π₯β22 +12+22 = 8.240 x 10β11 m
4. Monochromatic X-rays of wavelength 0.82 Γ undergo first order reflection from a crystal
of cubic lattice with lattice constant 3 Γ at a glancing angle of 7.855Β°. Identify the possible
planes which give rise to this reflection in terms of Miller indices. (π·ππ 11 & π½π’π β π½π’π16)
Given: n = 1 ; π = 7.855Β° ; a = 3 Γ = 3x10β10 m ; Ξ» = 0.82 Γ =0.82 x 10β10 m ; ( h k l ) = ?
Using nΞ» = 2dSinπ & d= π
ββ2 +π2 +π2
We get , ββ2 + π2 + π2 = 2π ππππ
ππ
= 2π₯3π₯10β10 πππ(7.855)
1π₯0.82 π₯ 10β10 = 1
Thus the possible planes ( h k l ) =( 1 0 0 ) ,(0 1 0 ) ,( 0 0 1)
(2 0 0) π± =
π
π, π² = β & π§ = β
Z Β½ 0 y X
(2 1 0) π± = βπ
π, , π² = π & π§ = β
Z Β½
0 1 Y
X
(1 3 2 ) π± = π, π² = βπ
π & π§ =
π
π
Z Β½
β1
3
0 Y 1 X
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 15
5. Calculate the glancing angle of the (1 1 0) plane of a simple cubic crystal (a=2.814 Γ )
corresponding to second order diffraction maximum for the x-rays of wavelength
0.710 Γ .
Given: n = 2 ; ( h k l ) = (1 1 0 ) ; a = 2.814Γ = 2.814x10β10 m ; Ξ» = 0.710
Γ =0.710 x 10β10 m ; π =?
Using nΞ» = 2dSinπ & d= π
ββ2 +π2 +π2
We get , π = π ππβ1( ππββ2 +π2 +π2
2π )
= π ππβ1(2π₯0.710 π₯ 10β10π₯β12 +12 +0
2π₯2.814π₯10β10 ) = 20.91Β°
6. Draw the following planes in the unit cube: i) ( 1 0 2 ) ii) ( 1 1 2 ). (π·ππ 2010)
Ans: i) ( 1 0 2 ) π₯ = β1, π¦ = β & π§ =1
2 & ii) (1 1 2 ) π₯ = β1, π¦ = 1 & π§ = β
1
2
7. A monochromatic X-ray beam of wavelength 1.5Γ undergoes second order Bragg
reflection from the plane (2 1 1) of a cubic crystal, at a glancing angle of 54.38Β°,calculate
the lattice constant. (π½π’π 2010)
Given: n = 2 ; ( h k l ) = (2 1 1 ) ; Ξ» = 1.5 Γ =1.5 x 10β10 m ; π = 54.38Β° ; π = ?
Using nΞ» = 2dSinπ & d= π
ββ2 +π2 +π2
We get , π = ππββ2 +π2 +π2
2ππππ
= 2π₯1.5 π₯ 10β10 π₯β22 +12 +12
2π₯πππ 54.38
= 4.52 x 10β10 π
( 1 0 2 ) π = βπ, π = β & π§ =π
π
Z
Β½ β1
0 Y
x
(1 1 2 ) π = βπ, π = π & π§ = βπ
π
Z β1
0 1 Y
X(β1
2)
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 16
8. In a calcite crystal, second order Braggβs reflections occur from the planes with d-spacing
3 Γ ,at a glancing angle of 24Β° .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 Γ =3x10β10 π , π = 24Β° , n = 2 ,Pd = ? and Ξ» = ?
Path difference,Pd = 2d Sinπ =2x3x10β10 xSin24Β° =2.44 x10β10 m
Also, Ξ» = ππ
π
= 2πππππ
π
= 2.44 π₯10β10
2
= 1.22 x10β10 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 =0.144x10β9 m , (h k l)=(1 1 0),
For FCC a =2β2 π =2β2 π₯0.144π₯10β9 π , d = ?
Using d= π
β(β2+π2+π2 )
= 2β2 π₯0.144π₯10β9
β(12+12+02 )
= 2.88 x 10β10 π
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 Γ .
Given: n = 2 ; ( h k l ) = (1 3 2 ) ; a = 3.81Γ = 3.81x10β10 m ; Ξ» = 0.058nm=0.058 x 10β9 m
; π =?
Using nΞ» = 2dSinπ & d= π
ββ2 +π2 +π2
We get , π = π ππβ1( ππββ2 +π2 +π2
2π )
= π ππβ1(2π₯0.058 π₯ ππβππ₯β12 +32 +22
2π₯3.81π₯10β10 )
= 34.72Β° ( also d = 1.018 X 10β10 m)
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 17
11. Draw the following planes in a cubic unit cell.
i) (2 0 0) ii) (Ξ Δͺ 0) iii) (Ξ 0 2) iV) (Ξ Ξ 2)
12. An X-ray beam of wavwlength 0.7 Γ undergoes first order Braggβs reflection
from the plane ( 302) of a cubic crystal at glancing angle 35Β° , πππππππππ πππ
lattice constant. ( 04 Marks)
Given: n = 1 ; ( h k l ) = ( 302 ) ; Ξ» = 0.7 Γ =0.7 x 10β10 m ; π = 35Β° ; π = ?
Using nΞ» = 2dSinπ & d= π
ββ2 +π2 +π2
We get , π = ππββ2 +π2 +π2
2ππππ
= 1π₯0.7 π₯ 10β10 π₯β32 +02 +22
2π₯πππ 35
= 2.200 x 10β10 π [ Note: d = 6.102x10β11m]
(102)βX = 1,Y = β & π = Β½ Z
Β½
0 Y
X 1
(112)βX = 1,Y = 1 & π = Β½
Z
Β½
0 1 Y
X 1
(Ξ Δͺ 0)βX = 1 ,Y = β1 & π = β
z
0 y
1
x
β1 0 Y
1
X
(200)βX = Β½ ,Y = β & π = β Z
0 Y 1/2
X
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 18
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 Γ . ( 04 Marks)
Given: Intercepts 3a:4b: βπ , a= π = π =2Γ =2 x 10β10m , (h k l ) =? ,d = ?
Coefficients of a,b and c are 3 ,4, β
Reciprocals of the coefficients are 1
3 ,
1
4 ,0
LCM 0f the denominators 3 & 4 is 12
Miller indices are 1
3π₯12 = 4 ,
1
4π₯12 = 3 , 0 x 12= 0
β΄ ( h k l ) = ( 4 3 0 )
π΄ππ π π€βππ π = π = π
We have d =π
ββ2+π2+π2 β΄ d =
2 π₯ 10β10
β42+32+02 = 4 x 10β11m
(102) X =1/1 =1, Y =1/0 = β 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 = β, Y =1/0 = β and Z =1/2
Z
Β½
0 Y
X
(011) X =1/0 = β, Y =1/1 =1and Z =1/1 = 1
Z 1
0 1 Y
X
REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY12/22) SEM-I/II Page 19
14. Draw the following planes in a cubic unit cell,
i) ( 0 0 I ) ii) ( I Δ« 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
Β½
0 1 y
X 1
(020) x =1/0= β, y =1/2 & z = 1/0 = β
Z
0 Β½ y
x
( I Δ« 0 ) x =1/1= 1, y =1/-1=β1 & z =1/0 =β
z
0 1 y
X 1
(001) x =1/0= β, y=1/0=β & z =1/1=1
Z 1
0 y
X
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
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)
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 2
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 = π½
π , 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(π) is the half the angle of cone of sound waves formed and is given by
π = πΊππβπ (π
π΄)
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, ππ£ = ππππ π‘πππ‘ or
β΄ π1π£1 = π2π£2 , Where π1, π2densities & π£1 , π£2 velocities.
2. Conservation of momentum states that the sum total momentum of the system always
remain constant.
Mathematically,π + ππ£2 = ππππ π‘πππ‘ ππ π1 + π1π£12 = π2 + π2π£2
2
Where π1, π2 pressures , π£1, π£2 velocities and π1, π2densities.
3. Conservation of energy states that the sum total energy of a system is always remains
constant.
Mathematically, β +π£2
2 = ππππ π‘πππ‘ ππ β1 +
π£12
2= β2 +
π£22
2
Where β1, β2 enthalpies and π£1, π£2 velocities.
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 3
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 π·π
π·π =
ππΈ
πΈ+ππ2 β
πΈβπ
πΈ+π β¦β¦.(1)
But temperature ratio, π»π
π»π = (
π
π2+
πΈβπ
π) (
ππΈ
πΈβπ π2 β 1) (
π(πΈβπ)
(πΈ+π)π) β¦β¦..(2)
Substituting for π2 ππππ πππ 1, ππ πππ 2, π€π πππ‘
π»π
π»π =
[π+πΈβπ
πΈ+π
π·ππ·π
]π·ππ·π
π·ππ·π
+πΈβπ
πΈ+π
β¦..(3) β΅ 4ab+(π β π)2 = (π + π)2
Also , density ratio, ππ
ππ =
π·ππ·ππ»ππ»π
β¦β¦.(4)
From eqns 1,3 &4,we get,
ππ
ππ
=
π·π
π·π
[π +πΈ β ππΈ + π
π·π
π·π]
π·π
π·ππ·π
π·π+
πΈ β ππΈ + π
= (
πΈ+π
πΈβπ
π·ππ·π
+π
πΈ+π
πΈβπ+
π·ππ·π
) β¦.(5)
eqns 3 & 5 are called RH relations/NS relations.
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 4
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 π2,temparatureπ2.
4. Two sensors π1 & π2 separated by a distance βπ fitted to driven section measures the
pressures π4,π5 and temperatures π4, π5 respectively.
Working :
1. Driver section is filled with gas at high pressure and Driven section is filled with gas of
low pressure π1 & π‘πππππππ‘π’ππ π1 .
2. Diaphram is ruptured to produce shock waves by pushing the piston and the rupture
pressure π2 & temperature is measured using sensor S.
3. The timeβπβ² taken by the shock wave to travel the distance βxβ is measured using CRO. also
the pressures π4,π5 and temperatures π4, π5 are measured using the sensors π1 & π2
respectively.
4. The speed of the shock waves is calculated using V = π₯
π‘ .
5. Then the match number of the shock waves is calculated using M = π
π ,where a is the
speed of sound at temperature π1.
6. Also the mach number M can be calculated using the RH relations π·π
π·π =
ππΈ
πΈ+ππ2 β
πΈβπ
πΈ+π , by finding π2, π1 for the gas of known πΎ.
Piston Diaphragm S π1( π4) π πππ πππ π2(π5)
π2 π1 x =7 cm
Driver section/gas Driven section/gas 2.9cm
Plunger
49 cm 51 cm
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 5
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 = πβππ ππ/ππ¬π/π
ππ ππΈ
Where m = mass, E = energy, h = Planckβs 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
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 6
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) ππΈ = 4ππ
β2 dE where m = mass,
h =Planckβs 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 = 2β2π1/2πΈβ1/2
β 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 πΈ1πΈ2πΈ3 πΈ
g(E)
0 πΈ1 πΈ2 πΈ3 πΈ
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 7
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 πΈ1 πΈ2 πΈ4 πΈ
Top Down method
Bulk material
Powder
Nanoparticals
Bottom Up method
Nanoparticals
Clusters /Groups
Atoms/molecules
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 8
Q: Explain Ball Mill method of synthesis/producing nanoparticles.
1. Ball mill method is a mechanical method based on β Top Down β 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
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 9
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 CNTβs..
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
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 10
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 ππ 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 ,SWCNTβs are
produced.
6. MWCNTβs are produced if the anode is not coated with catalytic agents.
7. Pure CNTβs can be obtained by using pure graphite rods.
Q:Expain the production of CNTS by Pyrolysis method.
π΅π πͺππ―π
1. The schematic diagram of Pyrolisis method is as shown in the diagram.
π»π πππ πππππ‘ π»π gas outlet
Chamber
Graphite anode
500 torr pressure 5 Ba
Graphite cathode
Furnace
Chamber
C Catalyst
Pressure gauge
π΅π πͺππ―π
Substrate on graphite sheet
β β β β β
β
β β β β β
β
β
β
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 11
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 MWCNTβs.
7. SWCNTβ
s are obtained when acetylene is replaced with methane or carbon monoxide at 1200
Q: Mention the properties of CNTSβ.
1. CNTβs are highly elastic.
2. Youngβs modulus of CNTs is about 9 times sronger than that of steel.
3. CNTβs exhibit large strength in tension.
4. CNTβs can be bent without breaking.
5. Electrical properties of CNTs ranges from semiconductor to good conductor.
6. CNTβs 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. CNTβs 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 CNTSβ.
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.
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 12
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 π·1.
5. Secondary electrons produced due to interaction with valence electrons are detected by
the detector π·2
6. and X-rays produced due to deep penetration are detected using the detector π·3
7. High resolution 3D image can be seen on the TV monitor to which π·1, π·2& π·3 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
π·1 back scattered electron detector
π·2 Secondary electron detector
π·3 X-rays detector
Evacuated chamber Specimen
Base
TV monitor
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 13
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 = π
π‘ ; M =
π
π ; Ξ» =
β
β2πππ or
1.228 π10β9
βπ ( 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=100x10β3 m,t =200ΞΌS=200π₯10β6 S,a=340 m/s ,M=?
Using, ,V=π
π‘ and M =
π
π
We get, M = π
ππ‘ =
100π₯10β3
340π₯200π₯10β6
=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=6.63x10β34 Js; m=9.1x10β31kg; e=1.6x10β19 C; Ξ» =?
Using Ξ» = β
β(2πππ)
= 6.63π₯10β34
β(2π₯9.1π₯10β31π₯1.6π₯10β19 π₯60π₯103) = 5.016 x 10β12 m
or Ξ» = 1.228 π10β9
βπ
= 1.228 π10β9
β(60π₯103) = 5.013 x 10β12 m
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 14
3. In a Reddy shock tube,it was found that, the time taken to travel between the two sensors
is 195 ΞΌs.If the distance between the two sensors is 100mm,find the Match number.(CBCS-
Jun/Jul16)
Given: x=100 mm=100x10β3 m,t =195ΞΌS=195π₯10β6 S,a=?(Not given) ,M=?
Using, ,ππ = π
π‘
= 100π₯1 0β3
195π₯10β6
= 512.8 m/s and
Also, M = ππ
π
= 512.8
π
4. Calculate the wavelength of an electron accelerated under a potential
difference of 100 V in scanning electron microscope.
Given: V=100V , h=6.63x10β34 Js; m=9.11x10β31kg; e=1.6x10β19 C; Ξ» =?
Using Ξ» = β
β( 2πππ)
=6.63π₯10β34
β( 2π₯9.11π₯10β31π₯1.6π₯10β19 π₯100)
= 1.228 x 10β10 m
or Ξ» = 1.228 π10β9
βπ
= 1.228 π10β9
β(100)
= 1.228 x 10β10 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=150x10β3 m,t = 0.3mS= 0.3π₯10β3 S, a=340 m/s ,M=?
Using, ,V = π
π‘ and M =
π
π
We get, M = π
ππ‘
= 150π₯10β3
340π₯0.3π₯10β3 = 1.471
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 15
6. Calculate the wavelength of an electron accelerated under a potential
difference of 100 V in scanning electron microscope.
Given: V=100V , h=6.63x10β34 Js; m=9.11x10β31kg; e=1.6x10β19 C; Ξ» =?
Using Ξ» = β
β(2πππ)
= 6.63π₯10β34
β(2π₯9.11π₯10β31π₯1.6π₯10β19 π₯100)
= 1.228 x 10β10 m
@@end@@
OR Ξ» = 1.228 π10β9
βπ
= 1.228 π10β9
β(100)
= 1.228 x 10β10 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 βs 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 Braggβs la a d e plai ho to erif Braggβs la a d Ide tif cr stals usi g Braggβs 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 SUCCESSβ¦β¦..
β SLOW AND STEADY WINS THE RACE β
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.
@@@@@