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SOLVED 60:40 CBCS

VTU DEC’17 QUESTION PAPER

Model Question Paper

Practice QUESTION Paper

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

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

17PHY12

First Semester B.E. Degree Examination,Dec-2017 Engineering Physics

Time: 3 hrs. Max. Marks: 100

𝐍𝐨𝐭𝐞: 𝟏. 𝐀𝐧𝐬𝐰𝐞𝐫 𝐚𝐧𝐲 𝐅𝐈𝐕𝐄 𝐟𝐮𝐥𝐥 𝐪𝐮𝐞𝐬𝐭𝐢𝐨𝐧𝐬, 𝐜𝐡𝐨𝐨𝐬𝐢𝐧𝐠 𝐎𝐍𝐄 𝐟𝐮𝐥𝐥 𝐪𝐮𝐞𝐬𝐭𝐢𝐨𝐧 𝐟𝐫𝐨𝐦 𝐞𝐚𝐜𝐡 𝐦𝐨𝐝𝐮𝐥𝐞.

2. Physical constants: Velocity of light C = 3x𝟏𝟎𝟖 m/s; Planck’s constant h = 6.63x𝟏𝟎−𝟑𝟒Js;

Mass of electron, 𝐦𝐞 = 9.1x𝟏𝟎−𝟑𝟏kg; Boltzmann constant ,k = 1.38x𝟏𝟎−𝟐𝟑J/k;

Avagadro number , 𝐍𝐀 = 6.02x𝟏𝟎𝟐𝟑 / mole.

Module-1 1. a) Write the assumptions of Planck’s law of radiation. Deduce Wein’s law and Rayleigh-Jeans

law from Planck’s law of radiation. (7 Marks)

b) Set up time independent one dimensional Schrodinger wave equation. (6 Marks)

c) What is Compton effect ? Explain its significance. (3 Marks)

d) An electron is bound in an one dimensional potential well of width 1 Å , but of infinite wall

height. Find its energy values in the ground state and also in the first excited states.(4 Marks)

OR

2. a) State Heisenberg’s uncertainty principle. Show that electrons cannot exist inside the

nucleus. (7 Marks)

b State de Broglie hypothesis and show that group velocity is equal to particle velocity.

c) Briefly explain three properties of wave function. (3 Marks)

d) Compute the de Broglie wavelength for an electron moving with one tenth part of

the velocity of light. (4 Marks)

Module-2 3. a) Explain Fermi energy and Fermi factor. Explain the variation of Fermi factor with

temperature. (7 Marks)

b) Derive the expression for electrical conductivity of an intrinsic semiconductor.

c) Write a note on Maglev vehicles. (4 Marks)

d) The electron concentration in a semiconductor is 5 x 1017𝑚−3. Calculate the conductivity

of the material if the drift velocity of electron is 350 m𝑠−1 in an electric field of 1000 V𝑚−1.

OR (4 Marks)

4. a ) Discuss the merits of quantum free electron theory. (6 Marks)

b) What is superconductivity ? Explain Type-I and Type-II superconductors. (6 Marks)

c) What is (i) mean collision time, (ii) drift velocity, (iii) Meissner effect ? (4 Marks)

d) Calculate the Fermi velocity and the mean free path for the conduction electrons

in silver, given that its Fermi energy is 5.5V and the relaxation time for electrons

is 3.83 x 10−14 S. (4 marks)

Module-3 5. a) Define angle of acceptance and numerical aperture.Obtain an expression for the

numerical aperture of an optical fiber. (7 Marks)

b) What is holography? Explain the principle of construction of hologram with suitable

ray diagram. (5 Marks)

c) Explain the processes of spontaneous emission and stimulated emission. (4 Marks)

d) A medium in thermal equilibrium at temperature 300K has two energy levels

with a wavelength separation of 1 µm. Find the ratio of population densities of

upper and lower levels. (4 Marks)

OR

6. a) Describe the construction of C𝑂2 laser and explain its working with the help of

energy level diagram. (6 Marks)

b) Discuss the three types of optical fibers with suitable diagrams. (6 Marks)

c) Mention four applications of LASER. (4 Marks)

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

of acceptance when it is in a medium of refractive index 1.33. (4 marks)

Module-4 7. a) Explain in brief the seven crystal systems with neat diagrams. (7 Marks)

b) Explain the crystal structure of diamond with neat diagram and calculate its

atomic packing factor. (6 Marks)

c) Define unit cell, primitive cell and Bravais lattice. (3 Marks)

d) Calculate the glancing angle for incidence of X-rays of wavelength 0.58 Å on the

plane(132) of NaCl which results in second order diffraction maxima taking the

lattice constant as 3.81 Å (4 Marks)

. OR

8. a) What are Miller indices ? Derive an expression for interplanar distance in terms

of Miller indices. (7 Marks)

b) Define co-ordination number and packing factor. Calculate the packing factor for

SCC and FCC structure. (6 Marks)

c) Derive Bragg’s law. (4 Marks)

d)Draw the following planes in a cubic unit cell: i) (1 1 1 ) ii) ( 1 0 1 ) iii) ( 0 1 1 ) (3 Marks)

Module-5 9. a) Describe the construction and working of Reddy’s shock tube. (6 Marks)

b) Discuss the variation of density of energy states for 3D,2D,1D and 0D structures.

c) Describe sol-gel method of producing nano particles. (5 Marks)

d) Mention any three applications of nano particles. (3 Marks)

OR

10 a)Describe the principle, construction and working of a scanning electron

microscope. (8 Marks)

b) Define: i) Mach number ii) Subsonic waves

iii) Supersonic waves iv) Ultrasonic waves. (4 Marks)

c) Explain pyrolysis method of obtaining carbon nanotubes. (4 Marks)

d) 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 100 microsecond. If the velocity of

sound under the same conditions is 340 ms−1, find the Mach number of the shock

wave. (4Marks)

**end**

SOLVED CBCS

ENGINEERING PHYSICS

DECEMBER-2017QUESTION PAPER

VTU MODEL QUESTION PAPER

PRACTICE QUESTION PAPER

C REDDAPPA SOLVED CBCS DEC-2017 VTU ENGINEERING PHYSICS QUESTION PAPER INTERLINE PUBLISHING.COM

17PHY12 VTU ENGINEERING PHYSICS Page 1

17PHY12

First Semester B.E. Degree Examination,Dec-2017 Engineering Physics Time: hrs. Max. Marks:

: . 𝐀 𝐰 𝐲 , . 2. Physical constants: Velocity of light C = 3x m/s; Planck s constant h = 6.63x − Js; Mass of electron, = 9.1x − kg;

Boltzmann constant ,k = 1.38x − J/k; Avagadro number , 𝐀 = 6.02x / mole.

Module-1 1. a) Write the assumptions of Planck s law of radiation. Deduce Wein s law and Rayleigh-Jeans

law from Planck s law of radiation. (7 Marks)

Assumptions of Planck s law of radiation 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(h i.e: E = n.h ,where n = Quantum number. 3. The oscillating particles absorb or emit energy in discrete units of h , only when they

undergo transitions between any two allowed energy states.

Planck s law of radiation is given by d = 𝛑 − d ……………

a) Wein s law:

For shorter wave lengths, is very small ,hence Uλ & e hλk are

Large i.e: e hλk so that e hλk − = e hλk ……….

from eqn 1 &2,we get Uλd = πλ hλk d

d = 𝐂 − −𝐂 d This is wein s law .where C = πhc & C =

b) Rayleigh-Jeans law: For longer wave lengths, is very large,

hence e hλk is very small ,Expanding e hλk as power series ,



we get , e hλk = 1+ λ + λ +………. = + λ , neglecting higher powers of λ ie: e hλk − = λ ………

from eqn 1 &3,we get Uλd = πλ hλk d : d = 𝛑

d , This is Rayleigh-Jeans law

b) Set up time independent one dimensional Schrodinger wave equation. (6 Marks)

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

freely in the positive direction of -direction is given by 𝚿 = Ae x−

= Ae x e− ,

Where Ae x represent the time independent part of the wave function and is

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

x = A ik e x and

x =A ik i e x

= i k Ae x ……. From eqns 1 & 2 we get

x = − πλ …….. = − & = 𝛑

But, de-broglie wave length = λ = = ½

= K EK = ½mv

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

λ = − ……

From eqns 3 & 4 ,we get x = − π −



x + π − = …… This is Schrodinger time independent equation for a particle For a free particle ,V=0

+ 𝜋ℎ =

c) What is Compton effect ? Explain its significance. (3 Marks)

The scattering of X-rays due to recoil of electrons is called Compton effect . Significance: Compton effect signifies the particle nature of X − rays photons/light

d) An electron is bound in an one dimensional potential well of width 1 Å , but of infinite wall height. Find its energy values in the ground state and also in the first excited states.(4 Marks)

Given: a = 1Å = 1x − m ; h=6.63x − Js; m = . x − kg ; E & E = ?

Using E = For ground state, n = E =

. x − x . x − x x − = 6.038x − J For 1st excited state, n =

E = . x − x . x − x x − = 2.415x − J

OR

2. a) State Heisenberg s uncertainty principle. Show that electrons cannot exist inside the nucleus. (7 Marks)

HUP states that the product of the uncertainty Δ in the position and the uncertainty ΔP in the momentum of a particle at any instant is equal to or greater than ( h/4𝛑)

i.e.: Δ ΔP ≥ ℎ𝜋 ,Where h is Planck s constant.

The significance of HUP is that, it is impossible to determine simultaneously both the position and momentum of the particle accurately at the same instant.



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

The minimum uncertainty in momentum ΔP ≥ π Δx

≥ . x − x . x −

ΔP ≥ 5.275 x 10-21 kg m/s = P(say)

Using E = mC , P = mV and m = √ −VC , the minimum energy E of the electron in the

nucleus is given by E = m C = o−vC = o− also P C = m v C = o−vC = o− E − P C = o ( − )− = 𝐂 + 𝐂

but the rest mass energy m c is very very small compared to P2c2,neglecting m c , we have E ≈ PC = 5.275 x − x 3 x J

= . x − x x . x − MeV

= 9.89 MeV = 10 MeV But the maximum energy of the electrons( -particle) emitted from the nucleus does not exceed 4MeV,hence electrons cannot exist in the nucleus.

b State de Broglie hypothesis and show that group velocity is equal to particle velocity. (6 Marks) de-Broglie hypothesis states that waves associated with material particles in motion. The waves are called de-Broglie/matter waves.

Consider a particle of mass m having particle velocity 𝒗 𝒂𝒓𝒕𝒊′ & 𝑖 ′𝒗 ′. We know that, 𝑔 = … … . .

but = 𝜋 & = ℎ ,we get = 𝜋 ℎ d = 𝜋ℎ …

Also from k = 𝜋𝜆 & = ℎ𝑃 we get k =

𝜋ℎ P

dk = 𝜋ℎ dP ……

From eqns 1,2 &3,we get 𝑔 = = 𝜋ℎ𝜋ℎ 𝑃 = 𝑃 …….



Also , from E = ½m = 𝑣 & P = m𝓥 , = 𝑃

dE = 𝑃 𝑖 ; 𝑃 = 𝑃 =

𝑣𝑃𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = 𝑃𝑎𝑟𝑡𝑖 ….

From eqns 4 & 5,we get 𝒗 = 𝒗 𝒂𝒓𝒕𝒊 Thus group velocity =Particle velocity

c) Briefly explain three properties of wave function. (3 Marks) Properties of the wave functions are :- (any three) 1. Wave function (𝚿) is single valued everywhere , 2. 𝚿 is finite everywhere 3. 𝚿 is continuous every where

4. First derivatives of 𝚿 are continuous every where 5. 𝛹 or 𝛹𝛹 * is called probability density.

6. 𝑖 𝑖 𝑖 𝑖 𝑖 𝑖 𝑖 𝑖 ∫ 𝛹 𝑉 =+∞−∞

d) Compute the de Broglie wavelength for an electron moving with one tenth part of the velocity of light. (4 Marks)

Given; V = = = . x ; h=6.63x − Js, m = m =9.1x − kg ;

C = 3 X m/s ; = ?

Using = V

= . x −. x − x . x = 2.429x − m.

Module-2 3. a) Explain Fermi energy and Fermi factor. Explain the variation of Fermi factor with

temperature. (7 Marks)

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

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

at thermal equilibrium.



It is given by the relation f(E) = + − / ,where E Fermi energy, E = energy of State, k = Boltzmann constant and T = temperature The variation of f(E) with temperature and energy is discussed below: When T = 0K

Case-1: If E < − 𝑖 − ,then − / 𝑻 = −∞ =

f(E) = + − / 𝑻 = + =1,

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

Case-2: If E > − 𝑖 + ,then − / 𝑻 = ∞ = ∞ f(E) = + − / 𝑻 = +∞ = ∞ =0,

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

Case-3: If E = − = ,then − / 𝑻 = indeterminate When T > 0K f(E)= + − / 𝑻

= − = ℎ − / 𝑻 = = f(E) = + − / 𝑻 = + = 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 .

b) Derive the expression for electrical conductivity of an intrinsic semiconductor. (5 Marks) In a semiconductor the net current is due to both electrons and holes . The current due to electrons is I = n eav ,where n = number density, a = area , v = drift velocity & = ℎ Also current due to holes is I = n eav But the total current I= I + I = ea(n v + n v

I = e (n v + n v

f(E) Note: = < =1000k

1 T = 0K

0.5 <

0 E



WKT, current density J = Ia J= e (n v + n v but from Ohm s law J=σE ie: σE = e n v + n v

σ = e n + n h

𝛔 = e( + ) where & mobilities of electrons and holes respectively. Note: 1. For an intrinsic semiconductors n = n = n called the density of intrinsic charge carriers. 2. For n-type semiconductor n n 𝛔 = e 3. For p-type semiconductor, n n 𝛔 = e


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

RW RW

Vehicle

Superconducting magnet

Aluminium guide way



d) The electron concentration in a semiconductor is 5 x − . Calculate the conductivity

of the material if the drift velocity of electron is 350 m − in an electric field of 1000 V − . (4 Marks)

Given: n =5 x m− , V = 350 ms− , E = 1000 Vm− . σ =?

Using σ = τ and V = τ

We get σ = V

= x x . x − x

= . /Ωm.

4. a ) Discuss the merits of quantum free electron theory. (6 Marks)

The merits of QFET are:

1.Specific heat: According to QFET the molar specific heat is given by 𝑣 = 𝐹

where k = Boltzmann constant and =Fermi energy. But 𝐹 = − 𝑣 = −

which agrees with experimental value. 2.Dependance of electrical resistivity ′𝝈′ on T (temperature):

We know that(WKT) 𝝈 = 𝝉∗ and 𝜏 = 𝜆𝑣𝐹 where =Fermi velocity, =wavelength 𝝈 = ∗ 𝝀𝒗 𝝈 ∝ 𝝀 … . .

Also it has been shown that ∝ 𝐴 ∝ 𝑟 ∝ 𝑇 ∝ 𝑇 … where A=area, r=radius & T=temperature. & ,we get 𝝈 ∝ 𝑻 which is the experimentally determined relation.

3.Dependance of ′𝝈′ ′ ′: WKT conductivity 𝝈 = ∗ 𝝀𝒗 , thus 𝝈 ℎ 𝝀𝒗 .With the decreases of

atomicity, n decreases and 𝝀𝒗 increases so that n . .𝝀𝒗 increases ,so that 𝝈 is more for

monoatomic metals than that of di&tri-atomic metals.This explains the conductivity of mono-atomic metals is more than that of di & tri-atomic metals. 4.Mean free path obtained from QFET is about 28.5 nm which is experimentally determined value.



b) What is superconductivity ? Explain Type-I and Type-II superconductors. (6 Marks)

Superconductivity is the phenomenon in which the resistivity/resistance of a substance becomes zero. Type-I

1) Type-I Superconductors(SC s) are also called soft SCs. 2) The magnetization of Type-I SC s increases with the increase of the applied magnetic field

and suddenly drops to zero at the magnetic field 𝐶 called the critical magnetic field as shown graphically.

3) Below 𝐶 SC becomes perfectly diamagnetic obeying Meissner effect and beyond 𝐶 SC changes to normal state.

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

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

2) The magnetization of Type-II SC s increases with the increase of the applied magnetic field and begin to decrease at the magnetic field 𝐶 called the lower critical field and finally becomes zero at the upper critical magnetic field 𝐶 .

−

Super

Conducting Normal

State State

O 𝐶 H

−

Super Intermediate Normal

conducting state state

state

O 𝐶 𝐶 𝐶 H



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

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

penetrate at 𝐶 penetrates at 𝐶 .

5) Thus Type-II SC s exists in three states namely SC state up to 𝐶 , intermediate state between 𝐶 & 𝐶 and normal state beyond 𝐶 .

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

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

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

indium as impurity to it.

c) What is (i) mean collision time, (ii) drift velocity, (iii) Meissner effect ? (4 Marks) Mean collision time(𝝉) is the average time elapsed between two consecutive collisions. Drift velocity(𝒗 ) is the average velocity with which the free electrons drift in a direction opposite to that of the applied electric field

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.

d) Calculate the Fermi velocity and the mean free path for the conduction electrons in silver, given that its Fermi energy is 5.5V and the relaxation time for electrons is 3.83 x − S. (4 marks) Given: E = . eV = . x . x − J ; τ = τ = 3.83 X − S ; m = . x − kg V = ? : = ?

Using ½ mV = E ,we get V = √ F

= √ x . x . x − . x −

= 1.391x m/s and

Also, =V . τ = 1.391x x 3.83 X − = . x − m



Module-3 5. a) Define angle of acceptance and numerical aperture.Obtain an expression for the

numerical aperture of an optical fiber. (7 Marks)

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.

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. Let , & be the RI of core, cladding and launch medium respectively. Also OA incident ray,AB refracted ray,BC totally reflected ray, 𝜃𝑖, 𝜃𝑟 & ∅ be the angles of incidence, refraction at A & angle of incidence at B respectively.

By snell s law at A, 𝑖 𝜃𝑖 = 𝑖 𝜃 𝑖 𝜃𝑖 = 𝑜 𝑖 𝜃𝑟 ……

But, from ∆ , 𝜃𝑟 = −∅ 𝑖 𝜃𝑟 = 𝑖 − ∅ =cos ∅ … . when 𝜃𝑖 𝑖 𝑖 = 𝜃 , , ℎ ∅ = ∅𝐶 Critical angle ……..

From eqns 1,2 &3 ,we get 𝑖 𝜃 = 𝑜 ∅ ……. Also, 𝑖 ∅ = and ∅ = √ − 𝑖 ∅ =

√ − …

From eqns 4 &5,we get 𝑖 𝜃 = √ −𝑜 or 𝜃 = 𝑖 − √ −

for air =

Launch medium ( )

Fibre axis

A

𝜃𝑖 O

𝛷 Core )

𝜃𝑟

D

B Cladding ( )



b) What is holography? Explain the principle of construction of hologram with suitable ray diagram. (5 Marks) 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. Construction of Hologram.

1. The schematic diagram for the construction of a Hologram is as shown in the diagram. 2. A laser beam of wavelength is made to fall on beam splitter, which split the beam in to

two beams. One beam which passes through splitter is called reference beam . The other reflected beam is made to fall on the object whose hologram is to be produced.

3. The beam reflected by the object is called spherical object beam . 4. The reference and object beams produce concentric circular rings called Gaber zones

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

5. On developing photographic plate we obtain hologram . 6. At each and every point on the hologram, complete information/details of the object are

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

of the object with less resolution.

c) Explain the processes of spontaneous emission and stimulated emission. (4 Marks) Let N1 & N2 be the population of the energy states E1 & E2 respectively so that (E2 −E1) = h & E2 >E1

Object

Reflected beam Object spherical beam

Laser beam ( )

Hologram

Beam splitter reference beam

𝑨∗ • 𝑨 •

⇒ h



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 h = E2 −E1). Mathematically, it is represented as 𝐀∗ → A + h Also, Rate of Spontaneous emission = A N where A = Einstein′s spontaneous emission coefficient.

Stimulated emission:

Stimulated emission is the phenomenon in which an atom (𝐀∗) in the excited state of energy E2 de-excite to the lower energy state E1 under the influence of an external photon h by emitting an identical photon of energy h = E2 −E1). Mathematically,it is represented as + 𝐀∗ → A + h Also, Rate of Stimulated emission= B N Uν,where Uν = energy density & B = Einstein′s Spontaneous emission coefficient.

d) A medium in thermal equilibrium at temperature 300K has two energy levels with a wavelength separation of 1 µm. Find the ratio of population densities of upper and lower levels. (4 Marks) Given: = µm = x − m, T =300K; h=6.63x − Js;

k=1.38x − J/K ; C=3x m/s ; = ?

Using = e− hCλk

= e−( . x − x x x − x . x − x )

= e− . = 1.364 x −

OR

𝑨∗ •

•

h h + h •A



6. a) Describe the construction of C laser and explain its working with the help of energy level diagram. (6 Marks)

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

𝑪 laser diagram

C gases outlet

Water outlet

-optical cavity(oc) (oc)

5 m

A electrodes C 2.5 cm Laser

BW BW=Brewster window

Water inlet Ba

Energy level diagram

Resonant transfer of energy C

= (meta stable state)

10.6 laser

Excitations by collission 9.6 laser

= 0 Ground state



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

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

upper & lower vibrational levels of CO2 molecules. 3) Ionisation takes place due to electric discharge when high DC voltage is applied between

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

C5 from their ground states 0 & C1 due to collision as follows: e+ → ∗

+e and e+𝑪 → 𝑪 ∗+e

where e& e are the energies of electron before and after collision. 5)

∗ molecule in excited level collide with CO2 molecules in their ground state C1 & excite it to metastable state C5 by resonant energy transfer as level C5 of CO2 is same as level 𝓥=1 of given by ∗ +𝑪 → 𝑪 ∗

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

place betweenC5 &C4 and C5 & C3. 7) The transitions/de-excitations takes place as follows:

C5 → C4 producing laser 10.6 (IR region) C5 → C3 producing laser 9.6 (IR region) C4 → C2

C3 → C2 Radiation less transitions C2 → C1

8) Due to high thermal conductivity of He, it removes heat from mixture and de-populate the lower states C3 &C2 quickly .

9) Laser beam is amplified by using optical resonators. 10) The laser output is 100kW for continuous mode and 10 kW in pulsed mode.

b) Discuss the three types of optical fibers with suitable diagrams. (6 Marks) The three types of optical fibres are:- 1) Single mode step index fibre(SMF

1. SMF has a core diameter − m of uniform RI and cladding diameter 60-70 m has uniform RI

Cladding

Core

Input output

Index profile Cross section profile Modes profile Pulse profile



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 < . . 5. Numerical aperture is < . . 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

2) Multimode step index fibre(MMF) Input output

1) MMF has a core diameter − m of uniform RI and cladding diameter 100-250 m has uniform RI.

2) The step index, cross section,modes and pulse profiles are as shown in the diagram. 3) MMF guides light in multi-modes as shown . 4) The V-number is > . . 5) Numerical aperture is . to . . 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

Cladding

Core

Input output


Cladding

Core

Input output




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

has uniform RI. 2) The graded index, cross section ,modes and pulse profiles are as shown in the diagram. 3) MMF guides light in multi-modes as shown in the diagram. 4) The V-number is > . . 5) Numerical aperture is . to . . 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.

c) Mention four applications of LASER. (4 Marks) The four applications of laser are:

1. laser welding.

2. laser cutting.

3. laser drilling.

4. Measurement of atmospheric pollutants(using LIDAR)

d) The angle of acceptance of an optical fiber is 30° when kept in air. Find the angle of acceptance when it is in a medium of refractive index 1.33. (4 marks) Given: θ =30° , n′ =1.33 , n = for air , θ′ =?

Using ′ 𝛉′ = Sin𝛉 = √n − n = constsnt

We get , θ′ = Sin− (n Sinθ n′⁄ = Sin− ( Sin ° .⁄ =22.08°



Module-4 7. a) Explain in brief the seven crystal systems with neat diagrams. (7 Marks)

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

≠ ≠ c = = = Types:SC/BaC/BCC/F 0 b Ex:KNO3, MgSO3 a

= = a = = ≠ Types:SC Ex: As, Sb, 0 a a

= ≠ = = , = c Types: SC Ex:Zn,Mg, 0 a a

≠ ≠ ≠ ≠ ≠ c Types:SC Ex: 5H2O CuSO4 0

a b

≠ ≠ = =90 ≠ c Types:SC/BaC Ex: CaSO4,2H2O b a 0

= = a = = = Types: SC/BCC/FCC a Ex: Copper/Gold a 0

= ≠ c = = = Types: SC/BCC Ex: SnO2,TiO2

0 a a



b) Explain the crystal structure of diamond with neat diagram and calculate its atomic packing factor. (6 Marks) Note: = 8 corner atoms, = 6 face atoms & = 4 body diagonal atoms

OR OR

1. Diamond is formed by two intervening FCC sub-lattices one of them is moved by 𝑎

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 = x 8+ x 6+4 =8

7. If R be the radius and a the lattice constant ,then we can show that

AB=2R=√𝑎 + 𝑎 + 𝑎

= √ 𝑎

= √ 𝑎 or

R= √ 𝒂

For Diamond n = 8 & R = √

Using APF = x πR = x

π √

O 1/2

O

1/2 O 1/2

O 1/2 O

3/4 1/4

1/4 3/4

B

2R

A



= x π √x x

= π√

= 0.34 or 34% c) Define unit cell, primitive cell and Bravais lattice. (3 Marks) Unit cellis the smallest block/geometrical figure from which the crystal can be built by repetition of it in three dimensions. Primitive cellis 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. Bravais lattice is the lattice in which all the lattice points are equivalent ,

each lattice point represent an identical set of one or more atoms/molecules. d) Calculate the glancing angle for incidence of X-rays of wavelength 0.58 Å on the plane(132) of NaCl which results in second order diffraction maxima taking the lattice constant as 3.81 Å. (4 Marks) Given: n = 2 ; ( h k l ) = (1 3 2 ) ; a = 3.81Å = 3.81x − m ; = . nm= . x − m ; θ =?

Using n = 2dSinθ & d= √ + + Note: Find, d = 1.018 X − m & then θ

We get , θ = sin− λ√ + +

= sin− x . x − x√ + + x . x −

= 34.72° 8. a) What are Miller indices ? Derive an expression for interplanar distance in terms

of Miller indices. (7 Marks)

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 and are denoted by ( h k l )



1. Consider the plane ABC of a crystal which intersect the X,Y & Z axes at A,B & C respectively.

2. Let ( h k l ) be the miller indices of the family of ABC planes.

3. Draw ON ⊥ to the plane ABC ,then ON= represent inter-planar distance/spacing.

Then OA= , OB= & OC = ,where , & are basis vectors

From right angle ∆ OAN, Cos θ = = / = ……. From ∆ OBN, Cos θ = = / = ……. and also, from ∆ OCN, Cos θ = = / = ……

For orthogonal co-ordinates, cos θ +cos θ + cos θ = ……

From eqns 1,2,3 &4 we get, + +

= 1

ie: d [ + + ] = 1

d = √ + +

For a cubic crystal = =

d= √ + +

Z

C

N B Y

𝜃 𝜃

0 𝜃

A X



b) Define co-ordination number and packing factor. Calculate the packing factor for SCC and FCC structure. (6 Marks) Co-ordination number of any atom in a crystal is the number of nearest equidistant

neighbouring atoms Surrounding that atom. 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 = xV V

For BCC , n = 2, R = √

Using, APF = x πR

= x π √

= x π √

= √ π

= 0.68 or 68%

For FCC, n = 4, R = √

Using, APF = x πR

= x π √

= x π √

= π√ = 0.74 or 74%

c) Derive Bragg s law. (4 Marks) Let the incident parallel x-rays of wavelength , incident on two atomic planes of Separation d at glancing angles θ and reflect along BC and EF respectively.

Incident rays -AD A C reflected rays-CF

Glancing angle

D 𝜃 𝜃 F

B

G 𝜃 𝜃 H d = Interplanar spacing

E



Draw BG & BH ⊥ to DE & EF respectively.

From the ∆ BGE & BHE we have GE=BE.Sinθ & = . 𝑖 𝜃 where BE = d ,interplanar distance. path difference, pd = GE + HE = d Sinθ + d Sinθ = d Sinθ … . Also for constructive interference pd= n …… from eqns & , n = 𝛉 ,

d) Draw the following planes in a cubic unit cell: i) (1 1 1 ) ii) ( 1 0 1 ) iii) ( 0 1 ) (3 Marks)

Module-5

9. a) Describe the construction and working of Reddy s shock tube. (6 Marks)

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.

Piston Diaphragm S )

x =7 cm

Driver section/gas Driven section/gas 2.9cm

Plunger

49 cm 51 cm

(111)x=y=z= =1

Z 1

0 1 y

X 1

(101)x= z= =1, y= =∞

Z 1

0 Y

X 1

(0 1)x= =∞, y= − =−1, z= =1

z 1

− 0 y

x



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 P ,temparatureT . 4. Two sensors S & S separated by a distance ∆ fitted to driven section measures the

pressures P ,P and temperatures T , T respectively.

Working :

1. Driver section is filled with gas at high pressure and Driven section is filled with gas of low pressure P & T .

2. Diaphram is ruptured to produce shock waves by pushing the piston and the rupture pressure P & temperature is measured using sensor S.

3. The time ′ taken by the shock wave to travel the distance x is measured using CRO. also the pressures P ,P and temperatures T , T are measured using the sensors S & S

respectively.

4. The speed of the shock waves is calculated using V = x .

5. Then the match number of the shock waves is calculated using M = V

,where a is the

speed of sound at temperature T . 6. Also the mach number M can be calculated using the RH relations = + M − −+ , by finding P , P for the gas of known γ.

b) Discuss the variation of density of energy states for 3D,2D,1D and 0D structures. (6 Marks) Variation of density of energy for various structures is as follows: 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 = √ 𝛑 / / dE

Where m = mass, E = energy, h = Planck s constant

g(E)

0 E



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.

2D structure/Quantum Film/Well: .

1. Quantum film/well is obtained when 3D structure reduced to nano scale in one dimension as shown in the diagram.

2. The density of states for quantum well is given by g(E) dE = π

dE where m = mass,

h =Planck s constant. 3. For each quantum state the density of states is constant. 4. The variation of g(E) with E is as shown in the graph. 5. The electrons are confined to 1 direction.

1D structure/Quantum Wire :

0

1. Quantum wire is obtained when 3D structure reduced to nano Scale in two dimensions as shown in the diagram.

2. The density of states for quantum well is given by g(E)dE = √ / − /

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

g(E)

0



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.

c) Describe sol-gel method of producing nano particles. (5 Marks)

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. 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. Advantages:

By Sol-gel method we can synthesis pure nanoparticles. Pure nanoparticles are obtained by this method. Nanoparticles obtained are of same size and shape.

g(E)

0

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



d) Mention any three applications of nano particles. (3 Marks) ( Any three of the following) Nano particles(matter) is used in:-

1. Scratchproof eye glasses. 2. 2. Crack resistant paints. 3. Anti graffiti coatings for walls.

4. Transparent sun screens. 5. Stain repellent fabrics.

6. Self cleaning windows. 7. Ceramic coatings for solar cells.

8. Gene theraphy. 9. To detect chemicals and biological substances in biochemical changes. 10. Stronger, lighter, cleaner, smarter surfaces and systems.

11.DNA probes. 12. Tissue engineering. 13.Nanoscale biochips. 14.Improving food packaging. 15. In tyres to improve adhesion to roads. 16. Separation and purification of biological molecules.

OR

17. a)Describe the principle, construction and working of a scanning electron microscope. (8 Marks)

Electron gun

To Vacuum pump

Magnetic condenser coil -1

Magnetic condenser coil-2

Scanning coil

Objective magnetic coil

back scattered electron detector

Secondary electron detector

X-rays detector

Evacuated chamber Specimen

Base

TV monitor



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 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 D . 5. Secondary electrons produced due to interaction with valence electrons are detected by

the detector D 6. and X-rays produced due to deep penetration are detected using the detector D 7. High resolution 3D image can be seen on the TV monitor to which D , D & D connected. 8. Biological specimens must be dehydrated and non conducting samples must be coated

with a thin conducting layer.

b) Define: i) Mach number ii) Subsonic waves iii) Supersonic waves iv) Ultrasonic waves. (4 Marks) 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 = , M is dimensionless quantity.

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 , aero planes etc produce subsonic waves. 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

Ultrasonic waves(Ultrasonics) are the Acoustic waves of frequency greater than 20 kHz. Ex: Bats/Dogs detect Ultrasonics .



c) Explain pyrolysis method of obtaining carbon nanotubes. (4 Marks) 𝐂

1. The schematic diagram of Pyrolisis method is as shown in the diagram. 2. It consists of a chamber in an electric furnace through which a mixture of nitrogen and

acetylene is passed and let out through the out let. 3. A substrate is placed on the quartz sheet inside the chamber. 4. The temperature in the chamber is maintained at about 700-800. 5. Due to high temperature acetylene breaks down in to carbon atoms. 6. When these atoms come near substrate they get attracted and deposited as carbon

nanotubes in the presence of catalysts which are MWCNT s. 7. SWCNT s are obtained when acetylene is replaced with methane or carbon monoxide at

1200

d) 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 100 microsecond. If the velocity of sound under the same conditions is 340 ms− , find the Mach number of the shock wave. (4Marks)

Given: x=100 mm=100x − m, t = S= x − S, a=340 ms− ,M=?

Using, V =

= x − x −

= 1000 m/s

Also, M = V

= = 2.941

**end**

Furnace

Chamber

Catalyst

pressure gauge

𝑪

Substrate on graphite sheet

⊙ ⊙ ⊙ ⊙ ⊙

⊙⊙ ⊙ ⊙ ⊙ ⊙ ⊙⊙⊙

C.REDDAPPA SOLVED VTU 60:40 CBCS ENGINEERING PHYSICS MODEL QUESTION PAPER INTERLINE PUBLISHING.COM

17PHY12/22 I/II SEM B.E Page 1

Model Question Paper with effect from 2017-18

CBCS Scheme 17PHY12/22 First/Second Semester B.E. Degree Examination

Engineering Physics

Time : 3 hrs Max. Marks: 100

Note: 1. Answer FIVE full questions, choosing one full question from each module.

2. Physical constants: Velocity of light c = 3 108m/s; h = 6.625 10

-34JS;

k = 1.38 10 -23

J/K; NA = 6.02 10 23

/Kmole; me = 9.1 10-31

kg; e = 1.6 10-19

C.

Module – 1

1 a. What is a black body? Discuss in brief Wien’s law and Rayleigh – jean’s law to explain black body spectrum.

Describe ultraviolet catastrophe. (8 marks)

Black body is the body which absorb all the radiations (Wavelengths) incident on it and it also give out all those radiations under suitable conditions.

Ex: Ferry black body radiator , metal ball coated with lamp black

Wein s law states that the energy density in the wave length interval λ & λ +dλ is 𝜆 dλ = − − /𝜆 dλ where & are constants. Wein s law explained only the shorter wavelength region of BBR spectra below and it failed to explain the longer wavelength region of the BBR spectra beyond . Rayleigh-Jean s law states that the energy density in the wave length interval λ &

λ +dλ is 𝜆 dλ = 𝜋𝜆 dλ ,where k = Boltsmann s constant. Rayleigh-Jean s law

explained the longer wavelength region of BBR spectra beyond and it failed explain the shorter wavelength region of BBR spectra below .

Ultraviolet catastrophe:

As per Rayleigh-Jeans law, Uλ → ∞ as λ → ,But experimentally observed that Uλ → as λ → . This failure of R-J s law beyond ultra-violet region is called ultraviolet catastrophe .



b. State and explain Heisenberg uncertainty principle. Discuss the significance of the principle. Prove that, using uncertainty principle, the electron emitted during -decay is not the preexisted electron in nucleus. (8 marks)

HUP states that the product of the uncertainty Δ in the position and the uncertainty ΔP in the momentum of a particle at any instant is equal to or greater than ( h/4𝛑) i.e.: Δ ΔP ≥ ℎ𝜋 ,Where h is Planck s constant.

The significance of HUP is that, it is impossible to determine simultaneously both the position and momentum of the particle accurately at the same instant.

To show that electrons do not present in the nucleus using HUP: Electron to be present in the nucleus, maximum uncertainty in position Δ =10-14 m (diameter) According to HUP,

The minimum uncertainty in momentum ΔP ≥ ℎ𝜋 𝛥

≥ . x − x . x − ΔP ≥ 5.275 x 10-21 kg m/s = P(say) Using E = mC , P = mV and m = √ −VC , The minimum energy E of the electron in the nucleus is calculated as follows:

= = 𝑜−𝑣𝐶 = 𝑜 6−𝑣 = = 𝑜𝑣−𝑣𝐶 = 𝑜 𝑣−𝑣 − = 𝑜 ( −𝑣 )−𝑣 = + , but the rest mass energy m c is very very small compared to P2c2, neglecting m c , we get, E ≈ PC = 5.275 x − x 3 x J

= . x − x x . x − MeV

= 9.89 MeV = 10 MeV But the maximum energy of the electrons(𝛃-particle) emitted from the nucleus does not exceed 4MeV,hence electrons do not present in the nucleus.



c. An electron is confined to a potential well of infinite height and width 5A0. Calculate the de Broglie wavelength when the electron is (i) in ground state and (ii) 3rd excited state. (4 marks)

Given: a=5Å = 5x − m,

Using KE of an electron, E= λ & energy of an electron in nth state E = a

We get , a = λ

λ= n For ground state ,n =1 λ= x x −

= 10 x − m

For 3rd excited state, n=4 , λ= x x −

= 6.25 x − m

OR

2 a. What are matter waves? Discuss their characteristics. (5 marks)

Matter (de-Broglie)waves are the waves associated with material particle in motion. Characteristics of matter waves: (Any four)

1. MW are non mechanical waves. 2. MW are associated with moving particles . 3. MW of microscopic particles can be measured . 4. MW are charge independent . 5. MW of macroscopic particles cannot be measured . 6. Different MW have different phase velocity. 7. Phase velocity of MW is greater than that of light. 8. Velocity of MW depends on the velocity of material particles.

b. Define Phase velocity and Group velocity. Obtain the relation between them and hence show that in a non-dispersion medium the group and phase velocities are same. (8 marks)

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

the wave travels and is given by vP = ω

where ω = angular velocity and k = wave constant. Group Velocity is defined as the velocity with which the resultant wave packet of

the group of waves travels and is given by v = ω

where dω = change in angular velocity and dk = change in wave constant. Relation between 𝒗 & 𝒗 for non-dispersive medium: Consider a particle of mass m having phase velocity 𝒗 and group velocity ′𝒗 ′. Let the wave have the wavelength λ , frequency ν, wave number k & angular velocity ω



We know that = 𝜔 …(1) and 𝑔 = 𝜔

….(2)

From eqns 1 & 2 , we get 𝑔 = 𝑉𝑃.

= + 𝑉𝑃

= + . 𝑉𝑃𝜆 𝜆 ……

Also, λ = 𝜋

𝜆 =

𝜋

=

− 𝜋 ……..

From eqns 3 & 4 ,we get 𝑔 = + . 𝑉𝑃𝜆 − 𝜋

= − 𝜋 𝑉𝑃𝜆 ie: 𝒗 = 𝒗 − 𝑉𝑃 λ = 𝜋

Note: . 𝑖 𝑖 𝑖

𝑉𝑃 = = ,Thus Group velocity is equal to Phase velocity.

c. Write a short note on properties of wave function. (3 marks)

1. Wave function (𝚿) is single valued everywhere , 2. 𝚿 is finite everywhere 3. 𝚿 is continuous every where

4. First derivatives of 𝚿 are continuous every where 5. I I or * is called probability density. 6. Probability of finding particle in space is given by ∫ I I dV =+∞−∞

d. Find the kinetic energy of an electron whose de Broglie wavelength is the same as that of a 100keV x-ray.

(4 marks)

Given: = 100 keV= 100x x1.6x − J , =λ , =9.1x − kg ; C =3 x m/s = ?

For x-ray , using , EX = CλX ,we get λX = CEX =. −x x . x − = 1.242x − m

For an electron, using, λ = √ Ee ,



we get E = ℎ 𝜆𝑒=. −. x − x . x − = 1.563 X − J

OR Using , = λ

ℎ𝑋 = ℎ√ 𝑒 = =

x x . x − x . x − x x

= 1.563 X − J

Module-2

3 a. Define the terms drift velocity, thermal velocity, mean free path and relaxation time. Explain why thermal velocity is not responsible for electric current in a conductor. (5 marks)

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

Thermal velocity (𝒗𝐓) is the velocity of the electrons due to temperature/heating. Mean free path λ is the average distance travelled by the electron between two consecutive collisions.

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

its velocity when the field is cut off. Because,the net flow of electrons across any cross section is zero as the electrons movie in different directions due to thermal velocity in the absence of external electric field.

b. Explain briefly the formation of cooper pairs and hence BCS theory. (5 marks)

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

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

− ) ( + q )

q

e e



3) An electron distorts the lattice due to electrostatic attraction when it approach a positive ion lattice, Smaller the mass of the ion distortion is more.

4) The oscillatory distorted lattice is quantized in terms of phonons . 5) This distorted lattice attract another electron by reducing its energy. 6) Thus the second electron attracts the first electron through lattice. This results in the pairing of two

electrons called Cooper Pair . The attraction between the electrons is maximum when the two electrons have equal and opposite spin and momentum.

7) The formation of cooper pair is considered as emission and absorption of the phonon by the two electrons as shown in the diagram.

8) The cooper pair formation begins at critical temperature TC and is completed at 0K. 9) Thus at 0K all the electrons are paired in to ordered cooper pairs, which can pass through lattice

without collisions when a potential difference is applied. This explains the zero resistivity of superconductors.

10) When the temperature increases the breaking of the cooper pairs begin beyond 0K and completed at TC as a result the superconductor attains normal state.

11. BCS theory explained the existence of energy gap in superconductors, Meissner effect ,Coherence length and Flux quantization, etc.

c. Discuss any three major failures of classical free electron theory. (6 marks)

Failures of CFET are:-

1.Specific heat :According to CFET the molar specific heat of electron gas at constant volume is given by 𝑣 = ,where R=universal gas constant. But experimentally determined 𝑣 𝑖 given by 𝑣 = − ,where T=absolute temperature. Thus CFET fail to explain the the dependence of 𝑣 T and numerical constant.

2.Dependance of electrical conductivity 𝝈 on T: According to CFET kinetic energy =

∝ √ … But mean collision time 𝜏 ∝ 𝑣 𝜏 ∝ √ … . ∝ √ Also from 𝜎 = 𝜏

,we get 𝜎 ∝ 𝜏 … .

From eqns 2 & 3 we get 𝝈 ∝ √ ,but experimentally it has been observed that 𝝈 ∝ 𝑻 , thus CFET

fail to explain the dependence of 𝜎 . 3.Dependance of ′𝝈′ 𝒕𝒓 𝒕𝒓 𝒕 n : 𝜎 = 𝜏

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

di-atomic and mono-atomic metals but from experimental results, it has been found that 𝝈 of mono-atomic metals of low n are more than that of di and tri-atomic metals of large n .Thus CFET fail to explain dependence of 𝝈 on concentration n.



4. Mean free path λ : Mean free path of Cu from CFET is given by λ= 𝜏=2.85nm. But the experimentally determined value of λ= . nm, which is times more than that λ of CFET.

d. For a metal having 6.5 x 10 28 free electrons per unit volume the relaxation time at room temperature 300K is 3.82 x 10 -14 second. Calculate its electrical resistivity using classical free electron theory. (4 marks)

Given: n = . x atoms.; τ = τ = . x − S, m = . x − kg ; e = . x − C ; = ? Using = τ0

= . x −. x x . x − x . x −

= 1.43x − m.

OR

4 a. What is a super conductor? Explain Meissner effect. (4 marks)

Superconductor is a conductor having zero resistivity /resistance.

Meissner s effect:

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

Is completely expelled out of the material at temperature called Critical temperature. 2) Below Critical temperature the material behave as perfect dia-magnet. 3) Consider a superconducting material placed in a magnetic field H, the magnetic lines penetrate through the conductor. 4) When the temperature of the conductor is cooled below the critical temperature TC,the magnetic flux inside the conductor is completely expelled out of the conductor. 5) Above TC ,B≠ and below TC ,B= . 6) Meissner effect is the principle of Maglev vehicles, Squid etc.

H H

T> 𝐓𝐂 T< 𝑻𝑪



b. Give the expressions for concentration of electrons and holes in an intrinsic semi-conductor. Obtain the expression for electrical conductivity of intrinsic semiconductor. (8 marks) Concentration of electrons, n = √ π e∗ / e(EF−EgkT ) and

Concentration of holes, n = √ π h∗ / e(−EFkT ) 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 = ,where = 𝑖 , = , = 𝑖 𝑖 & = ℎ also current due to holes is ℎ = ℎ ℎ But the total current I= + ℎ = ea( + ℎ ℎ

𝐼 = e ( + ℎ ℎ

, 𝑖 = 𝐼 J= e ( + ℎ ℎ but from Ohm s law J=𝜎 ie: 𝜎 = + ℎ ℎ 𝜎 = 𝑣𝑒 + ℎ 𝑣ℎ

𝝈 = e( + ) where & ℎ mobilities of electrons and holes respectively. For an intrinsic semiconductors = ℎ = 𝑖 called the density of intrinsic charge carriers. For n-type semiconductor ℎ 𝝈 = e For p-type semiconductor, ℎ 𝝈 = e

c. What is Fermi-Dirac statistics? Explain. (4 marks)

Fermi –Dirac statistics is the statistical rule applied to the distribution of identical, indistinguishable particles of spin called fermions like electrons, which obey Pauli s exclusion principle. The probability of occupation of a state by an electron is given by Fermi factor/Fermi- Dirac distribution function given by

f(E)= + − /𝑘𝑇 where = 𝑖 , E=energy of State, k=Boltzmann constant and

T= temperature. Case1. At T=0K & E < ,f(E)= 1 Case2. At T=0k & E > ,f(E)= 0 and Case3. At T> & E = ,f(E)= 0.5 Variation of f(E) with E is as shown in the graph.

f(E)

1 T = 0K

0.5 T>

0 E



d. A superconducting tin has a critical temperature at 3.7K in zero magnetic fields and a critical field of 0.0306 T at 0K. Find the critical field at TC. (4 marks)

Given : = . , , = 3.7K= , = ?

Using: = [ − 𝐶 ] = . [ − .. ]

= . [ − ] = . [ ] =

Module-3

5 a. State and explain the necessary conditions for laser (5 marks)

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 − s ie: their life is 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 = 𝜆

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



b. With a neat diagram explain the construction and working of semiconductor laser. (7 marks)

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

The schematic diagram of GaAs semi-conductor device is as shown in the diagram. 1. It consists of heavily doped n-region of GaAs doped with tellurium and p-region of GaAs

doped with zinc. 2. The upper and lower surfaces are metalized so that pn-junction is forward biased . 3. Two surfaces perpendicular to the Jn are polished so that they act as optical resonators and

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

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

like Gallium Arsenide (GaAs). 2. When GaAS diode is forward biased with voltage nearly equal to the energy gap voltage, electrons

from n-region & holes from p-region flow across the junction creating population inversion in the active jn region.

3. As the voltage is gradually increased due to forward biasing population inversion is achieved between the valence band and conduction band which in turn result in stimulated emission.

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

5. GaAs laser produce laser beam of wavelength 8870Å in IR region , GaAsP produce laser beam of 6500Å in visible region etc.

Semiconductor laser Energy level giagram Metalic coated(MC) surface P-type pn-Jn n-type

RS PS CB Ba p P Laser h+e

n-type VB Polished surfac(PS)

Laser Roughened surface(RS) MC

P-type

Pn-jn

n-type



c. Describe briefly the application of laser in welding and cutting. (4marks)

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. laser cutting:

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

/ inlet

Condenser lens

material

Laser beam

Condenser lens

Materials to be weld



d. The angle of acceptance of an optical fiber is 300 when kept in air. Find the angle of acceptance when it is in a medium of RI 1.33 (4 marks)

Given: 𝜃 =30° , ′ =1.33 , = 𝑖 , 𝜃′ =? Using ′ 𝑖 𝜃′ = Sin𝜃 =constant (√ − ) for a fibre

We get 𝜃′ = 𝑖 − 𝑜 𝑖 𝜃𝑜′𝑜

= 𝑖 − 𝑖 °. = 22.08°

OR

6 a. What is attenuation in optical fiber? Discuss the various loss factors in optical fiber communication.(7 marks)

Attenuation is the loss of power of light signal as it is guided along the fibre. 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 . 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.

b) 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.

b. Discuss the any three advantages and disadvantages of optical fiber communication system over conventional communication system. (6 marks) Advantages: 1. Optical fibres( system) are economical/cheap. 2. Optical fibres( system) are small in size,light weight,flexible ,mechanically strong. 3. Optical fibres( system) have large information carrying capacity as photons are used. 4. Optical fibres( system) are immune to electromagnetic and radio frequency interferences.

Disadvantages: 1. Optical fibres( system) are sensitive to temperature changes which leads to loss of signal. 2. Optical fibres( system) may break easily due to bending or accidents. 3. Re-establishing the connections of broken Optical fibres( system) is skillful work and

costly.



c. Mention the applications of holography (3 marks)

Holography are used in: 1. Holographic Interferometry-to study minor distortions of an object due to stress/vibrations etc 2. Holographic Diffraction Grating- gratings are produced with much more uniformity than in conventional gratings. 3. Acoustic Holography-used to construct 3D images of the internal parts of human body using ultrasound.

4. Holographic data storage and retrieval- used to store more informations. 5 Holography in chemistry-used to detect small concentrations of impurity atoms/molecules.

7 Holograms are used for security purposes.

b. Calculate on the basis of Einstein’s theory, the number of photons emitted per second by laser source emitting

light of wavelength 6328A0 with an optical power output 10mW. (4 marks)

Given: P= 10 mW = 10 x − w ,t= 1 S , C=3x m/s ; h=6.625x − Js, λ = Å = x − m ; n = ?

Using E= Pxt and E = n.Cλ n = λPC

= x − x x − x. x − x x

= 3.184 x photons.

Module – 4

8 a. Define lattice points, unit cell, Bravais lattice and primitive cell (4 marks)

The 3 dimensional periodic array(arrangement) of geometrical points in Space is called space lattice . Each point is called lattice point/site.

Bravais lattice is the lattice in which all the lattice points are equivalent , each lattice point represent an identical set of one or more atoms/molecules. Primitive cell-is the unit cell having lattice points only at the corners of it or is the unit cell containing only one lattice point in it.



b. Describe briefly the seven crystal systems (7 marks)

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

≠ ≠ c = = = Types:SC/BaC/BCC/F 0 b Ex:KNO3, MgSO3 a

= = a = = ≠ Types:SC Ex: As, Sb, 0 a a

= ≠ = = , = c Types: SC Ex:Zn,Mg, 0 a a

≠ ≠ ≠ ≠ ≠ c Types:SC Ex: 5H2O CuSO4 0 a b

≠ ≠ = =90 ≠ c Types:SC/BaC Ex: CaSO4,2H2O b a 0

= = a = = = Types: SC/BCC/FCC a Ex: Copper/Gold a 0

= ≠ c = = = Types: SC/BCC Ex: SnO2,TiO2

0 a a



c. Obtain an expression for the inter planar distance in a cubic crystal in terms of Miller indices. (6 marks)

1. Consider the plane ABC of a crystal which intersect the X,Y & Z axes at A,B & C respectively. 2. Let ( h k l ) be the miller indices of the family of ABC planes. 3. Draw ON ⊥ 𝑟 to the plane ABC ,then ON= represent inter-planar distance/spacing. Then

OA= ℎ , OB= &OC = ,where , & are basis vectors

From right ∆ OAN, Cos 𝜃 = = /ℎ = ℎ

……. From ∆ OBN, Cos 𝜃 = = / = ……. and also, from ∆ OCN, Cos 𝜃 = = / = ……

For orthogonal co-ordinates 𝜃 + 𝜃 + 𝜃 = ……

From eqns 1,2,3 &4 we get, ℎ + +

= 1

ie: [ ℎ + + ] = 1

d = √ + + For a cubic crystal = = d= √ + +

Z C N B Y 𝜃 𝜃 0 𝜃 A X



d. The grating space of calcite is 3.036A0 and for the first order Bragg reflection the glancing angle is 120. Find the path difference between the rays. (3 marks)

Given: n = 1 ; 𝜃 = ° ; d = 3.036Å = 3.036x − m ; Pd= nλ = ? Using nλ = 2dSin𝜃

Pd = 2x 3.036x − xSin °

= 1.262 x − m

OR

e. a. Discuss the allotropy of carbon with reference to diamond and graphite. (4 marks)

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 hardest having strong bonding and good insulator . 3. Graphite is very soft having strong bonding and good conductor.

b. Give a qualitative explanation of perovskite crystal structure (6 marks)

Perovskie is the common name for all the oxides of type AB ,where A & B are different metals. Ex: 𝑖 𝑖 𝑖 , 𝑖 ( calcium titanate) etc The common structure of perovskite is as follows: = 8 Ba/Ca atoms form are at the edges of the cell., = 6 atoms are at the face of the cell and = 1 𝑖 atom at the body centre of the cell. They exhibit superconductivity. They exhibit both piezo-electric & ferro-electric properties . They are used as dielectric in capacitors. They are used as piezo-electric in microphone. They are Type II superconductors have Perovskite structure.



c. Define coordination number and packing factor. Calculate the packing factor for BCC structure (6 marks)

Co-ordination number of any atom in a crystal is the number of nearest equidistant neighbouring atoms Surrounding that atom.

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 = a xV a aV

For BCC , n = 2, R = √ 𝐚

Using, APF = a x R

= a x √ a

= a x

√ a

= √ π

= 0.68 or 68%

d. Draw the crystal planes (1 2 0), (1 1 1), (1 0 1) and ( 00) in a cubic crystal. (4 marks)

(1 2 0) x = =1,y = , z = = ∞

z

o 1/2 y

1

x

(1 1 1) x = y = z = =1

z 1

o 1 y

1 x

(1 0 1) x = z = = 1,y = = ∞

z 1

o y

1

x

(00) x = − = −1,y = z = = ∞

z

−

O y

x



Module – 5

9 a. Define Mach number and hence distinguish between subsonic and supersonic waves. (4 marks)

Mach Number(M) is the ratio of the speed of an object (V) through a fluid to the

speed of sound(a) in the fluid at that point. Mathematically, M = 𝑽

.

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. 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

b. List the properties of shock waves. Give an example for weak and strong shock waves. (6 marks) (any four)

1. Shock waves(SWS) carry energy and propagate through a medium(solid/liquid/gas/plasma) and vaccum. 2. Across the shock waves there is always rapid changes in pressure, temperature, density of the flow. 3. SWS travels through most media at higher speed than other waves. 4. SWS dissipate energy relatively quickly with distance. 5. SWS are not conventional sound waves. 6. In SWS properties of the fluid(density, temperature, volume, pressure, mach number) change almost

instantaneously. 7. SWS can be normal, oblique and stationary waves. 8. SWS can change from nonlinear to linear over long distance. 9. SWS creates additional drag force on aircrafts with shocks. 10. Energy is preserved when SWS passes through matter but the energy which can be extracted as work

decreases and entropy increases. Weak shock waves are produced by burst of crakers / balloons, & Motor vehicles Strong shock waves are produced by Thunder bolt, Supersonic jets

c. With a neat diagram explain the construction and working of Reddy shock tube. (7 marks)



x =7 cm


Plunger 49 cm 51 cm



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 ,temparature . 4. Two sensors & separated by a distance ∆ fitted to driven section measures the

pressures , and temperatures , respectively.

Working : 1. Driver section is filled with gas at high pressure and Driven section is filled with gas of low

pressure & temperature T . 2. Diaphram is ruptured to produce shock waves by pushing the piston and the rupture pressure

& temperature is measured using sensor S. 3. The time 𝒕′ taken by the shock wave to travel the distance x is measured using CRO. also the

pressures , and temperatures , are measured using the sensors & respectively.

4. The speed of the shock waves is calculated using V = 𝑡 .

5. Then the match number of the shock waves is calculated using M = 𝑉

,where a is the speed of

sound at temperature . 6. Also the mach number M can be calculated using the RH relations = + − −+ , by finding , for the gas of known .

d. Discuss the advantages of Sol-Gel method. (3 marks) 1. By Sol-gel method we can synthesis pure nanoparticles. 2. Pure nanoparticles are obtained by this method. 3. Nanoparticles obtained are of same size and shape.

OR

10 a. With a neat diagram explain how the carbon nano tubes are synthesized using arc discharge method. (8 marks)

H gas inlet H gas outlet

Chamber

Graphite anode Ba

Graphite cathode

500 torr



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 are sealed. 3. Helium gas is circulated through the chamber at a pressure of 500 torrand a voltage

of 20-25V can be applied between the electrodes. 4. When a voltage of 20-25 V is applied between the electrodes, graphite evaporates and is

deposited on the cathode. 5. If the anode is coated with catalytic agents like Iron, cobalt or nickel ,SWCNT s are produced. 6. MWCNT s are produced if the anode is not coated with catalytic agents. 7. Pure CNT s can be obtained by using pure graphite rods.

b. Describe density of states for various quantum structures. (6 marks)

3D structure/Bulk metal :

1. The 3D structure is as shown in the diagram, which have all the 3 dimensions.

2. The density of states for 3D structure is given by g(E) dE = 𝟖√ 𝝅 / 𝑬 /

Where m = mass, E = energy, h = Planck s constant 3. The density of states g(E) increases with energy E is as shown in the graph 4. The electrons are not confined to any direction.

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.

g(E) 0 E

g(E)

0



2. The density of states for quantum well is given by g(E) = 𝜋ℎ dE where m = mass,

h =Planck s constant. 3. For each quantum state the density of states is constant. 4. The variation of g(E) with E is as shown in the graph. 5. The electrons are confined to 1 direction.

1D structure/Quantum Wire :

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 = √ / − /ℎ dE

where m = mass, E = energy, h = Planck const. 3. The variation of g(E) with E is as shown in the graph 4. The electrons are confined to 2 directions.

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. g(E) has discrete structure as shown.

g(E)

0

g(E)

0



c. Discuss the basics of conservation of mass, momentum and energy. (6 marks)

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, 𝜌 = or 𝜌 = 𝜌 , Where 𝜌 , 𝜌 densities & , velocities.

2. Conservation of momentum states that the sum total momentum of the system always remain constant.

Mathematically, + 𝜌 = + 𝜌 = + 𝜌 Where , pressures , , velocities and 𝜌 , 𝜌 densities.

3. Conservation of energy states that the sum total energy of a system is always remains constant.

Mathematically, ℎ + 𝑣 = ℎ + 𝑣 = ℎ + 𝑣

Where ℎ , ℎ enthalpies and , velocities.

@end@

C REDDAPPA SOLVED VTU ENGINEERING PHYSICS 60:40 CBCS PRACTICE QUESTION PAPER INTERLINE PUBLISHING.COM


Practice Model Question Paper CBCS SCHEME(60:40) 17PHY12/22

First/Second Semester B.E; Degree Examination

Engineering Physics Time: hrs. Max. Marks:

Note: . Answer any FIVE full questions, choosing ONE full question from each module. 2. Physical constants: Velocity of light C = 3x m/s; Planck’s constant h = 6.63x − Js;

Mass of electron m = 9.11x − kg; Boltzmann constant k = 1.38x − j/k;

Avagadro number NA = 6.02x /K mole.

Module-1

1. a) Reduce Planck’s law of radiation to Wein’s law and Rayleigh-Jeans law. (6 Marks)

Planck s law of radiation is given by Uλdλ = πλ cλ T − dλ ……………

a) Wein s law:

For shorter wave lengths, λ is very small ,hence Uλ & e cλ T are

Large i.e: e cλ T so that e cλ T − = e cλ T ……….

from eqn 1 &2,we get Uλdλ = πλ cλ T dλ

𝛌 dλ = 𝐂 𝛌− 𝐞−𝐂𝛌 dλ This is wein s law .where C = hc & C =

b) Rayleigh-Jeans law: For longer wave lengths, λ is very large,

hence e cλ T is very small ,Expanding e cλ T as power series ,

we get , e cλ T = 1+ λ + λ +………. = + λ , neglecting higher powers of λ ie: e cλ T − = λ ………

from eqn 1 &3,we get Uλdλ = πλ cλ T dλ 𝐞: 𝛌 dλ = 𝛌 dλ , This is Rayleigh-Jeans law



b) Set up 1D time independent Schrodinger equation for a free particle. (6 Marks)

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

freely in the positive direction of -direction is given by 𝚿 = Ae x− t

= Ae x e− t,

Where Ae x represent the time independent part of the wave function and is

represented by = Ae x …… differentiating eqn w.r.t twice we get

x = A ik e x and

x =A ik i e x

= i k Ae x …….(2) From eqns 1 & 2 we get

x = − πλ …….. = − & = 𝛌

But, de-broglie wave length λ = v λ = v = ½ v

= K EK = ½mv

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

λ = − ……

From eqns 3 & 4 ,we get x = − π −

x + π − = …… This is Schrodinger time independent equation for a particle For a free particle ,V=0

x + π =

This is Schrodinger time independent equation for free particle.



c) Show that group velocity is equal to particle velocity. (4 Marks)

Consider a particle of mass m having particle velocity 𝐯𝐏𝐚𝐫𝐭 𝐜 𝐞′ & group velocity ′𝐯 ′. We know that, v = … … . .

but = ν & E = hν ,we get = d = π dE …

Also from k = πλ & λ = P we get k =

πP

dk = π dP ……

From eqns 1,2 &3,we get v = = ππ P = P …….

Also , from E = ½mv = v & P = m𝓥 , we get E = P

dE = P dP ie; P = P =

vPart c e = vPart ….

d) Calculate the de-Broglie wavelength associated with an electron of energy 50keV. (4 Marks)

Given: E= 50keV= 5x x 1.6x − J , h=6.63x − Js; m=9.1x − kg; e=1.6x − C; λ =?

Using λ = √

= . x −√ x . x − x x x . x −

= 5.495 x − m

.

OR



2. a) Obtain eigen energy value relation and eigen functions for an electron in one dimensional potential well/Box of infinite height . (6 Marks)

a ∞ ∞ V=∞ V=0 Box/Well particle =0 =a Consider a particle of mass m moving by reflection at infinitely high walls of a Box/well of width a moving between =0 & =a Potential V=0 inside the Box/well and V=∞ outside the Box/well .one dimensional Schrodinger equation for particle is given by

x + π − = …… . For a particle inside the Box/well V=0 x + π = …. Putting

π = K ….. in equation , we get x + K = …… The general solution of the quadratic equation is of the form )=A sin(K ) +B Cos(K ) …… where A & B are constants determined from boundary conditions as follows : )=0 at =0 from eqn (5), 0=A x 0 +B x 1 B=0 also, )=0 at =a from eqn(4) 0=C Sin(Ka)+0xCos(Ka) 0= C Sin(Ka) Sin(Ka) = 0 as C≠ Sin(Ka) =0= Sin(n )

ie: Ka=n or K=πa …..

From eqns 3 & 6 ,we get π = π a

E = a or

In general En= a …

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 = a = Eo called end point /ground state energy .

n=2 gives E2 = a = 4 Eo called 1st excited state energy

n=3 gives E3 = a =9 Eo ,called 2nd excited state energy & so on

Substituting the values of B=0 & K= πa in eqn 5, we get

( ) = A Sin( πx a ……. this represents the permitted solutions To find A by normalization:

Applying the normalization condition ∫ I I dx =+∞−∞ to eqn 8 for x=0 & x=a, we get

∫ A Sin πx a dx =a A ∫ ½[ − Cos πx a ]dx =a

Sin θ)=½[1-Cos(2θ ]

½A [∫ dx − ∫ Cos πx a dxa ] =a

½A [x]a − [ a π Sin( πx a )]a = ∫ Cos mx dx =

x ½A [a − ] − [ a π Sin(2n ) − a π Sin(0)] =1

½A [a − ] − [ − ] =1

A a = A = /a or A =√ /𝐚 …

From eqns 8 &9, the normalized Eigen functions are given by 𝛙 ( )= √ /𝐚 Sin( 𝐱 𝐚 ) ….. n = 𝛙 I𝛙 I . n = 𝛙 I𝛙 I n = 𝛙 I𝛙 I =0 =a =0 =a Eigen functions 𝛙 , 𝛙 ,𝛙 , … . and their probability densities I𝛙 I , I𝛙 I , I𝛙 I …

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 [ ( )] Eigen Energy values are energy values for which Schrodinger equation



b) Explain Compton effect experiment and mention its significance. (6 Marks)

1. Compton experimental arrangement is as shown in the diagram. 2. X-ray photons of wave length λ from X-ray tube are collimated by passing through two

slits S1 &S2. 3. The collimated X-rays are made to fall on graphite target. 4. X-rays collide with the electrons of the target at rest and transfer part of their energy

resulting in the recoil of electrons in a direction making an angle ∅ and Scattered X −rays makes an angle θ respectively with the incident X-rays direction. 5. The intensity of the scattered X-rays in different directions are measured using Bragg s

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

found that scattered X-rays consists of two components namely unmodified component having the same wavelength λ as incident X-rays & modified component having slightly higher wavelength λ′

7. This scattering of X-rays due to recoil of electrons is called Compton effect . 8. The change in wavelength λ′ − λ is called Compton wavelength ,which depends only θ and independent of 𝝺. 9. Based on conservation of energy & momentum laws, considering the collision between X-

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

λ′ − λ = ∆λ = C (1−Cosθ where C = λ called Compton wavelength of electron = 0.02426 Å(constant)

10. ∆𝛌 varies from 0 for θ = ° to λ for θ = ° 11. Compton effect signifies the particle nature of X − rays light

Bragg’s spectrometer

Slits

X-rays Collimated X-rays 𝜃

∅

Graphite target Recoil electron



c) Find the expression for de-Broglie wavelength of an electron accelerated through a potential difference of ‘V ‘volts. (4 Marks)

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

by mv = eV , multiplying both Nr. and Dr. of LHS by m

we get, v = eV ie; m v = meV P = meV mv = P 𝐏 = √ 𝐞

But λ = 𝐏 𝛌 = √ 𝐞 = . 𝐱 −√ 𝐱 . 𝐱 − 𝐱 . 𝐱 −

= . 𝐱 − √ m

= .√ nm =

.√ Å

c) 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. (4 Marks)

Given: n=3(For 2rd excited state), a=0.18 nm =0.18 x − m,h=6.63x − Js; m=9.1x − kg & e=1.6x − Using E = a J OR E = a eV

= . x − x . x − x . x − = . x − x . x − x . x − x . x − = 1.677x − J = 104.83 eV = . x − . x −

= 104.81 eV

Module-2

3. a) Define Fermi Factor and discuss its dependence on energy and temperature. (5 Marks)

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) = +𝐞 − / ,where Fermi energy, E = energy of State, k = Boltzmann constant and T = temperature

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



When T = 0K

Case-1: If E < E − is − ve ,then 𝐞 − / = 𝐞−∞ =

f(E) = + − / = + =1,

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

Case-2: If E > − 𝑖 + ,then − / = ∞ = ∞ f(E) = + − / = +∞ = ∞ =0,

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

Case-3: If E = − = ,then − / = indeterminate When T > 0K f(E)= + − /

= − = ℎ − / = = f(E) = + − / = + = 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 .

b) Derive an expression for electrical conductivity and resistivity based on QFET. (6 Marks)

According to de-broglie hypothesis λ = ℎ𝑣 ,also λ = where symbols have usual

significance. = ℎ , where v & k are vectors

differentiating this w.r.t t ,

We get m = ….

If an electric field E is applied ,the electron experience a force m = − … & , , = −

= − ℎ Integration between the limits 0 to t ,we get ∫𝑡 = − ℎ ∫

k − = − ℎ −

𝜕 = − ℎ

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

f(E) Note: = < =1000k

1 T = 0K

0.5 <

0 E



applied electric field E in a time t as shown in the diagram.

If t = 𝜏 , ℎ 𝑖 𝑖 ℎ 𝑖 𝜕 𝑣 𝑖 ℎ is given

by 𝜕 𝑣 = − ℎ 𝜏 …. , Using mv =

ℎ

we get 𝜕 𝑣 = ℎ 𝜕 𝑣

𝜕 𝑣 = ℎ 𝜕 𝑣 …. From eqns 3&4,we get 𝜕 𝑣 = − 𝜏

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

ie: = − 𝜏 …

Also current density J = − …. 5&6, we get J = 𝜏 E

= 𝜏 ….

But from Ohm s law conductivity 𝜎 = ….. 7&8 ,we get 𝝈 = 𝝉 …(7),

where m is called effective electron mass usually denoted by ∗

Also resistivity = 𝜎 … & ,we get = 𝝉

c) Explain the resistance variation of a conductor with impurities and temperature or Write a note on Mathessien’s rule. (5 Marks)

1. Variation of resistivity of metals with temperature T is given by Matthiessen s rule = + 𝐏 (T)

Where = temperature independent resistivity due to impurities & imperfections

P (T)= temperature dependent resistivity due to att vibrations. 2. Variation of resistivity with temperature T is shown graphically.

E

0

𝜕

𝑖 0 T



At 0K, 𝐏 T = = , 3. For pure metals = ,but practically not possible to produce pure metals, ≠ 0 4. and at large T , is negligible compared to large P (T) = P (T)

d) Calculate the mobility of electrons in copper assuming that each atom contribute one free electron for conduction.Resistivity of copper = 1.7x − Ω , = . = . / . (4 Marks)

Given: = 1.7 x − -m ; M = 63.54 ; D = 8.96 x kg/ ; =6.025 x /kg mole ; = . − ; = . − ; = / , = ? ; = ? 𝑖 n = . 𝐴 =

x . x x. . x .

= 8.5x atoms.

= = . x x . x − x . x − = 4.325x − /vs

OR

4. a ) Explain the merits of QFET. (6 Marks)

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 𝑣 =

where k = Boltzmann constant and =Fermi energy. But = − 𝑣 = −

which agrees with experimental value. 2.Dependance of electrical resistivity ′𝝈′ on T (temperature):

We know that(WKT) 𝝈 = 𝝉∗ and 𝜏 = 𝜆𝑣 where =Fermi velocity, λ=wavelength 𝝈 = ∗ 𝝈 ∝ … . .

Also it has been shown that λ ∝ ∝ 𝑟 ∝ λ ∝ … where A=area, r=radius & T=temperature. & ,we get 𝝈 ∝ which is the experimentally determined relation.



3.Dependance of ′𝝈′ ′ ′: WKT conductivity 𝝈 = ∗ , thus 𝝈 ℎ .With the decreases

of atomicity, n decreases and increases so that n . .

increases ,so that 𝝈 is more

for monoatomic metals than that of di&tri-atomic metals.This explains the conductivity of mono-atomic metals is more than that of di & tri-atomic metals.

b) State Law of mass action and derive an expression for electrical conductivity of a semiconductor. (6 Marks)

Law of mass action states that the product of the concentration of charge carriers electrons ( )and holes( ℎ) in an intrinsic semiconductor is equal to the square of the intrinsic charge carriers( 𝑖 ) at any temperature.

ie: = . In a semiconductor the net current is due to both electrons and holes . The current due to electrons is = ,where = 𝑖 , = , = 𝑖 𝑖 & = ℎ Also current due to holes is ℎ = ℎ ℎ But the total current I= + ℎ = ea( + ℎ ℎ

= e ( + ℎ ℎ , 𝑖 = J= e ( + ℎ ℎ but from Ohm s law J=𝜎 ie: 𝜎 = + ℎ ℎ

𝜎 = 𝑣𝑒 + ℎ 𝑣ℎ

𝝈 = e( + ) where & ℎ mobilities of electrons and holes respectively. For an intrinsic semiconductors = ℎ = 𝑖 called the density of intrinsic charge carriers. For n-type semiconductor ℎ 𝝈 = e For p-type semiconductor, ℎ 𝝈 = e




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

d) Calculate the probability of an electron occupying an energy level 0.02 eV above the Fermi level at 300K. ( 4 marks)

Given: − = . = . . − J , k=1.38x − J/K and T=300K

Using f(E) = + − /𝑘𝑇

= + . 𝑥 . 𝑥 − / . 𝑥 − 𝑥

= + .

= + .

= .

=0.32 or 32 %

RW RW

Vehicle

Superconducting magnet

Aluminium guide way



Module-3

5. a) Obtained an expression for energy density of emitted radiations in terms of Einstein’s coefficients.

.

Consider a system of atoms in thermal equilibrium with radiation of energy density 𝜈 & frequency ν . Let N1 & N2 be the population of the energy states E1 & E2

respectively, where (E2 >E1 ) We know that, Rate of induced absorption = 𝜈 , where = 𝑖 𝑖 ′ 𝑖 𝑖 𝑖 𝑖 . Rate of Spontaneous emission = where A = Einstein′s coefficient of spontaneous emission and

Rate of Stimulated emission= 𝜈 , where B = Einstein′s coefficient of Stimulated emission . At thermal equilibrium,

Rate of induced absorption = Rate of Spontaneous emission + Rate of Stimulated emission

ie: 𝜈 = + 𝜈

𝜈 − =

𝑖 : 𝜈 = − dividing both Nr & Dr by ,we get

= [ − ] ……….

According to Boltzmann s law, = − / = ℎ / …….

From eqns 1 &2 , we get. 𝜈 = [ ℎ / − ] ……..(3)

But .the energy density given by Planck s law is 𝜈 = ℎ ℎ / − ……(4)

Comparing eqns 3 & 4 ,we get = or

= 𝜈 ( ℎ − and = Thus coefficient of stimulated absorption = coefficient of stimulated emission.

Thus energy density = [ /𝑲 – ]

Induced absorption Stimulated emission

• •

hν hν hν 2hν • • •

Spontaneous emission



b) Derive an expression for numerical aperture in an optical fibre. (5 Marks)

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 n , n & n be the RI of core, cladding and launch medium respectively. Also OA incident ray,AB refracted ray,BC totally reflected ray, θ , θr & ∅ be the angles of incidence, refraction at A & angle of incidence at B respectively.

By snell s law at A, no sin θi = n sin θr sin θ = sin θr ……

But, from ∆ ADB, θr = −∅ sin θr = sin − ∅ =cos ∅ … . when θ is maximum = θ , acceptance angle, then ∅ = ∅C Critical angle ……..

From eqns 1,2 &3 ,we get sin θ = cos ∅C ……. Also, sin ∅C = and cos ∅C = √ − sin ∅C =

√ −n …

From eqns 4 &5,we get sin θ = √n −n

θ = sin− √ −no

for air n =

Launch medium ( ) Fibre axis

A

𝜃𝑖 O

𝛷 Core )

𝜃𝑟

D

B Cladding ( )



c) Explain the construction and working of a Co2 laser. (6 Marks)

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

laser diagram

C gases outlet

Water outlet

-optical cavity(oc) (oc)

5 m

A electrodes C 2.5 cm Laser

BW BW=Brewster window

Water inlet Ba

Energy level diagram

Resonant transfer of energy C

= (meta stable state)

10.6 laser

Excitations by collission 9.6 laser

= 0 Ground state



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

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

upper & lower vibrational levels of CO2 molecules. 3) Ionisation takes place due to electric discharge when high DC voltage is applied between

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

C5 from their ground states 0 & C1 due to collision as follows: e+ → ∗

+e and e+ → ∗+e

where e& e are the energies of electron before and after collision. 5)

∗ molecule in excited level collide with CO2 molecules in their ground state C1 & excite it to metastable state C5 by resonant energy transfer as level C5 of CO2 is same as level 𝓥=1 of given by ∗ + → ∗

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

place betweenC5 &C4 and C5 & C3. 7) The transitions/de-excitations takes place as follows:

C5 → C4 producing laser 10.6 (IR region) C5 → C3 producing laser 9.6 (IR region) C4 → C2

C3 → C2 Radiation less transitions C2 → C1

8) Due to high thermal conductivity of He, it removes heat from mixture and de-populate the lower states C3 &C2 quickly .

9) Laser beam is amplified by using optical resonators. 10) The laser output is 100kW for continuous mode and 10 kW in pulsed mode.

d). The angle of acceptance of an optical fibre is 30° kept in air. Find the angle of acceptance when it is in a medium of refractive index 1.33. (4 Marks)

Given: 𝜃 =30° , ′ =4/3=1.333 , = 𝑖 , 𝜃′ =?

Using ′ 𝑖 𝜃′ = Sin𝜃 = √ − = constsnt

We get , 𝜃′ = 𝑖 − ( 𝑖 𝜃 ′⁄ = 𝑖 − ( 𝑖 ° .⁄ =22.03°

OR



6. a) What is holography? Explain the recording and reconstruction processes in holography with the help of suitable diagram. (5 Marks)

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. Construction/Recording of Hologram.

1. The schematic diagram for the construction of a Hologram is as shown in the diagram. 2. A laser beam of wavelength λ is made to fall on beam splitter, which split the beam in to

two beams. One beam which passes through splitter is called reference beam . The other reflected beam is made to fall on the object whose hologram is to be produced.

3. The beam reflected by the object is called spherical object beam . 4. The reference and object beams produce concentric circular rings called Gaber zones

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

5. On developing photographic plate we obtain hologram . 6. At each and every point on the hologram, complete information/details of the object are

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

of the object with less resolution. Re-construction of Hologram.

Object

Reflected beam Object spherical beam

Laser beam (λ) Hologram

Beam splitter reference beam

Hologram Observation direction

Laser beam (λ)

Object Virtual image Object Real image



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

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

3. The hologram act as diffraction grating. Due to diffraction and interference two images of the object are produced.

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

5. By changing the direction of the observation ,we can see all the details of the object originally hidden from view.

b)Explain the point to point communication using optical fiber with a block diagram. (5 Marks)

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. 1. 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) Explain different types of optical fibers. (6 Marks)

There are three types of optical fibres namely:- 1) Single mode step index fibre(SMF

1. SMF has a core diameter − of uniform RI and cladding diameter 60-70 has 𝑖

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 < . . 5. Numerical aperture is < . . 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

2) Multimode step index fibre(MMF) Input output

1) MMF has a core diameter − of uniform RI and cladding diameter 100-250 has 𝑖 .

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 > . . 5) Numerical aperture is . . .

Cladding

Core

Input output


Cladding

Core

Input output




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 − of variable RI and cladding diameter 100-250

has 𝑖 . 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 > . . 5) Numerical aperture is . . . 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.

d) The average output power of laser source emitting a laser beam of wavelength 6328Å is 10mW. Find the number of photons emitted per second by the laser source. (4 marks)

Given: P=10mW = 10x − ,t=1 s , C=3x m/s ; h=6.63x − Js, , λ = Å = x − m ; n = ?

Using E= Pxt and E =n.Cλ , we get n = λPtC

= x − x x − x. x − x x

= 3.18 x /m

Cladding

Core

Input output




Module-4

7 a) Explain the Diamond crystal structure with diagram and find its APF. (6 Marks)

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

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 = x 8+ x 6+4 =8

7. If R be the radius and a the lattice constant ,then we can show that

AB=2R=√ + + = √ = √ or R= √

For Diamond n = 8 & R = √

Using APF = a x R = a x

√ a

= a x √ ax x

= π√

= 0.34 or 34%

O 1/2

O

1/2 O 1/2

O 1/2 O

3/4 1/4

1/4 3/4

B

2R

A



b) Derive an expression for inter-planar distance. (6 Marks)

1. Consider the plane ABC of a crystal which intersect the X,Y & Z axes at A,B & C respectively.

2. Let ( h k l ) be the miller indices of the family of ABC planes. 3. Draw ON ⊥ 𝑟 to the plane ABC ,then ON= represent inter-planar distance/spacing.

Then OA= ℎ , OB= &OC = ,where , & are basis vectors

From right ∆ OAN, Cos 𝜃 = = /ℎ = ℎ

……. From ∆ OBN, Cos 𝜃 = = / = ……. and also, from ∆ OCN, Cos 𝜃 = = / = ……

For orthogonal co-ordinates 𝜃 + 𝜃 + 𝜃 = ……

From eqns 1,2,3 &4 we get, ℎ + +

= 1

ie: [ ℎ + + ] = 1

d = √ + +

For a cubic crystal = =

d= √ + +

Z

C

N B Y

𝜃 𝜃

0 𝜃

A X



c) Define Miller Indices and explain the procedure to find Miller Indices. (4 Marks)

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.

Procedure for finding miller indices.

1, Find the intercepts of the plane along X,Y & Z axes in terms of basis vectors , & . Ex: Let X,Y &Z axes intercepts are 3 ,2 &1 respectively.

2. Find the coefficients of ,b & .

ie: 3,2 & 1.

3. Find the reciprocals of the coefficients of basis vectors.

ie:The reciprocals are , ,

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 , , = , , =

d) The minimum order of Bragg’s reflection occurs at an angle of 20° in the plane (2 1 2 ).Find the wavelength of X-rays if lattice constant is 3.614Å. (4 Marks)

Given: n = 1 ; 𝜃 = 0° ; ( h k l ) = ( 2 1 2 ) ; a = 3.614Å = 3.614x − m ; λ = ?

Using nλ = 2dSin𝜃 & d= √ℎ + +

We get , λ = 𝑖 𝜃 √ℎ + +

= x . x − ° x√ + + = 8.240 x − m

OR



8 a) Describe Bragg’s spectrometer to verify Bragg’s law and to find the nature of crystals. (7 Marks)

The labeled schematic diagram of the Bragg s X-ray Spectrometer is as shown in the diagram. To verify Bragg s law:

1. Collimate the X-rays from X-ray tube by passing them through slits S1 & S2. 2. Allow the collimated X-rays to incident on the surface of a given crystal mounted vertically at the centre of a turn table at a glancing angle 𝜃

3. The position of the crystal is noted using the vernier V1 and horizontal circular scale.

4. Adjust the ionization chamber to receive the scattered X-rays and measure the ionization current (I)which is a measure of intensity of the X-rays using electrometer E.

5. Repeat the experiment as before by increasing the glancing angle 𝜽 gradually and note the corresponding ionization current each time. 6. Plot a graph of I v/s 𝜃′ ℎ ℎ note down the glancing angles 𝜃 , 𝜃 & 𝜃

for the 1st,2nd &3rd order diffraction respectively. 7 Then substituting for 𝜃 , 𝜃 & 𝜃 in the Bragg s equation 2d Sin𝜃 = , It will be found that ,Sin𝜽 : 𝜽 : 𝜽 = : : ,which verifies Bragg s law. To find the nature of the crystal:

1. Collimate the X-rays from X-ray tube by passing them through slits S1 & S2. 2. Allow the collimated X-rays to incident on the surface of a given crystal mounted

vertically at the centre of a turn table at a glancing angle 𝜃. The position of the 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

Vernier- Circular scale

Slits glancing angle-𝜃 Crystal target

X-rays Vernier-

Collimated X-rays

E -electrometer

Ionisation chamber (D) Ba

Intensity

0 𝜃

𝜃 𝜃 𝜃



ionization current (I) produced by the X-rays using electrometer E. 4. Adjust plane of the crystal to the incident X-rays and find the glancing angle 𝜃 for 1st order Spectrum. 5. Repeat the experiment as before for the crystal surfaces & and find

the corresponding 1st order glancing angles 𝜃 & 𝜃 . 6. Then using bragg s equation, 2d Sin𝜃 = , find = 𝜽 : 𝜽 : 𝜽

7. The nature of the crystal is identified from the ratios of 𝑖 𝜃 : 𝑖 𝜃 : 𝑖 𝜃

as follows:

For SC; 𝜽 : 𝜽 : 𝜽 = 1: √ : √

For BCC; 𝜽 : 𝜽 : 𝜽 = 1: √ : √

For FCC; 𝜽 : 𝜽 : 𝜽 = 1: √ : √

b) Find the APF of SC,BCC and FCC crystals. (6 Marks)

For SC, n =1, R = 𝐚

Using, APF = a x R

= a x a

= a x

a

= π = 0.52 or 52%

For BCC , n = 2, R = √ 𝐚

Using, APF = a x R

= a x √ a

= a x

√ a

= √ π

= 0.68 or 68%

For FCC, n = 4, R = a√

Using, APF = a x R

= a x a√

= a x

a√

= π√ = 0.74 or 74%



c) Explain lattice parameters with a diagram. (3 Marks) Lattice parameters are the Six quantities that describe the unit cell completely. The

lattice parameters are three basis vectors are → ,→ ,→ and three interfacial angles , & . e) Sketch the following planes in a cubic unit cell (1 0 0 ) ,( 0 2 0 ),( 1 3 ) and ( 0 1 2). (4 Marks)

Module-5

9 a) Explain the construction and working of Reddy shock tube to find mach number. (6 Marks)


Z

c

0 b y

a

x

(100) x=1/1=1,y=z=0/1=∞

Z

o Y

X 1

(020) x= z=0/1=∞,y= 1/2

z

o y

x

(1 ) x=1/1=1,y = − / , z =1/3

Z

O0o 1/3 y

1 o

X 1

(012) x= 0/1=∞,y= 1/1=1,z=1/2

z

1 y

x

O

1/2

o

-1/2

1

1/2 0



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 ,temparature . 4. Two sensors & separated by a distance ∆ fitted to driven section measures the

pressures , and temperatures , respectively.

Working :

1. Driver section is filled with gas at high pressure and Driven section is filled with gas of low pressure & .

2. Diaphram is ruptured to produce shock waves by pushing the piston and the rupture pressure & temperature is measured using sensor S.

3. The time ′ taken by the shock wave to travel the distance x is measured using CRO. also the pressures , and temperatures , are measured using the sensors &

respectively.

4. The speed of the shock waves is calculated using V = 𝑡 .

5. Then the match number of the shock waves is calculated using M = 𝑉

,where a is the

speed of sound at temperature . 6. Also the mach number M can be calculated using the RH relations =

𝜸𝜸+ − 𝜸−𝜸+ , by finding , for the gas of known .

b) Mention three properties and three uses of CNTs. (6 Marks)

Properties of CNTS ! (any three)

1. CNT s are highly elastic. 2. Young s modulus of CNTs is about 9 times sronger than that of steel. 3. CNT s exhibit large strength in tension. 4. CNT s can be bent without breaking. 5. Electrical properties of CNTs ranges from semiconductor to good conductor.


x =7 cm


Plunger

49 cm 51 cm



6. CNT s have low resistivity and low heat dissipation. 7. Electrical Conductivity of CNTs is maximum along the axis and very less along the

perpendicular direction. 8. CNT s exhibit magneto-resistance ie:their resistance decreses with increasing mag.field. 9. Thermal conductivity of CNTs is maximum along the axis and very less along the

perpendicular direction 10. CNTs have very high strength to weight ratio and have low density. 11. CNTs are chemically more inert compared to other forms of carbon.

Uses of CNTS ( any three)

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 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.

c) Explain the synthesis of nanoparticles by ball mill method. (4 Marks)

1. Ball mill method is a mechanical method based on Top Down approach to synthesis nanoparticles of metals and alloys on large scale.

2. It consists of a steel container filled with the material whose nanoparticles to be produced along with large number of heavy steel balls.

3. The container is capable of rotation about an axis inclined to the horizontal. 4. The steel balls rotate circularly about the axis and spin about their own axis. Due to these

motions of steel balls ,the material is continuously crushed and powdered.

Steel container Steel Balls

Nanopowder

Stand



5. As this process is continued finally we get powdered nanoparticles. Merits : By ball mill method we can produce nanoparticles on large scale economically. Demerits: 1. Nanoparticles produced are irregular in shape .

2.Nanoparticles produced are impure. 3. Nanoparticles produced are irregular in size.

d) Mention the uses of shock waves. (4 Marks) ( any four)

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 dry painting. 3. Sw are used in the extraction of sandal wood. 4. Sw are 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. 8. SW are used to heal broken bones quickly.

OR

10 a) Explain the principal, construction, working and uses of SEM. (7 Marks)

Principle:

1. SEM is a device used to produce very high resolving images based on the principle of wave nature of electrons.

Electron gun

To Vacuum pump

Magnetic condenser coil -1

Magnetic condenser coil-2

Scanning coil

Objective magnetic coil

back scattered electron detector

Secondary electron detector

X-rays detector

Evacuated chamber Specimen

Base

TV monitor



2. The resolving power of SEM is 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 . 5. Secondary electrons produced due to interaction with valence electrons are detected by

the detector 6. and X-rays produced due to deep penetration are detected using the detector D 7. High resolution 3D image can be seen on the TV monitor to which D , D & D connected. 8. Biological specimens must be dehydrated and non conducting samples must be coated

with a thin conducting layer. 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.

11 Explain the terms : Acoustic waves ,Infrasonic waves ,Audible waves and Ultrasonic waves. (4 Marks)

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)



f) Describe the pyrolysis method of synthesis of CNTs. (5 Marks) 𝐍 𝐂 𝐇

1. The schematic diagram of Pyrolisis method is as shown in the diagram. 2. It consists of a chamber in an electric furnace through which a mixture of nitrogen and

acetylene is passed and let out through the out let. 3. A substrate is placed on the quartz sheet inside the chamber. 4. The temperature in the chamber is maintained at about 700-800. 5. Due to high temperature acetylene breaks down in to carbon atoms. 6. When these atoms come near substrate they get attracted and deposited as carbon

nanotubes in the presence of catalysts which are MWCNT s. 7. SWCNT s are obtained when acetylene is replaced with methane or carbon monoxide at

1200 d) Describe the different structures of CNTs. (4 Marks)

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.

**end**

Furnace

Chamber

C Catalyst

Pressure gauge

Substrate on graphite sheet

⊙ ⊙ ⊙ ⊙ ⊙

⊙⊙ ⊙ ⊙ ⊙ ⊙ ⊙⊙⊙

Arm chair CNT Zig-Zag CNT Chiral CNT

Axis

solved 60:40 cbcs ]#) (17phy12/22) sem-i/iicbitkolar.edu.in/wp-content/uploads/2018/05/qps.pdfmodule...

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