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JRE SCHOOL OF Engineering
PREUNIVERSITY EXAMINATIONS MAY15
Subject Name Engineering Physics-II
Roll No. of Student
Date
02May 2014
All Branches
Subject Code
Max Marks
Max Duration
Time
NAS-201
50 Marks
2 Hr.
10.00AM-12.00Noon
Set- II
SECTION – A
(2x5=10)
NOTE: ATTEMPT ALL QUESTIONS
1. (a) Calculate coordination number of Body centered and face centered cube.
Coordination number of BCC:
Case I: In case of corner atom 4 body centered atoms in the bottom plane and 4 body
centered atoms in the top plane are available.
Case II: In case body centered atom 4 atoms on the top plane and 4 atoms on the bottom
plane are available. Therefore the coordination number in BCC is 8
Coordination number of FCC:
☼
8 atoms are in 8 corners of a cube and 6 atoms at all 6 faces.
☼
For any corner atom the nearest atom is face centered atom.
Hence for corner atom there are 4 face centered atoms in the same plane and 4 atoms
above and below the plane.
Therefore, the coordination number is 4 + 4 + 4 = 12 atoms
(b) Differentiate magnetic materials in terms of susceptibility.
For Diamagnetic materials magnetic susceptibility is negative. (  m
 1)
For Paramagnetic materials magnetic susceptibility is positive. (  m
 1)
For ferromagnetic materials magnetic susceptibility is positive and very high. (  m
 1 )
(c) Write the dimensions of magnitude of pointing vector.
S = E X H = Watt/ m2 or J/s m2.
(d) What is Meissner effect?
Meissner effect is the repulsion of all magnetic flux from the
interior of a super conductor. It occurs as a result of surface
current in the super conductor that produce a magnetic field
which exactly opposes the applied field inside the super
conductor .the super conductor behaves as if it were a perfect
diamagnetic with  =-1.
(e) Give important applications of nanotechnology.
Ans:-(1) Because of great mechanical properties of the carbon nano tube ,a variety of structure
have been proposed ranging from everyday items likes clothes, and sports gear to combat
jackets.
(2) Nano tube based transistors have been made that operate at room temp.
(3) The carbon nano tube is often used as a vessel for transporting drugs into the body..the
nano tubes allows for the drug dosage to hopefully be lowered by localizing its
distribution as well as significantly cut costs to pharmaceutical companies and their
consumers.
(4) Bulk carbon nano tubes have been used as composite fibers in polymers to improve the
mechanical, thermal and electrical properties of bulk product.
(5) Carbon nano tubes have also been implemented in nano electromechanical systems
including mechanical memory elements and nano scale electric motors.
SECTION – B
(5 x 3 = 15)
NOTE: ATTEMPT ANY THREE QUESTIONS
2. (a) X Rays of wavelength 0.3 Ao are incident on a crystal with a lattice spacing 0.5 Ao. Find
the angles at which second and third order Bragg’s diffraction maxima are observed.
Solution:- According to Bragg’s law of diffraction
2d sin θ = nλ
or sin θ =
n
2d
λ =0.30 A0 and d = 0.5 A0
For second order maxima ,n=2
sin θ =
2 x 0 .3
 0 .6
2 x 0.5
and
θ = sin-1 (0.6) =36.860
For third order maxima ,n=3
Sin θ=
3 x 0 .3
 0 .9
2 x 0.5
and
θ = sin-1 (0.9) =64.150
(b) Calculate the ratio between electronic and ionic polarizability for a material having
εr = 4.94 and n2 = 2.69.
According to clausius mossotti relation
 r  1 N  e   i 

-------------------(1)
r  2
3 0
Where αe and αi are the electronic and ionic polarizability for the given material.
In optical frequency region :
Therefore eqn (1) becomes
έr = n2
and αi = 0
n 2  1 N e

n 2  2 3 0
------------------(2)
Dividing eqn(1) by eqn(2), we get
 r 1 n2  2 e  i
.

 r  2 n2 1
e

(4.94  1) (2.69  2)
1 i 
.
 e (4.94  2) (2.69  1)
 i 3.94 4.69

x
1
 e 6.94 1.69
i
 1.576  1
e
i
 0.576
e
e
 1.736
i
(c) If the upper atmosphere of earth receives 1.38 K Wm-2 energy from the sun, what will be
peak value of electric and magnetic field at the layer?
Energy flux or poynting vector S is given by
S 
E rms 
E  xB
0
orS 
EB
0

E 2 rms
0c
E0
2
E0  2  0 c S
 0 = 4π x 10-7 Hm-1 c= 3x108 m/s and S = 1.38 x kwm-2
E0 =
B0=
2 x4 x3.14 x10 7 x3x108 x1380
 1.02kVm1
E0 1.02 x103

 3.4 x10 6 Wbm 2
8
c
3x10
(d) For a specimen, the critical fields are 0.176 T and 0.528 T at 14 K and 12 K resp. Find
transition temperature and critical fields at 0 K and 4.2 K.
  T 2 
Solution:--- we know that Hc (T) = Hc (0) 1  
 
  Tc  
Given At T=14K , Hc (T)=0.176T and at T = 12 K Hc (T)=
0.528T
  14  2 
There fore 0.176 = Hc(0) 1  
  --------------(1)
  Tc  
  12  2 
0.528 = Hc(0) 1  
  ----------------(2)
  Tc  
Equation (2) by (1) we get
2
 12 
1  
 Tc  = 0.528  3
2
0.176
 14 
1  
 Tc 
1
Tc =
 444  2

 = 14.9 K
 2 
Putting the value of Tc in eqn(1) we get
Hc(0) 1.512 Tesla
For T =4.2 K
  4.2  2 
Hc (T )  221  
 
  14.9  
Hc(T)= 1.51 x 0.6789 = 1.392 Tesla Ans.
(e) What would be mobility of electrons in copper if there are 9x109 valence electrons per m3
and conductivity of copper is 6 x 107 mho/m ?
The conductivity of metal is given by σ=neµ
or


ne
where n is the density of the electron in metal and µ its mobility.
σ= 6x 107 mho/m , n=9x109 /m3

6 x10 7
 4.16 x1016 m 3 / volt  sec
9
19
9 x10 x1.6 x10
Page 1 of 2
SECTION – C
(5 x 5 = 25)
NOTE: ATTEMPT ANY ONE PART FROM EACH QUESTIONS
Q.3 (a) Show that distance between successive planes of Miller indices (hkl) for cubic lattice
d hkl =
a
(h  k 2  l 2 )
2
Also find the interplanar spacing between successive planes in Simple Cubic Lattice.
As per notes given
(b) Explain the principle of x-ray diffraction. Derive Bragg’s law. How interplanar spacing
and lattice are constant determined using a Bragg’s spectrometer.
Bragg law identifies the angles of the incident radiation relative to the lattice planes for
which diffraction peaks occurs. W.L. Bragg considered crystals to be made up of
parallel planes of atoms. Incident waves are reflected specularly from parallel planes of
atoms in the crystal, with each plane is reflecting only a very small fraction of the
radiation, like a lightly silvered mirror.
These two x-ray beams travel slightly different distances. The difference in the distances
traveled is related to the distance between the adjacent layers.
Connecting the two beams with perpendicular lines shows the difference between the
top and the bottom beams.
The length DE is the same as EF, so the total distance traveled by the bottom wave is
expressed by: DE  d sin 
EF  d sin 
DE  EF  2d sin 
Or
n  2d sin 
Constructive interference of the radiation from successive planes occurs when the path
difference is an integral number of wavelenghts. This is the Bragg Law.
For a particular crystal Braggs reflections for three planes with spacing d 1, d2 d3 are obtained
for angles θ1,θ2 and θ3 for first order then
2 d1 Sin θ1, = 2 d2 Sin θ2 =2 d3 Sin θ3 = λ
Hence d1 : d2 :d3 = 1/ Sin θ1 = 1/ Sin θ2 = 1/ Sin θ3
Rewriting the Bragg law
d 
2  sin   
n
which makes n-th order diffraction off (hkl) planes of spacing ‘d’ look like first-order
diffraction off planes of spacing d/n.
Planes of this reduced spacing would have Miller indices (nh nk nl).
Q.4 (a) Define polarization. Explain different types of polarization in solids.
The induced electric moment per unit volume of the dielectric is called dielectric polarization (P).
The induced dipole moment μ is proportional to electric field acting on the molecule 𝐸
E
  E
or
(1)
If there are N molecules per unit volume then
Dielectric Polarization:
P  NE
P  N or
(2)
Types of Polarization:
A.
B.
C.
D.
Electronic Polarization
Ionic Polarization
Orientational Polarization
Space charge Polarization
A. Electronic Polarization [Pe]: In non-polar dielectric polarizability arises due to displacement of electrons
relative to nucleus by the action of electric field.
At equilibrium
Lorentz force = Coulomb force
 Ze 
1( Ze)(ch arg edisplaced )
4o d 2
The equation becomes
 ZeE 
 Ze  3Ze

4R 3 4R 3
3

We know Volume charge density
(3)
Ze(
1
4o
(4)
4d 3 
)
3
d2
Substituting 𝜌 from (4)
 3Ze
)
4R 3 )
3
4d 3 (
 ZeE 
Ze(
1
4o
d
4o R 3
Ze
d2
(5)
Hence induced dipole moment
  Zed  Ze
E
4o R 3 E
Ze
or
Where  e is called as Electronic Polarizability given by
Hence Electronic Polarization
  e E
 e  4o R 3
Pe  N e  N e E
(6)
(7)
B. Ionic Polarization [Pi]: Ionic polarization occurs in ionic crystals.
On applying electric field, positive and negative charges are further separated in opposite directions. This leads
to increase in separation (d).
The induced dipole moment due to ionic polarization
i   i E
Where
i
(8)
is ionic polarizability.
If there are N molecules per unit volume then Ionic polarization
Pi  N i  N i E
(9)
For most of the materials, ionic polarizability is less than electronic polarizability.
i 
1
10 e
C. Orientational Polarization [Po]: Orientational polarization occurs in polar molecules having permanent dipole
moment.
If there are N molecules per unit volume then Orientational polarization
Po  N o  N o E
(10)
Orientational polarization is temperature dependent and frequency dependent.
o 
 o2
3kT
(11)
D. Space charge Polarization [Ps]: Surface, grain boundaries may be charged, i.e. they contain dipoles which may
become oriented to some degree in an external field thus contributes to the polarization of the material.
The contribution from space charge polarization is negligible. Hence may be ignored
(b)What is local field? Derive Clausius Mossotti equation in dielectrics subjected to static
field.
Solution: The internal field is the electric field that acts at the site of any given atom of a
solid or liquid dielectric subject to an external electric field and is the resultant of the
applied field and the field due to all the surrounding dipoles.
When an electric field is applied, atoms of liquid and solid become electric dipoles which
produce an electric field. This field is different from the applied field. The total field at
atomic site is called local (internal) field or Lorentz field
Ei  E 
P
3 0
Clausius-Mossotti relation:
For non-polar dielectric, when placed in Electric field induces dipole moment
𝜇 = 𝛼 𝐸𝑖
Where 𝜇 is induced dipole moment,
𝛼 is polarizability and
𝐸𝑖 is local field.
Also as per definition of Polarization if N is the number of molecules per unit volume then
𝑃 = 𝑁 𝜇 = 𝑁 𝛼 𝐸𝑖
Also
Ei  E 
P
o
Ei  E 
{substituting
 
P
3 o

P 
P  N  E 

3 o 

Therefore Polarization
1
for crystals having cubic symmetry}
3
Or
P
N o
N
1
3 o
Using the definition of Polarization P:
P   o ( r  1) E   o E
Inserting it into the above equations above gives as final result the connection between the
polarizability (the microscopic quantity) and the relative dielectric constant (macroscopic
quantity)
N  r  1


E
3 o  r  1
 3
Also
 0
N

3 o   2 0
Named as specific Polarization of Dielectrics
Q.5 (a) Write the Maxwell’s equation in differential and integral form. Explain the physical
significance of each equation.
(I)
GAUSS LAW IN ELECTROSTATICS

s
1
E  .ds 
0
q
q   dv  volumech arg edensity
v
E

1
0 
.ds 

Displacement current .D = έ0 E
D
D
dv


.ds   dv

.dI   div.D  ds 


Using Gauss divergence Theorem
 div.D dv   dv

div.D   
..D  
It represents Gauss’s law. It is time independent or steady state equation. The flux of the lines
of electric force depends upon charge density. Charge acts as a source or sink for the lines of
electric force.
(II)
Gauss law in Magneto statics
(Magnetic mono pole does not exist.)
 B .ds  0
 div.B dv  0

Magnetic flux density

div.B   0
It expresses that isolated magnetic poles do not exist. It is time independent equation and state
those equal numbers of lines of magnetic induction enters and leave a given volume. There is
no source or sink for lines of magnetic force.
(III)
Faraday law of Electro-Magnetic Induction
B 
t
B 

t
  xE   
curlE 
It relates the electric field and magnetic induction. It is a time dependent or time varying
equation. It summarizes the well known Faraday’s maws or Lenz’s law of electromagnetic
induction. It relates the space variation of E with the time variation of B or H.
(IV)
Ampere’s Law---- If H→ is the magnetic field intensity and J→ is current density
H

.ds    J  .ds 
Stoke' sTheorem
curlH  .ds    J .ds
 curlH

 J
It relates the magnetic field B with electric displacement vector D and the current density J. It
is also time dependent equation. It represents Ampere’s law.
(b) Deduce pointing theorem for the flow of energy in an electromagnetic field. Give the
physical significance of each term.
Poynting Theorem:
Energy can be transported from one point (transmitter) to another point (receiver) by mean of
electromagnetic wave .the rate of such energy transportation can be obtained from Maxwell’s
equation.
Electric flux density j=σ E
and D= έE
H 
   (1)
t
E

   (2)
t
  xE    
xH  E 
Dotting both side of equation (2) with .E

.E  .(xH )  E 2  E
E
   (3)
t
but
.( AxB)  B.(xA)  A.(xB)
( A  HandB  E )
H .(xE)  .( HxE)  E 2  E
E
t
.
from
equation(1)
H
 
)
( H .H )
t
2 t
 H 2
1 2E

 ( ExH )  E 2   2
2 t
2 t
H .(xE)  H .( 
Applying volume integral both side
 1
 ( ExH )dv   t   2 E
2


1

H 2 dv   E 2 dv
v
2

Applying gauss div. theorem

1
 ( ExH )ds   t   2 E
v
Total power
2

1
H 2 
dv 

2

rate of decrease in energy
 E
2
dv
Ohmic power dissipated
Leaving the volume stored in electric and magnetic field
This equation reflect to as pointing‘s theorem. This various terms in the equation are
identified using energy conservation arguments for electromagnetic field. The first term on
R.H.S. is interpreted as the rate of decrease in energy stored in electric and magnetic field.
Second term is power dissipated due to the fact that the medium is conducting
σ ≠ 0. The
quantity
E  x H  is known as Poynting vectors in watt/m2.
P=
E x H 
Poynting theorem stated that the net power flowing out of a given volume v is equal to the time
rate of decreases in energy stored with in v minus the conduction losses.
Power out
Stored electrical engergy.
E
Ohmic
losses
Ohmic loss
σ
Stored in magnetic energy
Power in
Q.6 (a) Explain Type I and Type II superconductor. Write a note on high temperature
superconductors.
Type I and Type II superconductor:
The presence of a magnetic field causes the critical temperature of type I super conductor to
decrease as shown in fig. If the magnetic field exceeds a certain critical value Hc. which
depends on material and its temperature it super conductivity disappears altogether. Such
materials are super conductor only for value of T and Hc below their suspective curves and are
normal conductor for values of T and B above these curves. The critical field Hc would be
maximum at 0K.
Type I super conductor exist only in two states, normal and super conducting. Type II super
conductor, which discovered later on usually alloys, have normal super conductor and
intermediate state as well. Such materials have two critical magnetic field ,Hc 1 and Hc2 , a type
II super conductor behave just like type I counter part when H<Hc ,for an applied magnetic
field less than Hc1 .It is super conducting with no magnetic field in its interior. When H>Hc, a
type II super conductor exhibits normal behavior, again same as a type I super conductor.
However in applied field between Hc1 and Hc2 ,a type II super conductor is in a mixed state in
which it contains some magnetic flux but is super conducting. The stronger external magnetic
field,the more flux penetrates the materials up to the higher critical field Hc 2 .
A type II super conductor
behaves as though it consists of
filaments of normal and of
superconducting matter fixed
together. A magnetic field can
exist in the normal filament
,while the super conducting
filaments are diamagnetic and
resistanceless like type I super
conductor .Because of Hc2 can
be quick high, Type II super
conductors are used to make high field( up to 20 tesla) magnets for particle accelerators.
(b)What are nanotubes? Discuss its structure, creation, properties and applications.
Carbon nano tubes (CNT’s) are sheets of graphite (Graphite is an allotropic form of pure and
brittle form of carbon) up to make a tube i.e. carbon macro molecules in cylindrical form. The
nano tube dimensions are variable and can be as small as 0.4 in diameters.
Properties of Carbon nano tubes:



CNT’s are very light, flexible, thermally stable and durable and posses extra ordinary tensile
strength .CNT’s have tensile strength about 65Gpa.which is 50 times higher than sheet.
They also exhibit interesting electrical properties i.e. depending on the way the graphite
structure spirals around the tube .CNT’s be insulating, semiconducting or conducting.
Carbon nano tubes are very hydrophobic (i.e. they do not like water) and can easily be bind
to proteins. Because of this property they can serve as chemical and biological sensors.
Structure of carbon nano tubes:
The bonding in carbon nano tubes is sp2 the each atom joined to three neighbors similar to
those in graphite structure. The tubes can therefore be considered as rolled up graphite sheets.
The structure of a nano tubes can be specified by a chiral vector (n,m,) defines how the
graphite sheet is rolled up, where n and m are integers of vector equation
R = na1+ma2
The chiral vector can be understood. The value of n and m determines the chirality, or twist of
the nano tube .The chirality affects the conductance ,density, lattice, structure and other
proposed a single walled nano tube is considered metallic. If the value (n- m) is divisible by 3,
otherwise nano tube is semi conducting. Consequently which nanotube are formed with
random value of n and m we could expect that two-third of nano tubes would be semi
conducting ,while the rest would be metallic. Given the chiral vector (n,m,).the diameter of a
carbon nano tube can be determining using relation ship.
d= (n2 +m2 +nm)1/2 x 0.0783nm
Types of Carbon nano tubes: Depending upon the value of n and m the carbon nano tubes
have been divided into following three categories --


If m = 0 nano tubes are called “zig-zag’
If n=m the nano tubes are called ‘Arm chair’
For any other combination of n and m nano tubes are called ‘chiral’
Uses of Carbon Nano tubes:
i.
ii.
Because of great mechanical properties of the carbon nano tube ,a variety of structure have
been proposed ranging from everyday items likes clothes, and sports gear to combat
jackets.
Nano tube based transistors have been made that operate at room temp.
iii.
iv.
v.
vi.
The carbon nano tube is often used as a vessel for transporting drugs into the body..the
nano tubes allows for the drug dosage to hopefully be lowered by localizing its distribution
as well as significantly cut costs to pharmaceutical companies and their consumers.
Bulk carbon nano tubes have been used as composite fibers in polymers to improve the
mechanical, thermal and electrical properties of bulk product.
Carbon nano tubes have also been implemented in nano electromechanical systems
including mechanical memory elements and nano scale electric motors.
Other applications including conductive and high strength composites
energy storage and energy conversion devices,sensors,field emissive
display and radiation sources, hydrogen storage media and nano meter
sized semi conductor devices ,probes and inter connect.
SWNT: Single cylindrical wall. The structure of a SWNT can be visualized as a
layer of graphite, a single atom thick, called graphene, which is rolled into a
seamless cylinder. Most SWNT typically have a diameter of close to 1 nm. The
tube length, however, can be many thousands of times longer. SWNT are more
pliable yet harder to make than MWNT. They can be twisted, flattened, and
bent into small circles or around sharp bends without breaking. SWNT have
unique electronic and mechanical properties which can be used in numerous
applications, such as field-emission displays, nanocomposite materials, nanosensors, and logic
elements. These materials are on the leading-edge of electronic fabrication, and are expected to
play a major role in the next generation of miniaturized electronics.
MWNT: Multi-wall nanotubes can appear either in the form of a coaxial
assembly of SWNT similar to a coaxial cable, or as a single sheet of graphite
rolled into the shape of a scroll. The diameters of MWNT are typically in the
range of 5 nm to 50 nm. The interlayer distance in MWNT is close to the
distance between graphene layers in graphite. MWNT are easier to produce in
high volume quantities than SWNT. However, the structure of MWNT is less
well understood because of its greater complexity and variety. Regions of
structural imperfection may diminish its desirable material properties. The
challenge in producing SWNT on a large scale as compared to MWNT is
reflected in the prices of SWNT, which currently remain higher than MWNT.
SWNT, however, have a performance of up to ten times better, and are
outstanding for very specific applications.
Q.7 (a) Explain why ferromagnetic materials can be permanently magnetized
whereas paramagnetic ones cannot.
Ferromagnetism: When you think of magnetic materials, you probably think of iron, nickel or
magnetite. Unlike paramagnetic materials, the atomic moments in these materials exhibit very
strong interactions. These interactions are produced by electronic exchange forces and result in
a parallel or anti-parallel alignment of atomic moments. Exchange forces are very large,
equivalent to a field on the order of 1000 Tesla, or approximately a 100 million times the
strength of the earth's field.
The exchange force is a quantum mechanical phenomenon due to the relative orientation of the
spins of two electrons. The elements Fe, Ni, and Co and many of their alloys are typical
ferromagnetic materials.
Paramagnetism: This class of materials, some of the
atoms or ions in the material has a net magnetic
moment due to unpaired electrons in partially filled
orbital. One of the most important atoms with
unpaired electrons is iron. However, the individual
magnetic moments do not interact magnetically, and
like diamagnetism, the magnetization is zero when
the field is removed. In the presence of a field, there
is now a partial alignment of the atomic magnetic moments in the direction of the field,
resulting in a net positive magnetization and positive susceptibility.
In addition, the efficiency of the field in aligning the moments is opposed by the randomizing
effects of temperature. This results in a temperature dependent susceptibility, known as the
Curie Law.
(b) Explain Fermi-Dirac probability distribution function. Obtain an expression for the
electrical conductivity of an intrinsic and extrinsic semiconductor.
The Fermi Level is defined as the highest occupied molecular orbital in the valence band at 0 K,
so that there are many states available to accept electrons, if the case were a metal.
In metals and semimetals the Fermi level EF lies inside at least one band. In insulators and
semiconductors the Fermi level is inside a band gap; however, in semiconductors the bands are
near enough to the Fermi level to be thermally populated with electrons or holes. The Fermi–
Dirac distribution 𝑓(𝑒) gives the probability that (at thermodynamic equilibrium) an electron
will occupy a state having energy e. alternatively; it gives the average number of electrons that
will occupy that state given the restriction imposed by the Pauli exclusion principle.
1
𝑓(𝑒) =
1+ 𝑒
𝐸−𝐸𝑓
( 𝑘𝑇 )
(1)
Here, 𝐸𝑓 is energy in fermi level, T is the absolute temperature and k is Boltzmann's constant. If
there is a state at the Fermi level (E = 𝐸𝑓 ), then this state will have a 50% chance of being
occupied at any given time.
AS PER
GIVEN
NOTE’S
Semiconductors can be intrinsic or extrinsic. Intrinsic means that electrical
conductivity does not depend on impurities, thus intrinsic means pure. In extrinsic
semiconductors the conductivity depends on the concentration of impurities.
Conduction is by electrons and holes. In an electric field, electrons and holes move in opposite
direction because they have opposite charges. The current density is:
J = (nμn+ pμp) eE = σE
(n= electron and p= holes and σ is conductivity)
σ = (nμn +pμp) e
here p is the hole concentration and h the hole mobility
In an intrinsic semiconductor, a hole is produced by the promotion of each electron to the
conduction band. Thus for intrinsic semiconductor, n = p = ni
σi = (μn + μp) ni e
In semiconductor, electron- hole pairs are created. The mobility of electron is μe and that of
hole is μp.
The electron and holes move in opposite directions in the electric field as they are of opposite
sign.
For N-type semiconductor, n >> p, therefore,
σn ≈ neμn ≈ Ndeμn
Where Nd is concentration of donor atoms
For P-type semiconductor, p>>n, therefore
σp ≈ peμp ≈ NA e μp
NA is concentration of acceptor atoms.
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