PH 2161 ENGINEERING PHYSICS – II Second Semester question bank - Anna University

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Unit :I Lasers
Part A
1. What is Laser?
Laser is a device which produces a highly intense, monochromatic,collimated, coherent
beam of light . It stands for Light Amplification by Stimulated Emission of Radiation.
2. What are the characteristics of Laser radiation? Or State the properties of Laser beam.
Laser radiations has
(i) High Intensity
(ii) High Coherence
(iii) High Monochromatism
(iv) High directionality with less divergence
3. Define Stimulated (or) induced absorption.
An atom is taken from ground state E1 to the exited state E2 by a photon of
energy E2 − E1 = hυ . This is called stimulated absorption.
4. Define Spontaneous emission:
An atom in an excited energy level E2 while returning to ground level E1, spontaneously
it will emit a photon of energy 2 1 E − E = hυ . This emission is called spontaneous emission.
5.Define Stimulated (or) induced emission:
An atom in an excited energy level E2 is induced by a photon of energy 2 1 E − E = hυ ,
thereby the atom return to ground level by emitting two photons having energy 2 1 E − E = hυ .
Both the photons are in phase. This emission is called stimulated emission.

E2
E1
2 1 E − E = hυ
E2
E1
2 1 E − E = hυ 2 1 E − E = hυ
2 1 E − E = hυ
E2
E1
2 1 E − E = hυ
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6.What are Einstein’s relations?
In Einstein’s theory of spontaneous emission and stimulated emission, we have
B12=B21
A21/B21=8πhν3/c3
7. What are the three important requisites for laser action to take place?
(i) suitable active medium
(ii) population inversion
(iii) optical feed back
8. What is the principle of Laser?
The principle of Laser is Stimulated Emission. The number of photons emitted by
Stimulated emission process increases in cascade process resulting in amplification of
light.
9. What is meant by Active medium in laser?
The medium in which population inversion is achieved in laser technology is
called an active medium in laser.
10 .Mention the various types of lasers?
Types of lasers are
(v) Solid state lasers
(vi) Gas lasers
(vii) Liquid lasers
(viii) Dye lasers
(ix) Semiconductor lasers.
11.Define population inversion
It is a situation in which the number of atoms in the higher energy state is more
than the no of atoms in the lower energy state.
E2 (Excited State)
E1 (Ground State)
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12. Define pumping.
The method of achieving population inversion is called pumping.
13. What are the different methods of achieving population inversion? Or What are the
different types of pumping techniques?
a. Optical pumping
b. Electric discharge method
c. Inelastic collision between atoms
d. Direct conversion
14. What is active medium in CO2 laser?
It is a gas mixture consisting of CO2, molecular nitrogen and additives such as
Helium and water vapour.
15. What is heterojunction laser?
Heterojunction means that the material on the one side of the junction differs
from that on the other side of the junction. It is also called modern laser diode.
E.g.) Heterojunction is formed between Ga As and Ga Al As
16. What is homojunction laser?
Homojunction means that a single crystalline material forms a p-n junction. E.g.)
Gallium Arsenide (Ga As)
17. A diode emits green light of wavelength l=5511.11 Å. Find out the value of Eg?
Eg=hc/l joules
=hc/(1.602*10-19*l) eV
=2.25 ev
18. Mention some applications of laser.
(i) Used in fiber optic communication systems.
(ii) Used for welding, drilling, cutting, soldering purposes.
(iii) Used in CDs for storing and retrieving the data.
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(iv) Used for retinal correction, bloodless surgery, destroying kidney stones.
19. Give some applications of laser in medical field.
1) It is used to treat cancer and tumor in human body.
2) It is used to carry out micro surgery and bloodless operation
3) It is used to shatter kidney stones.
4) It is used for eye lens curvature correction.
20. Define holography
Holography is the process of recording the amplitude and phase of an object with
the help of lasers. It is a three-dimensional recording process.
21.List out the applications of holography.
(i) Three dimensional images produced by holograms are used in
educational and technical fields and also in advertising artistic
displays, etc
(ii) For imaging the objects.
22. What is the use of nitrogen and Helium in CO2 laser?
In CO2 laser nitrogen helps to increase the population of atoms in the upper level of CO2,
while helium helps to depopulate the atoms in the lower level of CO2, and also to cool the
discharge tube
23. What is Nd-YAG laser?
Nd-YAG is a neodymium-based laser.
Nd- Neodymium (rare earth element Nd3+)
YAG-Yttrium Aluminum Garnet, which is Y3Al5O12.
It is a four level solid-state laser.
24. Compare the characteristics of laser with ordinary thermal source of light.
Ordinary light Laser light
• Light emitted is not monochromatic
• Light emitted has no high degree of
• Light emitted is highly
monochromatic
• Light emitted has a high degree of
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coherence
• Emits light in all directions (not
directional)
• Light is not intense and bright
coherence
• Emits light only in one direction
(directional)
• Light is much intense and bright
25. Difference between spontaneous emission and stimulated emission :
Sl.No Spontaneous Emission Stimulated Emission
1.
2.
3.
4.
5.
6.
Without any stimulation photons are
immediately emitted.
The emitted photons move in all
directions. It is a random process.
It is an uncontrolable process.
Less intense.
Polychromatic radiation.
Chain reaction is not possible.
Photons are emitted by stimulation of the
external photons.
The emitted photons move in same
direction. It is not a random process.
It is a controlable process.
High intense.
Monochromatic radiation.
Chain reaction is possible.
26. Calculate the number of photons from green light of mercury (l=4961 A0) required to do
one joule of work.
Given l=4961 *10-10
We know E=hn
=hc/l
= 4,006*10-19 joules.
Number of photons required to do one joule of work =1/(4.006*10-19)
=2.4961*1018 /m3
27. What is the ratio of stimulated emission to spontaneous emission at a temperature of 2800C
for sodium D-line?
(Stimulated emission/ spontaneous emission)=1/(ehn/kt)
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= 1 /{exp[6.625*10-34*3*108/1.38*10-23*5535.890*10-7]-1}
= 6.264*10-26
28. A diode emits green light of wavelength l=5511.11 Å. Find out the value of Eg?
Eg=hc/l joules
=hc/(1.602*10-19*l) eV
=2.25 eV
29. What is meant by Optical pumping ?
In this method ground state atoms are excited with the help of photons emitted by the
external optical (light) source. The atoms absorb energy from the incident photons and raises to
excited state. Hence population inversion is achieved in higher energy state
E1
E2
Light photon
Light source
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PART: B
1) Explain the different modes of vibration in CO2 molecule.
There are three modes of vibration in CO2 molecule. They are
i) Symmetric stretching mode:
In this mode of vibration, the carbon atom is at rest in its position and both the oxygen
atoms simultaneously vibrate along the molecular axis. Due to this vibration the oxygen atoms
are either approaching or departing the fixed carbon atom.
ii) Asymmetric stretching mode:
In this mode of vibration, the carbon atom and both the oxygen atoms are simultaneously
vibrate along the molecular axis. Due to this vibration both the oxygen atoms are vibrating in one
direction and the carbon atom is vibrating in opposite to the direction of vibration of the oxygen
atom.
iii) Bending mode:
In this mode of vibration, the carbon atom and both the oxygen atoms are simultaneously
vibrate perpendicular to the molecular axis. When both the oxygen atoms are vibrating
downward (or upward), the carbon atom is vibrating upward (or downward).
C
O O
Molecular axis
O C O
Molecular axis
O C O
Molecular axis
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2. What does optical pumping mean? Explain the different methods of pumping process?
The process of raising more number of atoms to excited state by artificial means is
called optical pumping. The different methods of pumping process are,
a) Optical pumping;
Here, the atoms are excited with the help of photons emitted by an external optical
source. The atom absorbs energy from the photons and raises to excited state.
e.g.) Ruby laser, Nd-YAG laser
b) Direct electron excitation:
The electrons are accelerated to very high velocities by strong electric field and they
collide with gas atoms and these atoms are raised to excited state
e.g.) Argon lasers, CO2 laser
c) Inelastic atom-atom collision: Here, a combination of two types of gases is used, say A and B,
both having same or nearly coinciding excited states A* and B*
During electric discharge A atoms get excited due to collision with electrons. The excited A*
atoms now collide with B atoms so that B goes to excited state B*.
e.g.) Helium-Neon laser
e- + A=A*
A* +B= B*+A
d) Direct conversion:
Due to electrical energy applied in direct band gap semiconductor like GaAs etc., the
combination of electrons and holes takes place and electrical energy is converted to light energy
directly.
e.g.) Semi conductor laser
e) Chemical process:
Due to some chemical reactions the atoms may be raised to excited states.
e.g.)Dye laser
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3) Explain about the Industrial and medical applications of laser?
Industrial Applications:
A) Lasers in welding:
a) Very high welding rates are possible.(with 10KW CO2 laser 5mm thick stainless steel plates
can be welded at the rate of 10cm/sec)
b) Dissimilar metals can be welded.
c) It requires only very less heat affected zones.
d) Any extremely complex shaped contours can be welded using computers for controlling the
deflection of the laser beam.
e) Micro welding can be done with great ease
f) Being non-contact method, the work piece is not stressed
B) Laser heat treatment:
When laser light is focused onto a particular area for a long time, then that area alone will
be heated and the other areas will remain as such. This effect is called laser heat treatment.
C) Laser cutting
Laser is used as a tool to cut thin metal sheets by properly focusing the laser onto any particular
area to be cut for a longer time. Lasers cut through a wide variety of materials rapidly and
precisely.
Medical Applications:
a) Ophthalmologists use Argon ion lasers for welding retinal detachment.
b) For cataract removal lasers are used.
c) Laser scalpels are used for blood less surgery.
d) In laser angioplasty for removal of artery block Nd: YAG lasers are used.
e) In dermatology lasers arte used to remove freckles, acne, birthmarks and tattoo.
f) Lasers are used in breaking kidney stones and gallstones into smaller pieces.
g) Lasers are used in cancer diagnosis and therapy.
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4. Explain the Principle and working of Nd-YAG laser with neat diagram.
Nd-YAG laser is a four level solid-state laser. Nd-YAG stands for Neodymium Yttrium
Aluminium Garnet. It is a crystalline material.
Principle:
In Nd-YAG laser, Nd-YAG rod is the active material. Krypton flash tube is used to excite
the Nd3+
ions from ground state to excited state. Hence population inversion is achieved. During
the transition from higher energy level to lower energy level, a laser beam of wavelength 1.064
μm is emitted.
Working
i) When the krypton flash tube is switched on neodymium ions are excited from ground level E0
to excited energy level E3 and E4 by absorbing a photon of wave length 0.8μm and 0.73 μm
respectively.
ii) The excited neodymium ions from E3 and E4 make a radiation less transition and enter into E2
energy level.
iii) Large number of neodymium ions is collected in energy level E2. Hence population inversion
condition is achieved between the energy level E2 and E1.
iv) Few of neodymium ions in E2 spontaneously emit a photon of energy
E2 − E1 = hu thereby it returned to energy level E1. This emitted photon will trigger the
neodymium ions in E2 hence a chain of stimulated photons are emitted.
v) The generated photon travels back and forth between the plane mirrors and stimulates
more number of neodymium ions in E2 and grows in strength.
vi) After enough strength has been attained high intensity of laser beam of wavelength
1.064 μm is emitted through the partially reflecting mirror.
E2
E3
E1
E0
E4
0.73 μm
0.8 μm
Radiation less transition
Radiation less transition
1.064 μm
Laser Light
Nd3+ ions
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vii) Neodymium ions in E1 make rapid non-irradiative transition from E1 to E0.
5.Describe the working of a homo-junction semiconductor laser.
Principle:
When a pn junction is forward biased and if the forward bias voltage is equal to the
energy gap voltage Eg, the electrons and holes are injected into the junction called active region
and creates population inversion. Since GaAs has direct band gap there is large possibility for
recombination of electrons and holes, which leads to emission of photons. The photon thus
produced may interact with conduction band electrons thereby stimulating radiative
recombination creating coherent photons.
Working
When forward bias voltage is applied to p-n junction diode, valence band holes in P region and
conduction band electrons in N region are injected into p-n junction. Hence population inversion
is achieved in p-n junction. The injected holes and electrons are recombining with each other and
produce energy in the form of photon. The emitted photon reflected back and forth between the
polished faces of the junction and stimulates the recombination. Due to this process emitted
photons strength is increased. After gaining enough strength, it gives the laser beam of
wavelength 8400 Ǻ.
Conduction band
Valence band
p-n junction
P - Region N - Region
Photon E = hυ
Free holes
Free electrons
Forbidden energy gap (Eg)
Fermi energy level (EF)
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6.Describe the working of a hetero-junction LASER Diode.
Principle
The electron in conduction band recombines with a hole in the valance band. This
recombination of electron and hole produces energy in the form of light (photon). This photon
stimulates other charges and as a result stimulated emission take place which leads to LASER
light.
Working
The forward bias voltage is applied between the contact layers (1st and 5th). Due to the
forward bias voltage charge carriers are produced in the wide band gap layers (2nd and 4th). These
charge carriers are injected into the 3rd layer. When population inversion achieved, the charge
carriers recombines and produced spontaneous emitted photons. These photons stimulate other
charges and as a result stimulated emission take place which leads to LASER light emitted from
the active medium.
First layer
Second layer
Third layer
Fourth layer
Fifth layer
Top Contact Layer
Bottom Contact Layer
Forward Bias
Voltage
Ga As p-type
Ga Al As p-type
Ga As n-type
Ga Al As n-type
Ga As p-type
+
-
LASER
E = h υ
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Part C
1.Explain the Construction and Working of Nd-YAG laser with neat diagram.
Nd-YAG laser is a four level solid state laser. Nd-YAG stands for Neodymium Yttrium
Aluminium Garnet. It is a crystalline material.
Principle:
In Nd-YAG laser, Nd-YAG rod is the active material. Krypton flash tube is used to excite
the Nd3+
ions from ground state to excited state. Hence population inversion is achieved. During
the transition from higher energy level to lower energy level, a laser beam of wavelength 1.064
μm is emitted.
Construction
In Nd-YAG laser, YAG is a crystalline material, a small amount of Yttrium 3+ ions are
replaced by Nd 3+ ions. This crystal is cut into cylindrical rod about 5-10 cm long and 6-9 mm
diameter. This rod is used as the active material. A krypton flash tube is placed near the rod. It is
used as the optical pumping source. Both the rod and flash tube is kept inside the highly elliptical
reflecting cavity. The flash tube is controlled with the help of capacitor and battery. A pair of
reflecting plane mirror is placed at the end of the rod one is 100 % reflecting other is partially
reflecting. This arrangement forms an optical resonator. The ends of the rod are highly polished
and optically flat and parallel with each other. The system is cooled with air and water
circulation.
100 % Reflecting
Mirror
Partially Reflecting
Mirror
Highly Elliptical
Reflecting Cavity
Nd-YAG rod
Capacitor
Battery
Krypton flash tube
Laser
Light
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Working
i) When the krypton flash tube is switched on neodymium ions are excited from ground level E0
to excited energy level E3 and E4 by absorbing a photon of wave length 0.8μm and 0.73 μm
respectively.
ii) The excited neodymium ions from E3 and E4 make a radiation less transition and enter
into E2 energy level.
iii) Large number of neodymium ions is collected in energy level E2. Hence population
inversion condition is achieved between the energy level E2 and E1.
iv) Few of neodymium ions in E2 spontaneously emit a photon of energy
E2 − E1 = hu thereby it returned to energy level E1. This emitted photon will trigger the
neodymium ions in E2 hence a chain of stimulated photons are emitted.
v) The generated photon travels back and forth between the plane mirrors and stimulates
more number of neodymium ions in E2 and grows in strength.
vi) After enough strength has been attained high intensity of laser beam of wavelength
1.064 μm is emitted through the partially reflecting mirror.
E2
E3
E1
E0
E4
0.73 μm
0.8 μm
Radiation less transition
Radiation less transition
1.064 μm
Laser Light
Nd3+ ions
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vii) Neodymium ions in E1 make rapid non-irradiative transition from E1 to E0.
Characteristics of Nd-YAG laser
Type : It is a four level solid state laser.
Active medium &
Active material
Pumping method &
source
Optical resonator :
Power output : Power output is approximately 70 watt.
Nature of output : The output light is pulsed or continuous wave form.
Wavelength of
output light : The wavelength of output beam is 1.064 μm..
Advantages of Nd-YAG laser
i) Out put energy is very high.
ii) Repetition rate is very high.
iii) Population inversion is easily achieved.
iv) Excitation can be easily achieved by applying lower threshold energy. (Because YAG is a
crystalline material)
Disadvantages of Nd-YAG laser
The electron energy level structure of Nd3+ in YAG is complicated.
: Nd-YAG rod is the active medium and material.
: Optical pumping method is used. Krypton flash tube is used as pumping
source.
Pair of plane mirrors is used as optical resonator. One is 100 % reflecting
other is partially reflecting. Two end of the Nd-YAG rod is highly and
optically polished. One end is 100 % silvered other end is partially
silvered. It serves as optical resonator.
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2.Explain the construction and working of He-Ne laser with neat diagram.
He-Ne laser is the first gas laser.
Principle:
The active medium is the He-Ne gas. The gas laser medium is excited by electric
discharge method in which inelastic collision between atoms will result in population inversion.
Construction
He-Ne laser consists of a quartz discharge tube about 80 cm long and 1.5 cm diameter. The
discharge tube is filed with mixture of helium and neon gases with different partial pressures.
The gases are mixed under the pressure of 1 mm mercury of helium and 0.1 mm mercury of neon
in the ratio of 10 : 1. These mixture acts as active material. The ends of the discharge tube are
inclined at the polarizing angle. Hence the emitted laser light is plane polarized. This
arrangement is called Brewster window.
A radio frequency generator (RF generator) is connected to the discharge tube. It is used
to produce population inversion in active material. A pair of concave mirror is placed at the ends
of the discharge tube. One of the mirrors is 100 % reflecting and the other is partially reflecting.
These mirrors form an optical resonator.
Working
He Ne
F1
F2
F3
He*
He*
Ne*
Ne*
3.39 μm
1.15 μm
6328 Å
≈ 6000 Å
Radiation less transition
Energy transfer
through collision
E1
E2
E3
E4
E5
E6
RF Generator Brewster
window
Quartz discharge tube
100 % reflecting
concave mirror
Partially reflecting
concave mirror
Plane polarized
Mixture of He-Ne gas laser light
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When RF generator is switched on, electrons are generated inside the discharge tube. At
first the generated electrons collide with the helium atoms in ground state F1 thereby helium
atoms are excited to F2 and F3.
E4 and E6 are the excited energy level of the neon atoms which coincides with the excited
energy level of helium (F2 and F3). The excited helium atoms are collide with neon atoms in
ground state (E1) thereby neon atoms are excited to the energy levels E4 and E6. The helium
atom completely transfers its energy to the neon atom then it return to ground state (F1).
Three types of transitions constitute the laser beam, because population inversion
achieved between the energy levels E6→E5, E6→E3 and E4→E3.
E6→E5 transition : In this transition a laser beam of wavelength 3.39 μm is emitted. It is lying
in far infrared region.
E6→E3 transition : In this transition a laser beam of wavelength 6328 Å (red colour) is
emitted. It is lying in visible region.
E4→E3 transition : In this transition a laser beam of wavelength 1.15 μm is emitted. It is lying
in infrared region.
Laser beam of wavelength 6328 Å dominates the other two wavelengths. Therefore red
laser beam is emitted from the partially reflecting concave mirror.
*
1
e electron with higher kinetic energy
He atom in ground state
He atom in excited state
e electron with lesser kinetic energy
*
e + He ® He + e1
He* + Ne ® Ne* + He
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Characteristics of He - Ne laser
Type : It is a four level gaseous laser.
Active medium &
Active material
Pumping method
Optical resonator :
Power output : Power output is approximately 0.5 to 50 milli watt.
Nature of output : The output light is continuous wave form.
Wavelength of
output light : The wavelength of output beam is 6328 Å.
Advantages of He-Ne laser
i) Less heat is generated inside the discharge tube, therefore no need for cooling.
ii) It is operated in continuous mode.
iii) It has high stability of wavelength.
iv) Output of the laser can be tuned to a required wave length (±100 Ǻ).
v) It is not much expensive.
Disadvantages of He-Ne laser
i) Output power is moderate. (0.5 – 50 mill watt)
ii) Compare with solid state laser its size is large.
: The mixture of helium and neon gases is filled in the discharge tube.
Helium gas used as active medium and neon atom used as active
material
: Inelastic atom-atom collision method is used.
Pair of concave mirrors is used as optical resonator. One is 100 %
reflecting other is partially reflecting.
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3. Describe the construction and working of CO2 laser with neat diagram.
Principle
It is a four level molecular gas laser. The mixture of carbon di-oxide, nitrogen and helium
gases is used as an active material. Laser transition takes place between the vibrational states of
CO2 molecules. Inelastic atom-atom collision method is used to produce population inversion in
active material. A pair of concave mirror is used as an optical resonator. A laser beam of
wavelength 10.6 μm is emitted.
Construction
CO2 laser consists of a quartz discharge tube about 5 m long and 2.5 cm diameter. The discharge
tube is filled with mixture of CO2, N2 and helium gases with different partial pressures. In this
mixture CO2 act as active material, N2 act as active medium and helium act as coolant. The
ends of the discharge tube are fitted with alkali halide windows. Hence the emitted laser light is
plane polarized. This arrangement is called Brewster window.
The discharge tube is connected to the induction coil. It is used to produce population inversion
in active material. A pair of concave mirror is placed at the ends of the discharge tube. One of
100 % reflecting
concave mirror
Partially reflecting
concave mirror
laser light
CO2 N2 He
To induction coil
Vacuum
pump
NaCl
Brewster Window
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the mirrors is 100 % reflecting and the other is partially reflecting. These mirrors form an optical
resonator.
Working
When induction coil is switched on, electrons are generated inside the discharge tube. At
first the generated electrons collide with the nitrogen molecules in ground state F1 thereby
nitrogen molecules are excited to F5.
The excited N2 molecules are collided with CO2 molecules in ground state (E1) thereby
CO2 molecules are excited to the vibrational energy level E5. The N2 molecule completely
transfers its energy to the CO2 molecules then it return to ground state (F1).
Two types of transitions constitute the laser beam, because population inversion achieved
between the energy levels E5→E4 and E5→E3.
E5→E4 transition : In this transition a laser beam of wavelength 10.6 μm is emitted. It is lying
in infrared region.
*
e + N2 ® N2 + e1
* *
2 2 2 2 N + CO ® N + CO
N2 CO2
F1
F5
N2
*
10.6 μm
9.6 μm
Radiation less transition
Energy transfer
through collision
E1
E2
E3
E4
E5
CO2
*
Radiation less transition
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E5→E3 transition : In this transition a laser beam of wavelength 9.6 μm is emitted. It is lying
in infrared region.
Laser beam of wavelength 10.6 μm dominate the other wavelength. Therefore 10.6μm is
emitted from the partially reflecting concave mirror.
Characteristics of CO2 laser
Type : It is a four level gaseous laser.
Active medium &
Active material
Pumping method
Optical resonator :
Power output : Power output is approximately 10 k watt.
Nature of output : The output light is continuous or pulsed wave form.
Wavelength of
output light : The wavelength of output beam is 10.6μm.
Advantages of CO2 laser
i) Output power is very high
ii) The output power can be adjusted by adjusting the length of the tube.
iii) Efficiency is very high.
iv) Construction is simple.
Disadvantages of CO2 laser
i) Output power is affected by operating temperature.
ii) Corrosion problem occurred in the reflecting plates..
iii) Power output is very high; hence accidental exposure may crate damage to the eye.
: Mixture of CO2, N2 and helium gases is filled in a discharge tube.
Nitrogen molecules used as active medium and corban di-oxide used as
active material and helium used as coolant.
: Inelastic atom-atom collision method is used.
Pair of concave mirrors is used as optical resonator. One is 100 %
reflecting other is partially reflecting.
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iv) Contamination of carbon monoxide and oxygen will have some effect on the laser
action.
4. Describe the construction and working of homo-junction semiconductor laser.
Principle:
When a pn junction is forward biased and if the forward bias voltage is equal to the energy gap
voltage Eg, the electrons and holes are injected into the junction called active region and creates
population inversion. Since GaAs has direct band gap there is large possibility for recombination
of electrons and holes, which leads to emission of photons. The photon thus produced may
interact with conduction band electrons thereby stimulating radiative recombination creating
coherent photons.
Construction:
A homo-junction semiconductor laser diode is fabricated from same crystalline material
(Ga As : Gallium Arsenide). It is cut in the form of platelet about 0.5 mm thickness. It is used as
the active material. The platelets consist of two region called P and N and a junction is formed
in between the regions it is called p-n junction, it provides the active medium. In P region
majority charge carriers are holes and in N region majority charge carriers are electrons.
Electrical contact leads are fixed with top and bottom surface of the platelet and forward bias
voltage is applied through the leads. The end face of the junction is highly polished and parallel
with each other; it acts as the optical resonator.
P - Region
N - Region
+
- Laser beam
Contact lead
Contact lead
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Working:
When forward bias voltage is applied to p-n junction diode, valence band holes in P
region and conduction band electrons in N region are injected into p-n junction. Hence
population inversion is achieved in p-n junction. The injected holes and electrons are
recombining with each other and produce energy in the form of photon. The emitted photon
reflected back and forth between the polished faces of the junction and stimulates the
recombination. Due to this process emitted photons strength is increased. After gaining enough
strength, it gives the laser beam of wavelength 8400 Ǻ.
Characteristics of semiconductor laser
Type : Solid state semiconductor laser.
Active material & Active medium: The diode is made from a single crystal of gallium
arsenide; it is used as active material. P-N junction is used
as active medium.
Pumping method : Direct conversion method is used.
Optical resonator : End faces of the p-n junction are highly polished and
parallel with each other. It acts as the optical resonator.
Output power : Output power of this laser is about 1 milliwatt.
Nature of output : Continuous or pulsed wave form can be obtained.
Wavelength of output : The wavelength of output is 8400 Ǻ.
Conduction band
Valence band
p-n junction
P - Region N - Region
Photon E = hυ
Free holes
Free electrons
Forbidden energy gap (Eg)
Fermi energy level (EF)
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Advantages
1) Its cost is low.
2) It is very simple and exhibits high efficiency.
3) It can be operated at low power.
4) Output is easily modulated by controlling the junction current.
Disadvantages
1) The beam has very large divergence.
2) Purity of monochromatic is poor.
3) Output power is very low.
Applications
1) It is mostly used in fiber optic communication.
2) It can be used as a pain killer.
3) It is used in laser printers and CD writing and reading.
5.Describe the construction and working of a hetero-junction LASER Diode.
Principle
The electron in conduction band combines with a hole in the valance band. Hence the
recombination of electron and hole produces energy in the form of light (photon). This photon
stimulates other charges and as a result stimulated emission take place which leads to LASER
light.
Construction
First layer
Second layer
Third layer
Fourth layer
Fifth layer
Top Contact Layer
Bottom Contact Layer
Forward Bias
Voltage
Ga As p-type
Ga Al As p-type
Ga As n-type
Ga Al As n-type
Ga As p-type
+
-
LASER
E = h υ
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It consists of five layers. The third layer is made of Ga As – p-type that has a narrow
band gap. This layer is act as the active medium. Wide band gap layer is coated on the upper and
bottom of the third layer. The upper layer is second layer and it is made of Ga Al As – ptype
and fourth layer is Ga Al As – n-type. The junction of 2nd – 3rd layer and 3rd – 4th layer are
well polished hence it act as the optical resonator. A contact layer is made of Ga As – p-type is
formed on the second layer. All the four layers are grown on the fifth layer made of Ga As – ntype.
Working
The forward bias voltage is applied between the contact layers (1st and 5th). Due to the
forward bias voltage charge carriers are produced in the wide band gap layers (2nd and 4th). These
charge carriers are injected into the 3rd layer. When population inversion achieved, the charge
carriers recombines and produced spontaneous emitted photons. These photons stimulate other
charges and as a result stimulated emission take place which leads to LASER light emitted from
the active medium.
Characteristics
Type : It is a solid state semiconductor laser.
Active medium : p-n junction diode made from different layers
Pumping method : Direct conversion method
Optical resonator : Polished layer act as the optical resonators
Power output : The power output from this laser is 10 mW
Nature of the output : Continuous wave form
Wave length of output: It gives infrared radiation of wavelength 8000 Å
Advantages
i) The output power is high.
ii) It has high directionality and high coherence.
iii) Low threshold current density.
iv) Highly stable and longer lifetime.
Disadvantages
i) Cost is higher than homojunction laser.
ii) Difficult to grow different layers of p-n junction.
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6.Explain with a neat sketch, the construction and reconstruction of a hologram
using laser beam.
The method of producing three-dimensional image of an object due to the interference
phenomena of coherent light waves on a photographic plate is known as holography.
Principle
The light waves reflected from the object and directly from the source are interfering with
each other to produce the image of an object due to interference phenomena on the photographic
plate. The recorded photographic plate is called hologram. The hologram has no resemblance to
the object but it contains all the information about the object.
It consists of two-step processes
i) Construction of Hologram : Recording the image of an object on the photographic plate
due to interference phenomena.
ii) Reconstruction of Hologram : To display the recorded hologram.
Construction of Hologram
The light beam coming from the He-Ne laser source is made to pass through the beam
splitter. It splits into two beams. One is reflected by the beam splitter and is incident on the
photographic plate, this beam is called reference beam. The second beam is transmitted through
the beam splitter and is incident on the object, this beam is called object beam. When the object
beam is incident on the object, it is scattered in all directions and a part of scattered light is
incident on the photographic plate at the same area in which the reference beam strikes it.
These two waves are interfering with each other to produce the interference phenomena on the
photographic plate. The fringe width is very small as 1 μm.
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Reconstruction
The light beam coming from the He-Ne laser source is reflected by the plane
mirror and is made to pass through the hologram. This wave is called read out wave. This
wave interacts with the interference pattern on the hologram and produces two images. If
the hologram is transparent, two-dimensional real image is formed on the other side of the
hologram. The real image can be recorded on the photographic plate. When we look
through the hologram, a three-dimensional virtual image is formed.
Object beam
Reference
beam
Scattered
beam
Object
Beam splitter
Photographic
plate
He-Ne LASER Source
3D
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Unit 2 : Fibre Optics
Part A
1. What is an optical fibre?
An optical fibre is a dielectric waveguide used to transmit light signals from one place
to another.
2. What is the structure of an optical fibre?
Optical fiber consists of an inner cylinder made of glass or plastics called core and
it is surrounded by a cylindrical shell of glass or plastics called cladding. The refractive index of
core is greater than refractive index of cladding. The cladding is covered by a polyurethane
jacket which protects the fiber from moisture and abrasion.
3. What is the principle of fibre optic communication?
The principle of fibre optic communication is Total Internal reflection.
2. What are the conditions for total internal reflection?
i. The ray of light should travel from a denser to a rarer medium.
ii. The angle of incidence in the denser medium should be greater than the
critical angle of that medium.
3. Define acceptance angle
The maximum angle with which a ray of light can enter through one end of the fiber and
will be total internally reflected
Core
Cladding
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4. How are optical fibers classified?
Optical fibers are classified into three types
i) Based on the nature of the material
ii) Based on the modes
iii) Based on refractive index profile
5. Define Numerical aperture of a fibre.
Numerical Aperture of the fibre is the light gathering ability of a fibre.It is a
measure of the amount of light accepted by the fibre.
NA=sin θ0 =√n1
2-n2
2
6. Mention some applications of optical fibers
i) they are used in communication systems
ii) it is used foe signaling purposes
iii) it is used in television cables, submarine cables, etc
iv) it is used in an endoscope
8.What are the types of optical fibers based on the number of modes?
Optical fibers are classified depending on modes of propagation in the guide
1.Single Mode fiber
2.Multi Mode fibre
9. What are meridinal rays?
Meridinal rays are the ray s following Zig Zag path when they travel through fibre
and for every reflection it will cross the fiber axis.
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10. Distinguish between step index fiber and graded index fiber.
Sl.No. Step Index Fiber Graded Index Fiber
1
2
3
4
The difference in refractive indices
between core and cladding is obtained
in single step.
The path of propagation of light is
Zig-Zag.
Lower bandwidth
Distortion is more
The difference in refractive indices
between core and cladding is gradually
varying.
The path of propagation of light is
helical.
Higher bandwidth
Distortion is less
11.Define splicing
The technique of joining two optical fibres is called splicing .
12. Mention the steps involved in splicing
The techniques involved in splicing, basically consists of
i) Fusing the two fiber ends
ii) Bonding them together in an alignment structure.
13. What is single mode fibre?
In a fibre, if only one mode is transmitted through it, then it is said to be single
mode fibre.
14. What is multimode fibre?
If more than one mode is transmitted through optical fibers, then it is said to be
multi mode.
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15. Define attenuation
Attenuation (or fiber loss) is defined as the ratio of the output power Pout from a
fiber of length L to the optical input power Pin
16.What are the basic types of losses that occur in a fiber?
i) Absorption losses
ii) Scattering losses
iii) Radiative losses.
17. What are fibre optic sensors?
Sensor are the device used to measure or monitor quantities such as displacement,
pressure, temperature, flow rate, liquid level, chemical composition etc.
18.Explain the basic principle of a fiber optic sensor.
A fiber optic sensor consists of a light source coupled to an optical fiber and a
detector receives the signal-carrying light beam as it emerges from the fiber. The signal is
then electrically processed to get the desired output.
19. Mention the components involved in fibre optical system.
(i) Light source
(ii) Photo detector
(iii) Optical fibre transmission line.
.20. What are the advantages of fiber-optic communication?
i) Less cross-talk
ii) noise-free transmission
iii) economical
iv) longer life span
v) Easy maintenance.
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21. What are the properties of single mode fibre?
• Only one mode of light can be propagated.
• Core diameter is small
• The difference in refractive indices between core and cladding is small.
• No intermodal dispersion.
• Information can be carried to longer distances.
• Fabrication is difficult. Cost is high
22. What are active and passive sensors?
In active sensors the quantity to be measured acts directly on the light carrying
fibre itself and the fibre acting as a transducing element modifies the light passing
through it.
In passive sensors light reaches the transducing material through the fibre. The
quantity to be measured acting on the transducing material modifies the light. The
modified light again collected by the fibre and then carried to the detector.
23. What is meant by splicing in fibre optic?
A fibre splicing is a permanent or semi-permanent joint between 2 fibre and the
process of joining 2 fibres is called splicing.
24. What are the different types splicing in fibre optic?
1. Fusion splicing 2. Mechanical splicing
25. How is attenuation loss measured in an optical fibre?
Attenuation loss is measured in terms of decibel(dB) which is a logarithmic unit. The
loss of optical power is given by α = -10/L*log(Pout/Pin) dB/Km
Where Pout is the power of the output beam
Pin is the power of the input beam
L is the length of the fibre in Km
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26. What type of light source are to be used for fibre optic communication?
The light source used for optical fibre communication should meet the following
requirements
a. Monochromatic
b. intense
c. capable of being easily modulated
d. small and compact
e. durable and inexpensive
light emitting diodes (LEDs) and laser diodes (LDs) satisfy almost above requirements
27. What type of detectors are to be used for fibre optics communication?
The photo detectors used for optical communication should have
a. high quantum efficiency
b. adequate frequency response
c. low dark current
d. low signal dependent noise
PIN diodes and Avalanche diode satisfy the above requirements and hence are used for
fibre optics communication.
28. Give the application of fibre optics.
Optical fibres can be used
a. For communication purpose
b. As light pipe in endoscope application
c. As a part of fibre optic sensor used for different measurements
29.Write a note on endoscope.
Endoscope is a medical instrument. It has coherent optical bundle in the form of
imaging light pipe. This is inserted inside the human body to view and photograph the inner
organs.
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Part B
1. Explain with a neat diagram the Double crucible method of fibre drawing.
The double crucible method can be used to make both silica and halide glass fibers. The
technique is simple and straightforward. One glass rod is made from silica powders for the core
and one for the cladding. The rods are then used as feedstock for each of two concentric
crucibles. The inner contains the molten core, while the outer contains the cladding. In a
continuous process, the fiber is drawn from the molten state. The disadvantage of this method is
the possibility of introducing contaminants during the melting process. Figure illustrates the
double-crucible drawing process
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2.Derive the relation between numerical aperture and fractional index change and
explain Intermodal dispersion.
The sine of the acceptance angle of the fiber is called numerical aperture (NA).
i NA = sin θ
2 2
1 2
0
n n
NA
n

=
For air medium 0 n = 1 \ 2 2
1 2 NA = n − n (1)
The ratio of the refractive index difference between the core and cladding to the
refractive index of core is called fractional index change ( D).
Fractional index change
Refractive index difference between the core and cladding
refractive index of core
D =
1 2
1
n n
n

D =
1 1 2 D n = n − n (2)
From eqn. (1) ( ) ( ) 1 2 1 2 NA = n + n n − n (3)
Sub. eqn. (2) in eqn. (3) ( ) 1 2 1 NA = n + n D n
1 2 1 2 1 Assume n » n ; n + n » 2 n
2
1 \ NA = 2 D n
1 NA = n 2 D (4)
If we increase the fractional index change D, the light gathering capacity of fiber will be
increased. We can not increase Dto high value. If we increase to high value, it will produce
signal distortion. It is called intermodal dispersion.
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3.Distinguish between step index fiber and graded index fiber.
Sl.No. Step Index Fiber Graded Index Fiber
1
2
3
4
The difference in refractive indices between
core and cladding is obtained in single step.
The path of propagation of light is Zig-Zag.
Lower bandwidth
Distortion is more
The difference in refractive indices between core cladding is gradually varying.
The path of propagation of light is helical.
Higher bandwidth
Distortion is less
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4.Give differences between single mode and multimode fiber.
Sl.No. Single mode Fiber Multi mode Fiber
1
2
3
4
5
6
Only one mode of light can be
propagated.
Core diameter is small
The difference in refractive indices
between core and cladding is small.
No intermodal dispersion.
Information can be carried to longer
distances.
Fabrication is difficult. Cost is high.
More than one mode of light can be
propagated.
Core diameter is large
The difference in refractive indices
between core and cladding is large.
Intermodal dispersion is more.
Information can be carried to shorter
distances only.
Fabrication is easy. Cost is low.
5. Discuss the different types of Losses in optical fibres.
The optical power loss in a fibre is calculated using the formula
dB=-10log(Pout/Pin)
where Pout is the power coming out of the fibre
Pin is the power launched into the fibre
There are two main types of losses. They are
(i) Scattering losses and
(ii) Absorption losses
Scattering losses: The glass in optical fibre is an amorphous solid that is formed by allowing
the glass to cool from its molten state at high temperature until it freezes.While it is still plastic,
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the glass is drawn in the form of a very thin fibre under proper tension. During this process,
submicroscopic variations in the density of the glass are frozen into the fibre. Dopants added to
silica to modify the refractive index also cause fluctuation in refractive index.These
inhomogenities scatter a small portion of the light passing through the fibre, contributing for the
losses.
Absorption losses:
Three different mechanisms contribute to absorption losses in glass fibres. These are
ultraviolet absorption, infrared absorption, and ion resonance absorption.In pure fused silica,
absorption of ultraviolet radiation results in ionization of valence electrons into conduction band.
Thus there is loss of light due to ionization. Also during the fabrication of fibre,to change the
refractive index of the glass to any desired value, GeO2 is doped. This causes shift in the UV
absorption band towards longer wavelength region. The presence of impurities such as iron,
copper and chromium and OH- ions causes absorption losses in fibre.Hence extra care has to be
taken while purification of silicon.
6. Explain fibre optic communication systems.
Fibre optic communication systems comprise of the following units:
Transmitter consisting of
Light source and
Input signal modulator
Optical fibre that couples the modulated input signal from the transmitter with the receiver.
Receiver consisting of
Optical detector
Drive
circui
t
Light
sourc
e
Photo
detect
or
Signa
l
restor
ampli er
fier
Input
signal Output
signal
Optical
fiber
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Signal demodulator
Output signal display unit such as CRO
At first the information signal which is in analog form is converted into electrical signal.This
signal is fed into the transmitter.
Transmitter:
It consist of a drive circuit and a light source. The drive circuit converts electric input
signal into digital pulses. The light source converts digital pulses into optical pulses. The light
source may be an LED or Laser diode. Here electric pulses modulates the intensity of light
source and is then fed into the optical fibre.
Optical fibre:
Optical fibre acts as a wave guide to transmit the pulses towards the receiver by the
principle of Total internal reflection.
Receiver:
The photodetector in the receiver receives the optical pulses and converts it into the
electrical pulses. The amplifier amplifies the received signal.These signals are decoded into
analog signal. Thus original signal is obtained at the receiver end.
7. Write a note on total internal reflection
Total Internal Reflection:
Consider a ray of light traveling from core medium whose refractive index is 1 n to
cladding medium whose refractive index is 2 n . Let the refractive index of core is greater than
refractive index of cladding. ie., n1 > n2
When the incident ray incident at an angle θi on
the core-cladding interface, the light is refracted into the
cladding and makes an angle of refraction θr.
θr
θi
Cladding
Cladding
Core
Normal
Incident ray
Refracted ray
core cladding interface
1 2 n > n
1 n
2 n
2 n
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According to Snell’s law of refraction,
1 c n sin θ = n2 sin θr
r Where θ = 90 \ sin 90 = 1 o o
1 c 2 n sin θ = n
2 – 1 2
c c
1 1
n n
sin θ θ sin
n n
= \ =
The phenomenon of total internal reflection takes place when it satisfies the following
two conditions.
i) Light should travel from denser medium to rarer medium. ie., refractive index of core
( 1 n ) is greater than refractive index of cladding ( 2 n ).
ii) The angle of incidence on core ( i θ ) should be greater than the critical angle ( c θ ).
ie., i c θ > θ
When the angle of incidence θi increases, the
angle of refraction θr also increases. At a particular
angle of incidence the refracted ray passing through the
core-cladding interface. That angle of incidence is called
critical angle θc.
When the angle of incidence is further increased
(above the critical angle θc), the ray totally reflected by
the core cladding interface. This is called total internal
reflection. Hence the rays travel in core medium only.
θr = 90º
θc
Cladding
Cladding
Cladding
Cladding
Core
Core
Normal
Normal
Incident ray
Incident ray
Reflected ray
core cladding interface
core cladding interface
1 2 n > n
1 2 n > n
1 n
1 n
2 n
2 n
2 n
2 n
>θc
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Cladding
Cladding
Core
core - cladding interface
Fiber axis
Refractive index n2
θi
Refractive index n2
Refractive index n1
θ
θr
θc
90°
A
O
B C
D
A1
Normal
Fiber axis
Refractive
index
of air n0
Part C
1.Derive an expression for angle of acceptance and numerical aperture:
Consider a cylindrical fiber; it consists of core and cladding whose refractive indices are
‘ 1 n ’and ‘ 2 n ’ respectively. Let ‘ o n ’ be the refractive index of air in which the optical fiber is
placed.
Let ‘AO’ is the incident ray incident at a point ‘O’ on the core and make an angle θi with
the fiber axis. Then the ray is refracted along ‘OB’ and makes an angle θ with the fiber axis.
Further the ray proceeds to fall on the core-cladding interface at critical angle of incidence θc. At
this angle the ray ‘BC’ moves along core-cladding interface.
c From D OBD, θ + θ + 90 = 180 o o
c \ θ = 90 – θ o (1)
When the angle of incidence θi decreases the refractive angle θ also decreases but the
angle θc will be increased. Hence the ray is totally internally reflected by the core-cladding
interface. Thus, only those ray which passes within the acceptance cone will be totally internally
reflected.
Angle of acceptance:
The maximum angle within which a ray of light can enter through one end of the fiber
and still be totally internally reflected is called acceptance angle.
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Applying Snell’s law of refraction at a point ‘O’,
0 n sin θi = n1 sin θ
1
i
0
n
sin θ sin θ
n
=
1 2
i
0
n
sin θ 1 cos θ
n
\ = − (2)
Applying Snell’s law of refraction at a point ‘B’,
1 c 2 r n sin θ = n sin θ
c r We know θ = 90 – θ ; θ = 90 \ sin 90 = 1 o o o
( ) 1 2 n sin 90 – θ = n o
2
2 2 2
2
1 1
n n
cos θ ; cos θ
n n
= \ = (3)
Substitute eqn. (3) in eqn. (2)
\ eqn. (2) ⇒
2
1 2
i 2
0 1
n n
sin θ 1
n n
= −
2 2
1 1 2
i 2
0 1
n n n
sin θ
n n

= 1 2 2
1 2
0 1
n
n n
n n
= −
2 2
1 2
0
n n
n

=
2 2
– 1 1 2
i
0
n n
θ sin
n

= (4)
If the refractive index of air 0 n = 1, then the maximum value of acceptance angle is given as
– 1 2 2
i 1 2 θ = sin n − n (5)
Numerical aperture:
The sine of the acceptance angle of the fiber is called numerical aperture (NA).
i NA = sin θ
2 2
1 2
0
n n
NA
n

=
For air medium 0 n = 1 \ 2 2
1 2 NA = n − n (6)
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The sine angle of incidence is always less than the NA. This is the condition for
propagation of light within the fiber.
Discuss the different types of Optical fibers.
Types of Optical fibers:
Optical fibers are classified in to 3 main categories based on
(i) the material used
(ii) the number of modes
(iii) the refractive index profile.
(i)Based on the material used:
The fibers are classified into glass fibers and plastic fibers.
(a) Glass fibers: If the fibre is made up of a mixture of metal oxide and silica it is called glass
fibers.
(b)Plastic fibers: If the fibre is made up the plastic material it is called plastic fiber.They exhibit
greater attenuation than glass fibers.
(ii)Based on the number of modes:
Depending on the number of modes of vibration the fibres are classified into single mode
and multimode.
a) Single mode Fiber
• Only one mode of light can be propagated
• The difference in refractive indices between core and cladding is small.
• No intermodal dispersion.
• Information can be carried to longer distances.
• Fabrication is difficult. Cost is high.
b) Multi mode Fiber
• More than one mode of light can be propagated
• Core diameter is large
• Intermodal dispersion is more.
• Information can be carried to shorter distances only.
• Fabrication is easy. Cost is low
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(iii)Based on the refractive index profile:
Optical fibres are classified into two types based on the refractive index profile namely
step index and graded index fibre.
a) Step Index Fiber
• The difference in refractive indices between core and cladding is obtained in single step
• The path of propagation of light is Zig-Zag
• Lower bandwidth
• Distortion is more
Graded Index Fiber
• The difference in refractive indices between core and cladding is gradually varying
• The path of propagation of light is helical.
• Higher bandwidth
• Distortion is less
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3. Discuss the different types of losses in an optical fiber.
Attenuation
Attenuation in an optical fiber is caused by absorption, scattering, and bending losses.
Attenuation is the loss of optical power as light travels along the fiber. Signal attenuation is
defined as the ratio of optical input power (Pi) to the optical output power (Po). Optical input
power is the power injected into the fiber from an optical source. Optical output power is the
power received at the fiber end or optical detector. The following equation defines signal
attenuation as a unit of length:
Signal attenuation is a log relationship. Length (L) is expressed in kilometers. Therefore, the unit
of attenuation is decibels/kilometer (dB/km). Each mechanism of loss is influenced by fibermaterial
properties and fiber structure.
ABSORPTION. - Absorption is a major cause of signal loss in an optical fiber. Absorption is
defined as the portion of attenuation resulting from the conversion of optical power into another
energy form, such as heat. Absorption in optical fibers is explained by three factors:
• Imperfections in the atomic structure of the fiber material
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• The intrinsic or basic fiber-material properties
• The extrinsic (presence of impurities) fiber-material properties
Imperfections in the atomic structure induce absorption by the presence of missing molecules
or oxygen defects. Absorption is also induced by the diffusion of hydrogen molecules into the
glass fiber. Since intrinsic and extrinsic material properties are the main cause of absorption, they
are discussed further.
Intrinsic Absorption.
Intrinsic absorption is caused by basic fiber-material properties. If an optical fiber were
absolutely pure, with no imperfections or impurities, then all absorption would be intrinsic.
Intrinsic absorption sets the minimal level of absorption.
In fiber optics, silica (pure glass) fibers are used predominately. Silica fibers are used
because of their low intrinsic material absorption at the wavelengths of operation.
In silica glass, the wavelengths of operation range from 700 nanometers (nm) to 1600
nm. Figure shows the level of attenuation at the wavelengths of operation. This wavelength of
operation is between two intrinsic absorption regions. The first region is the ultraviolet region
(below 400-nm wavelength). The second region is the infrared region (above 2000-nm
wavelength).
Figure - Fiber losses.
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Intrinsic absorption in the ultraviolet region is caused by electronic absorption bands.
Basically, absorption occurs when a light particle (photon) interacts with an electron and excites
it to a higher energy level. The tail of the ultraviolet absorption band is shown in figure .
The main cause of intrinsic absorption in the infrared region is the characteristic
vibration frequency of atomic bonds. In silica glass, absorption is caused by the vibration of
silicon-oxygen (Si-O) bonds. The interaction between the vibrating bond and the electromagnetic
field of the optical signal causes intrinsic absorption. Light energy is transferred from the
electromagnetic field to the bond. The tail of the infrared absorption band is shown in figure .
Extrinsic Absorption.
Extrinsic absorption is caused by impurities introduced into the fiber material. Trace
metal impurities, such as iron, nickel, and chromium, are introduced into the fiber during
fabrication. Extrinsic absorption is caused by the electronic transition of these metal ions from
one energy level to another.
Extrinsic absorption also occurs when hydroxyl ions (OH-) are introduced into the fiber.
Water in silica glass forms a silicon-hydroxyl (Si-OH) bond. Theamount of water (OH-)
impurities present in a fiber should be less than a few parts per billion. Fiber attenuation caused
by extrinsic absorption is affected by the level of impurities (OH-) present in the fiber. If the
amount of impurities in a fiber is reduced, then fiber attenuation is reduced.
SCATTERING
Basically, scattering losses are caused by the interaction of light with density fluctuations
within a fiber. Density changes are produced when optical fibers are manufactured.
During manufacturing, regions of higher and lower molecular density areas, relative to
the average density of the fiber, are created. Light traveling through the fiber interacts with the
density areas as shown in figure .Light is then partially scattered in all directions.
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.
Figure: Light scattering
In commercial fibers operating between 700-nm and 1600-nm wavelength, the main source of
loss is called Rayleigh scattering. Rayleigh scattering is the main loss mechanism between the
ultraviolet and infrared regions as shown in figure. Rayleigh scattering occurs when the size of
the density fluctuation (fiber defect) is less than one-tenth of the operating wavelength of light.
Loss caused by Rayleigh scattering is proportional to the fourth power of the wavelength
(1/λ4). As the wavelength increases, the loss caused by Rayleigh scattering decreases.
BENDING LOSS
Bending the fiber also causes attenuation. Bending loss is classified according to the bend
radius of curvature: microbend loss or macrobend loss.
Microbends are small microscopic bends of the fiber axis that occur mainly when a fiber
is cabled. Macrobends are bends having a large radius of curvature relative to the fiber
diameter. Microbend and macrobend losses are very important loss mechanisms. Fiber loss
caused by microbending can still occur even if the fiber is cabled correctly. During installation, if
fibers are bent too sharply, macrobend losses will occur.
Microbend losses are caused by small discontinuities or imperfections in the fiber.
Uneven coating applications and improper cabling procedures increase microbend loss. External
forces are also a source of microbends. An external force deforms the cabled jacket surrounding
the fiber but causes only a small bend in the fiber. Microbends change the path that propagating
modes take, as shown in figure. Microbend loss increases attenuation because low-order modes
become coupled with high-order modes that are naturally lossy.
.
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Figure - Microbend loss
Macrobend losses are observed when a fiber bend's radius of curvature is large compared to the
fiber diameter.
These bends become a great source of loss when the radius of curvature is less than
several centimeters. Light propagating at the inner side of the bend travels a shorter distance than
that on the outer side. To maintain the phase of the light wave, the mode phase velocity must
increase. When the fiber bend is less than some critical radius, the mode phase velocity must
increase to a speed greater than the speed of light. However, it is impossible to exceed the speed
of light. This condition causes some of the light within the fiber to be converted to high-order
modes. These high-order modes are then lost or radiated out of the fiber.
Fiber sensitivity to bending losses can be reduced. If the refractive index of the core is
increased, then fiber sensitivity decreases. Sensitivity also decreases as the diameter of the
overall fiber increases. However, increases in the fiber core diameter increase fiber sensitivity.
Fibers with larger core size propagate more modes. These additional modes tend to be more
lossy.
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Unit 3: Conducting and Semiconducting materials
Part A
1. What are the postulates of classical free electron theory
i) In the absence of electric field, the free electrons in the metal moves freely in a metal
just like a gas molecule.
ii) In the presence of electric field, the electrons orient themselves along a particular
direction and move opposite to that of the field direction.
2. Mention any 3 drawbacks of classical free electron theory?
• The phenomena such as photo electric effect, Compton effect, black body radiation could not be
explained by this model
• Electrical conductivity of semi conductors or insulators could not be explained using this model
• Ferromagnetism could not be explained by this model
2. Define mobility of an electron
It is defined as the ratio of the drift velocity gained by the electron on the application of
an electric field.
3. What are the merits of classical free electron theory?
i) It verifies Ohm’s law.
ii) It explains the electrical and thermal conductivity of metals.
iii) It is used to derive Weidmann-Franz law
iv) Optical properties of metals can be explained using this theory.
4. What are the demerits of classical free electron theory?
(i) Classical theory says all electrons absorb energy, but quantum theory states that
only few electrons will absorb energy.
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(ii)This theory cannot explain Compton, photo-electric effect, paramagnetism,
ferromagnetism, etc.
(iii)The Lorentz number’s experimental and theoretical value does not match in
classical theory.
7. Define Fermi energy
The maximum or highest energy of the filled state at 0 K temperature. It is the maximum
energy that an electron in a metal can possess at 0 K temperature.
8.What is the importance of Fermi distribution function
It represents the probability of an electron occupying a given energy state.
9.Write fermi distribution function.
The fermi distribution function is given by
F(E)=1/1+exp(E-Ef)/KBT
Where KB Boltzmann constant
T is the absolute temperature
10.Define density of states What is its use?
Density of states is defined as the number of available energy states per unit
volume in an energy interval E and E + dE. It is used to calculate the number of charge
carriers per unit volume of the solid
11. State the properties of a semiconductor.
(i) At 0K they behave as insulators.
(ii) They have resistivity between that of conductors and insulators.
(iii) They have negative temperature coefficient of resistance.
(iv) The conductivity increases due to the temperature and impurities.
12. Define Intrinsic semiconductor.
A semiconductor in its purest form is called intrinsic semiconductor.
Eg: Ge, Si
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13. Define Extrinsic semiconductor.
A semiconductor having impurity is called extrinsic semiconductor.
Eg: Germanium doped with Phosphorous.
14. What is meant by doping?
The process of adding impurity to a pure semiconductor is said to doping.
15. What are compound semiconductors?
Semiconductors formed by combinations of equal atomic fractions of fifth and third
column or sixth and second column elements are called compound semiconductors.
Eg: GaAs, InP, MgO, CdS
16. Explain Hall effect
When a current carrying conductor is placed in a transverse magnetic field, an electric
field is produced inside the conductor in a direction normal to both the current and the
magnetic field. This effect is known as Hall Effect.
17. Mention some Applications of Hall effect
i) it is used to determine whether the material is p-type or n-type semiconductor
ii) it is used to find the carrier concentration
iii) it is used to find the mobility of charge carriers
iv) It is used to design magnetic flux meters on the basis of Hall voltage.
18. Define Fermi energy level and give its importance.
The maximum or highest energy level occupied by an electron at 0 K temperature.
It is the reference energy level which is used to separate the filled and vacant energy
levels.
19. What is meant by elemental semi conductors?
Elemental semiconductors are made from a single semiconducting element of
fourth group elements of the periodic table. Example: Germanium and silicon
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20. Difference between elemental and compound semiconductors
Sl.
No.
Elemental semiconductor Compound semiconductor
1.
2.
3.
4.
They are made of single element.
They are indirect band gap
semiconductors
During recombination energy is released
in the form of heat.
They are used for manufacture of
transistor, diode, FET, MOSFET etc.
They are made of more than one element.
They are direct band gap semiconductors.
During recombination energy is released
in the form of light.
They are used for manufacture of LED
and LASER diodes.
21. With increase of temperature the conductivity of semi conductor increases while that of
metals decreases. Give reasons
With increase of temperature more and more charge carriers are created and hence the
conductivity of semi conductor increases. In the case of metals with increase of temperature
The concentration of charge carriers remains the same. But due to increase of thermal energy the
electrons make frequent collision with lattice ions and hence the resistivity increases and
conductivity decreases.
22. How are N-type semiconductors produced?
When pentavalent impurity elements such as P, Ar, Sb is added to germanium or
silicon N-type semi conductors are produced
22. How are P-type semiconductors produced?
When trivalent impurity elements such as Al, Ga , In is added to germanium or silicon
P-type semi conductors are produced
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Part B
1. Discuss the Effect of temperature on Fermi Energy:
The maximum or highest energy level occupied by an electron at 0 K temperature is
called Fermi energy level. The maximum or highest energy of the filled state at 0 K temperature
is called Fermi energy.
Case : 1 Fermi energy of an electron at 0 K.
At T = 0 K , E < EF
(E E ) / kT ( ive) k(0)
1 1
F(E) 1
1 e 1 e − −
\ = = =
+ F + /
Carrier concentration of a metal at 0 K is n
C0 .
\ Eqn. (12) ⇒ ( )3/2 1/2
3
4
n 2m E dE
h
F
0
E
C
0
p
= ∫
( )3/2 1/2
3
4
2m E dE
h
EF
0
p
= ∫
( ) ( )
3/2
3/2 3/2 3/2
3 3
4 E 2 4
2m 2 m E
h 3 3 h
2
F
F
E
E
0
0
 
p   p =   =    
 
 
( )3/2 3/2
3
8
2m E
3 h
EF
0
p   =  
( )3/2 3/2
3 F
8
2 m E
3 h
p
=
( )3/2
3 F
8
n 2 m E
C0 3 h
p
= (13)
From Eqn. (13) ( ) ( )
2/3
2/3
3 F
8
n 2 m E
C0 3 h
 p 
=  
 
( )
( )
2/3 2/3 2/3 3
F 2/3
3
n 3 h 1
E n
8 8 2m
2 m
3 h
0
0
C
C
 
\ = =  
 p   p 
 
 
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2/3 2
F
3 n h
E
8 2m
C0
 
=  
 p 
(14)
Eqn.(14) is the expression for Fermi energy of an electron in a metal at 0 K temperature.
Fermi energy of an electron at non zero temperature
0
0
2
2
F F
F
kT
E E 1
12 E
     p  = −  
   
   
(17)
F0 E
is the Fermi energy of an electron at 0 k temperature
k is the Maxwell’s Boltzmann constant
2.Write the difference between elemental and compound semiconductors
Sl.
No.
Elemental semiconductor Compound semiconductor
1.
2.
3.
4.
They are made of single element.
They are indirect band gap
semiconductors
During recombination energy is
released in the form of heat.
They are used for manufacture of
transistor, diode, FET, MOSFET etc.
They are made of more than one
element.
They are direct band gap
semiconductors.
During recombination energy is
released in the form of light.
They are used for manufacture of LED
and LASER diodes.
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Si
Si Si Si
Si
Si
Si
Si
Si
Intrinsic semiconductor (Silicon)
At 0K temperature
EF
EV
EC
– ∞

At T = 0 K
Valence Band
Conduction Band
3. Explain Intrinsic Semiconductors.
A semiconductor in an extremely pure form is known as intrinsic semiconductor.
Example: Silicon and Germanium
In a crystal made up of silicon atoms, each atom is contributed by four valence electrons.
These electrons form a covalent bond with neighbouring silicon atoms and achieve stable
structure. At 0 K temperature the silicon semiconductor behaves as a perfect insulators, because
the covalent bonds are very strong. Therefore no free electrons are available in conduction band
and valence band is completely filled.
When temperature of an intrinsic semiconductor is raised, some of the electrons in
covalent bonds acquire sufficient energy to break the covalent bonds. Once the covalent bond is
broken, the electrons in those bonds are free and move to the conduction band. The motion of
these electron constitute electron current. When the electron leaves from the covalent bond it
creates a vacancy in valence band. This vacancy is called hole, it is ready to accept an electron
from any where in semiconductor. A combination of a free electron and a hole is called electronhole
pair. At any temperature, the number of free electron in conduction band is equal to number
of holes in valence band.
The energy band diagram is shown in fig. At 0 K temperature the maximum energy level
occupied by an electron is called Fermi level. In intrinsic semiconductors, the Fermi energy
level lies at the middle of the forbidden energy gap.
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Conduction through the Intrinsic Semiconductors
4. Explain about Extrinsic Semiconductors.
. A semiconductor in an impure form is called extrinsic semiconductors. Impure form
means adding some impurity into it. So that it’s electrical conductivity is increased.
The process of adding impurity to the pure semiconductor is called doping. Depending
upon the type of impurity added, extrinsic semiconductors are classified as two types. They are
a) N-type semiconductor b) P-type semiconductor
When an external electric field (V) is applied across an intrinsic semiconductor, the
current conduction through the intrinsic semiconductor is due to both the free electrons and
holes. Due to the external field conduction band free electrons are drifted towards the
positive polarity of the external field and the valence band holes are drifted towards the
negative polarity of the applied field
Free electrons
Holes
+ -
V
Intrinsic Semiconductor
Si
Si
Si
Si
Si
Si
Si
Si
Si
Hole Free
electron
Intrinsic semiconductor (Silicon)
At Room temperature
EF
EV
EC
– ∞

At T > 0 K
Valence Band
Conduction Band
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so it is called donor impurities. With the addition of pentavalent impurity, large numbers of free
electrons are available in the semiconductor. At 0 K all these free electrons form an energy level,
called donor energy level. The donor energy level is formed near and below the conduction band.
When the temperature is raised, the electrons in donor energy level acquire sufficient
energy and jumps to conduction band. Once an electron librated from donor energy level to
conduction band, it creates +ively ionized donor in donor energy level. At room temperature,
some of the covalent bonds are broken in valence band due to thermal energy supplied to the
Si
Si Si Si
Si
Si
Si
Si
Ar
Free
electron
N-type semiconductor At 0 K temperature
a) N-type semiconductor
When a pentavalent impurity (Ex. Antimony,
Arsenic, Bismuth ect.) is added to the tetravalent
intrinsic semiconductor, N-type semiconductor is
obtained.
Pentavalent impurity means the outermost orbit
has five valence electrons. Out of the five valence
electrons four electrons form a covalent bond with four
neighbouring intrinsic semiconductor atoms (Si or Ge).
The fifth electron is called as free electron. The
pentavalent impurity is ready to donate electrons to
the intrinsic semiconductor
E
eV Fermi energy level
Conduction band
Valence band
Free electrons
Holes
T > 0K
Donor energy level
+ively ionized donor
electron
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crystal. Once the covalent bond is broken, the electrons in those bonds are free and move to
conduction band. So that it creates a hole in that place. Hence in N-type semiconductor electrons
are majority carriers and holes are minority carriers.
The Fermi energy level is lying in between conduction band and donor energy level.
Conduction through the N-type Semiconductor
b) P-type semiconductor
When a trivalent impurity (Ex. Gallium, Indium, Boron ect.) is added to the tetravalent
intrinsic semiconductor, P-type semiconductor is obtained.
Trivalent impurity means the outermost orbit has three valence electrons. These three
valence electrons form covalent bonds with three neighbouring intrinsic semiconductor atoms (Si or
Ge). The fourth covalent bond is formed incomplete of one electron. The missing electron in that
bond is called hole. The trivalent impurities are ready to accept electrons from the intrinsic
semiconductor so it is called acceptor impurities. With the addition of trivalent impurity, large
numbers of holes are available in semiconductor. At 0 K all these holes form an energy level, called
acceptor energy level. The acceptor energy level is formed near and above the valence band. Hence
acceptor energy level formed near the valence band.
When an external electric field (V) is
applied across an N-type semiconductor, the
current conduction through the N-type
semiconductor is due to free electrons. Due to
the external field majority carrier free electrons
are drifted towards the positive polarity of the
external field and constitute large electric
current. The minority carrier holes are drifted
towards the negative polarity of the applied field
and constitute little current. This current is
practically neglected.
Free electrons
Holes
+ -
V
N-type Semiconductor
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When the temperature is raised, the electrons in covalent bond acquire sufficient energy
to break the covalent bonds. Once the covalent bond is broken, the electrons in those bonds are
free and move to acceptor energy level. So that it creates a hole in that place. Once an electron
reaches acceptor energy level, it creates -ively ionized acceptor in acceptor energy level. At
room temperature, more covalent bonds are broken due to the thermal energy supplied to the
crystal. In that time, few electrons may go to conduction band. Hence in P-type semiconductor
holes are majority carriers and electrons are minority carriers.
The Fermi energy level is lying in between valence band and acceptor energy level.
5. Determine Fermi energy and its variation with temperature in the case of an intrinsic
semiconductor.
In an intrinsic semiconductor the carrier concentration of electron in conduction band is
equal to carrier concentration of holes in valence band.
\ ne = nh
( ) ( )
* 3/ 2 * 3/ 2
e E E / kT h E E / kT
2 2
2 m kT 2 m kT
2 e 2 e
h h
− −  p   p 
  =  
   
F C V F
( ) ( ) ( ) ( )
3/ 2 3/ 2
* 3/ 2 E E / kT * 3/ 2 E E / kT
2 e 2 h
2 kT 2 kT
2 m e 2 m e
h h
 p  −  p  −
  =  
   
F C V F
( ) ( ) * 3/ 2 E / kT E / kT * 3/ 2 E / kT E / kT
e h m e e − m e e − F C = V F
* 3/ 2
E / kT E / kT h E / kT E / kT
*
e
m
e e e e
m
 
=  
 
F F V C
( )
* 3/ 2
E / kT h E E / kT
*
e
m
e e
m
+  
=  
 
F V C 2 (17)
Si
Si Si Si
Si
Si
Si
Si
In
Hole
P-type semiconductor At 0 K temperature
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Taking log on both sides of eqn.(17)
( ) ( )
* 3/ 2
E / kT h E E / kT
*
e
m
log e log e
m
+
   
=    
     
F V C 2
* 3/ 2
h
*e
2 E m E E
log
kT m kT
  +
=  
 
F V C
+
*
h
*
e
2 E 3 m E E
log
kT 2 m kT
  +
=  
 
F V C
+
*
h
*e
kT 3 m E E
E log
2 2 m kT
   + 
=    
   
V C
F
+
*
h
*
e
3 k T m E E
E log
4 m 2
  +
=  
 
V C
F
+ (18)
Case : 1
When T = 0 K then * *
h e m = m
E E
Eqn. (18) E
2
+
\ ⇒ = V C
F
The Fermi energy level is lying exactly at middle between the valence band and
conduction band.
Case : 2
When T is increasing above 0 K the mass of hole is slightly
greater than mass of an electron. Ie., * *
h e m > m . Therefore Fermi
energy level is slightly rises above the middle of energy gap.
Temperature
Valence Band
Conduction Band
EC
EV
EF (At T = 0 K)
EF (At T > 0 K)
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5. Determine hall coefficient for p-type semiconductor.
Let us consider a p-type semiconductor in the form of rectangular slab. Ix is the current
passes through the material along X direction and Bz is the magnetic field applied along Z
direction. As a result Hall voltage VH is developed along Y direction.
The direction of current is left to right along +ive X direction therefore the holes are
moving with constant velocity v along +ive X direction. When a magnetic field is applied along
Z direction, the hole will experience a downward force. Due to this force
holes are deflected downward and accumulate at bottom face of the specimen. This causes the
bottom face to be more +ive with respect to the top face; hence a potential difference is
developed across them. This potential difference causes an electric field called Hall electric field
EH along +ive Y direction. Due to this electric field, hole experiences an upward force.
Where, e is the charge of a hole
The downward force experienced by the
hole due to magnetic field = BZ e v (1)
Bzev
eEH
IX
Current
Top face
Bottom face
Hole direction
The upward force experienced by the
hole due to Hall electric field
= H e E (2)
X
Z
Y
IX
BZ
-
+
VH (Hall voltage)
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At equilibrium, From eqn.(1) & (2)
BZ e v = e EH
H Z \ E = B v (3)
The relation between current (Ix) and drift velocity (v) of a hole is given by
X I = p A e v (4)
Where
p is the number of holes per unit volume of the specimen
A is the cross sectional area of the specimen
X J
v
p e
\ =
(6)
Substitute eqn.(6) in (3)
X
H Z X Z
J 1
eqn. (3) E B J B
p e p e
⇒ = =
H
1
Where R
p e
+ =
It is called Hall coefficient or Hall constant. The +ive sign indicates the
direction of hall electric field is +ive Y direction.
H H X Z \ E = R J B (7)
H
H
X Z
E
R
J B
\ =
(8)
The current density acting along X
direction is
Jx = p e v (5)
X
X
I
J
A
Q =
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Part C
1. Define density of states in metals. Derive an expression for the same and hence evaluate
the carrier concentration in metals.
Density of state Z(E) dE is defined as the number of available energy states per unit
volume in an energy interval E and E + dE.
Let us consider a cubical metal with sides ‘a’. The number of energy level can be
obtained from various combinations of quantum numbers. A sphere is constructed with quantum
numbers nx, ny and nz in three dimensional space (n2 = nx
2 + ny
2 + nz
2). In this space each point
represents an energy level.
Let us construct a sphere of radius n in a space. The energy of this sphere is E. This
sphere is divided into many spheres. Each sphere represents a particular combination of quantum
numbers, particular energy value and particular radius. The energy of the next sphere is E + dE
having radius n + dn from its origin ‘O’.
The number of energy states within the sphere of energy E is equal to volume of the
sphere. But the principle quantum number nx, ny and nz can have only the positive integer value.
The quantum numbers have positive value only lying in 1/8th of the volume of the sphere.
nx
ny
nz
o
n
n+dn
E
E+dE
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Therefore number of energy states with in the sphere of energy E is equal to 1/8th of
volume of the sphere.
The number of available energy states
within the sphere of energy E ( 1)
The number of available energy states
within the sphere of energy E + dE
( ) 1 3
n dn
6
= p + (2)
The number of available energy states between
the energy interval E & E + dE
( )1 3 1 3
Y (E) dE n dn n
6 6
   
\ =  p +  −  p     
( )n dn 3 n3
6
p   =  + − 
n3 dn3 3 n2dn 3 n dn2 n3
6
p   =  + + + − 
dn3 3 n2dn 3 n dn2
6
p   =  + + 
The radial distance between the two energy levels ‘dn’ is very small therefore the higher order of
dn (ie., dn3 and dn2) terms are neglected..
Y (E) dE 3 n2dn
6
p   =  
n2dn
2
p
=
Y (E) dE n n dn
2
p
= (3)
The energy of the particle in three dimensional metal piece of sides ‘a’ is
2 2
2
n h
E
8ma
=
1 4 3
n
8 3
 
=  p   
( )1 4 3
n dn
8 3
 
=  p +   
= Eqn. (2) − Eqn. (1)
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2
2
2
8ma E
n
h
\ =
(4)
1
2 2
2
8ma E
n
h
 
=  
 
(5)
Differentiating eqn.(4) w.r.t E
\ Eqn. (4) ⇒
2
2
8m a
2n dn dE
h
=
2
2
8ma
n dn dE
2 h
\ =
(6)
Substitute eqn.(5) & (6) in (3)
\ Eqn. (3) ⇒
1
2 2 2
2 2
8m a E 8ma
Y (E) dE dE
2 h 2 h
p    
=    
   
1
2 2 2
1/ 2
2 2
8ma 8m a
E dE
4 h h
p    
=    
   
1
2 1 2
1/ 2
2
8ma
E dE
4 h
+ p  
=  
 
3
2 2
1/ 2
2
8ma
E dE
4 h
p  
=  
 
( )3/ 2 3 1/ 2
3 Y (E) dE 8m a E dE
4 h
p
=
(7)
Eqn.(7) is the number of energy states available in between energy interval E and E+dE
Density of energy states 3
Y (E) dE
Z (E) dE
Volume of the metal piece (a )
=
( )3/ 2 3 1/ 2
3
3
8m a E dE
Z (E) dE 4 h
a
p
=
( )3/ 2 1/ 2
3 Z (E) dE 8m E dE
4 h
p
\ =
( )3/2 1/2
3
2
or Z (E) dE 2m E dE
h
p
= (8)
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Density of electron:
Density of electron (dN) in between the energy interval E and E+dE is equal to number of
electron per unit volume in between the energy interval E and E+dE.
According to Pauli’s exclusion principle each energy state is occupied by two electrons
only. One is spin up and another one is spin down.
\ dN = 2 Z (E) dE
( )3/2 1/2
3
2
dN 2 2 m E dE
h
p
=
( )3/2 1/2
3
4
dN 2 m E dE
h
p
= (9)
Eqn. (9) is the expression for density of electron between the energy interval E and E+dE.
2.Derive an expression for Density of electron in a metal (or) Carrier concentration (nC
) in metals:
Density of state Z(E) dE is defined as the number of available energy states per unit volume in
an energy interval E and E + dE.
Let us consider a cubical metal with sides ‘a’. The number of energy level can be
obtained from various combinations of quantum numbers. A sphere is constructed with quantum
numbers nx, ny and nz in three dimensional space (n2 = nx
2 + ny
2 + nz
2). In this space each point
represents an energy level.
Let us construct a sphere of radius n in a space. The energy of this sphere is E. This
sphere is divided into many spheres. Each sphere represents a particular combination of quantum
numbers, particular energy value and particular radius. The energy of the next sphere is E + dE
having radius n + dn from its origin ‘O’.
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The number of energy states within the sphere of energy E is equal to volume of the
sphere. But the principle quantum number nx, ny and nz can have only the positive integer value.
The quantum numbers have positive value only lying in 1/8th of the volume of the sphere.
Therefore number of energy states with in the sphere of energy E is equal to 1/8th of
volume of the sphere.
The number of available energy states
within the sphere of energy E
1 3
n
6
= p (1)
The number of available energy states
within the sphere of energy E + dE
( )1 3
n dn
6
= p + (2)
The number of available energy states between
the energy interval E & E + dE
( )1 3 1 3
Y (E) dE n dn n
6 6
   
\ =  p +  −  p     
( )n dn 3 n3
6
p   =  + − 
1 4 3
n
8 3
 
=  p   
( )1 4 3
n dn
8 3
 
=  p +   
= Eqn. (2) − Eqn. (1)
nx
ny
nz
o
n
n+dn
E
E+dE
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n3 dn3 3 n2dn 3 n dn2 n3
6
p   =  + + + − 
dn3 3 n2dn 3 n dn2
6
p   =  + + 
The radial distance between the two energy levels ‘dn’ is very small therefore the higher order of
dn (ie., dn3 and dn2) terms are neglected..
Y (E) dE 3 n2dn
6
p   =  
n2dn
2
p
=
Y (E) dE n n dn
2
p
= (3)
The energy of the particle in three dimensional metal piece of sides ‘a’ is
2 2
2
n h
E
8ma
=
2
2
2
8ma E
n
h
\ =
(4)
1
2 2
2
8ma E
n
h
 
=  
 
(5)
Differentiating eqn.(4) w.r.t E
\ Eqn. (4) ⇒
2
2
8m a
2n dn dE
h
=
2
2
8ma
n dn dE
2 h
\ =
(6)
Substitute eqn.(5) & (6) in (3)
\ Eqn. (3) ⇒
1
2 2 2
2 2
8m a E 8ma
Y (E) dE dE
2 h 2 h
p    
=    
   
1
2 2 2
1/ 2
2 2
8ma 8m a
E dE
4 h h
p    
=    
   
1
2 1 2
1/ 2
2
8ma
E dE
4 h
+ p  
=  
 
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3
2 2
1/ 2
2
8ma
E dE
4 h
p  
=  
 
( )3/ 2 3 1/ 2
3 Y (E) dE 8m a E dE
4 h
p
=
(7)
Eqn.(7) is the number of energy states available in between energy interval E and E+dE
Density of energy states 3
Y (E) dE
Z (E) dE
Volume of themetal piece (a )
=
( )3/ 2 3 1/ 2
3
3
8m a E dE
Z (E) dE 4 h
a
p
=
( )3/ 2 1/ 2
3 Z (E) dE 8m E dE
4 h
p
\ =
( )3/2 1/2
3
2
or Z (E) dE 2m E dE
h
p
= (8)
Density of electron:
Density of electron (dN) in between the energy interval E and E+dE is equal to number of
electron per unit volume in between the energy interval E and E+dE.
According to Pauli’s exclusion principle each energy state is occupied by two electrons
only. One is spin up and another one is spin down.
\ dN = 2 Z (E) dE
( )3/2 1/2
3
2
dN 2 2 m E dE
h
p
=
( )3/2 1/2
3
4
dN 2 m E dE
h
p
= (9)
Carrier concentration in metals:
C n = ∫ dN F (E)
E = EF
E = 0
(10)
F(E) is the Fermi-Dirac distribution function.
(E EF ) / kT
1
F (E)
1 e −
=
+
(11)
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Substitute eqn.(9) & (11) in eqn. (10)
\ Eqn. (10) ⇒ ( ) ( )
3/2 1/2
3 E E /kT
4 1
n 2m E dE
h 1 e
F
F
E = E
C
E = 0

p
=
+
∫ (12)
Eqn.(12) is the expression for the density of electron in a metal at any temperature.
3. Derive expression for carrier concentration of electrons in intrinsic
semiconductor.
At 0 K temperature intrinsic semiconductor behaves as insulator. Ie., valence band is
completely filled and conduction band is completely empty. As the temperature increases
covalent bonds are broken hence electrons move from valence band to conduction band.
EV and EC represent the maximum and minimum energy of valence and conduction band
respectively. The energy of the electron in conduction band is varying between EC to ∞ and
the energy of holes in valence band is varying between – ∞ to EV The mass of the electron in
conduction band and mass of the hole in valence band is taken as me
* and mh
* respectively. At
any temperature number of electrons in conduction band is equal to number of holes in valence
band.
To determine carrier concentration of electrons in conduction band :-
The number of electrons available in the
conduction band between the energy
interval E and E + dE
( ) ( )
* 3/ 2 1/ 2
3 e
4
dN 2m E E dE
h
p
= − C
EF EF
EV
EV
EC
EC
– ∞
– ∞


At T = 0 K
At T > 0 K
Valence Band
Valence Band
Conduction Band
Conduction Band
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Density of electron in conduction band n dN F(E)
¥
= ∫
C
e
E
(1)
F(E) is the probability of electron occupying in a particular energy level in conduction
band.
(E E ) / kT
1
F(E)
1 e −
\ =
+ F
Substitute dN and F(E) values in eqn. (1)
( ) ( ) ( )
* 3/ 2 1/ 2
3 e E E / kT
4 1
Eqn. (1) n 2m E E dE
h 1 e
¥

p
\ ⇒ = −
+
∫ F
C
e C
E
(2)
The assumed energy level (E) in conduction band is greater than Fermi energy level (EF) ie.,
E EF \ E − EF kT > >>
(E E ) / kT e 1 − F >>
(E E ) / kT (E E ) / kT 1 e e − − \ + F » F (3)
Substitute eqn. (3) in eqn.(2)
( ) ( ) ( )
3/ 2 1/ 2 *
e 3 E E / kT
4 1
Eqn. (2) n 2 m E E dE
h e
¥

p
\ ⇒ = ∫ − F
C
e C
E
( ) ( ) ( ) 3/ 2 1/ 2 E E / kT *
e 3
4
n 2 m E E e dE
h
¥
p −
= ∫ − F
C
e C
E
( ) ( )
* 3/ 2 1/ 2 E / kT E / kT
3 e
4
2m E E e e dE
h
¥
p −
= ∫ − F
C
C
E
( ) ( )
* 3/ 2 E / kT 1/ 2 E / kT
3 e
4
n 2m e E E e dE
h
¥
p −
= F ∫ −
C
e C
E
(4)
Let us assume
E − EC = X (a)
\ E = X + EC
Differentiating both side
\ dE = dX (b)
Substitute eqn.(a), (b), (c) and (d) in eqn. (4)
To find the integral limits :-
Assume E = EC
\ Eqn. (a) ⇒ EC − EC = X
ie., X = 0 (c)
Assume E = ¥
\ Eqn. (a) ⇒ ¥ − EC = X
ie., X = ¥ (d)
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( * )3/ 2 E / kT 1/ 2 (X E ) / kT
3 e
4
Eqn. (4) n 2m e X e dX
h
¥
p − +
\ ⇒ = F ∫ C
e
0
( )* 3/ 2 E / kT 1/ 2 X/ kT E / kT
3 e
4
2m e X e e dX
h
¥
p − −
= F ∫ C
0
( )* 3/ 2 E / kT E / kT 1/ 2 X/ kT
3 e
4
2m e e X e dX
h
¥
p − −
= F C ∫
0
( ) ( ) * 3/ 2 E E / kT 1/ 2 X/ kT
3 e
4
n 2m e X e dX
h
¥
p − −
= F C ∫
e
0
If we solve the integral function in eqn. (5) we will get
( ) ( ) ( )3/ 2 1/ 2
* 3/ 2 E E / kT
3 e
4 kT
n 2m e
h 2
− p p
= F C
e
( )
* 3/ 2
e E E / kT
2
2 m kT
n 2 e
h
−  p 
=  
 
F C
e
(6)
Eqn.(6) is the carrier concentration of electron in conduction band of an intrinsic
semiconductor.
4. Derive an expression for carrier concentration of holes in valence band in an intrinsic
semi conductor :-
The number of holes available in the
valence band between the energy
interval E and E + dE
Density of hole in valence band n dN ( 1 F(E))
− ¥
= ∫ −
EV
h
(7)
1 – F(E) is the probability of hole occupying in a particular energy level in valence
band.
(E E ) / kT
1
1 F(E) 1
1 e −
\ − = −
+ F
( )
( )
E E / kT
E E / kT
1 e 1
1 e


+ −
=
+
F
F
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( )
( )
E E / kT
E E / kT
e
1 F(E)
1 e


− =
+
F
F
(8)
The assumed energy level (E) in valence band is less than Fermi energy level (EF) ie.,
E EF \ E − EF kT < < <
(E E ) / kT e 1 − F < <
(E E ) / kT 1 e 1 − \ + F »
\ Eqn. (8) ⇒
(E E ) / kT e
1 F(E)
1

− =
F
(E E ) / kT 1 F(E) e − \ − = F (9)
Substitute dN and 1 – F(E) values in eqn. (7)
( ) ( ) * 3/ 2 1/ 2 (E E ) / kT
3 h
4
Eqn. (7) n 2m E E e dE
h

− ¥
p
\ ⇒ = ∫ −
V
F
E
h V
( ) ( )
* 3/ 2 1/ 2 E / kT E / kT
3 h
4
2m E E e e dE
h

− ¥
p
= ∫ −
V
F
E
V
( ) ( )
* 3/ 2 E / kT 1/ 2 E / kT
3 h
4
n 2m e E E e dE
h

− ¥
p
= ∫ −
V
F
E
h V
(10)
Let us assume
EV − E = X (p)
\ E = EV − X
Differentiating both side
\ dE = − dX (q)
Substitute eqn.(p), (q), (r) and (s) in eqn. (10)
( ) ( ) ( ) ( )
* 3/ 2 E / kT 1/ 2 E X / kT
3 h
4
Eqn. (10) n 2m e X e dX
h
− −
¥
p
\ ⇒ = F ∫ V −
0
h
To find the integral limits :-
Assume E = − ¥
\ Eqn. (p) ⇒ EV + ¥ = X
ie., X = ¥ (r)
Assume E = EV
\ Eqn. (p) ⇒ EV − EV = X
ie., X = 0 (s)
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( ) ( ) * 3/ 2 E / kT 1/ 2 (E X) / kT
3 h
4
n 2m e X e dX
h
¥
p − −
= F ∫ V
h
0
( ) ( )
* 3/ 2 E / kT 1/ 2 E / kT X/ kT
3 h
4
2 m e X e e dX
h
¥
p − −
= F ∫ V
0
( ) ( )
* 3/ 2 E / kT E / kT 1/ 2 X/ kT
3 h
4
2 m e e X e dX
h
¥
p − −
= F V ∫
0
( ) ( ) ( )
* 3/ 2 E E / kT 1/ 2 X/ kT
3 h
4
n 2 m e X e dX
h
¥
p − −
= V F ∫
h
0
(11)
If we solve the integral function in eqn. (11) we will get
( ) ( ) ( )3/ 2 1/ 2
* 3/ 2 E E / kT
3 h
4 kT
n 2m e
h 2
− p p
= V F
h
( )
* 3/ 2
h E E / kT
2
2 m kT
n 2 e
h
−  p 
=  
 
V F
h
(12)
Eqn.(12) is the carrier concentration of hole in valence band of an intrinsic
semiconductor.
5.What is Hall effect? Derive an expression of Hall coefficient. Describe an
experimental setup for the measurement of Hall coefficient.
Hall Effect
When a current (Ix) carrying conductor or semiconductor is placed in a perpendicular
magnetic field (Bz), a voltage (Vy) is developed across the conductor or semiconductor in a
direction perpendicular to both the current and magnetic field. This effect is called Hall effect
and the generated voltage is called Hall voltage (VH).
The developed voltage causes an electric field it is called Hall electric field (EH).
Expression of Hall coefficient for n-type semiconductor
Let us consider an n-type semiconductor in the form of rectangular slab. Ix is the current
passes through the material along X direction and Bz is the magnetic field applied along Z
direction. As a result Hall voltYa ge VH is developed along Y direction.
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The direction of current is left to right along +ive X direction therefore the electrons are
moving with constant velocity v along –ive X direction (ie., from right to left). When a magnetic
field is applied along Z direction, the electron will experience a downward force. Due to this
force electrons are deflected downward and accumulate at bottom face of the specimen.
This causes the bottom face to be more –ive with respect to the top face, hence a potential
difference is developed across them. This potential difference causes an electric field called Hall
electric field EH along –ive Y direction. Due to this electric field, electron experiences an upward
force.
The downward force experienced by the
electron due to magnetic field = BZ e v (1)
Bzev
eEH
IX
Current
Top face
Bottom face
electron direction
X
Z
IX
BZ
+
-
VH (Hall voltage)
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Where, e is the charge of an electron
At equilibrium, From eqn.(1) & (2)
BZ e v = e EH
H Z \ E = B v (3)
The relation between current (Ix) and drift velocity (v) of an electron is given by
X I = − n A e v (4)
Where
n is the number of electrons per unit volume of the specimen
A is the cross sectional area of the specimen
And the –ive sign indicates the charge of electron is –ive.
X J
v
n e
\ = −
(6)
Substitute eqn.(6) in (3)
X
H Z X Z
J 1
eqn. (3) E B J B
n e n e
⇒ = − = −
H
1
Where R
n e
− =
It is called Hall coefficient or Hall constant. The –ive sign indicates
the direction of hall electric field is –ive Y direction.
X
X
I
J
A
Q =
The upward force experienced by the
electron due to Hall electric field = H e E (2)
The current density acting along X
direction is
= − n e v
(5)
Jx
Bzev
eEH
IX
Current
Top face
Bottom face
Hole direction
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H \ E = RH JX BZ (7)
H
H
X Z
E
R
J B
\ =
(8)
Experimental setup
A semiconductor material of thickness t, breadth b and length l is taken in the form of
rectangular slab. Ix is the current passes along X direction through a battery and Bz is the
magnetic field applied along Z direction (ie., perpendicular to the plane of paper). As a result
Hall voltage VH is developed along Y direction. The developed voltage is measured by fixing
two probes at the centers of bottom and top faces of the sample. This voltage causes an electric
field called Hall electric field EH.
b
l
t
mA Rh
+
-
VH (Hall voltage)
+ -
Key
Battery
IX
BZ
Y
X
Z
Y
I
X
B
Z
-
+
VH (Hall
voltage)
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Hall electric field H
H
V
E
t
=
\ VH = EH t (9)
Substitute eqn. (8) in (9)
H H X Z eqn. (9) ⇒ V = R J B t (10)
Current density X
X
I
J
Area of the sample
=
Area of the sample = breadth x thickness of the sample (b t)
X
X
I
J
b t
\ =
(11)
Substitute eqn. (11) in (10)
X
H H Z
I
eqn. (10) V R B t
b t
 
⇒ =  
 
H X Z
H
R I B
V
b
=
H
H
X Z
V b
R
I B
\ =
(12)
Mobility of charge carriers
Hall coefficient of n-type semiconductor is given by
H
1
R
n e
= −
The above expression is valid only for conductors where the velocity is taken as the drift
velocity. But for semiconductor velocity is taken as average velocity. Therefore Hall coefficient
RH is modified as
H
3 1
R
8 n e
p  
= −  
 
H
1.18
R
n e
= −
(13)
H
1.18
n e
R
\ = − (14)
Electrical conductivity of n-type semiconductor is
e e s = n e μ
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Therefore mobility of electron e
e n e
s
μ =
(15)
Sub. Eqn.(14) in (15)
e
e
H
eqn. (15)
1.18
R
s
⇒ μ =
 
− 
 
e H
e
R
1.18
s
μ = − H
H
X Z
V b
R
I B
=
Q
H
e
X Z
e
V b
I B
1.18
 
s  
 
\ μ = −
e H
e
X Z
V b
1.18 I B
s
μ = −
Similarly
h H
h
X Z
V b
1.18 I B
s
μ =
The mobility of electron in n-type
semiconductor is
(16)
The mobility of hole in p-type
semiconductor is
(17)
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UNIT 4 Magnetic , superconducting and dielectric materials
PART A
1. What is meant by magnetic materials? Give example.
The materials which can be easily magnetized by keeping it in an external magnetic
field are called magnetic materials.
Eg: Iron, Ferrites, etc
2. Define Magnetic dipole moment.
A system having two opposite magnetic poles separated by a distance‘d’ is called
as a magnetic dipole. If ‘m’ is the magnetic pole strength and ‘l’ is the length of the
magnet, then its dipole moment is given by
M=ml
3.Define magnetic field intensity(H).
It is defined as the force experienced by a unit North Pole placed at the given
point in a magnetic field.
4..Define magnetization (or) Intensity of magnetization (I)
It is the process of converting a nonmagnetic material into a magnetic material. It
is also defined as the magnetic moment per unit volume.
5. Define magnetic flux density (or) magnetic induction
It is defined as the number of magnetic lines of force passing normally through
unit area of cross section.
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6.Define magnetic permeability
It is defined as the ratio between magnetic flux density (B) and the magnetic field
intensity (H). it is the measure of degree at which the lines of force can penetrate through
the material.
7.Define magnetic susceptibility
It is the measure of the ease with which the specimen can be magnetized by the
magnetizing force. It is the ratio between I and H.
8. What do you understand by the term magnetic domains?
Magnetic domains are small regions in a ferromagnetic material where all the dipoles
are aligned in the same direction.
9. What is Bohr magneton?
The orbital magnetic moment and spin magnetic moment of an electron in an atom
can be expressed in terms of atomic unit of magnetic moment called Bohr magneton.
1 Bohr magneton=eh/4πm=9.27x10-24 Am2
10. Define hysteresis
When a ferromagnetic material is made to undergo a cycle of magnetization, the
intensity of magnetization and the magnetic flux density lags behind the applied magnetic
field.This process is known as hysteresis.
11.What are the four types of energy involved in the growth of magnetic domains?
The four types of energies involved in the growth of magnetic domains
are
i) Exchange energy
ii) Anisotropy energy
iii) Domain wall energy
iv) Magneto-strictive energy
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12. What is meant by reversible and irreversible domains
When the external magnetic field applied to a domain is increased, it starts
expanding. Now when the external magnetic field is removed, if the domain returns to its
original position it is called reversible domains and if the domain doesn’t returns to its
original position it is known as irreversible domains.
13. What are soft and hard magnetic materials?
The materials which can be easily magnetized and demagnetized are called soft magnetic
materials. The materials which are very difficult to magnetize and demagnetize are called
hard magnetic materials.
14. Define retentivity
Even when the applied field is zero(or)removed, the material still acquires some magnetic
induction which is known as residual magnetism (or) retentivity
15. Define coercivity
To remove the residual magnetism in a magnetic material, the magnetic field
strength has to be reversed during a hysteresis cycle and this phenomenon is known as
coercivity.
16. Define hysteresis loss
It is the loss of energy in taking a ferromagnetic material through a complete
cycle of magnetization and the area enclosed is called hysteresis loop.
17. Define critical magnetic field.
It is the minimum magnetic field that is required to destroy the superconducting
property.
18. Define transition (or) critical temperature
The temperature at which a normal conductor loses its retentivity and becomes a
super conductor is known as transition temperature or critical temperature.
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19. Explain Meissner Effect.
When the superconducting material is placed in a magnetic field, under the
condition that T ≤ Tc and H ≤ Hc the flux lines are excluded from the material. Thus the
material exhibits perfect diamagnetism. This phenomenon is known as Meissner effect.
20. Define critical current
The minimum current that can be passed in a sample without destroying its
superconductivity is called critical current(ic).
ic = 2πrHc
21. Define persistant current
When d.c current of large magnitude is once induced in a superconducting ring
then the current persists in the ring even after the removal of the field. This is known as
persistant current.
22. Define super conductivity
The process of conducting the electrical current with zero resistance is called super
conductivity.
PART B
1. Differentiate between hard and soft magnetic materials .
Soft Materials
Hard Materials
They can be easily magnetized and
demagnetized
They cannot be easily magnetized and
demagnetized
Hysteresis loop is very narrow.
Hysteresis loop is very broad
Susceptibility and permeability are high.
Susceptibility and permeability are low.
Loop area is less and hence hysteresis loss is
minimum
Loop area is large and hence hysteresis loss is
maximum.
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Retentivity and coercivity are small Retentivity and coercivity are large.
They have low Eddy current loss.
They have high Eddy current loss.
Movement of domain wall is easy and hence even
for a small applied field large magnetization
occurs.
Movement of domain wall is not easy and hence
large field is required for magnetization
These materials are free from irregularities like
strain or impurities.
These materials have large amount of
irregularities strain or impurities.
They are used to make electro magnets.
Eg. Iron –silicon alloys, Ferrous
nickel alloys, ferrites and garnets.
They are used to make permanent magnets.
Eg. Iron –nickel aluminium alloys
with certain amont of cobalt called ALNICO
alloy, copper nickel iron alloys, copper nickel
cobalt alloys.
2. Explain hysteresis.
When a ferro magnetic material is made to undergo through a cycle of magnetization
the variation of B with respect to H can be represented by a closed hysteresis loop or
curve that is it refers to the lagging of magnetization behind the magnetizing field.
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If a magnetizing field H is applied to a ferromagnetic material and if H is
increased to H max the material acquires magnetism . So the magnetic induction also
increases represented by ‘ oa’ in the fig. Now if the magnetic field is decreased from
Hmax to zero, the magnetic induction will not fall rapidly to zero, but falls to “b”
rather than zero. This shows that even when the applied field is zero or removed, the
material still acquires some magnetic induction “ob” which is so called retentivity.
Now to remove this residual magnetism the magnetic field strength is
reversed and increased to –Hmax represented as “oc” so called co-ercivity and hence
we get the curve “bcd” . Then the reverse field –H is reduced to zero and the
corresponding curve “de” is obtained and by further increasing H to H max the
curve”efa “ is obtained.
3.Give the applications of soft and hard magnetic materials
Soft magnetic materials:
1. They are used in a wide variety of machines in daily uses such as power transformers
, output transformers , motors, generators etc. electrical steel are used as core material
for these machines.
2. Nickel iron alloy and soft ferrites are used in the magnetic amplifiers , saturable core
devices, computers etc.
3. They are also used in switching circuits , micro wave isolaters and matrix storage of
computers.
Hard Materials
1. They are mainly used to make permanent magnets.
2. They are used for the storage of information.Magnetic taps and discs for the storage
and reproduction of audio , video and digital sequences.
4.Explain the properties of superconductors
1.Electrical Resistance:
Electrical Resistance of a superconducting material is very less of
the order of 10-5 ohm-cm.
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2.Magnetic property:
When super conducting materials are subjected to very large value of magnetic field , super
conducting property is destroyed. The field required to destroy super conductingproperty is
called as critical magnetic field.
3.Dia magnetic property-Messiner effect
When the super conducting material is placed in a magnetic field of flux density B the
magnetic lines of force penetrates through the material. When the material is cooled below
it’s transition temperature, the magnetic lines of force are expelled out from the material.
This effect is called Meissner Effect.
4.Effect of electric current:
When a large value of AC current is applied to a super conducting material it induces some
magnetic field in the material. And because of this magnetic field the super conducting
property of the material is destroyed.
5.Persistent Current
when DC current of large magnetitude is once induced in the super conducting ring then
the current persist in the ring even after the removal of the field. This is known as persistent
current.
6.Thermal property:
1. The entropy and specific heat decreases at transition temperature.
2. The thermal conductivity of type I super conductor is low.
3. The thermo-electric effect disappears in the super conducting state.
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7. Isotope Effect:
The transition temperature varies due to the presence of isotopes.
5.Explain about SQUID
SQUID is the acronym for the Super conducting Quantum Interference Device. It is an
ultra sensitive instrument for the measurement of very weak magnetic fields of the order of
10 –14 tesla.
Principle:
Small change in magnetic field produces variation in the flux quantum.
Explanation:
It consist of a super conducting ring which can have magnetic fields of quantum values(1, 2,
3……..) of flux placed in between two josephson junction when the magnetic field is applied
perpendicular to the plane of the ring current is induced at the two josephson junctions and
produces interference pattern. The induced current flows around the ring so that the magnetic
flux in the ring can have quantum values of flux which corresponds to the values of magnetic
fields applied.
Therefore SQUIDS are used to detect the variation in very minute magnetic signals in terms
of Quantum flux. They are used as the storage devices for magnetic flux.
6.What do you meant by magnetic levitation? Explain
We know that super conducting material exhibits meissner effect because of this
nature super conducting materials strongly repel external magnets. It leads to a levitation
effect.
When a magnet is placed over a super conductor the magnet floats. This effect is known
as magnetic levitation.
Diamagnetic property of a super conductor namely rejection of magnetic flux lines is
the basis of magnetic levitation. A super conducting material can be suspended in air against
a repulsive force from a permanent magnet. This magnetic levitation effect can be used for
high speed transportation such as super fast trains without frictional loss.
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7. Give the applications of superconductors
Engineering applications:
1. Since there is no loss in power (zero resistivity) super conductors can be used for the
transmission of power over very large distances.
2. Since the super conducting property can be easily destroyed it can be used in switching
devices.
3. Since the variation in small voltages causes large constant current it can be used in very
sensitive electrical instruments. eg galvano meter
4. Since the current in the super conducting ring can flow with out any change in its value
(persistant current) it can be used as the memory or storage element in computers.
5. Since the size of the specimen can be reduced to about 10-4 cm it can be used to
manufacture electrical generators and transformers in small sizes with high efficiency.
Apart from this they are used to design cryotron, josephson devices, SQUID, magnetic
levitated trains, modulators, rectifiers, commutators etc…
Medical Applications
1. super conducting materials are used in NMR imaging equipments which is used for
scanning purposes.
2. They are applied in the detection of brain wave activity such as brain tumour,
defective cells etc
3.Discuss about type I superconductors and type II superconductors
Type I super conductors:
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When a super conductors is kept in the magnetic field and if the field is increased the
super conductors becomes a normal conductor abruptly at critical magnetic field. His type of
materials are named as type I super conductors.
Below the critical field the specimen excludes all the magnetic line of forces and exhibit
complete meissner effect. Hence they are perfect dia magnets. They have only one critical
magnetic field. The maximum known critical field for type I super conductors is of the order of
0.1 T
Type II super conductors:
The material which looses super conducting property gradually due to increase in magnetic
field are called Type II super conductors. They do not show complete meissner effect. They do
not behave as perfect diamagnet. When the super conductor is kept in the magnetic field and if
the field is increased below the lower critical field HC1 the material exhibit complete meissner
effect, and above HC1 the magnetization decreases and hence the magnetic flux starts penetrating
through the material. The specimen is said to be in a mixed state between HC1 and HC2. Above
HC2 (upper critical field) it becomes a normal conductor.
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Part – C
1. Explain the domain theory of ferromagnetic materials
Ferro magnetic materials, though spontaneously magnetized do
not show macroscopically observable magnetization. In order to explain the above Weiss put
forward domain theory.
According to this theory the entire ferromagnetic volume splits into a large number of small
regions where all the magnetic moments are aligned in a same direction is called Domains. A
magnetic domain has definite boundaries.
In the absence of an external magnetic field the magnetic moment vectors of separate
domains are oriented in all possible directions so that the net magnetic moment of the entire body
equals to zero.When an external field is applied, in the initial stages of magnetization in the
material, the domain having moments parallel to the magnetic field increases in area. In the final
saturation stage, the other domains are rotated parallel to the field.
Processes of domain magnetization:
There are two possible ways to align a random domain structure by applying an external
magnetic field
H
H H
a) H=0 b) By wall movement c) by rotation of
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1.By the motion of domain walls
By the motion of domain walls, ie by increase in the volume of domains that are
favorably oriented with respect to the magnetizing field at the cost of those that are unfavorably
oriented as shown in fig(b)
2.By rotation of domains
By rotation of domains ie by the rotation of the direction of magnetization of domain
along the direction of field as shown in fig (c)
In a weak magnetizing field the magnetization of the specimen is due to the motion of
domain walls and in stronger fields that is due to the rotation of domains.
Energies involved in the domain growth:
To study the domain structure clearly we must know the 4 types of energy involved in the
process of domain growth.
1. Exchange energy: It is the energy associated with quantum mechanical coupling that align
individual atomic dipoles within a single domain .It arises from the interaction of electron spins.
It depends upon the interatomic distance.
2. Crystal anisotropic energy: Crystal anisotropic energy arises from the difference of energy
required for magnetization along any two different direction in a single crystal. Thus the energy
required for magnetization is a function of crystal orientation.
3. Domain wall energy: It is the sum of contribution from the exchange and crystalline
anisotropy energy in the domain wall region. The boundary between the two domain is known as
Bloch wall.
4. Magnetostritive energy: The change in length along the direction of magnetization of a
domain solid refers to magnetostriction. The change in length is independent of the sign of the
field and may be either decrease or increase depending upon the nature of the material. Thus the
Magnetostritive energy is the energy is due to the mechanical stresses generated by
magnetostriction in the domains.
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2.Explain about BCS theory of superconductivity
The theoretical explanation of super conductivity is first given by Bardeen Copper and
Schrieffer in 1957 and it is called BCS theory.
The BCS theory explains most of the phenomena associated with super conductivity such
as zero resistivity, isotope effect etc. This theory involves the electron interaction through
phonon as inter mediators.
The electrons traveling in a solid interact with lattice vibrations by the virtue of
electroststic forces between them. This interaction is called electron –electron inter actions,
which leads to scattering of electrons and hence causes a change in the electrical resistivity. The
resistivity is sensitive with temperature particularly in the low temperature region since the no. of
phonons increases with temperature.
It is assumed from the BCS theory that the electron –phonon interaction produces an
attractive interaction between two electrons.
For eg an electron of Wave vector k emits a virtual phonon which is absorbed by an electron k’ .
Thus k is scattered as k-q and k’+q. Once the phonon energy exceeds the electronic energy the
interaction becomes attractive interaction. Therefore the two electrons interacting attractively in
the phonon field are called copper pair.
The energy of the pair of electrons in the bound structure is less in the free state. The
electron- lattice-electron interaction is stronger when the temperature is less than critical
temperature. The copper pair is completed at T=0K and is completely broken when it reaches the
critical temperature Tc.
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3.Discuss about type I superconductors and type II superconductors
Type I super conductors:
When a super conductors is kept in the magnetic field and if the field is increased the super
conductors becomes a normal conductor abruptly at critical magnetic field. His type of materials
are named as type I super conductors.
Below the critical field the specimen excludes all the magnetic line of forces and exhibit
complete meissner effect. Hence they are perfect dia magnets. They have only one critical
magnetic field. The maximum known critical field for type I super conductors is of the order of
0.1 T
Type II super conductors:
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He materials which looses super conducting property gradually due to increase in magnetic
field are called Type II super conductors. They do not show complete meissner effect. They do
not behave as perfect diamagnet. When the super conductor is kept in the magnetic field and if
the field is increased below the lower critical field HC1 the material exhibit complete meissner
effect, and above HC1 the magnetization decreases and hence the magnetic flux starts penetrating
through the material. The specimen is said to be in a mixed state between HC1 and HC2. Above
HC2 (upper critical field) it becomes a normal conductor.
4.Give the properties and applications of superconductors
1. Electrical Resistance:
Electrical Resistance of a superconducting material is very less of the order of 10-5 ohmcm.
2. Magnetic property:
When super conducting materials are subjected to very large value of magnetic field ,
super conducting property is destroyed. The field required to destroy super conductingproperty is
called as critical magnetic field.
3. Dia magnetic property-Messiner effect
When the super conducting material is placed in a magnetic field of flux density B the
magnetic lines of force penetrates through the material. When the material is cooled below it’s
transition temperature, the magnetic lines of force are expelled out from the material. This effect
is called Meissner Effect.
4.Effect of electric current:
When a large value of AC current is applied to a super conducting material it induces
some magnetic field in the material. And because of this magnetic field the super conducting
property of the material is destroyed.
5.Persistent Current
When DC current of large magnetitude is once induced in the super conducting ring
then the current persist in the ring even after the removal of the field. This is known as persistent
current.
6.Thermal property:
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4. The entropy and specific heat decreases at transition temperature.
5. The thermal conductivity of type I super conductor is low.
6. The thermo-electric effect disappears in the super conducting state.
7. Isotope Effect:
The transition temperature varies due to he presence of isotopes.
Engineering applications:
1.Since there is no loss in power (zero resistivity) super conductors can be used for the
transmission of power over very large distances.
2.Since the super conducting property can be easily destroyed it can be used in switching
devices.
3.Since the variation in small voltages causes large constant current it can be used in very
sensitive electrical instruments.eg galvano meter
4.Since the current in the super conducting ring can flow with out any change in its
value(persistant current) it can be used as the memory or storage element in computers.
5.since the size of the specimen can be reduced to about 10-4 cm it can be used to
manufacture electrical generators and transformers in small sizes with high efficiency.
Apart from this they are used to design cryotron, josephson devices, SQUID, magnetic
levitated trains, modulators, rectifiers, commutators etc…
Medical Applications
1.super conducting materials are used in NMR imaging equipments which is used for
scanning purposes.
2.They are applied in the detection of brain wave activity such as brain tumour, defective
cells etc
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UNIT 5 Modern Engineering Materials
PART A
1. What are metallic glasses?
Metallic glasses are metal alloys which have non crystalline or amorphous
structure and exhibit the property of both metals and glasses.
2. What do you mean by the term “quenching”?
Quenching is a technique used to form metallic glasses. Quenching means extremely
rapid cooling of a molten liquid which results in the irregular arrangement of atom .
3. Give some properties of metallic glasses
(i) They have high corrosion resistance.
(ii) The ferromagnetic properties of metallic glasses have received a great deal of
attention, probably because of the possibility that these materials can be
used as transformer cores
(iii)High rupture, strength and toughness
(iv) Electrical resistivity is high in amorphous phase of metglasses
4. What is meant by glass transition temperature?
The temperature at which liquid like atomic configuration can be frozen into a solid is
said to be glass transition temperature.
5. List out a few applications of metallic glasses.
i) Some of the met glasses can behave as superconductors.
ii) They are used in the cores of high power transformers
iii) As they have high corrosion resistance they are used in reactor vessels, marine
cables, surgical clips, orthopaedical implants , etc.
iv) They are used to make computer memories, magneto-resistance sensors etc.
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6. What are nano phase materials?
Materials with grain size of the order of 1-100 nm are known as nanophase materials.
7. What are methods of synthesizing nanophase materials?
i) top down approach where bulk materials are broken into nanosizes and
ii) bottom-up approach in which nano materials are made by building atom by atom.
8. Mention some properties of nanomaterials
i) Size of grains controls the mechanical, electrical, optical, chemical,
semiconducting and magnetic properties.
ii) These materials are very strong. the strength of the material is inversely
proportional to the grain size.
iii) The melting point of nanophase material is reduced by reducing the grain size.
iv) Undergoes super elastic properties even at lower temperatures.
v) Magnetic moment is increased by decreasing its material size.
9.Give some uses of nanophase materials.
vi) used as ceramic capacitors to store elkectrical energy
vii) used in current controlling devices
viii) magnetic devices made from these materials are used in RAM,
READ/WRITE HEAD, sensors etc
ix) they are used to make semiconductor lasers.
x) They are used in power generation
10. What are Shape memory alloys?
Shape memory alloys are metal alloys which have the ability to return back to their
original shape when subjected to some appropriate thermal procedures.
11. What do you understand by “Martensite” and “Austenite” phases?
The crystal structure of SMA at lower temperature is said to be Martensite phase and the
crystal structure of SMA at higher temperature is said to be Austenite phase.
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12. Explain transformation temperature
Shape memory alloys have the ability to switch from a temporary shape to a parent shape
above a certain temperature called as transformation temperature.
13. Define Pseudoelasticity.
Pseudoelasticity occurs in some types of SMA in which the change in its shape will occur
even without change in its temperature.
14. How are SMA’s classified
i) Materials which regain the shape only upon heating are referred to as one-way shape
memory.
ii) Materials that take up their own shape not only upon heating but also upon cooling
are referred as two way shape memory.
15. List out some properties of SMA.
(i) The transformation occurs over a range of temperature
(ii) They exhibit pseudoelastic or superplastic property
(iii) They exhibit hysteresis curve during cooling and heating process.
16. Mention some uses of shape memory alloys.
i) It is used as a blood-clot filter
ii) They are used to make glass frames
iii) They are used in the opening and closing of valves.
iv) They are used in controlling and preventing cracks
v) They are used to correct irregularities in the teeth
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Part B
1. What are nanomaterials? Explain how the physical properties vary with
the geometry?
• Materials having grain size of the order of 10-9 m are called nano materials.When the
grain size is reduced there will be a great change in the properties.
• Starting from the bulk the first effect of reducing particle size is to create more surface
sites.This in turn changes the surface pressure and results in a change in the interparticle
spacing.
• It is important to note that the inter particle spacing decreases with size. This is due to
competition between long range electronic forces and short range core-core repulsion.
• Nano materials have very high strength and super hardness. It is because of the cluster of
grains in nano materials and are mostly free from dislocations, they become stronger than
conventional metals.
Other properties include
(i) Interparticle spacing is very less in nano materials
(ii) High strength and super hardness
(iii) Melting point will be very less
(iv) Energy bands becomes narrower
(v) Ionisation potential will be higher.
(vi) They are capable of storing hydrogen
(vii) They exhibit spontaneous magnetization
2. Give the charactercities and applications of shape memory alloys?
Shape memory effect: Certain metallic alloys like alloy of gold and cadmium exhibit a plastic
nature when they are cooled to lower temperature. They return to their original dimensional
configuration during heating at high temperature. This effect is called shape memory effect.
Pseudoelasticity: When a metallic object of a given shape is cooled from a given temperature
T2 to a lower temperature T1 it deforms and changes its shape. Upon reheating from T1 to T2
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the shape change accomplished at T1 is recovered so that the object returns to its original
configuration. This is known as pseudoelasticity.
Applications :
(i) It is used as a blood clot filter.
(ii) It is used as a circuit edge connector.
(iii) It is used to make glass frames.
(iv) It is used in controlling and preventing cracks.
(v) It is used to correct the irregularities in teeth.
3. Explain about shape memory alloys.
The group of metallic alloys which demonstrates the ability to return to the original shape
or size when subjected to appropriate thermal procedure (heating or cooling) are called shape
memory alloys.
Shape memory alloys are also called as shape shifters, active materials ,smart materials,
intelligent materials and adaptive materials.
The two phases which occur in SMA are martensite and Austenite. Martensite is
relatively soft and easily deformed phase which exist at low temperature. Austenite is a high
temperature phase having a crystal structure with a high degree of symmetry. Mechanism
involved in SMA is reversible martensite transformation. Stress and temperature have a large
influence on the martensitic transformation.
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4. Explain melt spinning method of preparing metallic glasses.
The process involved in the formation of metallic glass is by melt spinning technique.
This device consists of a disc made of copper. A refractory tube having fine nozzle at the bottom
of the tube is placed just over the spinning disk. An induction heater is affixed with the refractory
tube which melts the alloy under Helium atmosphere. Melt can be ejected through the nozzle and
the ejection rate may be increased by increasing the gas pressure. Thus the glassy alloy ribbon is
formed over the rotating roller. Thickness of the ribbon may be varied by increasing or
decreasing the speed of the rotating roller.
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5. Explain super-elasticity and Hysteresis in S M A.
Super elasticity:
SMA show a superelastic behaviour if deformed at a temperature which is slightly above
their transformation temperature. This effect is caused by the stress induced deformation of
Martensite above its normal temperature. Because it has been formed above its normal
temperature, the martensite reverts immediately to undeformed Austenite as soon as the stress is
removed. This process provides a very springy rubber like elasticity in these alloys.
Hysteresis:
When a shape memory alloy is subject to a constant tensile load for a particular temperature the
transformation of Martensite to Austenite occurs over a range of temperature and not on a single
temperature.
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When the material is cooled from the state of Austenite phase(A) at a particular
temperature , it transforms to Martensite phase(B) due to elongation . Correspondingly there is a
change in shape.
When the material is heated from martensite state (B) at a particular temperature it
regains its original shape at Austenite Phase(A).The transformation does not occur at a single
temperature, but over a range of temperatures that vary with each alloy system. The difference
between the transition temperature upon heating and cooling is called Hysteresis.
Part C
1. Give a detailed account of metallic glasses their method of production, properties and
application.
The process involved in the formation of metallic glass is by melt spinning technique.
This device consist of a disc made of copper. A refractory tube having fine nozzle at the bottom
of the tube is placed just over the spinning disk. An induction heater is affixed with the refractory
tube which melts the alloy under Helium atmosphere. Melt can be ejected through the nozzle and
the ejection rate may be increased by increasing the gas pressure. Thus the glassy alloy ribbon is
formed over the rotating roller. Thickness of the ribbon may be varied by increasing or
decreasing the speed of the rotating roller.
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Properties:
Mechanical properties:
(i) The random ordering in metallic glasses gives very high corrosion resistance.
(ii) The absence of point defects, dislocation, result in a metallic glass with high strength
than metals and alloys.
(iii) They are highly ductile.
Electrical Properties:
(i) The electrical resistivity of metallic glasses is high.
(ii) Due to high resistivity the eddy current loss is very small.
Magnetic properties:
(i) The metallic glasses have both hard and soft magnetic properties.
(ii) They exhibit high saturation magnetization.
Applications:
(i) Metallic glasses are used in cores of high power transformers.
(ii) It is used to make different kinds of springs.
(iii) It is used in leads of tape recorder, magnetic shields.
(iv) It is used to make computer memories, magnetoresistance sensors, etc.
(v) It is used in reactor vessels, surgical clips , marine cables etc.
(vi) It is used for producing high magnetic fields.
It can be used for simple filament winding to reinforce pressure vessels.
2. Describe the ball milling method used to produce nanomaterials. Explain the properties
of nanomaterials.
Ball Milling Method:
In ball milling also called Mechanical crushing small balls are allowed to rotate around
the inside of a drum and then fall on a solid with gravity force and crush the solid in to nano
crystallites. Ball milling can be used to prepare a wide range of elementals and oxide
powders . For eg. Iron with grain sizes of 10-30 nm can be formed. A variety of intermetallic
compounds based on Nickel and Aluminium can also be formed using this method.
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Fig: Ball milling method
Variation of Physical Properties with geometry.
Starting from the bulk the first effect of reducing particle size is to create more surface sites. This
in turn changes the surface pressure and results in a change in the interparticle spacing. It is
important to note that the interparticle spacing decreases with size. This is due to competition
between long range electronic forces and short range core-core repulsion.
Nano materials have very high strength and super hardness. It is because of the cluster of grains
in nano materials and are mostly free from dislocations, they become stronger than conventional
metals.
Electron affinities and Chemical properties.
The Variations in interparticle spacing and geometry also result in the variation of electronic
properties with size. The electronic bands in metals become narrower when the size is reduced
from bulk and hence changes the value of ionization potential. The ionization potentials at small
sizes are higher than that for the bulk.
The large surface to volume ratio, the variation in geometry and the electronic structure
have a strong effect on catalytic properties.
Magnetic properties:
Nano particles of even non magnetic solids exhibit totally a new type of magnetic
properties. In small particles a large fraction of the atoms is deposited at the surface. These
atoms have lower coordination numbers than the interior atoms.When investigated it is found
that local magnetic moment is primarily determined by the local coordination numbers.
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As the coordination number decreases, the magnetic moment increases with the atomic
value. That is small particles are more magnetic than the bulk material.
3. Describe the Chemical vapour deposition technique production of nanomaterials.
Mention the applications.
Chemical vapour deposition technique:
In this method nanoparticles are deposited from the gas phase. Material is heated to form
a gas and then allowed to deposit on a solid surface usually under vacuum condition. The
deposition may be either physical or chemical. Nano powders of oxides and carbides of
metals can be formed if vapours of carbon or oxygen are present with the metal.
Chemical vapour deposition can also be used to grow surfaces. If the object to be coated
is introduced inside the chemical vapour, the atoms or molecules coated may react with the
substrate atoms or molecules. The way the atoms or molecules grow on the surface of the
substrate depends on the alignment of the atoms or molecules of the substrate.
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Applications:
(i) Nano particles are used as fillers, paints, magnetic recording media, ferro fluids,
drugs, rocket propellent, fuel additives etc in dispersed state.
(ii) Nano particles are used as catalyst, electrodes of solar cells and fuel cells,
sensors,adsorbents, synthetic bone, self cleaning glass, etc in consolidated state.
(iii) Nano particles are used in the synthesis of flexible and dense ceramics, insulators,
harder metals etc in very dense phase.
(iv) Nano particles are used as quantum electronic devices, photonic crystals , DNA chips,
Biosensors etc in ordered assembly form.
4. Explain the Characteristics and applications of S M A.
Shape memory effect:
Certain metallic alloys like alloy of gold and cadmium exhibit a plastic nature when they
are cooled to lower temperature. They return to their original dimensional configuration during
heating at high temperature. This effect is called shape memory effect.
Pseudoelasticity: When a metallic object of a given shape is cooled from a given temperature T2
to a lower temperature T1 it deforms and changes its shape. Upon reheating from T1 to T2 the
shape change accomplished at T1 is recovered so that the object returns to its original
configuration. This is known as pseudoelasticity.
Super elasticity:
SMA show a superelastic behaviour if deformed at a temperature which is slightly above
their transformation temperature. This effect is caused by the stress induced deformation of
Martensite above its normal temperature. Because it has been formed above its normal
temperature, the martensite reverts immediately to undeformed Austenite as soon as the stress is
removed. This process provides a very springy rubber like elasticity in these alloys.
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Hysteresis:
When a shape memory alloy is subject to a constant tensile load for a particular
temperature range the transformation of Martensite to Austenite occurs on heating.
When the material is cooled from the state of Austenite phase(A) at a particular
temperature , it transforms to Martensite phase(B) due to elongation . Correspondingly there is a
change in shape.
When the material is heated from martensite state (B) at a particular temperature it
regains its original shape at Austenite Phase(A).The transformation does not occur at a single
temperature, but over a range of temperatures that vary with each alloy system. The difference
between the transition temperature upon heating and cooling is called Hysteresis.
Applications:
(i) It is used as a blood clot filter.
(ii) It is used as a circuit edge connector.
(iii) It is used to make glass frames.
(iv) It is used in controlling and preventing cracks.
(v) It is used to correct the irregularities in teeth
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5. Give an account of the properties and applications of metallic glasses.
Properties:
Mechanical properties:
(iv) The random ordering in metallic glasses gives very high corrosion resistance.
(v) The absence of point defects, dislocation, result in a metallic glass with high strength
than metals and alloys.
(vi) They are highly ductile.
Electrical Properties:
(iii) The electrical resistivity of metallic glasses is high.
(iv) Due to high resistivity the eddy current loss is very small.
Magnetic properties:
(iii) The metallic glasses have both hard and soft magnetic properties.
(iv) They exhibit high saturation magnetization.
Applications:
(vii) Metallic glasses are used in cores of high power transformers.
(viii) It is used to make different kinds of springs.
(ix) It is used in leads of tape recorder, magnetic shields.
(x) It is used to make computer memories, magnetoresistance sensors, etc.
(xi) It is used in reactor vessels, surgical clips , marine cables etc.
(xii) It is used for producing high magnetic fields.
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