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DEPARTMENT OF PHYSICS RV COLLEGE OF ENGINEERING

ENGINEERING PHYSICS NOTES-2017-18

COURSE CODE: 16PH12/22

Semester: I/II ENGINEERING PHYSICS( Theory and practice)

Course Code: 16PH12/16PH22 CIE Marks:100+50=150

Hrs/Week: L:T:P:S 4:0:2:0 SEE Marks:100+50=150

Credits: 05 SEE ( Theory)Duration: 3 Hrs

SEE (Practice)Duration: 3 Hrs

SYLLABUS:

UNIT-I: LASERS AND OPTICAL FIBERS

Basic principles of Laser: Absorption and emissions, Einstein’s coefficients. Energy density in terms of Einstein coefficients, conditions and requisites of laser. Types of lasers: Helium -Neon Laser, Semiconductor diode Laser. Characteristics of laser beam. Industrial applications of lasers: laser cutting, welding and drilling, measurements of pollutants in atmosphere. Principle of Optical fibers: propagation mechanism, condition for propagation, acceptance angle and numerical aperture. Modes of propagation, types of optical fibers. Attenuation: Absorption, scattering and radiation loss, attenuation coefficient. Application of optical fiber in point to point communication, advantages of optical fiber communication over electrical mode of communication.

08 h

UNIT-II: QUANTUM MECHANICS

Black body radiation spectrum, Laws of black body radiation spectrum, Plank’s quantum theory, Review of Photoelectric effect and Compton effect. Wave –particle duality, de-Broglie hypothesis. Matter waves: properties of matter waves, wave packet, group velocity, phase velocity and their relations. Application of matter waves: Scanning Electron Microscope (SEM) – construction and working. Uncertainty principle: illustrations-Non-confinement of electron inside the nucleus and broadening of spectral lines. Setting up of one dimensional time independent Schrodinger’s wave equation-wave function, physical significance of wave function, Eigen function, Eigen values. Application of Schrodinger’s wave equation: Free particle, Particle in a one dimensional potential well of infinite depth. Problems.

11 h

UNIT-III: OSCILLATIONS AND WAVES

Simple Harmonic Motion, Characteristics of Simple harmonic motion. Un damped / Free vibrations, differential equations of un damped / free vibrations and solutions. Examples of Simple harmonic oscillators a) Spring and Mass system, b) Torsional Pendulum. Damped vibrations: Differential equations of damped vibrations and solutions. Forced vibrations: Differential equations of forced vibrations and solutions, Resonance. Examples of forced vibrations- LCR circuits. Problems.

09h

UNIT-IV: ELECTRICAL CONDUCTIVITY IN METALS AND SEMICONDUCTORS Review of Classical free electron theory, Quantum free electron theory. Fermi energy and Fermi factor in metals, variation of Fermi factor with temperature. Density of states and carrier concentration in metals. Hall effect-Determination of number and sign of charge carriers. Band theory of solids, (qualitative approach). Intrinsic semiconductors: carrier concentration, concept of effective mass (qualitative), derivation of electron and hole concentration, intrinsic carrier concentration, Fermi level in intrinsic semiconductors, Expression for the energy gap of intrinsic semiconductors. Extrinsic semiconductors: Types of extrinsic semiconductors, doping methods (qualitative). Variation of carrier concentration in extrinsic semiconductors with temperature, variation of Fermi level in extrinsic semiconductors with temperature and impurity concentration. Hall effect in semiconductors.

10h

UNIT-V: DIELECTRICS AND THERMAL CONDUCTIVITY Dielectrics: Electric dipole, Dipole moment, Field due to electric dipole at a point in a plane. Polarization of dielectric materials: Types of polarizations, frequency dependence of polarization mechanisms, dielectric loss. Internal field in solids: for one dimensional infinite array of dipoles (Lorentz field), Clausius - Mossotti equation. Thermal conductivity: conduction of heat in solids, steady state, coefficient of thermal conductivity, thermal conductivity of a good conductor by Searle’s method and thermal conductivity of a poor conductor by Lee’s and Charlton’s method.

09h

CONTENTS:

• UNIT I : LASERS AND OPTICAL FIBERS

• UNIT II : QUANTUM MECHANICS

• UNIT III : OSCILLATIONS AND WAVES • UNIT IV : ELECTRICAL CONDUCTIVITY IN METALS AND

SEMICONDUCTORS • UNIT V : DIELECTRICS AND THERMAL CONDUCTIVITY

UNIT- I: LASERS AND OPTICAL FIBERS

1


LASER

Introduction:

LASER is an acronym for Light Amplification by Stimulated Emission of Radiation. Laser device produces a beam of coherent, monochromatic, intense and directional light. Hence laser light is highly organized when compared with the ordinary light. This is because the waves of a laser beam move in phase with each other, travel in a narrow path in one direction. In the case of an ordinary light it spreads out, travels in different directions and hence it is incoherent. On account of the special properties, lasers are the most versatile and exploited tools in different fields such as Engineering, Medicine, Defense, Entertainment, Communication etc., Other common applications of lasers include reading the bar code, cutting and welding metals, displays in light shows, playing music, printing documents, guiding missile to its target and so on.

Basic principles: Interaction of Radiation with Matter Production of laser light is a consequence of interaction of radiation with matter under

appropriate conditions. The interaction of radiation with matter leads to transition of the quantum system such as an atom or a molecule of the matter from one quantum energy state to another quantum state.

A material medium is composed of identical atoms or molecules each of which is characterized by a set of discrete allowed energy levels. An atom can move from one energy state to another when it receives or releases an amount of energy equal to the energy difference between those two states, which is termed as a quantum jump or transition.

Consider a two energy level system with energies E1 and E2 of an atom. E1 is the energy of lower energy state and E2 is the energy of excited state. The energy levels E1 and E2 are identical to all the atoms in the medium. The radiation (either absorbed or emitted) may be viewed as a stream of photons of energy (E2-E1) = hν, interacting with the material. These interactions lead to any one of the following Induced Absorption of radiation

Spontaneous emission of radiation

Stimulated emission of radiation


2

Induced Absorption:

Excitation of atoms by the absorption of photons is called induced absorption.

An atom in the lower energy state E1absorbs the incident photon of energy (E1-E2) and goes to

the excited state E2. This transition is known as absorption. For each transition made by an

atom one photon disappears from the incident beam.

for an atom A,

A + hν A* (excited state)

The number of absorption transitions per second per unit volume occurring in the material at

any instant of time will be proportional to

(i) The Number of atoms in the ground state N1

(ii) Energy density of the incident radiation (Uν)

Rate of induced absorption = B12UνN1

where B12 is proportionality constant which gives the probability of absorptions and it is

called Einstein co-efficient of absorption. Since the number of atoms in the lower energy

state is greater, the material absorbs more number of the incident photons.

Spontaneous Emission:

An atom which is at higher energy state E2 is unstable spontaneously returns to the lower

energy state E1 on its own during which a single photon of energy (E2-E1) = hυ is emitted, the

process is known as spontaneous emission.

This spontaneous transition can be expressed as

A * A + hυ

E1

Incident photon = hν= E2- E1

E2

E1

E2

Before

Emitted photon= hν=(E2-E1) E2

E1 After


3

The number of spontaneous transitions per second, per unit volume depends on the number of

atoms N2 in the excited state.

Therefore, the rate of spontaneous emission = A21N2

Where A21 is proportionality constant which gives the probability of spontaneous

emission and it is called Einstein co-efficient of spontaneous emission of radiation.

The process has no control from outside. The instant of transition, directions of emission of

photons, phases of the photons and their polarization states are random quantities. There will

not be any correlation among the parameters of the innumerable photons emitted

spontaneously by the assembly of atoms in the medium. Therefore the light generated by the

source will be incoherent (ex: light emitted from conventional sources).

Stimulated Emission:

Emission of photons by an atomic system with an external influence is called stimulated

emission. A mechanism of forced emission was first predicted by Einstein in 1916 in which

an atom in the excited state need not wait for the spontaneous emission to take place. A

photon of energy hυ = (E2-E1), can induce the excited atom to make downward transition and

emit light. Thus, the interaction of a photon with an excited atom triggers it to drop down to

the ground state (lower energy) by emitting a photon. The process is known as induced or

stimulated emission of radiation.

A* + hυ A + 2 hυ

The number of stimulated transitions per sec per unit volume in the material is proportional to

(i) The Number of atoms in the excited state N2

(ii) Energy density of the incident radiation (Uν)

Rate of stimulated emission= B21UνN2

Where B21 is proportionality constant which gives the probability of stimulated

emissions and it is called Einstein co-efficient of induced (stimulated) emission.

The process of stimulated emission has the following properties.

(i) The emitted photon is identical to the incident photon in all respects. (It has the same

frequency; it will be in phase and will travel in the same direction and will be in the same

state of polarization).

Stimulated photon Stimulating photon

E2

E1 Before After


4

(ii) The process can be controlled externally.

(iii) Stimulated emission is responsible for laser.

Some basic definitions

1. Atomic system It is a system of atoms or molecules having discrete energy levels.

2. Active medium It is the material medium composed of atoms or ions or molecules supports the basic interaction of radiation with matter in thermal equilibrium condition.

3. Energy density The energy density Uν refers to the total energy in the radiation field per unit volume

per unit frequency due to photons. It is given by the Plank’s distribution law

𝑈𝜈 =8𝜋ℎ𝜈3

𝑐3 1

𝑒ℎ𝜈𝑘𝑇 − 1

4. Population

It is the number density (the number of atoms per unit volume) of atoms in a given energy state.

5. Boltzmann factor

It is the ratio between the population of atoms in the higher energy state to the lower energy state under thermal equilibrium. If N2 is the number density of atoms in the energy state E2 and N1 is the number density of atoms in the ground state then According to Boltzmann condition N1 > N2

And N2N1

= e− hν kT 6. Population inversion

It is the condition such that the number of atoms in the higher energy (N2) state is

greater than the number of atoms in the ground state (N1). i.e, N2 > N1

If N2 > N1, it is non-equilibrium condition and it is called population inversion.

Expression for energy density of incident radiation in terms of Einstein coefficients:

Consider an atomic system interacting with radiation field of energy density Uγ. Let E1 and E2

be two energy states of atomic system (E2 > E1). Let us consider atoms are to be in thermal

equilibrium with radiation field, which means that the energy density Uγ is constant in spite of

the interaction that is taking place between itself and the incident radiation. This is possible

only if the number of photons absorbed by the system per second is equal to the number of

photons it emits per second by both the stimulated and spontaneous emission processes.


5

We know that

The rate of induced absorption = B12UνN1,

The rate of spontaneous emission = A21N2

The rate of stimulated emission = B21UνN2

N1 and N2 are the number of atoms in the energy state E1 and E2 respectively, B12, A21 and B21

are the Einstein coefficients for induced absorption, spontaneous emission and stimulated

emission respectively.

At thermal equilibrium,

Rate of induced absorption = Rate of spontaneous emission + Rate of stimulated

emission

B12N1Uγ = A21N2 + B21N2Uγ

or Uγ (B12N1 – B21N2) = A21N2

221112

221

NBNBNAU−

=γ

By rearranging the above equation, we get

−=

1

1

221

11221

21

NBNBB

AUγ --------- (1)

In a state of thermal equilibrium, the populations of energy levels E2 and E1 are fixed by the

Boltzmann factor. The population ratio is given by,

)12(

12 kT

EE

eNN −

−=

2 1

2 1 1

1

2

E E hkT kT

hkT

N N e N e

N eN

γ

γ

− − − = =

∴ =

Since (E2-E1 = hν)

∴ Equation (1) becomes,

=−1

21

1221

21 1

kTh

eBBB

AU γγ ------ (2)


6

According to Planck's law of black body radiation, the equation for U γ is,

Uν = 8πhν3

c3 1

ehνkT−1

------- (3)

Now comparing the equation (2) and (3) term by term on the basis of positional identity we have

3

3

21

21 8ch

BA γπ

= and 121

12 =BB

or 2112 BB =

This implies that the probability of induced absorption is equal to the probability of stimulated emission. Due to this identity the subscripts could be dropped, and A21 and B21 can be simply represented as A and B and equation (3) can be rewritten.

∴ At thermal equilibrium the equation for energy density is

1][e

1BAUγ

kThγ

−

=

(Think: Even though the probability of induced absorption is equal to the probability of stimulated emission, the rate of induced absorption is not equal to rate of stimulated emission. Why?) Conditions for light amplification:- Conditions for laser emission can be studied by taking the ratios of rate of stimulated emission to spontaneous emission and rate of stimulated emission to absorption. At thermal equilibrium,

21

21

221

221

tan AUB

NAUNB

EmissioneousSponEmissionStimulated νν ==

Since, B21/A21 = C3 /8πhν3 = constant, This suggests that in order to enhance the number of stimulated transitions the radiation density Uν should be made high.

112

221

112

221

NBNB

UNBUNB

orptionInducedabsEmissionStimulated

==ν

ν

(B21/B12 = 1)

The stimulated emission will be larger than the absorption only when N2>N1. If N2>N1 the stimulated emission dominates the absorption otherwise the medium will absorb the energy. This condition of N2>N1 is known as inverted population state or population inversion.


7

Requisites of a laser system: The essential components of a laser are An active medium to support population inversion. Pumping mechanism to excite the atoms to higher energy levels. Population inversion Metastable state An optical cavity or optical resonator.

Active medium: It is the material medium composed of atoms or ions or molecules in which the laser action is made to take place, which can be a solid or liquid or even a gas. In this only a few atoms of the medium (of particular species) are responsible for stimulated emission. They are called active centers and the remaining medium simply supports the active centers. Pumping Mechanism: To achieve the population inversion in the active medium, the atoms are to be raised to the excited state. It requires energy to be supplied to the system. The process of supplying energy to the medium with a view to transfer the atoms to higher energy state is called pumping. Important pumping mechanisms are

a) Optical pumping: It employs a suitable light source for excitation of desired atoms. This method is adopted in solid state lasers (ex: Ruby laser and Nd:YAG laser).

b) Electric discharge: In this process an electric field causes ionization in the medium and raises it to the excited state. This technique is used in gas lasers (ex: Ar+ laser).

c) Inelastic atom-atom collision : In this method a combination of two types of gases are used, say A and B. During electric discharge A atoms get excited and they now collide with B atoms so that B goes to excited state. This technique is used in gas lasers (ex: He-Ne laser).

d) Direct conversion : In this process electrical energy is directly converted into light energy. This technique is used in semiconductor lasers (ex: GaAs laser).

Population Inversion and Meta stable state:

In order to increase stimulated emission it is essential that N2>N1 i.e., the number of atoms in the excited state must be greater than the number of atoms in the ground state. Even if the population is more in the excited state, there will be a competition between stimulated and spontaneous emission. The possibility of spontaneous emission can be reduce by using


8

intermediate state where the life time of atom will be little longer (10-6 to 10-3 s ) compared to excited state (10-9

s). This intermediate state is called metastable state and it depends upon the nature of atomic species used in the active medium. Principle of pumping scheme: Consider three energy levels E1, E2 and E3 of a quantum system of which the level E2 is metastable state. Let the atoms be excited from E1 to E3 state by supply of appropriate energy. Then the atom from the E3 state undergoes downward transition to either E1or E2 states rapidly. Once the atoms undergo downward transitions to level E2 they tend to stay, for a long interval of time, because of which the population of E2 increases rapidly. Transition from E2 to E1 being very slow, in a short period of time the number of atoms in the level E2 is far greater than the level E1. Thus Population inversion has been achieved betweenE1 and E2 . The transition from meta stable state to ground state is the lasing transition. It occurs in between upper lasing E2 and lower lasing level E1.

Optical resonator: • An optical resonator generally consists of two plane mirrors, with the active material

placed in between them. One of the mirrors is semitransparent while the other one is 100% reflecting. The mirrors are set normal to the axis of the active medium and parallel to each other.

• The optical resonant cavity provides the selectivity of photon states by confining the possible direction of photon propagation, as a result lasing action occurs in this direction.

• The distance between the mirrors is an important parameter as it chooses the wavelength of the photons. Suppose a photon is traveling between two reflectors, it undergoes reflection at the mirror kept at the other end .the reflected wave superposes on the incident wave and forms stationary wave such that the length L of the cavity is given by

𝐿 = 𝑛 λ2 Hence λ = 2L

𝑛

Where, L is the distance between the mirrors. λ is the wavelength of the photon, n is the integral multiple of half wavelength

E2

E3

E1

Non-radiative transition

Meta stable state (Upper lasing

level)

Lasing transition

(Lower lasing level)


9

The wavelengths satisfying the above condition are only amplified. Hence the cavity is also called resonant cavity. The main Role of the optical resonator is to

• Provide positive feedback of photons into the active medium to sustain stimulated emission and hence laser acts as a generator of light.

• Select the direction of stimulated photons which are travelling parallel to the axis of optical resonator and normal to the plane of mirrors are to be amplified. Hence laser light is highly directional.

• Builds up the photon density (Uν) to a very high value through repeated reflections of photons by mirrors and confines them within the active medium.

• Selects and amplifies only certain frequencies of stimulated photons which are to be highly monochromatic and gives out the laser light through the partial reflector after satisfying threshold condition.

Helium-Neon (He-Ne) Laser:

The Helium – Neon laser is a gas laser that produces a continuous laser beam. It is widely used in surgery, communication, printing and scanning.

Fig : Schematic diagram of He- Ne Laser

Construction: He-Ne laser system consists of a narrow long quarts tube (1m long and 10cm diameter) with two electrodes and the tube is filled with He-Ne gas mixture (10:1 ratio) at a pressure of 1mm (torr). The tube ends are fitted with Brewster’s quartz window, the light travelling parallel to the axis incident at polarizing angle (tanθB=n). Two mirrors one fully silvered (m1) and the other partially silvered (m2) are located outside the ends of the tube such that they are perpendicular to the axis of the tube and perfectly parallel to each other. Mirrors along with

Active medium

100%

Ref

lect

ing

mirr

or

Sem

i tra

nspa

rent

mirr

or


10

the quartz tube form the laser cavity. The entire arrangement is placed in an enclosure. Electrodes are connected to a powerful d.c. source. Working: When a d.c. voltage (at 1000 V) is applied, the electric field ionizes some atoms of the gas mixture, due to this the electron are released and are accelerated towards the anode and helium and neon ions are accelerated towards cathode. Due to the smaller mass electrons acquire very high velocity. The free electrons while moving towards anode collide with atoms in their way. Collisions are more with the Helium (as Helium and Neon are in 10:1 ratio) and helium atoms are excited to the levels 3S and 1S level which are meta stable state of helium. This is electrical pumping. This collisions are of first kind i.e.,

He + e1 He* + e2 Where He: Helium in ground state, He*: Helium in excited state, e2 is lesser with

energy less than that of e1 When current is continuously passed through the discharge tube more and more helium atoms are excited to state 3S and 1S. The energy levels 3S2 (20.66 eV) and 2S2 (19.78 eV) of neon atoms are very close to the helium levels 1S and 3S.

Collisions of 2nd type takes place helium and neon atoms. In this process Neon atoms are excited to 3S2 and 2S2 levels and the Helium atoms to ground state. This is called resonance transfer of energy. He* + Ne ---------------- He + Ne*


11

This is the pumping mechanism in the He-Ne laser and Ne atoms are active centers and population inversion sets with lower energy states. It is to be noted that there may be a chance of quick transition from 2S2 state to the ground state since it is radiatively connected to the ground state. Such case possible when the probability of decay (2S2 state) to the ground state exceeds that to the 2p4 levels which occurs between S and P levels at very low pressures. Since we are applying a high pressure of the order of a mm Hg, the transitions to the ground state undergo complete resonance trapping instead of escaping from the gas i.e, every time a photons is emitted is simply absorbed by another atom in the ground state and ends up in the excited 2S2 state through resonance energy transfer. Hence population increases at 2S2 levels through continuous process. This leads to increase the life time of S levels to ~100 ns when compared to the life time of P4 levels of the order of 20 ns. Therefore a favorable lifetime ratio for producing the required population inversion satisfies the lasing action in Ne atoms.

There are three types of laser transitions which are as follows, 3S --------- 3P transmission with 3361.2 nm (3.39μm) which are in IR region 3S ----------2P transition with 632.8 nm (0.632 μm) which are in visible region 2S --------- 2P transition with 1152.3 nm (1.15 μm) which are in IR region 2P --------- 1S transition is spontaneous. The 1S level is a meta stable state and it should be quickly depopulated. To facilitate this the tube is made narrow and 1S to ground state transition takes place due to collision with the walls of the quartz tube. The Infra- red (IR) radiations of 3.39 μm and 1.15 μm are absorbed by quartz window and 632.8 nm component is amplified in the resonance cavity and comes out through the partially silvered mirror m2.

Semiconductor diode laser:

Semiconductor lasers are very popular in the modern world. The semiconductor lasers are built using highly doped direct band gap semiconductors like Gallium arsenide (GaAs), Gallium arsenide phosphide (GaAsP) materials (Direct band gap semiconductors have high probability of recombination of holes and electrons). They are continuous laser.

Fig: Schematic construction of semiconductor laser


12

Construction:

A semiconductor laser in its simplest form consists of highly doped GaAs p-n junction. The junction is forward biased with d.c. supply. The p-n junction forms the active region. The resonant cavity is obtained by cleaving (polished) the front and back faces of the semiconducting material The cleaved/polished faces are perfectly parallel and flat. The back face is fully reflecting and front face is partially reflecting.

Working:

The energy band diagram of p-n junction consists of valence band (V) and conduction band (C ) separated by an energy gap Eg. If the semiconductor is assumed to be at absolute zero temperature (T=0K), the valence band is completely filled state and the conduction band is completely empty The hatched area represented in figure ‘b’ (before inversion) shows that the valence band is in a completely filled state of energy. In operation, when the p-n junction is heavily forward biased with a large current, it raises the electrons from the valence band to the conduction band, but this is an unstable state and within a short time (10-13s) the electrons in the conduction band drop to the lowest level in that band itself. At the same time, the electrons near the top of the valence band will drop to the lowest unoccupied levels, leaving behind holes. As a result the lowest level of conduction band is full of electrons (indicated in fig ‘c’) while the top of the valence band is full of holes (indicated by removed hatch in fig ‘c’), which is an indication of population inversion between valence band and conduction band. The narrow depletion region becomes the active region in semiconductor.

Now, at some instant, one of the excited electrons from the conduction band falls back into the valence band to recombine with a hole, and the energy associated with this recombination is emitted as a photon of light. This photon as it moves, stimulates the recombination of another excited electron with a hole and release another photon. These two photons are in


13

phase with each other and have the same wavelength, travel together in the same direction and get reflected at the end face. As they travel back they stimulate more electron hole recombination with the release of additional photons and become a part of the monochromatic and coherent laser beam. This beam gets resonated by traveling back and forth and finally leaves through the partially reflecting face. Since the current is being passed continuously (through forward bias), more electrons get excited, rise to the conduction band and new holes generate in the valence band. This situation maintains the population inversion, while the recombination of electron-hole pairs continuous with the generation of laser beam. Note: At very low current, the p n junction acts like an LED Characteristics of Laser beam • Directionality: The design of the resonant cavity, especially the orientation of the

mirrors to the cavity axis ensures that laser output is limited to only a specific direction. Since laser emits photons in a particular direction, the divergence is less when compared the other ordinary sources.

• Monochromacity: The laser beam is characterized by a high degree of mono-chromaticity (single wavelength or frequency) than any other conventional monochromatic sources of light. Ordinary light spreads over a wide range of frequencies, whereas laser contains only one frequency. The spectral bandwidth is comparatively very less when compared to ordinary light. Hence the degree of mono chromaticity is very high in lasers.

• Coherence: The degree of coherence of a laser beam is very high than the other sources. The light from laser source consists of wave trains that are in identical in phase. Laser radiation has high degree of special (with respect to Space) and temporal (with respective to time) coherence.

• High Intensity: The laser beam is highly intense. Since wave trains are added in phase and hence amplitudes are added. Laser light emits as a narrow beam and its energy is concentrated in a small region. Since all the energy is concentrated in the particular focus point, it is highly intense and bright. When laser beam is focused on a surface, the energy incident is of the order of millions of joules.

• Focus ability: Since laser is highly monochromatic, it can be focused very well by a lens. It is so sharp the diameter of the spot will be close to the wavelength of the focused light. It can be focused to a very small area 0.7µm2. Since even laser is not ideally monochromatic the spot diameter in actual cases will be 100 to 150 times larger than the wavelength.

Applications of Lasers Applications of lasers are wide spread over various scientific disciplines like Physics, Chemistry, Biology, Medicine, Communication, Holography, Material processing, Industry etc., In industry lasers are mainly used in cutting, welding and drilling processes.


14

Welding:

In the welding process the laser beam is focused on to a spot to be welded. Due to heat generation, the material melts over a tiny area on which beam focused. As a result an impurity such as oxides floats upon the surface. On cooling the material within becomes homogenous solid structure which makes weld strong. In the laser welding process No destruction occurs in the regions around the welded portion. Since it is a contact less process

no foreign material comes in contact in to the welded joint. Further since the heat-affected zone is small, laser welding is ideal in areas such as microelectronics. CO2 lasers are the most popular ones in this particular application.

Advantages: • High quality and speed • Weld dissimilar metals • No need of filler • Low heat affected Zone • Welding can be done in inaccessible areas

Cutting: Laser cutting is generally done assisted by gas blowing. In this process a jet of gas is issued through nozzle right at the spot where laser beam is focused. The combustion of the gas burns the metal thus reducing the laser power requirement for cutting. The tiny splinters along with the molten part of metal will be blown away by oxygen jet. The blowing action increases the depth and speed of the cutting process

Advantages:

• The quality of cutting is very high. • Complicated 3D cutting profiles is possible. • There will be no thermal damage and chemical change, when cutting is done to regions around cut area. • High edge quality. • No wear and tear of cutting tool. • Minimal amount of mechanical distortion.

Laser Light

Welding Spot

Laser Light


15

Drilling:

Drilling of holes is achieved by subjecting materials to laser pulses of duration 10-4to 10-3

second. The intense heat generated over a short duration by the pulses evaporates the material

locally, thus leaving a hole.

Advantages:

• In conventional methods tools wear out while drilling whereas the problem doesn't

exist with laser setup.

• Drilling can be achieved at any angle.

• Very fine holes of diameter 0.2 to 0.5mm could be drilled with a laser beam

• Very hard materials or brittle materials could be subjected to laser drilling since there

is no mechanical stress with a laser beam.

ND-YAG laser is used to drill holes in metals, where as CO2 laser is used in case of metallic

and non -metallic materials.

Measurement of pollutant in atmosphere:

There are various types of pollutants in the atmosphere, they are oxides of nitrogen, carbon

monoxide, sulphur dioxide etc.. In conventional technique the type and the concentration of

pollutants in the atmosphere are determined by the chemical analysis. However this is not a

real time data. This limitation can be overcome by using laser which yields a real time data.

In the measurement of pollutant, laser is made use of the way RADAR system is used.

Hence it is often referred to as LIDAR which means Light Detection And Ranging. A LIDAR

can be employed to measure distance, altitude & angular coordinates of the object.

In LIDAR

• Ruby laser is used as transmitting source which sends the laser beam through the

desired region of the atmosphere.

• The transmitted light is reflected by the mirrors and reaches a receiver.

• The receiving part consists of concave mirror collects scattered light. The mirror

focuses the light on to a photo-detector which converts the light energy into electrical

energy. A narrowband filter is used to cuts off extraneous light and background noise.

Then the electrical signal is fed to a computer a data processor, which gives

information regarding distance, dimensions of the object etc.


16

Two methods can be employed to know the composition of the pollutants

(i) Absorption technique (ii) Raman back scattering

Absorption technique:

When laser beam is made to pass through the atmosphere then molecules can either absorb

light of certain frequencies or scatter light of certain frequencies. Depending upon the

characteristic absorption pattern, the composition of the pollutant molecules can be

determined.

Raman back scattering:

In this method laser light is passed through the sample, and the spectrum of the transmitted

light is obtained. Since laser is highly monochromatic we expect to see only one line in the

spectrum but due to Raman scattering, in the spectrum not one but several other lines of weak

frequencies can be seen symmetrically. Additional spectral lines are called side bands and

they are formed when the oscillating frequencies of the molecules of the gas are added to or

subtracted from the incident's light's frequency. Different gases produce different side bands,

the shift in frequencies are termed as Raman shifts. Thus by observing Raman spectra of the

back-scattered light in the gas sample one can get the information about the composition of

the pollutants.


17

OPTICAL FIBERS Optical fibers are the light guides used in optical communications as wave-guides. They are thin, cylindrical, transparent flexible dielectric fibers. They are able to guide visible and infrared light over long distances. The working structure of optical fiber consists of three layers. Core- the inner cylindrical layer which is made of glass or plastic. Cladding- which envelops the inner core. It is made of the same material of the core but of lesser refractive index than core. The core and the cladding layers are enclosed in a polyurethane jacket called sheath which safeguards the working structure of fiber against chemical reactions, mechanical abrasion and crushing etc.

Propagation mechanism in Optical fiber:

In optical fibers light waves can be guided through it, hence are called light guides. The cladding in an optical fiber always has a lower refractive index (RI) than that of the core. The light signal which enters into the core can strike the interface of the core and the cladding at angles greater than critical angle of incidence because of the ray geometry. The light signal undergoes multiple total internal reflections within the fiber core. Since each reflection is a total internal reflection, the signal sustains its strength and also confines itself completely within the core during propagation. Thus, the optical fiber functions as a wave guide.

Numerical Aperture and Ray Propagation in the Fiber: Consider an optical fiber consists of core and cladding material placed in air medium. Let n0, n1 and n2 be the refractive indices of surrounding air medium, core and cladding material respectively. The RI of cladding is always lesser than that of core material (n2 < n1) so that the light rays propagate through the fiber. Let us consider the special case of ray which suffers critical incident at the core cladding interface. The ray travels along AO entering into the core at an angle of θ0 with respect to the fiber axis. Let it be refracted along OB at an angle θ1 in the core and further proceed to fall at critical angle of incidence (= 90-θ1) at B on the interface of core and cladding. Since it is critical angle of incidence, the refracted ray grazes along core and cladding interface.


18

It is clear from the figure that a ray that enters at an angle of incidence less than θ0 at O, will have to be incident at an angle greater than the critical angle at the core-cladding interface, and gets total internal reflection in the core material. When OA is rotated around the fiber axis keeping θ0 same, it describes a conical surface, those rays which are funneled into the fiber within this cone will only be totally internally reflected and propagate through the fiber. The cone is called acceptance cone.

The angle θ0 is called the wave guide acceptance angle or the acceptance cone half-angle which is the maximum angle from the axis of optical fiber at which light ray may enter the fiber so that it will propagate in core by total internal reflection. Sin θ0 is called the numerical aperture (N.A.) of the fiber. It determines the light gathering ability of the fiber and purely depends on the refractive indices of core, cladding and surrounding medium. Let n0, n1 and n2 be the refractive indices of surrounding medium, core and cladding respectively for the given optical fiber. By applying the Snell’s law at O, 𝑛0𝑠𝑖𝑛𝜃0 = 𝑛1𝑠𝑖𝑛𝜃1 𝑠𝑖𝑛𝜃0 = 𝑛1

𝑛0𝑠𝑖𝑛𝜃1 ------------ (1)

At the point Bon the core and cladding interface, the angle of incidence = 90-θ1 Applying Snell’s law at B 𝑛1 sin(90 − 𝜃0) = 𝑛2𝑠𝑖𝑛90 or 𝑛1𝑐𝑜𝑠𝜃1 = 𝑛2 𝑐𝑜𝑠𝜃1 = 𝑛2

𝑛1 ------------------- (2)

From equation (1) 𝑠𝑖𝑛𝜃0 = 𝑛1

𝑛0𝑠𝑖𝑛𝜃1 = 𝑛1

𝑛0(1 − 𝑐𝑜𝑠2𝜃1)

𝑠𝑖𝑛𝜃0 = 𝑛1𝑛01 − 𝑛22

𝑛12= 𝑛1

𝑛0𝑥𝑛12−𝑛22

𝑛12=

𝑛12−𝑛22

𝑛0

If the medium surrounding the fiber is air, then n0 = 1, Therefore, 𝑠𝑖𝑛𝜃0 = 𝑛12 − 𝑛22 sinθ0 = Numerical aperture: NA Therefore, 𝑁𝐴 = 𝑠𝑖𝑛𝜃0 = 𝑛12 − 𝑛22 If θi is the angle of incidence of an incident ray, then the ray will be able to propagate, If θi < θ0 or if sin θi < sin θ0 or sin θi < NA Fraction Index Change (∆): The fractional index change ∆ is the ratio of the difference in the refractive indices between the core and the cladding to the refractive index of core of an optical fiber. It is also known as relative core clad index difference, denoted by ∆. If n1 and n2 are the refractive indices of core and cladding, then,


19

∆ = 1

21 )(n

nn −

Relation between NA and ∆ We have RI change Δ = n1−n2

n1

Or Δn1 = (n1 − n2) -----------(1) We know that,

∆+

−+

−=

121

2121

22

21

)(

))((

..

nnn

nnnn

nnAN

Since n1~ n2 , (n1+ n2) = 2n1

Therefore, ∆=

∆=

2..

2..

1

21

nAN

nAN

Though an increase in the value of ∆ increases NA and thus enhances the light gathering capacity of the fiber, we cannot increase ∆ to a very large value, since it leads to a phenomenon called “intermodal dispersion” which causes signal distortion. Modes of Propagation: The possible number of paths of light in an optical fiber determines the number of modes available in it. It also determines the number of independent paths for light that a fiber can support for its propagation without interference and mixing. We may have a single mode fiber supporting only one signal at a time or multimode fiber supporting many rays at a time. Such number of modes supported for propagation in the fiber is determined by a parameter called V-number. If the surrounding medium is air then the V- number is given by 𝑉 = 𝜋𝑑

𝜆(𝑛12 − 𝑛22)

or ( )dV NAπλ

=

where d is the core diameter, n1 is the refractive index of core, n2 is the refractive index of the cladding and λ is the wavelength of the light propagating through the fiber.

𝑉 = 𝜋 2𝑟𝜆𝑛12 − 𝑛22 = (𝑟)2𝜋

𝜆 𝑛12 − 𝑛22 = 𝑟.𝑘 𝑛12 − 𝑛22 .

Where k is the propagation constant (k= 2π/λ) If fiber is surrounded by a medium of refractive index n0, then

2 21 2

0

)n ndVn

πλ

−=

For single mode fiber, the number of modes supported by step index fiber is ≈ 2

2V

For graded index fiber, the number of modes are ~ V2/4


20

Types of Optical fibers: The optical fibers are classified under 3 categories. They are

a) Step index Single mode fiber (SMF) b) Step index multi mode fiber (MMF) c) Graded index Multi Mode Fiber (GRIN)

This classification is done depending on the refractive index profile and the number of modes that the fiber can guide. Refractive Index Profile (RI): Generally in any types of optical fiber, the refractive index of cladding material is always constant and it has uniform value throughout the fiber. But in case of core material, the refractive index may either remain constant or subjected to variation in a particular way. This variation of RI of core and cladding materials with respect to the radial distance from the axis of the fiber is called refractive index profile. This can be represented as follows,

Intermodal dispersion: In a single mode fiber, a single light ray (called single mode) propagates through the fiber. Since other modes are not interfering with this mode, intermodal dispersion is zero in single mode fiber. In case of multimode fiber, intermodal dispersion is maximum. Different modes are propagating with different distance hence there is a time delay in the propagation Due to this, there is a possibility of dispersion in their path and losses their information. This is called intermodal

a) Step index Single mode fiber (SMF): A single mode fiber has a core material of uniform refractive index (RI) value. Similarly cladding also has a material of uniform RI but of lesser value. This results in a sudden increase in the value of RI from cladding to core. Thus its RI profile takes the shape of a step. The diameter value of the core is about 8 to 10 µm and external diameter of cladding is 60 to 70 µm. Because of its narrow core, it can guide just a single mode as shown in Figure. Hence it is called single mode fiber. Single mode fibers are most extensively used ones and they constitute 80% of all the fibers that are manufactured in the world today. They need lasers as the source of light. Though less expensive, it is very difficult to splice them (joining of optical fibers). Since single mode is propagating through the fiber, intermodal dispersion is zero in this fiber. They find particular application in submarine cable system.

60 to 70 µm 8 to 10 µm

RI profile of Step index fiber

X axis: Radial distance from the centre of the fiber


21

Fig: Step Index Single Mode Fiber b) Step index multimode fiber (MMF): The geometry of a step-index multimode fiber is as shown in below figure. It’s construction is similar to that of a single mode fiber but for the difference that, its core has a much larger diameter by the virtue of which it will be able to support propagation of large number of modes as shown in the figure. Its refractive index profile is also similar to that of a single mode fiber but with larger plane regions for the core. The step-index multimode fiber can accept either diode laser or LED (light emitting diode) as source of light. It is the least expensive of all. Since multi modes are propagating through this fiber with different paths, intermodal dispersion is maximum this fiber. Its typical application is in data links which has lower bandwidth requirements.

Cladding

Core

light ray

100 to 250 µm

50 to 100 µm

refractive index profile

Radial distance

Ray propagation

RI profile

Radial distance

Input and output pulse


22

Fig. Step index multimode fiber

c) Graded index multimode fiber (GRIN) Graded index multimode fiber is also denoted as GRIN. The geometry of the GRIN multimode fiber is same as that of step index multimode fiber. Its core material has a special feature that its refractive index value decreases in the radially outward direction from the axis and becomes equal to that of the cladding at the core-cladding interface. But the RI of the cladding remains uniform. Its RI profile is also shown in figure. Either a diode laser or LED can be the source for the GRIN multimode fiber. It is most expensive of all. Its splicing could be done with some difficult. Since multi modes are propagating with different velocity and meets at nodes position, intermodal dispersion is minimized in this fiber. Its typical application is in the telephone trunk between central offices.

Core

Cladding

100 to 250 µm 50 to 100 µm

RI profile

Radial distance

Fig. Graded index multimode fiber




23

Differences between single and multimode fibers: Single mode fiber Multi mode fiber • Only one mode can be propagated • Smaller core diameter • Low dispersion of signal • Can carry information to longer

distances • Launching of light and connecting two

fibers are difficult

• Allows large number of modes for light to pass through it

• Larger core diameter • More dispersion of signal • Information can be carried to shorter

distances only • Launching of light and connecting of

fibers is easy Differences between step and graded index fibers: Step index fiber Graded index fiber • Refractive index of core is uniform • Propagation of light is in the form of

meridional rays • Step index fibers has lower bandwidth • Distortion is more (in multimode) • No. of modes for propagation

Nstep = V2/2

• Refractive index of core is not uniform • Propagation of light is in the form of skew

rays • Graded index fibers has higher bandwidth • Distortion is less • No. of modes for propagation Ngrad = V2/4

Attenuation in optical fibers: The total energy loss suffered by the signal due to the transmission of light in the fiber is called attenuation. The important factors contributing to the attenuation in optical fiber are

i) Absorption loss ii) Scattering loss iii) Bending loss iv) Intermodal dispersion loss and v) coupling loss.

Attenuation is measured in terms of attenuation co-efficient and it is the loss per unit length. It is denoted by symbol α. Mathematically attenuation of the fiber is given by,

L

InPoutP

10log10

α

×

−= dB/km

Where Pout and Pin are the power output and power input respectively, and L is the length of the fiber in km. Therefore, Loss in the optical fiber = α x L


24

1. Absorption loss: There are two types of absorption; one is absorption by impurities and the other intrinsic absorption. In the case of first type, the type of impurities is generally transition metal ions such as iron, chromium, cobalt and copper. During signal propagation when photons interact with these impurities, the electron absorbs the photons and get excited to higher energy level. Later these electrons give up their absorbed energy either as heat energy or light energy. The re-emission of light energy is of no use since it will usually be in a different wavelength or at least in different phase with respect to the signal. The other impurity which would cause significant absorption loss is the OH- (Hydroxyl) ion, which enters into the fiber constitution at the time of fiber fabrication. In the second type i.e., intrinsic absorption it is the absorption by the fiber itself, or it is the absorption that takes place in the material assuming that there are no impurities and the material is free of all inhomogeneities and this sets the lowest limit on absorption for a given material. 2. Scattering loss: The signal power loss occurs due to the scattering of light energy due to the obstructions caused by imperfections and defects, which are of molecular size, present in the body of the fiber itself. The scattering of light by the obstructions is inversely proportional to the fourth power of the wavelength of the light transmitted through the fiber. Such a scattering is called Rayleigh scattering. The loss due to the scattering can be minimized by using the optical source of large wavelength.

3. Bending losses (radiation losses): There are two types of bending losses in optical fiber a) macroscopic and b)microscopic bending loss.

a) Macroscopic bends: Macroscopic bends occurs due the wrapping of fiber on a spool or turning it around a corner. The loss will be negligible for small bends but increases rapidly until the bending reaches a certain critical radius of curvature. If the fiber is too bend, then there is possibility of escaping the light ray through cladding material without undergoing any total internal reflection at core-cladding interface.

b) Microscopic bends: This type of bends occurs due to repetitive small-scale fluctuations in the linearity of the fiber axis. Due to non-uniformities in the manufacturing of the fiber or by non-uniform lateral pressures created during the cabling of the fiber. The microscopic bends cause irregular reflections at core-cladding interface and some of them reflects back or leak through the fiber.


25

This loss could be minimized by extruding a compressible sheath over the fiber which can withstand the stresses while keeping the fiber relatively straight.

4. Coupling losses: Coupling losses occur when the ends of the fibers are connected. At the junction of coupling, air film may exist or joint may be inclined or may be mismatched and they can be minimized by following the technique called splicing. Applications of Optical Fibers: Point-to-point Communication The use of optical fibers in the field of communication has revolutionized the modern world. An optical fiber acts as the channel of communication (like electrical wires), but transmits the information in the form of optical waves. A simple p to p communication system using optical fibers is illustrated in the figure. The main components of p to p communication is

1) An optical transmitter, i.e., the light source to transmit the signals/pulses 2) The communication medium (channel) i.e., optical fiber 3) An optical receiver, usually a photo cell or a light detector, to convert light pulses back

into electrical signal.


26

The information in the form of voice or video to be transmitted will be in an analog electric signal format. This analog signal at first converted into digital electric (binary) signals in the form of electrical pulses using a Coder or converter and fed into the optical transmitter which converts digital electric signals into optic signals. An optical fiber can receive and transmit signals only in the form of optical pulses. The function of the light source is to work as an efficient transducer to convert the input electrical signals into suitable light pulses. An LED or laser is used as the light source for this purpose. Laser is more efficient because of its monochromatic and coherent nature. Hence semiconductor lasers are used for their compact size and higher efficiency. The electrical signal is fed to the semiconductor laser system, and gets modulated to generate an equivalent digital sequence of pulses, which turn the laser on and off. This forms a series of optical pulses representing the input information, which is coupled into the optical fiber cable at an incidence angle less than that of acceptance cone half angle of the fiber.

Next the light pulses inside the fiber undergo total internal reflection and reach the other end of the cable. Good quality optical fibers with less attenuation to be chosen to receive good signals at the receiver end.

The final step in the communication system is to receive the optical signals at the end of the optical fiber and convert them into equivalent electrical signals. Semiconductor photodiodes are used as optical receivers. A typical optical receiver is made of a reverse biased junction, in which the received light pulses create electron-hole charge carriers. These carriers, in turn, create an electric field and induce a photocurrent in the external circuit in the form of electrical digital pulses. These digital pulses are amplified and re-gain their original form using suitable amplifier and shaper. The electrical digital pulses are further decoded into an analogues electrical signal and converted into the usable form like audio or video etc.,

As the signal propagates through the fiber it is subjected to two types of degradation. Namely attenuation and delay distortion. Attenuation is the reduction in the strength of the signal because power loss due to absorption and scattering of photons. Delay distortion is the reduction in the quality of the signal because of the spreading of pulses with time. These effects cause continuous degradation of the signal as the light propagates and may reach a limiting stage beyond which it may not be retrieve information from the light signal. At this stage repeater is needed in the transmission path. An optical repeater consists of a receiver and a transmitter arranged adjacently. The receiver section converts the optical signal into corresponding electrical signal. Further the electrical signal is amplified and recast in the original form and it is sent into an optical transmitter section where the electrical signal is again converted back to optical signal and then fed into an optical fiber.

Finally at the receiving end the optical signal from the fiber is fed into a photo detector where the signal is converted to pulses of electric current which is then fed to decoder which converts the sequence of binary data stream into an analog signal which will be the same information which was there at the transmitting end.

Reciever Amplifier Transmitter


27

Advantages over conventional communication: Advantages of optical fibers over conventional cables are 1) Large Bandwidth: Optical fibers have a wider bandwidth (when compared to

conventional copper cables). This helps in transmitting voice, video and data on a single line and at very fast rates (1014 bps as compared to about 104 bps in ordinary communication line)

2) Electromagnetic Interference (EMI): EMI and disturbance in the transmission is a very common phenomenon in ordinary copper cables. However, the optical fiber cables are free from EMI, since Electromagnetic radiation has no effect on the optical wave. Hence, there is no need to provide specially shielded conditions for the optical fiber.

3) Low attenuation: Compared to metallic cables, optical fibers have a low attenuation level (as they are relatively independent of frequency). The loss in optical fibers is very low, of the order of 0.1 to 0.5 dB/km of transmission.

4) Electrical Hazards: Since, optical fibers carry only the light signals, there are no problems of short-circuiting and shock hazards.

5) Security: Unlike electrical transmission lines, there is no signal radiation around the optical fiber, hence the transmission is secure. The tapping of the light waves, if done, leads to a loss of signal and can be easily detected.

6) Optical fiber cables are small in size, light weight and have a long life. Disadvantages The disadvantages in the communication systems using optical fibers are

• Fiber loss is more at the joints if the joints do not match (the joining of the two ends of the separate fibers are called splicing)

• Attenuation loss is large as the length of the fiber increases. • Repeaters are required at regular interval of lengths to amplify the weak signal in long

distance communication. • Sever bends will increase the loss of the fiber. Hence, the fiber should be laid straight

as far as possible and avoid severe bends. Note:

• Point to Point haul communication system is employed in telephone trunk lines. This system of communication covers the distances 10 km and more. Long-haul communication has been employed in telephone connection in the large cities of New York and Los Angeles. The use of single mode optical fibers has reduced the cost of installation of telephone lines and maintenance, and increased the data rate.

• Local Area Network (LAN) Communication system uses optical fibers to link the computer-oriented communication within a range of 1 or 2 km.

• Community Antenna Television (CATV) makes use of optical fibers for distribution of signal to the local users by receiving a multichannel signal from a common antenna.


28

1. Medical application Endoscopes are used in the medical field for image processing and retrieving the image to find out the damaged part of the internal organs of the human body. It consists of bundle of optical fibers of large core diameter whose ends are arranged in the same sequence. Endoscope is inserted to the inaccessible damaged part of the human body. When light is passed through the optical bundle the reflected light received by the optical fibers forms the image of the inaccessible part on the monitor. Hence, the damage caused at that part can be estimated and also it can be treated. 2. Industry Optical fibers are used in the design of Boroscopes, which are used to inspect the inaccessible machinery parts. The working principle of boroscope is same as that of endoscopes.

3. Domestic Optical fiber bundles are used to illuminate the interior places where the sunlight has no access to reach. It can also be used to illuminate the interior of the house with the sunlight or the incandescent bulb by properly coupling the fiber bundles and the source of light. They are also used in interior decorating articles.

Sl. No SHORT ANSWERE QUESTIONS CO’s 1. Define the terms:

a. Spontaneous emission. b. Stimulated Emission. c. Active medium. d. Population inversion.

1

2. Explain any one of the industrial applications of laser. 2 3. Give any two differences between the laser light and ordinary light. 2

Sl. No LONG ANSWSER QUESTIONS

4. Explain the requisites of a laser system. 2 5. With energy level diagram of He-Ne gas, explain the working of He-Ne

laser. 3

6. Discuss the conditions required for laser action. 2 7. Write a note on measurement of pollutants in atmosphere using laser. 2 8. Explain the three processes which take place when radiation interacts

with matter. 2

9. Explain the terms stimulated emission and population inversion. Obtain an expression for energy density of photons in terms of Einstein’s co-efficient.

2&3

10. Explain the characteristics of a laser beam. 11. Explain the working principle of a semiconductor laser using band

diagram and discuss its advantages.


29

S. No Questions CO

1 Give reason Optical fibers are immune to electromagnetic interference. Intermodal dispersion is minimum in GRIN compared to MMSI fiber. Repeaters are used in the path of optical fibers in point to point communication system.

1

2 Explain the terms a. Acceptance angle b. Cone of acceptance c. Numerical aperture d. Modes of propagation e. Attenuation.

1

3 Explain attenuation losses in optical fibers. 2 4 Write any two advantages of optical fiber communication over normal

communication. 2

5 Explain propagation mechanism in optical fibers. 2 6 Distinguish between step and graded index fibers. 2

LONG ANSWER QUESTIONS S. No Questions CO’s 1 With the help of ray diagram, explain the working principle of optical fibers. 1&2 2 Derive an expression for acceptance angle of an optical fiber in terms of

refractive indices. 1&2

3 What is numerical aperture? Obtain an expression for numerical aperture in terms of refractive indices of core and cladding and arrive at the condition for propagation.

4 Explain the terms 1) modes of propagation and 2) types of optical fibers 1 5 What is attenuation? Explain the different losses in optical fibers. 2 6 With the help of a block diagram explain point to point communication. 2 7 Discuss the advantages and disadvantages of an optical communication

system. 1


30

PROBLEMS: 1. The ratio of population of two energy levels out of which one corresponds to metastable

state is 1.059x10-30. Find the wavelength of light emitted at 330K. 30

1

2 10059.1 −= xNN

, T=330K, λ=?

Constants h=6.63x10-34Js, K= 1.38x10-23J/K, C= 3x108m/s Using the relation for Boltzmann’s factor

kThce

kThe

NN

λν −=

−=

1

2

λ = 632.8nm.

2. Calculate the ratio of i) Einstein Coefficients, ii) Stimulated to spontaneous emissions, for a system at 300K in which radiations of

wavelength 1.39µm are emitted.

1533

3

21

21 102.688 −=== xhch

BA

λπνπ

Since B12 = B21 we can write 15

21

21 102.6 −= xBA

We have

Rate of stimulated emission/Rate of spontaneous emission= νν U

AB

NAUNB

21

21

221

212 =

But

−=

1

183

3

kTh

echU νννπ

−=

1

121

21

kTh

eBAU νν

Therefore rate of stimulated emission/rate of spontaneous emission=

−1

121

21

21

21

kTh

eBAx

AB

ν

==

−1

1

kTh

eν =10-15

3. Calculate on the basis of Einstein’s theory, the number of photons emitted per second by a He-Ne laser source emitting light of wavelength 6328Ao with an optical power of 10mW.

tnhc

tEn

tEP

λ∆=

∆==

Hence hc

Ptn λ∆=

= 3.182x1016


31

4. Calculate the numerical aperture, relative RI difference, V- number and number of modes in an optical fibre of core diameter 50µm. Core and cladding Refractive indices 1.41 and 1.40 at λ= 820nm.

(NA)2 = (n12- n2

2) 1

21

nnn −

=∆ λ

π )( 22

21 nnd

V−

=

NA= 0.1676 = 0.007 =32 Hence no of modes=V2/2 =512.

5. An optical fibre has clad of RI 1.50 and NA 0.39. Find the RI of core and the acceptance angle.

(NA)2 = n12- n2

2 θo= sin-1 0.39 (0.39)2= n1

2 - (1.50)2 = 22.96o n1 = 1.54

6. The NA of an OF is 0.2 when surrounded by air. Determine the RI of its core. Given The RI of cladding as 1.59. Also find the acceptance angle when it is in a medium of RI 1.33.

(NA)2 = (n12 – n2

2) θo = 8.64o n1= 1.60

7. A glass clad fibre is made with core glass of RI 1.5 and cladding is doped to give a

fractional index difference of 0.0005. Determine a) The cladding index. b) The critical internal reflection angle. c) The external critical acceptance angle d) The numerical aperture

8. The attenuation of light in an optical fibre is estimated at 2.2dB/km. What fractional initial intensity remains after 2km & 6km?

L = 2 : Pout/Pin = 36.3% L=6 : Pout/Pin = 4.79%

9. Find the attenuation in an optical fibre of length 500m, when a light signal of power 100mW emerges out of the fibre with a power 90mW.

α = L

Pp

in

out )(log10 10−

α = 0.915dB/km 10. A semiconductor laser emits green light of 551 nm. Find out the value of its band gap. Eg = hc/𝜆 = 2.25 eV


32

11. The probability of spontaneous transition is given as 0.08. in a laser action which results with the radiation of 632.8 nm wavelength. Calculate the probability of stimulated emission. (Ans. 1.22x1013)

12. Calculate the critical angle if the refractive indices of optical fiber are 1.5 & 1.48.

(Ans. θc = 80.63o) 13. The optical fiber power after propagating through a fiber of 1.5 km length is reduced

to 25% of its original value. Compute the fiber loss in dB/km. (Ans. 4 dB/km)

UNIT- II: QUANTUM MECHANICS

33

UNIT –II: QUANTUM MECHANICS

Introduction:

At the beginning of the 20th century, Newton’s laws of motion were able to successfully describe the motion of the particles in classical mechanics (the world of large, heavy and slow bodies) and Maxwell’s equations explained phenomena in classical electromagnetism. However the classical theory does not hold in the region of atomic dimensions. It could not explain the stability of atoms, energy distribution in the black body radiation spectrum, origin of discrete spectra of atoms, etc. It also fails to explain the large number of observed phenomena like photoelectric effect, Compton Effect, Raman Effect, Quantum Hall effect, superconductivity etc. The insufficiency of classical mechanics led to the development of quantum mechanics. Quantum mechanics gives the description of motion and interaction of particles in the small scale atomic system where the discrete nature of the physical world becomes important. With the application of quantum mechanics, most of the outstanding problems have been solved. Black Body Radiation

Black-body radiation is electromagnetic radiation emitted by a black body held at constant, uniform temperature. Black Body Spectrum: It is a graph showing the variation of the energy of the black body radiations as a function of their wavelengths or frequencies. The energy distribution in the black body spectrum is explained by Wien’s distribution law in the lower wavelength region and Rayleigh Jeans law explains the energy distribution in the larger wavelength region.

Wien’s law:

λλλ λ

λ deAdE TB−

−= 5

Rayleigh Jeans law λλπλλ dTkdE 48 −=

Fig : Blackbody Radiation spectrum


34

Neither, Wien’s law nor Rayleigh- jean’s law could explain the energy distribution in the entire blackbody spectrum. The energy distribution in the entire blackbody spectrum was successfully explained by Max. Planck by quantum Theory.

Planck’s quantum theory The energy distribution in the black body radiation spectrum was successfully

explained by Max Planck in the year 1900. According to Planck’s quantum theory thermal energy is not emitted or absorbed continuously, but it is emitted or absorbed in discrete quantities called quanta. Each quantum has an energy ‘hν’ where h is the Planck’s constant and υ is the frequency of the radiation. Applying the Planck’s quantum theory an expression for the energy distribution in the black body spectrum was obtained and it is called Planck’s formula. The Planck’s formula is as follows

λλπλ

λλ d

e

hcdEkT

hc1

18)(5−

=

Where k is the Boltzmann’s constant; h- Planck’s constant and c is the velocity of light, λ is the wavelength of the black-body radiation and ω is the angular frequency of the radiation. Photoelectric effect:

When the light of a suitable wavelength shines on certain materials, then electrons are spontaneously emitted from the surface of material. It can be observed in any material but most readily and steadily in metals and good conductors. This phenomenon is known as the photoelectric effect.

The materials that exhibit photoelectric effect are called photosensitive materials and the emitted electrons are called photoelectrons. Heinrich Hertz first observed this phenomenon in 1887.

The electrons are emitted only when the photons reach or exceed a threshold frequency (energy) and below that threshold, no electrons are emitted from the metal regardless of the light intensity or the length of time of exposure to the light. To explain this phenomenon, Albert Einstein proposed that light be seen as a collection of discrete bundles of energy (photons), each with energy hυ, where υ is the frequency of the light that is being quantized and h is known as the Planck constant. Einstein’s photoelectric equation:

Einstein, in 1905, proposed that the light energy is localized in small packets similar to the Planck’s idea of quanta, and named such packets as photons. According to Einstein, in photoelectric effect one photon is completely absorbed by one electron, which thereby gains the quantum of energy and may be emitted from the metal.

Thus the photon energy is used in the following two parts: i). A part of its energy is used to free the electron from the atoms of the metal surface. This energy is known as a photoelectric work function of metal (Wo) ii) The other part is used in giving kinetic energy (½ mv2) to the electron.

Thus 212oh W mvν = +

where ‘v’ is the velocity of the emitted electron.


35

θ

Φ

This equation is known as Einstein’s photoelectric equation. When the photon’s energy is of such a value that it can just liberate the electron from metal, then the kinetic energy of the electron will be zero. Then the above equation reduces to

oo Wh =ν , where oν is called the threshold frequency. Threshold frequency is defined as the minimum frequency which can cause photoelectric emission. Below this frequency no emission of electron takes place. Compton Effect: When a monochromatic beam of high frequency radiation (X – rays, γ – rays, etc.) is scattered by a substance, then the scattered radiation contains two components - one having a lower frequency or greater wavelength called as modified radiation and the other having the same frequency or wavelength called as unmodified radiation. This phenomenon is known as Compton effect and was discovered by Prof. A.H. Compton in 1921. The process of recoiling of electron and scattering of photon is as shown in the following figure: Φ----- is the recoil angle θ------ is the scattering angle.

Schematic diagram of Compton Effect

According to the quantum concept of radiation, the radiation is constituted by energy packets called photons. The energy of photon is hν, where h is Planck’s constant and ν is the frequency of radiation. The photons move with velocity of light c, possess momentum hν/c and obey all the laws of conservation of energy and momentum. According to Compton, the phenomenon of scattering is due to an elastic collision between two particles, the photon of incident radiation and the electron of the scatterer. When the photon of energy hν collides with the electron of the scatterer at rest, it transfers some energy to the electron, i.e., it loses the energy. The scattered photon will therefore have a smaller energy and consequently a lower frequency (ν’) or greater wavelength (λ’) than that of the incident photon. The observed change in wavelength of the scattered radiation is known as Compton Shift. In the scattering process, the electron gains kinetic energy and thus recoils with a velocity v. The change in wavelength of the photon scattered through an angle θ is given by

'

0

( ) [1 cos ]hm c

λ λ λ ϑ∆ = − = −

Electron at rest Incident photon

Scattered photon

Recoil electron


36

Wave and particle duality of radiation: To understand the wave and particle duality, it is necessary to know what a particle is and what a wave is. A particle is a localized mass and it is specified by its mass, velocity, momentum, energy, etc. In contrast a wave is a spread out disturbance. A wave is characterised by its wavelength, frequency, velocity, amplitude, intensity, etc. It is hard to think mass being associated with a wave. Considering the above facts, it appears difficult to accept the conflicting ideas that radiation has a wave particle duality. However this acceptance is essential because the radiation exhibits phenomena like interference, diffraction, polarization, etc., and shows the wave nature and it also exhibits the particle nature in the black-body radiation effect, photoelectric effect, Compton Effect etc. Radiation, thus, sometimes behave as a wave and at some other time as a particle, this is the wave particle duality of radiation. De Broglie’s concept of matter waves: Louis de-Broglie in 1924 extended the wave particle dualism of radiation to fundamental entities of physics, such as electrons, protons, neutrons, atoms, molecules, etc. de-Broglie put a bold suggestion that like radiation, matter also has dual characteristic, at a time when there was absolutely no experimental evidence for wavelike properties of matter waves. de Broglie Hypothesis of matter waves is as follows. • In nature energy manifests itself in two forms, namely matter and radiation. • Nature loves symmetry. • As radiation can act like both wave and a particle, material particles (like electrons,

protons, etc.) in motion should exhibit the property of waves. These waves due to moving matter are called matter waves or de-Broglie waves or pilot waves. Wavelength of matter waves:

The concept of matter waves is well understood by combining Planck’s quantum theory and Einstein’s theory. Consider a photon of energy E, frequency γ and wavelength λ.

By Planck’s theory λ

γ hchE ==

By Einstein’s mass-energy relation 2mcE = By equating and rearranging the above equations, we get

ph

mch

hcmc

==

=

λ

λ2

Where, p is the momentum of the photon and h is a Planck’s constant. Now consider a particle of mass m moving with a velocity v and momentum p. According to the do-Broglie hypothesis matter also has a dual nature. Hence the wavelength λ of matter waves is given by


37

ph

mvh

==λ

This is the equation for the de-Broglie wavelength. de Broglie wavelength of an electron Consider an electron of mass ‘m’ accelerated from rest by an electric potential V. The electrical work done (eV) is equal to the kinetic energy E gained by the electron.

mEmv

mEvm

mvE

eVE

2

221 Also

22

2

=

=∴

=

=

Therefore wavelength of electron wave

2

h hp mv

hmE

λ

λ

= =

∴ =

But eVE =

22

hh memeV V

λ∴ = =

Substituting the values of e, m and h, we get

12.28

2oh A

me=

12.28 oAV

λ∴ =

Note: Instead of a an electron, if a particle of charge ‘q’ is accelerated through a potential difference V, then

mqVh

2=λ

Properties of matter waves and how they are different from electromagnetic waves: 1. Lighter the particle, greater would be the wavelength of matter waves associated with it. 2. Smaller the velocity of the particle, greater would be the wavelength. 3. For p = 0, λ is infinity ie., the wave becomes indeterminate. This means that matter waves

are associated with moving particles only. 4. Matter waves are produced by charged or uncharged particles in motion. Whereas

electromagnetic waves are produced only by a moving charged particle. Hence matter waves are non – electromagnetic waves.


38

5. In an isotropic medium the wavelength of an electromagnetic wave is a constant, whereas wavelength of a matter wave changes with the velocity of the particle. Hence matter waves are non- electromagnetic waves.

6. A particle is a localized mass and a wave is a spread out disturbance. So, the wave nature of matter introduces a certain uncertainty in the position of the particle.

7. Matter waves are probability waves because waves represent the probability of finding a particle in space.

Wave packet: A wave packet refers to the case where two (or more) waves exist simultaneously. A wave packet is often referred to as a wave group. This situation is permitted by the principle of superposition.

In physics, a wave packet (or wave train) is a short "burst" or "envelope" of localized wave action that travels as a unit. A wave packet can be analyzed into, or can be synthesized from, an infinite set of component sinusoidal waves of different wave numbers, with phases and amplitudes such that they interfere constructively only over a small region of space, and destructively elsewhere. Each component wave function, and hence the wave packet, are solutions of a wave equation. Depending on the wave equation, the wave packet's profile may remain constant (no dispersion) or it may change (dispersion) while propagating.

Quantum mechanics ascribes a special significance to the wave packet. It is interpreted as probability amplitude. The modulus square of the probability amplitude describes the probability density that a particle in a particular state will be measured to have a given position or momentum. The equation that describes the evolution of this wave packet is the Schrödinger equation. It can be Time independent or Time dependent.

A monochromatic wave has velocity called the phase velocity given by υλω==

kvp where ω

is the angular frequency, λπ2

=k is the wave number, υ is the frequency. However, if we

have a compound wave(wave packet) that is composed of individual waves with a range of frequencies, each individual wave has its phase velocity, but the amplitudes of the waves add up to produce a wave packet which has a velocity all of its own. This velocity is called the group velocity and is usually different from the individual phase velocities of the waves that

make up the packet. It is given by the expression kk

vg ∂∂

=∆∆

=ωω

Schematic diagram of Wave Packet formed by superposition of waves


39

Relation between group velocity and phase velocity: kv

kv pp ==>= ωω

)2()2()2(

)(2k

vv

kv

vkv

kvkkv

kv p

pp

pp

pp

gπ

λλπλ

λλπω −

∂

∂+=

∂∂

∂

∂+=

∂

∂+=

∂

∂=

∂∂

=

λλλ

ππ

λλπ

∂

∂−=

−∂

∂+= p

pp

pg

vv

vvv 2

2 )4

2()2(

Relation between group velocity and particle velocity:

We know that ( )1.........2πωυ hhE == while

( )2...........22 ππλhk

khhp === (Where E, p is the Energy, momentum of the

particle). Thus we can show that the group velocity is given by:

particleparticle

g vm

mvmp

pmp

pE

kv ===

∂∂

=∂∂

=∂∂

=)2/( 2ω

Thus we have shown that for a free particle, the group velocity equals the particle

velocity, while the phase velocity is half the group velocity.

Relation between group velocity, phase velocity and light velocity:

gparticleparticlep v

cv

cmv

mcpE

kv

222

=====ω

Thus we get the relation 2. cvv pg =

Application of matter waves:

Scanning Electron Microscope The wave nature of electrons has been practically applied in diverse fields, of these the

most important is in the development of scanning electron microscope, having magnification and resolving power much higher than the optical microscope. Principle: The smallest distance resolved by a microscope is called Limit of resolution (Δx) and the resolving power of a microscope is the reciprocal of the limit of resolution.

2 sin

x λµ θ

∆ =

Resolving Power = 1/Δx = 2μsinθ/λ

where 2 sinµ θ is called the numerical aperture of the objective. Thus shorter the wavelength(λ) of the radiation, greater is the resolving power.


40

It is this resolving power that sets a limit to the magnification of the microscope, because a mere increase in the magnifying power without a corresponding increase in the resolving power would only produce large blurred image. Optical microscope uses a light beam of wavelength ≈ 600 nm so that the smallest distance resolvable ≈ 300 nm(Assuming numerical aperture = 2). Hence, if instead of light waves we could use electron beam the resolving power can be increased by more than 106 fold. Electrons are known to exhibit wave characteristics under suitable conditions, the wavelength of electron wave is governed by de Broglie relation.

2hmeV

λ =

The magnification can be increased by making use of de Broglie waves associated with electrons, accelerated to approximately 75 kV which has a wavelength of 4.48x10-12 m which is roughly 105 times shorter than visible radiation. In optical microscopes, ‘light waves’ produce the image and magnification is obtained by ‘optical lenses’. An electron microscope generally uses electron waves and the electron beam is focused by suitable magnetic lenses. Magnetic lens:

The magnetic fields which are axially symmetric have a focusing effect on the electron beam passing through them. The axially symmetric magnetic fields are produced by short solenoids. Such solenoids are called magnetic lens. The focal length of such magnetic lens depends on the quality of coil and current. By adjusting the current through the coils the intensity of the magnetic field and the focal length can be adjusted. It is widely used in electron microscopes.

Scanning Electron Microscopy (SEM) The SEM uses electrons instead of light to form an image. SEM is a powerful and commonly used instrument, in both academia and industry for imaging the surfaces of almost any material with a resolution down to about 1 nm. Most of the scanning electron microscopes have magnification ranges from X20 to X100000. SEM image gives a characteristic three-dimensional appearance and useful for adjudging the surface structure of the sample. Working Principle of the SEM:

A beam of electrons are produced by heating a thermionic filament. And it is operated with an acceleration voltage of 1 KV to 40 KV for the electrons. The electron beam follows a vertical path and it makes its way through a set of magnetic lenses which focuses and directs the beam towards the sample. These electrons interact with the sample to a depth of ~1μm and generate a number of different types of signals, which are emitted from the area of the specimen where the electron beam is impinging. The induced signals are detected and the intensity of one of the signals (at a time) is amplified and used as the intensity of a pixel on the image on the computer screen. The electron beam, then moves to the next position on the sample and the detected intensity gives the intensity in the second pixel and so on. This


41

produces an image with depth-of-field which is usually 300-600 times better than that of an optical microscope, and also enables a three-dimensional image to be obtained. Construction:

A schematic of SEM is shown in the following figure.

Fig . Basic construction of SEM.

Electrons are thermionically emitted from a tungsten filament (cathode) are drawn to an anode and focused by two successive magnetic lenses into a beam with a very fine spot size that is typically 10Å in diameter. Pairs of scanning coils located at the objective (magnetic) lens deflect the beam either linearly or in raster fashion over a rectangular area of the specimen surface. Upon impinging on the specimen, the primary electrons decelerate and several processes such as elastic scattering viz., forward scattering and backscattering of the incoming electrons and inelastic scattering viz., generation of secondary electrons, Auger electrons, Bremsstrahlung, characteristic x-rays, electron-hole pairs (in insulators and semiconductors), long-wavelength electromagnetic radiation etc. A number of different types of signals, which are emitted from the area of the specimen where the electron beam is impinging, are as shown in fig.

Fig. Different types of signals produced when high-energy electrons impinges on a material.


42

Various SEM techniques are differentiated on the basis of what is subsequently detected and imaged, and the principle images produced in the SEM are of three types:

• Secondary electron images, • Backscattered electron images and • Elemental X-ray maps.

The secondary electrons are the electrons which are generated from the surface of the specimen when it is bombarded by high energy primary electrons. These electrons are collected by detector which creates a pattern of light and dark areas on a computer screen corresponding to the emission of secondary electrons from the specimen. As the number of electrons produced at any given point can be related directly to the topography of the specimen with respect to the detector, the patterns created on the viewing screen represent the surface topography of the specimen. Backscattered electrons are the high energy electrons that are elastically scattered and essentially possess the same energy as the incident or primary electrons. The probability of backscattering increases with the atomic number of the sample material. Therefore the primary electrons arriving at a given detector position can be used to yield images containing information on both topology and atomic composition. An additional electron interaction in the SEM is that the primary electron collides with and ejects a core electron from an atom in the sample. The excited atom will decay to its ground state by emitting either a characteristic X-ray photon or an Auger electron. By analyzing energies of characteristic of the x-ray photon the atoms can be identified. Further the concentration of atoms in the specimen can be determined by counting the number of X-rays emitted. Applications: SEM is used in a wide range of fields as a tool for observing surfaces at the nanometer level. It is an indispensable instrument for not only basic research but also for highly integrated semiconductor devices and new advanced materials. The SEM is used in the investigation of surface unevenness of the material (topography), the shape and size of the particle making up the object (Morphology), heterogeneity of the materials to visualize various mineral components in their distinct growth forms and their relation in terms of overall micro fabric and texture (composition) and to visualize the arrangement of atom in an object (crystallographic information). Besides surface topographic studies the SEM can also be used for determining the chemical composition of a material, its fluorescent properties, and the formation of magnetic domains and so on. Heisenberg’s uncertainty principle Heisenberg’s uncertainty principle is a direct consequence of the dual nature of matter. In classical mechanics, a moving particle at any instant has a fixed position in space and a definite momentum which can be determined if the initial values are known (we can know the future if we know the present) In wave mechanics a moving particle is described in terms of a wave group or wave packet. According to Max Born’s probability interpretation the particle may be present


43

anywhere inside the wave packet. When the wave packet is large, the momentum can be fixed, but there is a large uncertainty in its position. On the other hand, if the wave packet is small the position of the particle may be fixed, but the particle will spread rapidly and hence the momentum (or velocity) becomes indeterminate. In this way certainty in momentum involves uncertainty in position and the certainty in position involves uncertainty in momentum. Hence it is impossible to know within the wave packet where the particle is and what is its exact momentum. (We cannot know the future because we cannot know the present). Thus we have Heisenberg’s uncertainty principle. According to the Heisenberg’s uncertainty principle “It is impossible to specify precisely and simultaneously certain pairs of physical quantities like position and momentum that describe the behavior of an atomic system”. Qualitatively, this principle states that in any simultaneous measurement the product of the magnitudes of the uncertainties of the pairs of physical quantities is equal to or greater than h/4π (or it is of the order of h) Considering the pair of physical quantities such as position and momentum, we have ΔpΔx ≥ h/4π ...….1 Where Δp and Δx are the uncertainties in determining the momentum and the position of the particle. Similarly, we have other canonical forms as ΔEΔt ≥ h/4π ……2 ΔJΔθ ≥ h/4π ……3 Where ΔE and Δt are uncertainties in determining energy and time while ΔJ and Δθ are uncertainties in determining angular momentum and angular position. Illustration of Heisenberg’s uncertainty principle: 1. Nonexistence of electrons in the atomic nucleus: If an electron is to exist inside the nucleus, then the maximum uncertainty in its position ∆x must not exceed the size of the nucleus. The diameter of the nucleus is of the order of 10-14m. i.e,. ∆x ≤ 10-14m. Heisenberg’s Uncertainty principle states that,

Therefore Uncertainty in momentum is, -34

14

20 1

h46.625 x 104 3.14 100.5275 10

x

x

x

Px

P

P kgms

π

−

− −

∆ ≥∆

∆ ≥× ×

∆ ≥ ×

This is the minimum uncertainty in the momentum of the electron. The momentum of the electron is greater than the uncertainty in the momentum. Thus

( )1.........................4πhpx x ≥∆∆

( )1......................../.105275.0 20 skgmpx−×≥


44

If the momentum is as per the equation (1) then the velocity of the electron is much greater than the velocity of light. Hence we have to consider the relativistic equation for the energy of the electron. We now show the derivation of this equation as done by Einstein.

According to the theory of relativity the energy equivalence E, of mass m is given by,

where ‘mo’ and ‘m’ is the rest mass and relativistic mass of the particle and ‘v’ its velocity. Squaring the above equation we get,

The momentum of the particle is given by

Multiplying by c2, we get

Subtracting equation (6) from equation (3), we have

Thus we get the equation ( )9.........................220

22 cmcpE +=

Where E is the relativistic energy of the particle.

Now from equation (8), 2 2 2 2 4 2 2 2 2

o oE p c m c c p m c= + = +

Since, the rest mass of the electron mo=9.11 x 10-31kg. Now by using the condition (9) in the above equation, we can say that, in order that the electron to exist within the nucleus, its energy E must be such that, the energy E of the electron in the nucleus is approximately is

MeVE

JxE89.9

10582454.1 12

≥≥ −

( )8.........................420

222 cmcpE +=( )7.........................)()(

22

)2242

0222

vcvccmcpE

−−

=−

)2....(....................1

.

2

2

202

cv

cmmcE

−

==

)3...(.....................1

22

620

2

2

4202

vccm

cvcm

E−

=−

=

)4....(....................1

.

2

2

0

cv

vmmvp

−

== )5....(....................22

22202

vccvm

p−

=

)6....(....................22

422022

vccvm

cp−

=


45

In order that an electron may exist inside the nucleus, its energy must be greater than or equal to 9.89 MeV. But, the experimental investigations on beta decay shows that the kinetic energy of the beta particles (electrons) is of the order of 3 to 4 MeV. This clearly indicates that electrons cannot exist within the nucleus. 2. Broadening of spectral lines

When an atom absorbs a photon, it rises to the excited state and it will stay in the excited state for certain time called the lifetime. Lifetime of atoms in the excited levels is of the order of 10-8s. When the atom comes to the ground state it emits a photon of energy exactly equal to the energy difference between the two levels as shown in the figure .

λ

Fig. Line width for emitted photons The energy of the emitted photon is given by

.................(1)hcE hνλ

= =.

where h is a Planck’s constant, ν is the frequency, c is the velocity of light and λ is the wavelength. Thus for the first set of diagrams, where there is only one sharply defined excited state, with ΔE = 0, there is only a single wavelength of the photons emitted and there would be no broadening of the spectral lines. However, in nature, there is always a finite amount of uncertainty in the energy of the lines (ΔE) that would lead to a finite broadening in the wavelength of the photons emitted as shown in the second set of diagrams. Differentiating equation (1) with respect to wavelength (λ), we get

2λλ∆

−=∆Εhc

ΔE

E =hν E =hν

E =hν E =hν

λ In

tens

ity Δλ= 0

λ

Inte

nsity

Δλ = FWHM


46

2

.................(2)hcE λλ∆

∆ =

According to Heisenberg’s uncertainty principle, the finite lifetime Δt of the excited state means there will be an uncertainty in the energy of the emitted photon is given by

t

h∆

≥∆Επ4

Substituting for ΔE from (2) and applying the condition of minimum uncertainty, we get

=∆2λλhc

th∆π4

or 2

4 c tλλπ

∆ ≥∆

This shows that for a finite lifetime of the excited state, the measured value of the emitted photon wavelength will have a spread of wavelengths around the mean value λ. This uncertainty in the measured value of wavelength demands for very narrow spread, the lifetime of the excited state must be very high (of the order of 10-3s). Such excited levels are called Metastable states. This concept is adopted in the production of highly monochromatic laser light. Schrödinger’s Wave Equation:

In1926 Schrödinger starting with de-Broglie equation (λ = h/mv) developed it into an important mathematical theory called wave mechanics which found a remarkable success in explaining the behavior of the atomic system and their interaction with electromagnetic radiation and other particles. In water waves, the quantity that varies periodically is the height of water surface. In sound waves it is pressure. In light waves, electric and magnetic fields vary. The quantity whose variation gives matter waves is called wave function (ψ). The value of wave function associated with a moving body at a particular point x in space at a time t is related to the likelihood of finding the body there at a time. A wave function ψ(x,t) that describes a particle with certain uncertainty in position, moving in a positive x-direction with precisely known momentum and kinetic energy may assume any one of the following forms: Sin( ωt - kx) , cos( ωt - kx), ei(ωt – kx), e-i(ωt – kx) or some linear combinations of them. Schrödinger wave equation is the wave equation of which the wave functions are the solutions. It cannot be derived from any basic principles, but can be arrived at, by using the de-Broglie hypothesis in conjunction with the classical wave equation. Time Independent one dimensional Schrödinger wave equation (TISE): In many situations the potential energy of the particle does not depend on time explicitly; the force that acts on it, and hence the potential energy vary with the position of the particle only. The Schrödinger wave equation for such a particle is time independent wave equation. Let ψ(x,t) be the wave function of the matter wave associate with a particle of mass m moving with a velocity v. The differential equation of the wave motion is as follows.


47

)1.(..........1

2

2

22

2

tvx ∂∂

=∂∂ ψψ

The solution of the Eq.(1) as a periodic displacement of time t is ψ(x,t) =ψ0(x) e-iωt …..(2) Where ψ0(x) is the amplitude of the matter wave. Differentiating Eq.2 partially twice w.r.t. to t, we get

ωψ i

t−=

∂∂ ψ0(x) e-iωt

222

2

ωψ it

=∂∂ ψ0(x) e-iωt

2

2

2

ωψ−=

∂∂

t ψ0(x) e-iωt

=

∂∂

2

2

tψ - 2ω ψ …… ...(3)

Substituting Eq.3 in Eq.1 2 2

2 2x vψ ω ψ∂

= −∂

…(4)

We have 2

22

22

2 42λπ

λπω

=

== k

v

Substituting this in Eq4, we get

)5.(..........42

2

2

2

ψλπψ

−=∂∂

x

)6.(..........04

2

2

2

2

=+∂∂ ψ

λπψ

x Substituting the wavelength of the matter waves λ=h/mv in Eq.6 we get

)7.(..........04 2

222

2

2

=+∂∂ ψπψ

hvm

x If E and U are the total and potential energies of the particle respectively, then the kinetic energy Ek of the particle

UEmvEk −== 2

21

)(222 UEmvm −=∴ Substituting this in Eq.7, we get

)8.(..........0)(82

2

2

2

=−+∂∂ ψπψ UE

hm

x Hence ψ is a function of x alone and is independent of time. This equation is called the Schrödinger time- independent one dimensional wave equation.


48

Physical significance of the wave function: The wave function ψ(x, t) is the solution of Schrödinger wave equation. It gives a quantum-mechanically complete description of the behavior of a moving particle. The wave function ψ cannot be measured directly by any physical experiment. However, for a given ψ, knowledge of usual dynamic variables, such as position, momentum, kinetic energy, etc., of the particle is obtained by performing suitable mathematical operations on it. The most important property of ψ is that it gives a measure of the probability of finding a particle at a particular position.

ψ is also called the probability amplitude. In general ψ is a

complex quantity, whereas the probability must be real and positive. Therefore a term called probability density is defined. The probability density P (x,t) is a product of the wave function ψ and its complex conjugate ψ*.

*( , )P x t ψψ∴ = =

2),( txψ

Normalization of wave function If ψ is a wave function associated with a particle, then τψ d2 is the probability of finding the particle in a small volume dτ. If it is certain that the particle is present in a volume τ then the total probability in the volume τ is unity i.e., ∫ =

ττψ 12d . This is the normalization

condition. In one dimension the normalization condition is ∫ =

xdx 12ψ

Note: When the particle is bound to a limited region the probability of finding the particle at infinity is zero i.e., ∞=at x *ψψ is zero. Properties of wave function: The wave function ψ should satisfy the following properties to describe the characteristics of matter waves. 1. ψ must be a solution of Schrödinger wave equation. 2. The wave function ψ should be continuous and single valued everywhere. Because it is

related to the probability of finding a particle at a given position at a given time, which will have only one value.

3. The first derivative of ψ with respect to its variables should be continuous and single valued everywhere, because ψ is finite.

4. Ψ must be normalized so that ψ must go to 0 as x ±∞→ , so that τψ d∫ 2 over all the

space be a finite constant.

Eigen functions and Eigen values: The Schrödinger wave equation is a second order partial differential equation; it will have many mathematically possible solutions (ψ). All mathematically possible solutions are not physically acceptable solutions. The physically acceptable solutions are called Eigen functions (ψ).


49

The acceptable wave functions ψ has to satisfy the following conditions: 1. ψ is single valued. 2. ψ and its first derivative with respect to its variable are continuous everywhere. 3. ψ is finite everywhere. Once the Eigen functions are known, they can be used in Schrödinger wave equation to evaluate the physically measurable quantities like energy, momentum etc., these values are

called Eigen values. In an operator equation λψψ =^O where

^O is an operator for the

physical quantity and ψ is an Eigen function and λ is the Eigen value. For example:

ψψ EH =∧

Where H is the total energy (Hamiltonian) operator, ψ is the Eigen function and E is the total energy in the system. Similarly other eigen value equations are :

ψψ pP =∧

Where P stands for the momentum operator and p stands for momentum eigen value. Similarly we have

ψψ xx =∧

Where x stands for position operator and position eigen value. Applications of Schrodinger’s wave equation: 1. For a Particle in an one-dimensional potential well of infinite depth (Particle in a box) Consider a potential energy landscape extending from x = 0 to x = a, where it is throughout constant and we therefore set it to zero for our convenience. Outside this region, the potential energy U is infinity. This type of landscape can be represented by the figure below. Consider a particle of mass ‘m’ moving freely in x- direction in the region from x=0 to x=a. Outside this region potential energy ‘U’ is infinity and within this region U=0. As the particle cannot exist outside this region, due to infinitely high potential energy, we take ψ = 0 outside this region/box. Inside the box U = 0, hence the Schrodinger’s equation is given by,

)1....(..........0

08

22

2

2

2

2

2

=Ψ+∂Ψ∂

=Ψ+∂Ψ∂

kx

Eh

mx

π

where, )2..(..........82

22

hEmk π

=

Discussion of the solution The solution of the above equation is given by )3(..........sincos kxBkxA +=Ψ where A & B are constants which depending on the boundary conditions of the well. Now apply boundary conditions for this,

V=0

x x=0 x=a

U=


50

Condition: I at x =0, U = ∞, thus the particle is not present. Hence ψ = 0. Substituting the condition I in the equation 4, we get A =0 and B≠ 0. (If B is also zero for all values of x, ψ is zero. This means that the particle is not present in the well.) Now the equation 3 can be written as )4(..........sin kxB=Ψ Condition: II at x =a, U = ∞, thus the particle is not present. Hence ψ = 0 Substituting the condition II in equation 5 we get 0= B sin(ka)

2

222

0sin,0 Since

ank

ank

nkaka

B

π

ππ

=

=

=∴=≠

where, n = 1,2,3…………….

Substitute the value of k2 in equation (3).

)5......(..........

8

8

2

22

2

22

2

2

mahnE

an

hEm

=

=ππ

The equation (5) gives energy values or Eigen value of the particle in the well. When n=0, from eqn (4) for all values of x between x = 0 to x = a, ψn = 0. This would imply that the particle is not present inside the box, which is not true. Hence the lowest physically allowed value of ‘n’ for this problem is 1. ∴ The lowest energy corresponds to ‘n’=1 is called the zero-point energy or Ground state

energy. 2

2

int 8mahE pozero =−

All the states of n > 1 are called excited states. To evaluate B in equation (3), one has to perform normalization of wave function. Normalization of wave function:

Consider the equation, xkxBa

nBsin sin πψ ==

The total probability of finding the particle inside the well is unity because there is only one particle within the box.

∫ =Ψa

dx0

2 1

2 2

0

sin 1a nB xdx

aπ

=∫

But ( )θθ 2cos121sin 2 −=


51

aB

aB

nnaaB

axn

naxB

dxa

xndxB

dxa

xnB

a

a a

a

2

12

102sin22

12sin22

12cos2

12cos12

2

20

2

0 0

2

0

2

=∴

=∴

=

−−

=

−

=

−

=

−

∫ ∫

∫

ππ

ππ

π

π

Thus the normalized wave function of a particle in a one-dimensional box is given by,

xa

nan

=Ψπsin2 where, n=1,2,3……………

This equation gives the Eigen functions of the particle in the box. The Eigen functions for n=1,2,3.. are as follows.

xaa

xaa

xaa

=Ψ

=Ψ

=Ψ

π

π

π

3sin2

2sin2

sin2

3

2

1

Since the particle in a box is a quantum mechanical problem we need to evaluate the most probable location of the particle in a box and its energies at different permitted state. Let us consider first three cases Case (1): n=1

This is the ground state and the particle is normally found in this state. For n=1, the Eigen function is

xaa

=Ψπsin2

1

At x = 0 , a ψ = 0 and |ψ|2 = 0

At x = a/2, ψ = a2 and |ψ|2 = 2/a


52

The plot of Ψ1, the wave function/amplitude and 2

1Ψ the probability density versus ‘x’ is as

shown above. From the figure, it indicates the probability of finding the particle at different locations inside the box. This means that in the ground state the particle cannot be found at the walls of the box and the probability of finding the particle is maximum at the central region. The Energy in the ground

state is given by 2

2

1 8mahE = .

Case 2: n =2 This is the first excited state. The Eigen function for this state is given by

xaa

=Ψπ2sin2

2

At x = 0 , a/2, a ψ = 0 and |ψ|2 = 0

At x = a/4 ψ = a2 and |ψ|2 = 2/a

At x = 3a/4 ψ = -a2 and |ψ|2 = 2/a

These facts are seen in the following plot. This means that in the first excited state the particle cannot be observed either at the walls or at the center. The energy is 12 4EE = . Thus the energy in the first excited state is 4 times the zero point energy.

Ψ1

x=0 x=a/2 x=a

√(2/a)

U→ ∞

|Ψ1|2

x=0 x=a/2 x=a

2/a

U→ ∞

3a/4 |Ψ2|2

2/a

U→ ∞

Ψ2

√(2/a)

U→ ∞

a/4

3a/4 a/4


53

Case 3: n =3 This is the second excited state and the Eigen function for this state is given by

xaa

=Ψπ3sin2

3

At x = 0 , a/3, 2a/3,a ψ = 0 and |ψ|2 = 0

At x = a/6, 5a/6 ψ = a2 and |ψ|2 = 2/a

At x = a/2 ψ = -a2 and |ψ|2 = 2/a

The energy corresponds to second excited state is given by 13 9EE = . Note: The particle in an infinite well/box, closely resembles the real life problem of electrons trapped in the attractive force field of the positively charged nucleus. The energy levels in which the electrons can reside are quantized as we know! Further there is not one potential well but an array of such wells which are stacked periodically, with the periodicity given by the details of the material that we are considering. The height of the barrier is not infinity in real life as otherwise it would be impossible to get free/itinerant electrons as seen in metals. 2. Free Particle: Free particle means, it is not under the influence of any kind of field or force. Thus, it has zero potential energy, i.e., U = 0 over the entire region. Hence Schrodinger’s equation becomes,

ψπ

π

Exm

h

Eh

mx

=

∂Ψ∂

−

=Ψ+∂Ψ∂

2

2

2

2

2

2

2

2

8

08

∂∂

− 2

2

2

2

8 xmhπ

is nothing but the Kinetic energy operator which acts on ψ to give the total

energy E times ψ, as the potential energy is zero throughout. Further there is no restriction on the value of momentum or energy as there are no boundary conditions setting ψ = 0 anywhere. Hence all energy values are allowed, and there is no quantization of energy levels.

Ψ3

√(2/a)

U→ ∞

a/6 5a/6

a/3

a/2

2a/3

|Ψ3|2

2/a

U→ ∞

a/6 5a/6

a/3

a/2

2a/3


54

Q.No Sample Questions CO 1. What is Wien’s law and Rayleigh Jeans law and discuss the limitations. How

does Planck’s law extrapolate between the two? 2

2. State De Broglie hypothesis. 1 3. What is wave function? Give its physical significance and properties. 1 4. With a neat labeled diagram, explain the construction and working of Scanning

Electron Microscope. 1&2

5. State Heisenberg’s uncertainty principle. By applying Heisenberg’s uncertainty principle, illustrate the broadening of spectral lines.

1&2

6. State Heisenberg’s uncertainty principle. Using Heisenberg’s uncertainty principle, prove the non-existence of electron within the nucleus of an atom.

1&2

7. Using Heisenberg’s uncertainty principle explain the broadening of spectral lines.

2

8. What are Eigen functions? Mention their properties. 1 9. Setup time independent one-dimensional Schrodinger’s wave equation for a

matter wave. 1

10. Apply the time independent Schrodinger’s wave equation to find the solutions for a particle in an infinite potential well of width ‘a’. Hence obtain normalized wave function

1&2

11. Solve the Schrodinger’s wave equation for a free particle. 1&2 P No. Problems CO

1. Calculate the de Broglie wavelength associated with a proton moving with a velocity equal to (1/20)th of the velocity of light. To be found: de Broglie wavelength, λ Solution:

3414

27 8

6.625 10 2.64 101.67 10 (1/ 20) 3 10

h mmv

λ−

−

−

×= = = ×

× × × ×

3

2. An electron and a proton are accelerated through the same potential difference. Find the ratio of their de Broglie wavelengths. To be found: Ratio of de Broglie wavelength, λ Solution: De Broglie Wavelength, λ

2hmE

λ = , 1m

λα∴

For electron, 1e

emλ α For proton, 1

p

pmλ α

Ratio of De Broglie Wavelengths, pe

p e

m

mλλ

∴ =

3


55

3. Compare the energy of a photon with that of a neutron when both are associated with wavelength of 1Ao. Given that the mass of neutron is 1.67 ×10-27kg. To be found: Comparison of energy of photon with that of neutron Solution: Energy of neutron,

2 2 34 221

2 10 2 27

(6.625 10 ) 13.1 102 2 2 (1 10 ) 1.67 10n

P hE Jm mλ

−−

− −

×= = = = ×

× × × ×=0.082eV

Energy of photon, 34 8

1610

6.625 10 3 10 19.89 101 10p

hcE Jλ

−−

−

× × ×= = = ×

×=12421.9eV

Ratio of energies, 51.5148 10p

n

EE

= ×

3

4. An electron has a speed of 4.8 x 105 m/s accurate to 0.012 %. With what accuracy with which its position can be located. To be found: Uncertainty in position, Δx Solution:

Uncertainty principle is given by,4hx pπ

∆ ∆ ≥

Uncertainty in speed, Δv = 4.8 x 105 x 0.012100

= 57.6m/s

Uncertainty in position,34

631

6.625 10 1 104 4 3.14 9.1 10 57.6

hx mm vπ

−−

−

×∆ = = = ×

∆ × × × ×

3

5. The inherent uncertainty in the measurement of time spent by Iridium-191 nuclei in the excited state is found to be 1.4x10-10 s. Estimate the uncertainty that results in its energy in the excited state. To be found: Uncertainty in energy, ΔE

Solution: Uncertainty principle is given by,4hE tπ

∆ ∆ ≥

34

2510

6.625 10 3.77 104 4 3.14 1.4 10

hE Jtπ

−−

−

×∴∆ ≥ ≥ ≥ ×

∆ × × ×

3

6. The position and momentum of 1 keV electron are simultaneously determined and if its position is located within 1Å. What is the percentage of uncertainty in its momentum? To be found: Percentage of uncertainty in momentum of electron, Δp

Solution: Uncertainty principle is given by,4hx pπ

∆ ∆ ≥

Uncertainty in momentum,

34

2410

6.625 10 0.53 10 . /4 4 3.14 1 10

hp kg m sxπ

−−

−

×∆ = = = ×

∆ × × ×

Momentum, p = 16 232 2 9.1 10 31 1 10 1.707 10 /mE kgm s− −= × × − × × = ×

Percentage of uncertainty in momentum of electron, 100 3.1pp∆

× =

3


56

7. Show that the energy Eigen value of a particle in second excited state is equal to 9 times the zero point energy. To be found: Energy Eigen value for second excited state is equal to 9 times the zero point energy.

Solution: Energy Eigen value equation is given by,2 2

28n hEma

=

n=1, zero-point energy state, 2

1 28hEma

=

n=3, second excited state, 2

3 12

9 98

hE Ema

= =

3

8. An electron is bound in a one-dimensional potential well of width 1Å, but of infinite height. Find the energy value for the electron in the ground state. To be found: Energy Eigen value Solution:

Energy Eigen value equation is given by, 2 2

28n hEma

=

For n=1, ground state energy,

( )( )

234

1 231 10

6.625 1037.65

8 9.1 10 10E eV

−

− −

×= =

× × ×

3

9. An electron is bound in one dimensional potential well of infinite potential of width 0.12 nm. Find the energy values in the ground state and also the first two exited state. To be found: Energy Eigen value Solution:

Energy Eigen value equation is given by, 2 2

28n hEma

=

For n=1, ground state energy, ( )

( )

234

1 231 9

6.625 1026.16

8 9.1 10 0.12 10E eV

−

− −

×= =

× × × ×

For n=2,first excited state, For n=3,second excited state,

2 14 104.61E E eV= = 3 19 235.44E E eV= =

3

10. An electron is trapped in a potential well of a width 0.5nm. If a transition takes place from the first excited state to the ground state find the wavelength of the photon emitted. To be found: Wavelength of the photon emitted, λ Solution: For n=1, ground state energy,

( )( )

23419

1 231 9

6.625 102.41 10 1.507

8 9.1 10 0.5 10E J eV

−

−

− −

×= = × =

× × × ×

3


57

For n=2,first excited state

192 14 9.64 10 6.025E E J eV−= = × =

Energy difference,

192 1 – 7.23 10 4.518E E E J eV−∆ = = × =

Wavelength of the photon emitted , 34

1

8

9

6.625 10 3 17.

0 223 10

74.8hc nmE J

λ−

−×× × ×

= = =∆

Exercise:

1. Taking Planck’s law as the starting point try to derive Wien’s law and Rayleigh –Jeans law in the two limits in which they are valid.

2. How does the concept of superposition of waves leading to a wave packet naturally reconcile the dual wave and particle nature of matter?

3. Is there any connection between the position – momentum and energy-time uncertainty relationship? Discuss.

4. Is the derivation for particle in a box done above valid for highly energetic relativistic particles? Discuss.


58

UNIT –III: OSCILLATIONS AND WAVES

59

UNIT – III: OSCILLATIONS AND WAVES Introduction: Motion of bodies can be broadly classified into three categories:

[1] Translational motion [2] Rotational motion [3] Vibrational / Oscillatory motion

Translational motion: When the position of a body varies linearly with time, such a motion is termed as translational motion. Example: A car moving on a straight road, a ball moving on the ground. Rotational motion: When a body as a whole does not change its position linearly with time but rotates about its axis, this motion is said to be rotational motion. Example: rotation of earth about its axis, rotation of a fly wheel on ball bearings. Vibrational /Oscillatory motion: When a body executes back and forth motion which repeats over and again about a mean position, then the body is said to have Vibrational/oscillatory motion. If such motion repeats in a regular interval of time then it is called Periodic motion or Harmonic motion and the body executing such motion is called harmonic oscillator. In harmonic motion there is a linear relation between force acting on the body and displacement produced. Example: bob of a pendulum clock, motion of prongs of tuning fork, motion of balance wheel of a watch, the up and down motion of a mass attached to a spring. Note: 1. If there is no linear relation between the force and displacement then the motion is called

un harmonic motion. Many systems are un harmonic in nature. 2. In some oscillatory systems the bodies may be at rest but the physical properties of the

system may undergo changes in oscillatory manner Examples: Variation of pressure in sound waves, variation of electric and magnetic fields in electromagnetic waves.

Parameters of an oscillatory system

1. Mean position: The position of the oscillating body at rest.

2. Amplitude: The amplitude of an SHM is the maximum displacement of the body from its mean position.

3. Time Period: The time interval during which the oscillation repeats itself is called the time period. It is denoted by T and its unit in seconds.

DisplacementPeriod = T= 2π Acceleration

4. Frequency: The number of oscillations that a body completes in one second is called

the frequency of periodic motion. It is the reciprocal of the time period T and it is denoted by n .

1Frequency= n =T


60

5. Phase: It is the physical quantity that expresses the instantaneous position and direction of motion of an oscillating system. If the SHM is represented by y=A sin (ωt+φ ), together with ω as the angular frequency. The quantity (ωt+φ ) of the sine function is called the total phase of the motion at time ‘t’ and ‘φ ’ is the initial phase or epoch.

All periodic motions are not vibratory or oscillatory. In this chapter we shall study the simplest vibratory motion along one dimension called Simple Harmonic Motion (SHM) which usually occurs in mechanical systems. SIMPLE HARMONIC MOTION (SHM) A body is said to be undergoing Simple Harmonic Motion (SHM) when the acceleration of the body is always proportional to its displacement and is directed towards its equilibrium or mean position. A particle or a system which executes simple harmonic motion is called Simple Harmonic Oscillator. Examples of simple harmonic motion are motion of the bob of a simple pendulum, motion of a point mass fastened to spring, motion of prongs of tuning fork etc., TYPES OF SHM Simple harmonic motion can be broadly classified in to two types, namely linear simple harmonic motion and angular simple harmonic motion. Linear Simple Harmonic Motion: If the body executing SHM has a linear acceleration then the motion of the body is linear simple harmonic motion. Examples: motion of simple pendulum, the motion of a point mass tied with a spring etc., Angular Simple Harmonic Motion: If the body executing SHM has an angular acceleration then the motion of the body is angular simple harmonic motion. Examples: Oscillations of a torsional pendulum

(1) UN DAMPED OR FREE VIBRATIONS If a body oscillates without the influence of any external force then the oscillations are called Free oscillations or un damped oscillations. In free oscillations the body oscillates with its natural frequency and the amplitude remains constant (Fig.1)

Figure 1: Free Vibrations

In practice it is not possible to eliminate friction completely .Actually the amplitude of the vibrating body decreases to zero as a result of friction. Hence, practical examples for free oscillation/vibrations are those in which the friction in the system is negligibly small.


61

DIFFERENTIAL EQUATION OF FREE OSCILLATIONS Consider a block of mass m suspended from a rigid support through a mass less spring. The restoring force produced in the spring obeys Hooke’s Law. Hooke’s Law: “The restoring force produced in a system is proportional to the displacement”. When the mass is displaced through a distance ‘y’ then the restoring force (F) produced is

F = − k y --------- (1)

The constant k is called force constant or spring constant. It is a measure of the stiffness of the spring. The negative sign in equation (1) indicates that the restoring force is in the direction of its equilibrium position.

Figure 2: A one-dimensional harmonic oscillator.

In equilibrium condition the linear restoring force (F) in magnitude is equal to the weight ‘mg’ of the hanging mass, shown in figure 2.

i.e, F = mgmg = ky

mgand k = (2)y

∴

−− − − − − −

Now, if the load is displaced down through a distance ‘y’ from its equilibrium position and released then it executes oscillatory motion.

2

2dy d yVelocity of the body = v = and Acceleration of the body = a =dt d t

From Newton’s second law of motion for the block we can write

2

2

2

2

2

2

2

22

2

d yF = ma = m (3)dt

from equation (1) substitue F= -ky

d y-ky = mdt

d y kor + y = 0

Whereωis the angular natural frequency.

dt mkSubstitue =ω , m

i.e., y=0----------(4)d ydt

ω

−− − − − − − − − − −

+


62

Equation (4) is the general differential equation for the free oscillator. The mathematical solution of the equation (4), y(t) represents the position as a function of time t. Let y(t) = A sin (ωt +φ ) be the solution. y(t) = Asin(ωt +φ ) -------------------- (5) Where A is the amplitude and (ωt +φ ) is called the phase, φ is the initial phase (i.e., phase at t=0). Period of the oscillator (T)

At equilibrium, magnitude of restoring force F equal to the weight ‘

We have F = ma, also F = ky( )is

ma = kykyAcceleration = a = m

DisplacementPer

mg’of the hanging mass

iod = T = 2πAcceleration

yT=2π k

.∴

2m 1 2π= 2π = 2π = y k ωω

m

2π m Period of the oscillator = T= =2πω k

∴

Although the position and velocity of the oscillator are continuously changing, the total energy E remains constant and is given by

2 21 1 (6)2 2

E mv ky= + − − − − − − − −

The two terms in (6) are the kinetic and potential energies, respectively. VELOCITY AND ACCELERATION We can find velocity from the expression for displacement, y =A sin (ωt+φ ) Velocity in SHM The rate of change of displacement is the velocity of the vibrating particle. By differentiating the above expression with time, we get


63

( )

2

2

2

2 2

dyv = = dt

yv = Aω 1-sin (ωt+ ) As sin(ωt+ )=A

yv

Aωcos ωt

= Aω 1-A

v=ω A y

+

-

φ φ

φ

∴ This is the expression for velocity of the particle at any displacement ‘y’. The maximum velocity is obtained by substituting y = 0, ∴ vmax = Aω Since y = 0 corresponds to its mean position, the particle has maximum velocity when it is at its mean position. At the maximum displacement, i.e., at the extreme position y = A of the particle, the velocity is zero. Acceleration in SHM The rate of change of velocity is the acceleration of the vibrating particle. Differentiating velocity expression with respect to time t, we get acceleration

( )2

2

2

2

2

2Aω sin ωd ya= = -dtd

t+

ya= =t

y -ωd

φ

The above equation gives acceleration of the oscillating particle at any displacement. This equation is the standard equation of SHM. Since y=0 corresponds to its mean position, the particle has minimum acceleration. At the maximum displacement, i.e., at the extreme position y=A, maximum acceleration is 2 Aω− Graphical representation of displacement, velocity and acceleration

( )

( )

( )2

22

y = Asin ωt+

dyvelocity v=

Let the displace

= Aωcos ωt+

ment of a particle executing SHM be

dtd yand Accleration a= =-Aω sin ωt+dt

φ

φ

φ

∴


64

Table 1: Displacement, Velocity and Acceleration for various values of time ‘t’

Time ωt Displacement

( )Asin ωt φ+ Velocity

( )cos ωtAω φ+ Acceleration

( )2sin ωtAω φ− +

t = 0 0 0 +Aω 0 Tt = 4

2π +A 0 2Aω−

Tt = 2

π 0 - Aω 0 3Tt = 4

32π -A 0 2Aω+

t = T 2π 0 + Aω 0

The variation of displacement, velocity and acceleration with time‘t’ is shown graphically:

Energy in simple harmonic motion The total energy (E) of an oscillating particle is equal to the sum of its kinetic energy and potential energy under ideal conditions (no friction). The velocity of a particle executing SHM at a position where its displacement is y from its mean position is 2 2v = ω A -y

Kinetic energy: Kinetic energy of the particle of mass m is

22 2

2 2 2

1K = m ω A -y21K = mω A -y -- (1)2

− − − − − − − − − −


65

Potential energy: Let F be the restoring force and dy be displacement against the restoring force. Let dw be the work done by the force during the small displacement dy.

dw = -F.dy = -(-ky) dy = ky dy

∴ Total work done for the displacement y is given by

( )

y

0y

2 2

0

2 2

W = dW = kydy

W = mω ydy k=mω

1 w = mω y (2)2

∴ −−− − − − − − − −

∫ ∫

∫

This work done is stored in the body as potential energy 2 21U = mω y2

Total energy 2 2 2 2 2 2 21 1 1mω A - y + mω y = m ω A2 2 2

E K U = + =

Thus we find that the total energy of a particle executing simple harmonic motion is 2 21 mω A2

Special cases: (i) When the particle is at the mean position y = 0, from eqn (1) the kinetic energy is maximum and from eqn. (2) the potential energy is zero. Hence the total energy is wholly kinetic

2 2max

1E = K = m ω A2

(ii) When the particle is at the extreme position y = +A, from eqn. (1) the kinetic energy is zero and from eqn. (2) the Potential energy is maximum. Hence the total energy is wholly potential.

2 2max

1E = U = m ω A2

The variation of energy of an oscillating particle with the displacement can be represented in a graph as shown in the Figure 3.

Figure 3. Variation of KE and PE with respect to displacement


66

DIFFERENTIAL EQUATION OF ANGULAR SHM AND ITS SOLUTION EXAMPLE: TORSIONAL PENDULUM Simple harmonic motion can also be angular. In this case, the restoring torque required for producing SHM is directly proportional to the angular displacement and is directed towards its mean position.

A heavy metal disc suspended by means of a wire from a rigid support, constitutes a Torsional Pendulum (Fig. 4).When the wire is twisted by an angle θ, a restoring torque/couple acts on it, tending it to return to its mean position.

Restoring torque or couple (τ) = − C θ ……….. (1)

Where, C is the torsional constant. It is equal to the moment of the couple required to produce unit angular displacement. Its unit is Nmrad−1. The negative sign indicates that the torque is acting in the opposite direction to the angular displacement.

The angular acceleration produced by the restoring couple in the wire, α = 2

2

dtd θ and I be the

moment of inertia of the metal disc about the axis of the wire. According to Newton’s second law, the equation of torque or couple acting on it is given by

τ = I α = I 2

2 (2)ddtθ− − − − − −

Therefore, on equating (1) and (2) we have, I = -Cθ

We get angular acceleration, = CI

− θ ,

22 2

2,C dSubstituting wehaveI dt

θω ω θ= = −

The above relation shows that the angular acceleration is proportional to angular displacement and is always directed towards its mean position. Therefore, the motion of the disc is always simple harmonic motion (SHM).

2

22 0 (3)d

dtθ ω θ+ = − − − − − − − − − − − −

This is the differential equation of angular simple harmonic motion of the Torsional Pendulum. Therefore, the time period and frequency of oscillator is given by relation,

Period 2π I CT= = 2π Where ω=ω C I

and Frequency 1 12

CnT Iπ

= =

Note: For a given wire of length l, radius r and rigidity modulus η 4

Torsionalconstant of the wire 2

rCl

πη=

2

2

dtd θ

2

2

dtd θ

Figure 4: Torsional Pendulum


67

(2) DAMPED VIBRATIONS Free vibrations are oscillations in which the friction/resistance considered is zero or negligible. Therefore, the body will keep on vibrating indefinitely with respect to time. In real sense if a body set into vibrations, its amplitude will be continuously decreasing due to friction/resistance and so the vibrations will die after some time. Such vibrations are called damped vibrations (Fig. 5).

Figure 5: Damped vibrations

DIFFERENTIAL EQUATION OF DAMPED VIBRATIONS AND SOLUTION A block of mass m connected to the free end of a spring is partially immersed in a liquid (Fig. 6) and is subjected to small vibrations. The damping force encountered is more when the block moves in the liquid and hence the amplitude of vibration decreases with time and ultimately stops vibrating; this is due to energy loss from viscous forces. Let y to be the displacement of the body from the equilibrium state at any instant of time ‘t’, then dy/dt is the instantaneous velocity.

Figure 6: Spring and mass system The two forces acting on the body at this instant are: i. A restoring force which is proportional to displacement and acts in the opposite direction,

it may be written as Frestoring = −ky

ii. A frictional or damping force is directly proportional to the velocity and is oppositely

directed to the motion, it may be written as

damping

dyF rdt

= −


68

The net force on an oscillator subjected to a linear damping force that is linear in velocity is simply the sum is thus given by

restoring

2

d

22

2

2

2

amp

2

2

ingdy= -ky -rdt

d yBut by Newton's law of motion, F = mdt

d ywheremis themassof thebod

F= F

y and is the accleration of the body.dt

d y dyThen , m =-ky-rdtdt

d y r dy kor + + y=0-----------(1)m dt mdt

+ F

Let r

m=2b and 2k

mω= , then the above equation takes the form,

22

2 2 y=0d y dybdtdt

ω+ + ------------------(2)

This is the differential equation of second order. In order to solve this equation, we assume its solution as

α ty = Ae --------------------(3) Where A and α are constants. Differentiating equation (3) with respect to time t, we have

2α t 2 α t

2dy d = Aαe and = Aα ed t d t

y

By substituting these values in equation (2), we have

( )

2 2

2 2

2 A =0or

2 0

t t t

t

A e bA e e

Ae b

α α α

α

α α ω

α α ω

+ +

+ + =

For the above equation to be satisfied, either y=0, or ( )2 22 0bα α ω+ + = Since y=0, corresponds to a trivial solution, one has to consider the solution

( )2 22 0bα α ω+ + = The standard solution of the above quadratic equation has two roots, it is given by,

2 2 2

2 2

4 4 42

b b

b b

ωα

α ω

− ± −=

=− ± −

Therefore, the general solution of equation (2) is given by 2 2 2 2( ) ( ) (4)b b t b b ty C e D eω ω− + − − − −

= + − − − − − Where C and D are constants, the actual solution depends upon whether 2 2b ω> , 2 2b ω= and

2 2b ω< .


69

Case 1: Heavy damping ( 2 2b ω> ) In this case, 2 2b ω− is real and less than b, therefore in equation (4) both the exponents are negative. It means that the displacement ‘y’ of the particle decreases continuously with time. That is, the particle when once displaced returns to its equilibrium position quite slowly without performing any oscillations (Fig.7). Such a motion is called ‘overdamped’ or ‘aperiodic’ motion. This type of motion is shown by a pendulum moving in thick oil or by a dead beat moving coil galvanometer.

Figure 7: Heavy damping or overdamped motion Case 2: Critical damping ( 2 2b ω= ) By substituting 2 2b ω= in equation (4) the solution does not satisfy equation (2). Hence we

consider the case when 2 2b ω− is not zero but is a very small quantity β. The equation (4) then can be written as

[ ][ ]

( ) ( )

( )sinceβissmall,wecan approximate,

1 1 ,on thebasisof exponentialseriesexpansion(1 ) (1 )

( ) ( ) which is theproduct of terms

b t b t

bt t t

t t

bt

bt

y C e Dey e C e De

e t and e ty e C t D t

y e C D t C D

β β

β β

β ββ ββ β

β

− + − −

− −

−

−

= +

= +

= + = −

= + + −

= + + −

As ‘t’ is present in bte− and also in the term ( )t C Dβ − , both of them contribute to the variation of y with respect to time. But by the virtue of having t in the exponent, the term bte− predominantly contributes to the equation. Only for small values of t, the term ( )t C Dβ −contributes in a magnitude comparable to that of bte− . Therefore, though y decreases throughout with increase of t, the decrement is slow in the beginning, and then decreases rapidly to approach the value zero. i.e., the body attains equilibrium position (Fig.8). Such damping motion of a body is called critical damping. This type of motion is exhibited by many pointer instruments such as voltmeter, ammeter...etc in which the pointer moves to the correct position and comes to rest without any oscillation.

Figure 8: Critical damping motion.


70

Case 3: Low damping ( 2 2b ω< )

This is the actual case of damped harmonic oscillator. In this case 2 2b ω− is imaginary. Let

us write

[ ]

2 2 2 2

2 2

(-b + iβ )t (-b - iβ )t

-bt iβ t -iβ t

-bt

-bt

b -ω =i ω -b =iβ

where β = ω -b andi= -1.Then,equation (4) now becomesy = Ce +Dey = e (Ce +De )y = e (C(cosβ t+isinβ t)+D(cosβ t-isinβ t))y = e (C+D)cosβ t+i(C-D)sinβ t)RewritingC+D=Asi

′ ′

′ ′

′

′

′ ′ ′ ′

′ ′

[ ]-bt

-bt

n and i(C-D)=Acos ,

WhereAand areconstants.y = e Asin cosβ t+Acos sinβ t

y= Ae sin(β t+ )

φ φ

φφ φ

φ

′ ′

′

This equation sin( )bty Ae tβ φ− ′= + represents the damped harmonic oscillations. The oscillations are not simple harmonic because the amplitude ( btAe− ) is not constant and decreases with time (t). However, the decay of amplitude depends upon the damping factor b. This motion is known as under damped motion (Fig.9). The motion of pendulum in air and the motion of ballistic coil galvanometer are few of the examples of this case.

Figure 9: Under damped motion

The time period of damped harmonic oscillator is given by

2 2 2

2

2π 2π 2πT= =β ω -b k r-

m 4m′

The frequency of damped harmonic oscillator is given by

2

2

1 1n =2 4

k rT m mπ= −


71

(3) FORCED VIBRATIONS In the case of damped vibrations, the amplitude of vibrations decrease with the time exponentially due to dissipation of energy and the body eventually comes to a rest. When a body experiences vibrations due to the influence of an external driving force the body can continue its vibration without coming to a rest. Such vibrations are called forced vibrations.

For example: When a tuning fork is struck on a rubber pad and its stem is placed on a table, the table is set in vibrations with the frequency of the fork. These oscillations of the table are the forced oscillations/vibrations. So forced vibrations can also be defined as the vibrations in which the body vibrates with frequency other than natural frequency of the body, and they are due to applied external periodic force. DIFFERENTIAL EQUATIONS OF FORCED VIBRATIONS AND SOLUTIONS Suppose a particle of mass m is connected to a spring. When it is displaced and released it starts oscillating about a mean position. The particle is driven by external periodic force

. The oscillations experiences different kinds of forces viz,

1) A restoring force proportional to the displacement but oppositely directed, is given by Frestoring = -ky, where k is known as force constant.

2) A frictional or damping force proportional to velocity but oppositely directed, is given by

F damping = - dyrdt

, where r is the frictional force/unit velocity.

3) The applied external periodic force is represented by , where Fo is the maximum value of the force and ωd is the angular frequency of the driving frequency.

The total force acting on the particle is given by,

net external restore damp

2

2

2

o2

F = F + F

r

+F

dy= F sinω t - r - kyd dt

d yBy Newton's second law of motion F= m .dtd y dyHence,m =F sinω t - r - kyd dtdt

The differentialequation of forced vibrations is given by

2d y dy k+ + y=2 m dt mdt

o

o

2 o

22

d

o

2

Fr kSubstitute = 2b, = ω and = f, then equation (1) becomesm m m

d y dy+2b +ω y=f sinω t-

Where b is the damping coefficient,

----------(2)

F is the amplitude of the e

dtdt

F sinω td -----------(1)m

d

xternal driving force and ω is the angular frequencyof theexternalforce.

( sin )o dF tω

( sin )o dF tω


72

The solution of the above equation (2) is given by dy= Asin(ω t ) (3)φ− − − − − − − − −

Where, A is the amplitude of the forced vibrations. By differentiating equation (3) twice with respect to time t, we get

22

2cos( ) sin( )d d d ddy d yA t and A tdt dt

ω ω φ ω ω φ= − =− −

By substituting the values of y, dy/dt and d2y/dt2 in equation (2), we get

2 2

22

sin ( ) 2 cos ( ) sin ( ) sin ( )

or

( )sin ( ) 2 cos ( ) sin ( )cos cos ( )sin (4)

d d d d d d

d d d d dd

A t b A t A t f t

A t b A t f t f t

ω ω φ ω ω φ ω ω φ ω φ φ

ω ω ω φ ω ω φ ω φ φ ω φ φ

− − + − + − = − +

− − + − = − + − − − − − −

If equation (4) holds good for all values of t, then the coefficients of sin (ωdt- φ) and cos (ωdt- φ) must be equal on both sides, then

2 2( ) cos (5)2 sin (6)

d

d

A fand b A f

ω ω φω φ− = − − − − − − − − −

= − − − − − − − − − By squaring and adding equations (5) and (6), we have

2 2 2 2 2 2 2 2

2 2 2 2 2

( ) 4

(7)( ) 4

d d

d d

A b A f

fAb

ω ω ω

ω ω ω

− + =

= − − − − − − − −− +

By dividing equation (6) by (5), we get

12 2 2 22 2tan , Phase tan (8)d d

d d

b bω ωφ φω ω ω ω

− = = − − − − − − − −

Substituting the value of A in equation (3) we get

2 2 2 2 2sin( )

( ) 4d

d d

fy tb

ω φω ω ω

= −− +

--------------- (9)

This is the solution of the differential equation of the Forced Harmonic Oscillator. From equations (7) and (8), it is clear that the amplitude and phase of the forced oscillations depend upon 2 2( )dω ω− , i.e., these depend upon the driving frequency (ωd) and the natural frequency (ω) of the oscillator. We shall study the behaviour of amplitude and phase in three different stages of frequencies i.e, low frequency, resonant frequency and high frequency.


73

Case I: When driving frequency is low i.e., ( )dω ω<< . In this case, amplitude of the vibrations is given by

( ) ( )

2 2 2 2 2

2 22 2 2 2d d

2 2d d

2o o o2

1 12 2

( ) 4

As ω ω, ω -ω ω =ω

and 4b ω 0 ω 0

f F m F F kA= = = f= and ω =k m k m mω

2tan tan (0) 0

d d

d

d

fAb

band

ω ω ω

ωφω ω

− −

=− +

→

= = = −

This shows that the amplitude of the vibration is independent of frequency of the driving force and is dependent on the magnitude of the driving force and the force-constant (k). In such a case the force and displacement are always in phase. Case II: When dω ω= i.e., frequency of the driving force is equal to the natural frequency of the body. This frequency is called resonant frequency. In this case, amplitude of vibrations is given by

2 2 2 2 2

1 1 12 2

( ) 4

2 ( )

2 2tan tan tan ( )0 2

d d

o o

d d d

d d

d

fAb

f F m FAb r m r

b band

ω ω ω

ω ω ω

ω ω πφω ω

− − −

=− +

= = =

= = = ∞ = −

Under this situation, the amplitude of the vibrations become maximum and is inversely proportional to the damping coefficient. For small damping, the amplitude is large and for large damping, the amplitude is small. The displacement lags behind the force by a phase π/2. Case III: When ( )dω ω>> i.e., the frequency of force is greater than the natural frequency of the body. In this case, amplitude of the vibrations is given by

2 2 2 2 2d d

2 2 2 4d d d

4 2 2 o od d d 2 2 2

d d d

-1 -1 -1d2 2

dd

fA =(ω -ω ) +4b ω

ω ω (ω -ω ) ωf F m Fsinceω is very large,ω 4b ω , A= = =

ω ω mω

2bω 2band = tan = tan - =tan (-0)=πωω -ω

φ

∴

>>> ∴

This shows that amplitude depends on the mass and continuously decreases as the driving frequency ωd is increased and phase difference towards π.


74

RESONANCE: If we bring a vibrating tuning fork near another stationary tuning fork of the same natural frequency as that of vibrating tuning fork, we find that stationary tuning fork also starts vibrating. This phenomenon is known as Resonance. Resonance is a phenomenon in which a body vibrates with its natural frequency with maximum amplitude under the influence of an external vibration with the same frequency. Theory of resonant vibrations: (a) Condition of amplitude resonance. In case of forced vibrations, the expression for amplitude A and phase φ is given by,

2 2 2 2 2( ) 4d d

fAbω ω ω

=− +

and 12 22tan d

d

bωφω ω

− = −

The amplitude expression shows the variation with the frequency of the driving force ωd. For a particular value of ωd, the amplitude becomes maximum. The phenomenon of amplitude becoming a maximum is known as amplitude resonance. The amplitude is maximum when

2 2 2 2 2( ) 4d dbω ω ω− + is minimum. If the damping is small i.e., b is small, the condition of

maximum amplitude reduced to max 2 d

fAbω

= .

(b) Sharpness of the resonance. We have seen that the amplitude of the forced oscillations is maximum when the frequency of the applied force is at resonant frequency. If the frequency changes from this value, the amplitude falls. When the fall in amplitude for a small change from the resonance condition is very large, the resonance is said to be sharp and if the fall in amplitude is small, the resonance is termed as flat. Thus the term sharpness of resonance can be defined as the rate of fall in amplitude, with respect to the change in forcing frequency on either side of the resonant frequency.

Figure 10: Effect of damping on sharpness of resonance


75

Figure 10, shows the variation of amplitude with forcing frequency for different amounts of damping. Curve (1) shows the variation of amplitude when there is no damping i.e., b=0. In this case the amplitude is infinite at dω ω= . This case is never realized in practice due to friction/dissipation forces, as a slight damping factor is always present. Curves (2) and (3) show the variation of amplitude with respect to low and high damping. It can be seen that the resonant peak moves towards the left as the damping factor is increased. It is also observed that the value of amplitude, which is different for different values of b (damping), diminishes as the value of b increases. This indicates that the smaller is the damping, sharper is the resonance or large is the damping, flatter is the resonance. EXAMPLE FOR FORCED VIBRATIONS: SERIES LCR CIRCUIT An L-C-R circuit fed by an alternating emf is a classic example for a forced harmonic oscillator. Consider an electric circuit containing an inductance L, capacitance C and resistance R in series as shown in the figure 11. An alternating emf has been applied to a circuit is represented by sino dE tω .

Figure 11: Series LCR circuit.

Let q be the charge on the capacitor at any instant and I be the current in the circuit at any

instant. The potential difference across the capacitor is qC

, the back emf due to self inductance

in the inductor is dILdt

and, the potential drop across the resistor is IR. The sum of voltages

across the three LCR elements illustrated must be equal the voltage supplied by the source element. Hence the voltage equation at any instant is given by,

L R C S

o d

2

d o2

2

d o d2

2d o

d2

2

2

V +V +V = V (t)

dI qL +IR+ = E sinω tdt C

Differentiating,weget

d I dI 1 dqL +R + =ω E cosdt C dtdt

dq d I dI Ibut = I, L +R + =ω E cosω tdt dt Cdt

d I R dI 1 ω E+ + I= cosω tL dt LC Ldt

d I R dI 1+ +L dt LCdt

dtω

∴

d od

ω E πI= sin ω t+ (1)L 2

− − − −


76

This is the differential equation of the forced oscillations in the electrical circuit. It is similar to equation of motion of a mechanical oscillator driven by an external force.

oFm

2d y dy 2+2b +ω y= sinω td2 dtdt --------------------------(2)

The explicit and precise connection with the mechanical oscillation equation is given below:

Displacement : y I

Velocity : dydt

dIdt

Damping constant : 2b R/L

Natural frequency : ω 1/√LC

External force : F(t) Eo

And Fo/m ωdEo/m

Forced oscillations (refer previous section) LCR Circuit application Starting with equation from forced vibrations, we have

oFm

2d y dy 2+2b +ω y= sinω td2 dtdt

Amplitude of mechanical oscillations/ vibrations is given by

2 2 2 2 2( ) 4o

d d

F mAbω ω ω

=− +

The equivalent equation for LCR circuit is given by

2d o

d2d I R dI 1 ω E π+ + I= sin ω t+

L dt LC L 2dt

Amplitude of current Io is given by

0 2 2 2 2 2

22 2

2

2 22 2

2

( ) 4

1 4 ,

1

do

d d

do

o

d d

ELI

b

RSubstitute for and bLC L

we get

ELI

RLC L

ω

ω ω ω

ω

ω

ω ω

=− +

= =

= − +


77

( )

2

22

2 22

22 22

2 2

In thedenominator,multiply and divde the1term by and ,we get

LC

1

1

d d

do

o

dd d

d

do

o

dd d

d

by L

ELI

L RLC L L

ELI

RLCL L

ω ω

ω

ω ω ωω

ω

ω ω ωω

= − +

=

− +

( ) ( )

22

2 2 22

1

1 (Capacitive reactance)

(Inductive reactance)

do

o

dd

d

Cd

d L

o oo

L CC L

ELI

L RL C

substitute XC

and L X

E EIX X RX X R

ω

ω ωω

ω

ω

=

− +

=

=

= = − +− +

The solution of the equation (1) for the current at any instant in the circuit is of the form

2

2

sin( ) sin( ) (3)1

od o d

dd

EI t I t

R LC

ω φ ω φ

ωω

= − = − − − − − − −

+ −

Electrical Impedance: The ratio of amplitudes of alternating emf and current in ac circuit is

called the electrical impedance of the circuit. o oo

o

E E. ., Z = II Z

i e ∴ =

The amplitude of the current is o oo 2

2

E EI = =Z1R + ω L-

ω Cdd

Impedance of the circuit Z = 2

2 1R + ω L- (4).ω Cd

d

− − − − − − − − −


78

The quantity 1ω L-ω Cd

d

is the net reactance of the circuit which is the difference between

the inductive reactance L(X )=ω Ld and the capacitive reactance C1(X )=

ω C.d.

Equation (3) shows that the current I lags in phase with the applied emf sino dE tω by an

angle φ and is given by 1 1

1

tan tand

d L CL

C X XR R

ωωφ − −

− − = =

The following three cases arise:

1. When XL>XC , φ is positive, that is, the current lags behind the emf. Circuit is inductive.

2. When XL<XC , φ is negative, that is, the current leads behind the emf. Circuit is capacitive.

3. When XL=XC , φ is zero, that is, the current is in phase with the emf. Circuit is resistive.

Electrical Resonance: According to equation (4), the current have its maximum amplitude when

1 1 10 0L C d d dd d

X X or L or L orC C LC

ω ω ωω ω

− = − = = =

Where ωd is the angular frequency of the applied emf, while ω = 1LC

is the (angular)

natural frequency of the circuit. Hence, the maximum amplitude of the current oscillations occurs when the frequency of the applied emf is exactly equal to the natural (un-damped) frequency of the electrical circuit. This is the condition of electrical resonance.

Sharpness of resonance and Bandwidth: When an alternating emf is applied to an LCR circuit, electrical oscillations occur in the circuit with the frequency ωd is equal to the applied emf. The amplitude of these oscillations (current amplitude) in the circuit is given by

22

= , Where is theimpedanceof thecircuit.1

o oo

dd

E EI ZZ

R LC

ωω

=

+ −


79

At resonance, when the frequency ωd of the applied emf is equal to the natural frequency 1LC

ω = of the circuit, the current amplitude Io is maximum and is equal to Eo/R. Thus at

resonance the impedance Z of the circuit is R. At other values of ωd, the current amplitude Io is smaller and the impedance Z is larger than R.

Figure 12: Graphical plot of current vs frequency. The variation of the current amplitude Io with respect to applied emf frequency ωd is shown in the figure 12. Io attains maximum value (Eo/R) when ωd has resonant value ω and decreases as ωd changes from ω. The rapidity with which the current falls from its resonant value (Eo/R) with change in applied frequency is known as the sharpness of resonance. It is measured by the ratio of the resonant frequency ω to the difference of two frequencies ω1 and ω2 taken at

12

of the resonant (ω) value.

i.e., Sharpness of resonance (Q) =2 1

ωω ω−

,

ω1 and ω2 are known as the half power frequencies. The difference of half power frequencies, ω1 - ω2 is known as “band-width”. The smaller is the bandwidth, the sharper is the resonance.

Eo/

ω1 ω ω2


80

SI No

Sample Questions (one mark each) CO

1. What is simple harmonic motion? 1 2. Write the general equation representing SHM. 1 3. List any two characteristics of SHM. 1 4. A particle executes a S.H.M. of period 10 seconds and amplitude of 2 meter.

Calculate its maximum velocity. 3

5. Hydrogen atom has a mass of 1.68x10-27 kg, when attached to a certain massive molecule it oscillates as a classical oscillator with a frequency1014 cycles per second and with amplitude of 10-10 m. Calculate the acceleration of the oscillator.

3

6. A body executes S.H.M such that its velocity at the mean position is 4cm/s and its amplitude is 2cm. Calculate its angular velocity.

3

7. What is free vibration? 1 8. What is damped vibration? 1 9. Give any two examples for damped vibration. 1 10. What is forced vibration? 1 11. Given two vibrating bodies what is the condition for obtaining resonance? 2 12. Explain why a loaded bus is more comfortable than an empty bus? 2 13. What is a simple harmonic oscillator? 1 14. What is torsional oscillation? 1 15. Displacement of a particle of mass 10g executes SHM given by 𝑥 =

15𝑠𝑖𝑛 2𝜋𝑡𝑇

+ 𝜑 and its displacement at t=0 is 3cm where the amplitude isn15cm. Calculate the initial phase of the particle.

3

16. Name the two forces acting on a system executing damped vibration. 2 17. How critical damping is beneficiary in automobiles? 1 18. What is restoring force? 1

Descriptive Questions 1. Every SHM is periodic motion but every periodic motion need not be SHM.

Why? Support your answer with an example. 2

2. What is the phase difference between (i) velocity and acceleration (ii) acceleration and displacement of a particle executing SHM?

1

3. Show graphically the variation of displacement, velocity and acceleration of a particle executing SHM.

2

4. Setup the differential equation for SHM. 1 5. Define the terms (i) time period (ii) frequency and (iii) angular frequency of

oscillations. 1

6. What is phase of SHM? Explain the term phase difference. 1 7. Derive an expression for the time period of a body when it executes angular

SHM. 2

8. Explain the oscillations of a mass attached to a horizontal spring. Hence deduce an expression for its time period.

1

9. Distinguish between linear and angular harmonic oscillator? 2 10. What are damped vibrations? Establish the differential equation of motion 2


81

for a damped harmonic oscillator and obtain an expression for displacement. Discuss the case of heavy damping, critical damping and low damping.

11. What is damping? On what factors the damping depends? 1 12. What is forced vibration? Give an example. 1 13. What do you mean by forced harmonic vibrations? Discuss the vibrations of

a system executing simple harmonic motion when subjected to an external force.

2

14. What is driven harmonic oscillator? How does it differ from simple and damped harmonic oscillator?

2

15. What is resonance? Explain the sharpness of resonance. 2 16. Illustrate an example to show that resonance is disastrous sometimes. 2

SI No

Problems CO

1. A particle executes SHM of period 31.4 second and amplitude 5cm. Calculate its maximum velocity and maximum acceleration. Solution: The maximum velocity at y =0 in 2 2v = ω A -y , ∴ vmax= Aω

max

2 2 0.231.4

0.2 5 1.0 / sec 0.01 /

radianT

v cm m s

π πω = = =

∴ = × = =

At the maximum displacement, i.e., at the extreme position x=A, maximum acceleration is - 2 Aω

2 20.2 5 0.002 /a m s= − × = −

3

2. A circular plate of mass 4kg and diameter 0.10 metre is suspended by a wire which passes through its centre. Find the period of angular oscillations for small displacement if the torque required per unit twist of the wire is 4 x 10-3 N-m/radian. Solution:

Moment of Inertia2 2

24(0.05) 0.0052 2

MRI kgm= = =

Time period is given by 30.0052 2 3.144 10

ITC x

π −= == × × =7.021s.

3


82

3. A mass of 6kg stretches a spring 0.3m from its equilibrium position. The mass is removed and another body of mass 1kg is hanged from the same spring. What would be the period of motion if the spring is now stretched and released? Solution:

F mg 6×9.8 m 1F=ky, k= = = =196N/m and T=2π =2×3.14 =0.45sy y 0.3 k 196

3

4. The particle performing SHM has a mass 2.5g and frequency of vibration 10Hz. It is oscillating with amplitude of 2cm.Calculate the total energy of the particle. Given :Frequency ω=2πn=20πs-1, Amplitude A=2cm=0.02m, mass m=2.5 x 10-3kg Solution: Total energy, = ( ) ( )2 22 2 -3 -31 1E= mω A = 2.5 10 20π 0.02 =1.97 10 J

2 2× × × ×

3

5. A vibrating system of natural frequency 500 cycles/sec, is forced to vibrate with a periodic force/unit mass of amplitude 100 x 10-5 N/kg in the presence of a damping/ unit mass of 0.01 x 10-3 rad/s. Calculate the maximum amplitude of vibration of the system. Given :Natural frequency = 500 cycles/sec, Amplitude of the force / unit mass , Fo/m=100 x 10-5 N/kg, Damping coefficient , r/m = 0.01 x 10-3 rad/s Solution: Maximum amplitude of vibration

5

3

2 ( )

100 10 0.0318 .0.01 10 2 500

Themaximumamplitudeof vibration of the system is 0.0318meter

of F mAb r m

A m

ω ω

π

−

−

= =

×∴ = =

× × × ×

∴

3

6. A circuit has an inductance of 1 π henry and resistance 100Ω. An A. C. supply of 50 cycles is applied to it. Calculate the reactance and impedance offered by the circuit. Solution: The inductive reactance is XL= ωdL =2πnL Here ωd=2πn=2π x 50= 100π rad/sec and L=1 π Henry. XL=ωdL=2πnL =2π x 50x 1 π =100Ω.

The impedance is 2 2 2 2100 100 141.4LZ R X= + = + = Ω

3

7. A series LCR circuit has L=1mH, C=0.1µF and R=10Ω.Calculate the resonant frequency of the circuit. Solution: The resonant angular frequency of the circuit is given by

1LC

ω =

Here L=1mH and C=0.1µF 5

3 7

1 10 /10 10

rad sx

ω− −

= =

3

UNIT –IV: ELECTRICAL PROPERTIES OF METALS AND SEMICONDUCTORS

83

UNIT- IV: ELECTRICAL PROPERTIES OF METALS AND SEMICONDUCTORS

Introduction: Conducting materials play a vital role in Engineering. It is very essential to know the electrical properties of materials for specific application of the materials. The properties of metals such as electrical conduction, thermal conduction, specific heat etc., are due to the free electrons or conduction electrons in metals. The first theory to explain the electrical conductivity of metals is Classical free electron theory and it was proposed by Drude in the year1900 and later developed and refined by Lorentz. Hence classical free electron theory is known as Drude-Lorentz theory.

Assumptions of Classical Free Electron Theory: 1. A metal is imagined as a three dimensional ordered network of positive ions with the

outermost electrons of the metallic atoms freely moving about the solid. The electric current in a metal, due to an applied field, is a consequence of drift velocity of the free electrons in a direction opposite to the direction of the field.

2. The free electrons are treated as equivalent to gas molecules and thus assumed to obey the laws of kinetic theory of gases. As per kinetic theory of gases, in the absence of the field

the energy associated with each electron at a temperature T is 32

kT , where k is

Boltzmann constant. It is related to the kinetic energy through the relation 23 1

2 2 thkT mv=

where vth is the thermal velocity of the electrons. 3. The electric potential due to the ionic core (lattice) is taken to be essentially constant

throughout the metal. 4. The attraction between the free electrons and the lattice ions and the repulsion between the

electrons are considered insignificant. Drift Velocity Initially the electrons in the metal which are in thermal equilibrium will move in random directions and often collide with ions with no net displacement. When electric field is applied, the equilibrium condition is disturbed and there will be net displacement in randomly moving free electron’s positions, with time in a direction opposite to the direction of the field. This displacement per unit time is called drift velocity which will be constant for the free electrons in the steady state. This accounts for the current in the direction of the field.


84

If ‘E’ is the electric field applied to the metal, ‘τ ’ is mean collision time, then drift velocity for conduction electron in a metal is given by

deEvmτ=

where ‘e’ and ‘m’ are charge and mass of electron respectively. Current density (J): It is the current per unit area of cross section of an imaginary plane held normal to the direction of current in a current carrying conductor. i.e. J = I/A where A is the area of cross section. Electric Field (E): Electric field across homogeneous conductor is defined as the potential drop per unit length of the conductor. If ‘L’ is the length of a conductor of uniform cross section and uniform material composition and ‘V’ is the potential difference between its two ends, then electric field ‘E’ is given by E = V/L Mean Free Path (λ ): It is the average distance travelled by the conduction electrons between successive collisions with lattice ions. Mean Collision Time (τ ): It is the average time that elapses between two consecutive collisions of an electron with the lattice ions. Relation between v, τ and λ : If ‘v’ is the total velocity of the electrons, then the mean collision time ‘τ ’ is given by

vλτ =

Resistivity ( ρ ): For a material of uniform cross section, the resistance ‘R’ is directly proportional to length ‘L’ and inversely proportional to area of cross section ‘A’


85

i.e. LR

ALRA

α

ρ =

where ‘ ρ ’ is called resistivity. It is the property of the material and gives the measure of opposition offered by the material during the current flow in it.

RAL

ρ∴ =

Conductivity (σ ): It is reciprocal of resistivity. It is a physical property that characterizes conducting ability of a material.

1 LRA

σρ

= =

Relation between J, σ and E: From ohms law

V=IRAlI ρ.=

ρ1.

lV

AI=

JAI= and σ

ρ=

1

EJ σ= Expression for electric current in a conductor: (I)

dI nev A= n - Number of free electrons in unit volume of the conductor

dv - Drift velocity of electrons A - Area of cross section of the conductor e – Charge of an electron

Conductivity:

dneAVI = m

EeVdτ

=

I=dVA

mneAE

mne

mEeneA .)())((

22 τττ==

RIV

IAlV

IAl

nemV

=

=

=

.

..2

ρτ


86

mne

nem

τρ

σ

τρ

2

2

1==

=

Mobility of electrons: Mobility of electrons (µ ) is defined as the magnitude of drift velocity acquired by the electron in unit field.

i.e. 1dv eE eE

E E m mτµ = = =

Mobility in terms of Electrical conductivity: We have equation from ohm’s law,

µσ

σ

neAE

AnevAEI

EJEJ

d ====

=

Hence µσ ne= Thermal conductivity of Metals: It is defined as the ratio of flow of heat across a unit cross section of a conductor with a unit temperature gradient between the opposite faces perpendicular to the direction of the heat flow.

It is given by

where negative sign shows that the heat energy flows from the hot end to the cold end. Wiedemann-Franz Law: This law states that the ratio of the thermal conductivity to the electrical conductivity is

directly proportional to the absolute temperature. TK ασ

or tconsT

K tan=σ

, This constant L is

known as Lorentz number. Failures of classical free electron theory: Although electrical and thermal conductivity in metals can be explained successfully through classical free electron theory, it failed to account for many other experimental facts such as specific heat, temperature dependence of ‘σ ’ & dependence of electrical conductivity on electron concentration.

1. The molar specific heat of a gas at constant volume is RCv23

= , where R is a universal

constant. But the experimental value of electronic specific heat is Cv=10-4RT which the classical theory could not explain. Also the experimental value shows that the electronic

Adxd

tQ

Kθ

−=


87

specific heat is temperature dependent, whereas the classical free electron theory says that it is temperature independent.

2. The electrical conductivity of a metal is inversely proportional to temperature. According to classical free electron theory, electrical conductivity is inversely proportional to the

square root of temperature, i.e. T1

∝σ .

3. Electrical conductivity is given as m

ne2τ=σ

According to classical electron theory electrical conductivity is directly proportional to the electron concentration. But monovalent metals like copper found to have high electrical conductivity than the divalent & trivalent metals like Zinc and Aluminium. Hence CFET fails to explain the observation.

4. Though metals are expected to exhibit negative Hall co-efficient since the charge carriers in them are electrons, some metals like zinc have positive Hall co-efficient. The free electron theory could not explain the positive Hall co-efficient of metals.

Assumptions of quantum free electron theory: The main assumptions of quantum free electron theory are 1. The energy values of free electrons are quantized. The allowed energy values are realized

in terms of a set of energy levels. 2. The distribution of electrons in the various allowed energy levels follows Pauli’s

exclusion principle. 3. Distribution of electrons in energy states obey Fermi-Dirac statistics. 4. The free electrons travel in a constant potential inside the metal but stay confined within

its boundaries. 5. The attraction between the free electrons and lattice ions, the repulsion between the

electrons themselves are ignored. Fermi level and Fermi energy: If we assume the number of electrons per unit volume as ne then these electrons should be accommodated in the various energy levels. At absolute zero temperature, the electrons occupy the lowest available energy levels. The highest occupied level in metals at zero Kelvin is called as the Fermi level and the corresponding energy value of that level is called as the Fermi energy, it is denoted by EF. Thus at 0K all levels up to the Fermi level are occupied while the levels above it are vacant.

The dotted level is the Fermi level. Levels from Eo up to EF are occupied while levels above EF are empty.

EF

E0 Energy band


88

Fermi-Dirac statistics: Under thermal equilibrium the free electrons are distributed in various available energy states. The distribution of electrons among the energy levels follows statistical rule known as Fermi-Dirac statistics.

Fermi-Dirac statistics is applicable to fermions. Fermions are indistinguishable particles with half integral spin. Since electron has half spin they obey Fermi-Dirac statistics and they are called Fermions.

Fermi factor represents the probability that a quantum state with energy E is occupied

by an electron, is given by Fermi-Dirac distribution function,

−

+=

kTEE

EfFexp1

1)(

Where k is the Boltzmann’s constant, T is the temperature in Kelvin, E is the energy and EF is the Fermi energy.

Dependence of Fermi factor on temperature: The dependence of Fermi factor on temperature at T=0K is given in the figure. Case 1: the probability of occupation for E < EF at T = 0K Substituting the value of T = 0K in the Fermi function we get

( ) ( ) 1

1 1 1 11 0 1FE E

kTf E

ee−∞− +

= = = =+ +

fE)=1 implies that all the levels below EF are occupied by electrons. Case 2: the probability of occupation for E>EF at T = 0K. Substituting the value of T = 0K in the Fermi function, we get

( ) ( ) 1

1 1 1 01FE E

kTf E

ee∞− +

= = = =+ ∞

.

This shows that all energy levels above EF are vacant.

0

At T=0K

f (E) 1

EF

E


89

Case 3: probability of occupation at temperature > 0K

1. At ordinary temperatures, though the value of probability remain 1 for E<< EF, it starts

decreasing from 1 as the values of E become closer to EF. 2. The value of f(E) become ½ at E=EF. This is because at E = EF

( ) ( ) 01

1 1 1 11 1 1 2FE E

kTf E

ee− +

= = = =+ +

3. For values just beyond EF, f(E)>0 4. Further above E > EF, the probability value falls off to zero rapidly. It implies that the probability of occupancy of Fermi level at any temperature other than 0K is 0.5 Hence Fermi level is defined as the energy level at which the probability of electron occupancy is half. Also, Fermi energy, EF is the average energy possessed by the electrons which participate in conduction process in conductors at temperatures above 0K. Density of states g (E): The permitted energy levels for electrons in a solid will be in terms of bands. Each energy band spread over an energy range of few eV. The number of energy levels in each band will be extremely large and hence the energy values appear to be virtually continuous over the band spread. Each energy level consists of two states and each state accommodates only one electron. Therefore, energy level can be occupied by two electrons only, having opposite directions of spin. The exact dependence of density of energy states on the energy is realized through a function denoted as g (E) and is known as density of states function. It is defined as, the number of available states per unit volume per unit energy interval. The number of states lying in the range of energies between E and E+dE is given by

dEEmh

dEEg 2/12/33 )2(4)( π

= .

Where E is the kinetic energy of the electron in the energy level E.


90

Carrier concentration in metals and Fermi energy at 0K Number of free electrons /unit volume which possess energy in the range E and E+dE is given by N (E) dE = g(E)× dE× f(E) The number of free electrons/unit volume of the material, i.e., n is equal to the total number of electrons that are distributed in various energy levels upto EF. Thus we have

∫=

=FE

E

dEENn0

)(

∫=

=FE

E

dEEfEgn0

)()(

But, f (E) =1, at T= 0K

∫=

=FE

E

dExEgn0

1)(

g(E) dE is given by, dEEmh

dEEg 2/12/33 )2(4)( π

=

dEEmh

nFE

E∫=

=0

2/12/33 )2(4π

2/32/33 )(

32)2(4

FEmh

n π=

2/32/33 )(

3228

FExmh

n π=

2/32/33 )()2(

38

FEmh

n π=

This is the equation of concentration of electrons in a metal at 0K.

Expression for the Fermi energy at 0K is given by 3/22

)3(8 π

nm

hEF =

3/2BnEF =

Where B= 3/22

)3)(8

(πm

h is a constant=5.85x10-38J.

Success of Quantum Free Electron theory 1. The theory could successfully explain the specific heat capacity of metals. 2. It could also explain temperature dependence of electrical conductivity. 3. It explained the dependence of electrical conductivity on electron concentration. 4. It also explained photoelectric effect, Compton effect, Black body radiation, Zeeman

effect etc.,


91

HALL EFFECT: When a transverse magnetic field ‘B’ is applied perpendicular to current carrying conductor, a potential difference is developed across the specimen in a direction perpendicular to both current and the magnetic field. This phenomenon is called the Hall effect. The Voltage so developed is called Hall voltage. Hall effect helps to i) determine the sign of charge carrier in the material 2) determine the charge carrier concentration and iii) determine the mobility of charge carrier, if conductivity of material is known. Hall effect measurement showed that the negative charge carrier electrons are responsible for conduction in metal and it also shows that there exist two types of charge carriers in semiconductor. BAND THEORY OF SOLIDS: The energy band structure of a solid determines whether it is a conductor, an insulator, or a semiconductor The electron of an isolated atom has certain definite energies such as 1s ,2s, 3p, 3s, etc. Between two consecutive allowed values of energy there is forbidden gap. As we bring together large number of identical atoms to form a solid, significant changes take place in the energy levels. The energy levels of each atom will interact with the other identical atoms. The wave functions of each atom will overlap and as a result the energy levels of each atom are distributed slightly and split into a number of levels corresponding to the number of atoms. The split energy levels are very close to each other and they form a narrow band known as energy band. The range of energies possessed by electrons in a solid is known as energy band. The energy band formed by the energy levels of the valence electrons is called valence band. The energy band immediately above the valence band where the conduction electrons are present is called conduction band. The separation between the upper level of valence band and the bottom level of conduction band is known as forbidden energy gap, Eg.

The forbidden energy gap is a measure of the bondage of valence electrons to the atom. The greater the energy gap more tightly valence electrons are bound. When energy is supplied, electrons from the valence band jump to the conduction band and thereby the material starts conducting. SEMICONDUCTORS Pure semiconductors are the materials having electrical conductivity greater than that of insulators but significantly lower than that of a conductor at room temperature. They have conductivity in the range of 10-4

to 104 S/m. The interesting feature about semiconductors is that they are bipolar and current is transported by two types of charge carriers of opposite sign namely electrons and holes. The number of carriers can be drastically enhanced by doping the


92

semiconductor with suitable impurities. The doped semiconductor which exhibits higher conductivity is called an extrinsic semiconductor. The conductivity of an extrinsic semiconductor depends on the doping level which is amenable to control. The current transportation in extrinsic semiconductor occurs through two different processes namely drift and diffusion. Pure semiconductors are of relatively less importance whereas extrinsic semiconductors are widely used in fabricating devices. These devices are more generally known as solid-state electronic devices. INTRINSIC SEMICONDUCTORS A semiconductor in an extremely pure form is known as an intrinsic semiconductor. Intrinsic carriers in pure semiconductors

At room temperature in pure semiconductors, a single event of breaking of bonds leads to two carriers; namely electron and hole. The electron and hole are created as a pair & the phenomenon is called electron-hole pair generation. At any temperature T the number of electrons generated will be equal to the number of holes generated. If ‘n’ denotes number density of electrons in the conduction band & ‘p’ is the number of holes in the valence band then n = p = ni where, ‘ni’ is called intrinsic concentration or the intrinsic density After the generation, the carriers move independently; the electrons move in the conduction band & the holes move in the valence band. The motion of these two carriers is random in their respective band as long as no external field is applied. Concept of Effective Mass of the Electron and Holes: Consider an isolated electron of mass m and charge –e in an electric field of strength E. The electric force acting on it is –eE. The electron gets accelerated, then -eE = ma. However, an electron within a crystal is in a periodic potential due to positive ion cores. The neighboring ions and electrons in the crystal do exert some force on the electron in a crystal. Then ma= -eE + force due to neighboring ions and electrons. Since the latter force is not known quantitatively, we can write the above equation as me*a= -eE or me*= -eE/a where me* is called the effective mass of the electron within the crystal. Thus it is inferred that the effective mass of an electron depends on its location in the energy band. Electrons near the bottom of the conduction band have an effective mass which is almost equal to the effective mass of a free electron. Electrons near the bottom of the valence band have negative effective mass. The removal of an electron with a negative effective mass is identical to creating a particle of positive mass. Thus hole is given the status of particle with positive effective mass mh

*. Carrier concentration in intrinsic semiconductor The actual number of electron in the conduction band is given by

of the band

( ) ( )c

top

cE

n f E g E dE= ∫ (1)


93

Since F-D function describes the probability of occupancy of energy state. Under thermal equilibrium condition, the electron concentration obtained from eqn. (1) is the equilibrium concentration. As f(E) rapidly approaches zero for higher energies, the integral in eqn. (1) can be re-written as

( ) ( )c

cE

n f E g E dE∞

= ∫

dEEmh

Eg ec21

23

*3 )()2(4)( π

= Where E is the kinetic energy of the electron.

E Ec

Conduction band In the above fig. the bottom edge of conduction band EC corresponds to the potential energy of an electron at rest in conduction band. Therefore the quantity (E – EC) represents the kinetic energy of conduction level electron at high energy level.

dEEEmh

Eg cec21

23

*3 )()2(4)( −=π (2)

dE

kTEE

EEmh

nF

ce

Ec

−

+

−= ∫

∞

)(exp1

)()2(4 21

23

*3π (3)

As E > EF : 1>>−kT

EE F

e : kTEE

kTEE FF

ee−−

≅+1

Therefore kTEE

kTEE

F

Fe

e

)(

1

1 −−

− ≅+

Using this eqn in eqn. (3) we get

dEeEEmh

n kTEE

Ece

F

c

)(21

23

*3 )()2(4 −−∞

∫ −=π

dEeEEemh

n kTEE

Ec

kTEE

e

c

c

cF )(21)(

23

*3 )()2(4 −−∞−

∫ −=π

Let E-Ec = x then dx = dE Lower limit when E=Ec x = Ec – Ec = 0 Upper limit when E=∞ x = ∞ - Ec = ∞

Therefore dxexemh

n axkTEE

e

CF−

∞−

∫=0

21)(

23

*3 )2(4π (4)


94

The integral is similar to standard integral. The solution of eqn.(4) is given by

( )

1/ 2

0

3 2 ( )* 3/ 23

, where a=1/kT2

4 = 2 ( ) 2

F c

ax

E E kTe

x e dxa a

n m e KTh

π

π π

∞−

−

=

∴

∫

Rearranging the term we get 3/ 2*

( )2

22 C FE E kTem kTn eh

π − − =

(5)

3/ 2*

C 2

2 N 2 em kTLeth

π =

( )CN C FE E kTn e− −∴ = (6)

Nc is temperature-dependent material constant known as effective density of states in the conduction band. Expression for hole concentration in valence band

If f (E) is the probability for occupancy of an energy state at E by an electron, then probability that energy state is vacant is given by [1- f(E)]. Since hole represents the unoccupied state in valence band, the probability for occupancy of state at E by a hole is equal to probability of absence of electron at that level. The hole concentration in valence band is therefore given by

[ ]vE

1 ( ) ( )vbottomband

p f E g E dE= −∫ (7)

Ev

E Valence band

1- f(E ) rapidly approaches to zero for lower energy levels, the above equation rewritten as

[ ]vE

1 ( ) ( )vp f E g E dE−∞

= −∫

( )[ ] ( ) ( ) dEEEmh

Efp vh

Ev

21

23

*3 241 −−= ∫

∞−

π

Now ( )

−

+=

−

+−=−

−

kTEEe

e

kTEEe

EfF

kTEE

F

F

11

111 (8)


95

For E<EF (E-Ev) is negative. Therefore 0≅−kT

EEe F

Therefore 11 ≅

−

+kT

EEe F and equation 8 reduces to 1- f(E)= kTEEF

e)( −−

( )( )

dEeEEmh

p kTEE

vh

E F −−

∞−

−= ∫ 21

23

*3 )(24π

( ) dEeEEem

hp kT

EEE

vkT

EE

h

vvvF

−

−−

∞−

−

−

∫ −=21

23

*3 )2(4π

Let Ev-E=x then -dE = dx or dE = -dX

∞=∞+=−∞−=

xExEx

LowerLt

v

v )(

00

==−=

=

xEEx

EEUpperLt

vv

v

( ) )()2(4 21

023

*3 dEeEEem

hp kT

EE

vkT

EE

h

vvF

−−=

−

−−

∞

−

−

∫π

( ) dEeEEemh

p kTEE

vkT

EE

h

vvF

−

−−∞

−

−

∫ −=21

0

23

*3 )2(4π

Above equation is of the standard form aa

dxex ax

20

21 π

=−∞

∫ where Ev-E= x and a=kT1

23

23

*3 )(

2)2(4 kTem

hp kT

EE

h

vF ππ

−

−=

kTEE

hvF

eh

kTmp)(2

3

2

*22−

−

=

π

3/ 2*

v 2

2 N 2 hm kTLeth

π =

where Nv is temperature-dependent material constant known as effective density of states in the valence band.

kTEE

v

vF

eNp)( −−

= Fermi level in intrinsic semiconductor In an intrinsic semiconductor electron and hole concentrations are equal. Therefore n = p

kTEE

VkT

EE

c

VFFC

eNeN)()( −−−−

= Taking logarithm on both side and rearranging the term, we get


96

=>kT

EENN

kTEE VF

C

VFC )(ln)( −−−

=

−−

Multiplying by kT throughout

=> VFC

VFC EE

NNkTEE +−

=+− ln

=> VCC

VF EE

NNkTE ++

= ln2

=>

+

+

=C

VVCF N

NkTEEE ln21

2

Substituting the values of NV and NC and after simplification we get *h*e

m3 ln2 4 m

C vF

E EE kT + = +

(1)

As kT is small and the effective mass *em and *

hm do not differ much, the second term in the eqn. (1) may be ignored. If *

em = *hm , then we get

2C v

FE EE + =

(2)

we can write eqn. (2) as

2 2C v v v C v

F vE E E E E EE E+ + − − = = +

2

C v g

gF v

but E E EE

E E

− =

∴ = +

If top of the valence band Ev is taken as zero level, then 2

gF

EE =

Thus Fermi level in the intrinsic semiconductor lies at the centre of the energy gap as shown below:

k

Eg

Ev

Ec

E


97

INTRINSIC DENSITY, ni In an intrinsic semiconductor at T=0K, the electron concentration in the conduction band is identical to hole concentration in the valence band. n=p=ni From this, we get np=ni

2

kT

EE

vkT

EE

Ci

VFFc

eNeNn)()(

2−

−−

−=

kT

EE

VC

Vc

eNN)(

)(−

−=

But Ec-Ev =Eg

kTE

VCigeNNn /2 )( −=

kTE

VCigeNNn 2/2/1)( −=

Substituting the values of Nc and Nv we get,

kTEhei

geTmmh

kn 2/2/34/3**2/32 )(]2[2 −=π

The following important points may be inferred from the above relation 1. The intrinsic density is independent of Fermi level position. 2. The intrinsic density is a function of band-gap Eg, which represents the energy needed

to break a bond. 3. The intrinsic density strongly depends on the temperature. The contribution of

temperature increase to ni is mostly due to the exponential term and only to a marginal extent due to the term T3/2.

Expression for the band gap of a Semiconductor: The band gap is the energy separation between the conduction band and the valence band of a semiconducting material. The conductivity of an intrinsic semiconductor is given by hiei enen µµσ +=

)( heien µµσ += Substituting the value of ni, we get

)(2

exp2243

2

**23

2 heghe e

kTE

mmmx

hkTm µµπσ +

−

=

The above equation can be written as

−=

kTE

A g

2expσ

Where )(2243

2

**23

2 hehe e

mmmx

hkTmA µµπ

+

=

As ρ

σ 1= :

=

kTE

B g

2expρ


98

We know that l

RA=ρ then

=

kTE

AlBR g

2exp

=

kTE

CR g

2exp where

ABlC =

Taking log on both sides kTE

CR g

2lnln +=

Therefore )ln(ln2

CRkTEg −=

The band gap is given by )ln(ln2 CRkTEg −=

kTE

CR g

2lnln += is of the form cmxy += : By taking ln R in the y-axis and

T1 in the x-axis,

if a graph is plotted, a straight line is obtained as shown in below figure.

Slope kTE

m g

2=

Therefore kTmEg )(=

By finding the slope of the straight line, the band gap of the semiconductor is determined using the relation, Eg = 2k x slope of the straight line drawn between ln R and 1/T. Extrinsic semiconductor

The intrinsic semiconductor has low conductivity which is not amenable to control. However a judicious introduction of impurity atoms in a intrinsic semiconductor produces useful modification of its electrical conductivity. The method of introduction of controlled quantity of impurity into an intrinsic semiconductor is called doping. The impurity added is called dopant. The semiconductor doped with impurity atoms is called extrinsic semiconductor. There are two types of extrinsic semiconductor namely p-type & n-type which are produced depending on the group of impurity atoms.

n-type semiconductors are produced when pure semiconductors are doped with pentavalent impurity atoms such Phosphorous, Arsenic etc.

p-type semiconductors are produced when pure semiconductors are doped with trivalent impurity atoms such as Aluminum, Boron etc.

L

M N

ln C

ln R

1/T


99

Temperature variation of carrier concentration in extrinsic semiconductor The dependence of electron concentration on temperature for n-type semiconductor is as shown in the figure below.

At 0K the donor levels are filled which means that all the donor electrons are bound to the donor atoms. At low temperature, corresponding to region- I, there is no enough energy to ionize all the donors and not at all enough to break covalent bond. As temperature increases, the donor atoms get ionized and donor atoms go into the conduction band. The region-I is known as ionization region. Occasionally a covalent band maybe broken out, but number of such events will be insignificantly small. At about 100K all donor atoms are ionized, once all electrons from donor level are excited into conduction band, any further temperature increase does not create additional electrons and the curve levels off. The region (region-II) is called depletion region. In the depletion region the electron concentration in the conduction band is nearly identical to the concentration of dopant atom. If ND is donor concentration then nn = ND (depletion region) where nn – electron concentration in n-type As temperature grows further, electron transitions from valence band to conduction band increases. At high temperature (region-III) the number of electron transition becomes so large that the intrinsic electron concentration exceeds the electron concentration due to donor. This region is therefore called intrinsic region. In intrinsic region, nn = ni

Similarly in p-type semiconductor, the acceptor levels are vacant at 0K & valence band is full. As temperature increases in the ionization region, the electrons from the valence band jump into acceptor level. However the electrons do not acquire enough energy to jump into conduction band levels. At the temperature Ts, the acceptor levels are saturated with electrons. The region- II lying between Ts (saturation temperature) and Ti is called the saturation region. In case of p-type materials within this temperature interval the hole concentration remains constant as thermal energy is not yet sufficient to cause electron

I II III

Depletion region

Intr

insic

regi

on

T

n

Ioni

zatio

n re

gion

Td Ti


100

transition from valence band to conduction band. In the saturation region, the hole concentration is equal to the acceptor impurity concentration. Thus pp = NA With increase of temperature beyond T, electron transition due to intrinsic process commence & hole concentration due to intrinsic process far exceeds that due to impurity atom. In region-III, pp = ni

Fermi level in extrinsic semiconductor

The carrier concentration in extrinsic semiconductors varies with temperature as discussed earlier. It follows that the probability of occupancy of respective bands & position of Fermi level varies with temperature.

In n-type semiconductor, in low temperature region the electron in the conductor band is only due to the transition of electrons from donor levels. Therefore Fermi level lies between the donor level ED & the bottom edge of conduction band.

As temperature increases the donor level gradually gets depleted & the Fermi level shifts downward. At the temperature of depletion Td, the Fermi level coincides with the donor level ED

i.e. EFn = ED.

As temperature increases further above Td, the Fermi level shifts downward approximately in linear fashion, though electron concentration in the conduction band

remains constant. This is in accordance with the relationD

cDcF N

NkTEEEn

ln22

−+

= .

At temperature Ti, where intrinsic process contributes to electron concentration significantly, the Fermi level approaches the intrinsic value EFi = Eg/2. With further increase in temperature the behavior of extrinsic semiconductor transitions into that of an intrinsic type & Fermi level stays at EFi .Thus

EFn = EFi= Eg/2. N-type semiconductor


101

P-type semiconductor In case of p-type semiconductor the Fermi level EFp rises with increasing temperature from below the acceptor level to intrinsic level EFi as shown in fig2.

2A v

FpE EE +

∴ = (ionization region)

As temperature increases further above Ts, the Fermi level shifts downward approximately in linear fashion, though hole concentration in the valence band remains constant. This is in

accordance with the relationA

vAvF N

NkTEEEp

ln22

++

= .

EFp = EA (at T=Ts) and EFp = Eg/2.

Fig 2.

Effect of variation of impurity concentration: The addition of donor impurity to an intrinsic semiconductor leads to the formation of discrete donor level below the bottom edge of conduction band. At low impurity concentrations the impurity atom are spaced far apart & do not interact with each other. With an increase in the impurity concentration the impurity atom separation in the crystal decreases & they tend to interact. Consequently the donor level also undergoes splitting & form energy band below the conduction band. The larger the doping concentration, the broader is the impurity band & at one stage it overlaps with the conduction band.. The broadness of donor levels into a band is accompanied by a decrease in the width of forbidden gap & also the upward displacement of Fermi level. The Fermi level moves closer & closer to the conduction band with increasing impurity concentration & finally moves into the conduction band as donor band overlaps the conduction band. In similar way, in p-type semiconductor, the acceptor level broadens and forms into a band with increasing impurity concentration which ultimately overlaps the valence band. The Fermi level moves down closer to the valence band and finally at high impurity concentration it will shift in to valence band.


102

n-type semiconductor

HALL EFFECT IN SEMICONDUCTORS: Let us consider a rectangular plate of p-type semiconductor. When potential difference is applied across its ends, a current ‘I’ flows through it along x direction. If holes are majority charge carriers in p-type semiconductors then the current is given by I = pAevd (1)

Where p- concentration of holes A- Area of cross section of end face e- charge on the hole vd- drift velocity of holes

therefore current density J = I/A = pevd (2) Any plane perpendicular to current flow direction is an equipotential surface. Therefore potential difference between front and rare faces is zero. If magnetic field is applied normal to crystal surface and also to the current flow, a transverse potential difference is produced between the faces F & F/. It is called Hall voltage VH.


103

Before the application of magnetic field B, the holes move in an orderly eay parallel to faces F & F/. On the application of magnetic field B, the holes experience a sideway deflection due to the Lorentz force FC. The magnitude of this force is given by FL = e B vd Because of this force, holes are deflected towards the front dace F and pile up there. Initially the material is electrically neutral everywhere. However as holes pile up on the front side, a corresponding equivalent negative charge is left on the rare dace F/. As a result an electric field is produced across F & F/.The direction of electric field will be from front face to rare face. It is such that it opposes the further pile up of holes on the front face F. A condition of equilibrium is reached when FE due to transverse electric field EH balances the Lorentz force. The transverse electric field EH is known as Hall field. In equilibrium condition FE = FH

FE = e EH = e(VH/w) (3) Where w- width of semiconductor plate From eqn (2) vd= Jx/Pe

Therefore xL

BJF = p

(4)

Equating (3) and (4) we get H x

xH

eV BJ = w p

wBJ wBIVpe peA

⇒ = =

If ‘t’ is the thickness of the semiconductor plate, A=wt. Then above equation reduces to

HBIVpet

=

Hall field per unit current density per unit magnetic field is called Hall co-efficient RH Thus H H x

Hx x x

E V w wBJRBJ BJ wpeBJ

= = =

H1Rpe

∴ = (6)

Substitute (6) in (5) we get

HBIVtHR= (7)

HV tBIHR = (8)

The Hall voltage is a real voltage & can be measured with a voltmeter with the direction of magnetic field & current depicted in this fig, the sign of Hall voltage is +ve.

For n-type semiconductor Hall voltage will be –ve, when the direction of current is same as in the fig. Therefore by knowing the sign of Hall voltage the type of semiconductor & the sign of the majority charge carriers will be known.


104

The carrier concentration is given by

H

1pR e H

BIV te

= =

In case of n-type semiconductor

H

1nR e H

BIV te

= − = −

SOLVED NUMERICALS: 1. What is the probability of a level lying 0.01 eV below the Fermi level not being occupied

by electrons at T = 300K? Solution: Probability not being occupied by electrons = 1-f(E) = 1- (e(E-E

F)/k

BT + 1)-1

= 1 - (e 0.01/0.026 + 1)-1 = 1/(1.47 + 1) = 0.405

2. Find the temperature at which there is 1% occupancy probability of a state 0.5 eV above Fermi energy. Solution: f(E) = 0.01 = 1/[e (E-E

F)/k

BT + 1] for E-EF= 0.5 eV

Solving we get 0.01 = 1/[e5797/T + 1]. Thus e5797/T = 1/0.01 - 1 = 99. Taking log we get, T = 5797/4.595 = 1261.1 K

3. The effective mass of holes in a material is 4 times that of electrons. At what temperature would the Fermi energy be shifted by 10% from the middle of the forbidden energy gap? Given band gap = 1 eV. Solution: EF = (EC + EV)/2 + (3kT/4) log(mh/me) Fermi level is shifted by 10% = 0.1 eV. Originally Fermi energy was 0.5 eV above EV . Now it is 0.5+0.1 = 0.6 eV above EF. (EV + 0.6) eV = (EC+EV)/2 + (3kT/4)log(4) ----(1) And (EV + 0.5) eV = (EC+EV)/2 -----------(2) Subtracting 2 from 1 we get 0.1 eV = (3kT/4)log(4) 0.1602x10-19J = (3x1.38x10-23xT/4)0.6021.

T = 1116K 4. For an intrinsic semiconductor with gap width Eg=0.7eV. Calculate the concentration of

intrinsic charge carriers at 300K assuming that me*=mo (rest mass of electron).

ni =33.49x1018 /m3

kTEgi e

hmkTn 2/

2/3

222 −

=π


105

5. Calculate the free electron concentration, mobility and drift velocity of electrons in an Aluminium wire of length 5m and resistance 60m-ohm, if it carries a current of 15A assuming that each Aluminium atom contributes 3 free electron for conduction. Solution:

htatomicweigxDensityxatomxNectronspernooffreeeln A

310=

98.26107.210025.63 326 xxxxn =

3291081.1 −= mxn .

enρµ 1= = 29198 1081.110107.2

1xxxx −−

.1027.1 1123 −−−= svmxµ

EVd=µ

smxxxxxLIREVd /103.2

510601510279.1 4

33−

−−

====µµ

6. The Fermi level in silver is 5.5eV at 0K. Calculate the number of free electrons per unit volume and the probability occupation for electrons with energy 5.6eV in silver at the same temperature. Solution:

3/23/22 3

80n

mhEF

=

π

.1084.5 328 −= mxn

7. Calculate the probability of an electron occupying an energy level 0.o2eV above the Fermi level at 200K and 400K in a material.

kTEEe

EfF−

+=

1

1)(

2001038.1106.102.0

23

19

1

1)(xx

xx

e

Ef−

−

+

= =0.24

f(E) =0.36 at 400K. 8. A semiconducting material 12 mm long, 5 mm wide and 1 mm thick has a magnetic flux

density of 0.5 Wb/m2 applied perpendicular to the largest faces. A current of 20 mA flows through the length of the sample, and the corresponding voltage measured across its width is 37µV. Find the Hall coefficient of the semiconductor. Solution:

Hall coefficient RH = Ey/JxBz = -1/ne. Since Ey = Vy/w, RH = Vy/wJxBz

Thus RH = (37x10-6x10-3)/(20x10-3x0.5) = 3.7x10-6 m3C-1


106

Electrical Properties of matter S. No Questions 1. An electron is accelerated by an electric field of 4V/cm, is found to have mobility 8x

10-3 m2 / Vs. What is its drift velocity? 2. How many valence electrons will a donor impurity has in a n-type semiconductor? 3. With increase in temperature, how does resistance of a pure semiconductor vary? 4. What is a hole in context of semiconductors? 5. In Hall effect experiment what is the polarity of Hall voltage for a n-type

semiconductor? 6. What will be the Fermi velocity of an electron in copper if Fermi energy (EF ) = 6 eV ? 7. At 300K, if probability for occupancy of an energy state E by an electron is 0.75,

calculate probability for occupancy of the same state by a hole? 8. Write any two assumptions of Drude-Lorentz theory? 9. Sketch the graph of Fermi factor f(E) verses E for the case E=EF at at T> 0K in metals. 10. State density of states in metals. 11. Write an expression for density of states in metals. 12. Sketch the variation of fermi level with temperature for n type semiconductor. 13. What are Fermions? 14. Outline the phenomenon of Hall effect in materials. 15. For silicon semiconductor with band gap 1.12eV, determine the position of the Fermi

level at 300K if me*=0.12mo and mh

*=0.28mo. 16. Distinguish between intrinsic and extrinsic semiconductors. 17. Find the probability that energy level at 0.2 eV below Fermi level being occupied at

temperature 1000K? 18. What is the value of Fermi function when E=Ef at T>0K? 19. What is the effect of increase of impurity concentration on band gap in extrinsic

semiconductors? 20. Mention any two demerits of classical free electron theory. 21. Find the probability of a level lying 0.01 eV below the Fermi level being occupied by

electrons at T = 0K? 22. What is the magnitude of Lorentz force in Hall effect experiment? 23. With neat sketch, show the Fermi level position in p-type semiconductor. 24. Give the expression for Ohm’s law in terms of J,σ and E. 25. What is Fermi factor in Fermi Dirac distribution? 26. Find the relaxation time of conduction electrons in a metal if its resistivity is 2.5x10-8

Ωm and it has 5x1028 conduction electrons/m3 27. Sketch the position of Fermi level at 0K in a band diagram of a n-type semiconductor,

at low doping 28. Find the Fermi velocity of conduction electron if the Fermi energy of silver is 8eV 29. Determine the probability of occupancy of an energy level situated 0.05eV above the

Fermi energy at temperature of 0K


107

30. Write an expression for carrier concentration of p-type semiconductor. 31. Write any two postulates of classical free electron theory of solids. 32. Give the relation between electrical conductivity and mobility of charge carriers in a

conductor. 33. If a system is composed of indistinguishable, half integral spin particles and obeys

Pauli exclusion principle, then what statistics is obeyed by the system? 34. What is the value of the Fermi factor for metals at room temperature? 35. Plot variation of Fermi factor with temperature in a metal. 36. Distinguish between free electron theory and band theory of solids in terms of

influence of lattice on the electron moving in a metal. 37. Give expression for Fermi level at 0 k in an intrinsic semiconductor. 38. Find the temperature at which there is 1% probability that a state with energy 0.5 eV

above Fermi energy is occupied? 39. What is Hall Effect? 40. Sketch the variation in the energy of the Fermi level in a ‘n’ type semi-conductor as a

function of temperature? 41. Describe in words Wiedemenn-Franz Law 42. What is the formula for intrinsic carrier density (ni)? 43. A wire of diameter 0.2 meter contains 1028 free electrons per cubic meter. For an

electric current of 10A, calculate the drift velocity for free electrons in the wire? 44. The fermi level in an intrinsic semi-conductor is at .25 eV. What is the width of the

band gap? 45. Electrical conductivity of Cu is 5.78×107Ω-1m-1. If the free electron density of Cu is

8.46×1028m-3. Find the mobility of electrons? 46. The fermi energy for an intrinsic semiconductor is at 5 eV. At 0K, calculate the

probability of occupation of electrons at E= 5.5eV? 47. A sample of silicon is doped with 107 phosphorous atoms/cm3. Find the Hall voltage, if

the sample is 100µm thick, Ix=1 mA and Bz= 10-5 Wb/m2? 48. Write any one drawback of classical free electron theory? 49. Write the relation for specific heat of a metal as per quantum free electron theory 50. Write the condition at which the value of f(E) = 1 at 0° K. 51. Mention any two assumptions of quantum free electron theory. 52. Find the relaxation time of conduction electrons in a metal of resistivity 1.54×10-8 Ωm.

If the metal has 5.8×1028 conduction electrons per m3. 53. Why is that only the electrons near the Fermi level contribute to electrical

conductivity? 54. Find the probability with which an energy level 0.02 eV below Fermi level will be

occupied at room temperature of 300K. 55. A copper strip of 2.0 cm wide and 1.0 mm thick is placed in a magnetic field of 15000

gauss. If a current of 200 A is setup in the strip with the Hall voltage appears across the strip is found to be 0.18 V. calculate the Hall coefficient.

56. Which statistical rule is obeyed by electrons in quantum free electron theory? 57. Where does the Fermi level lie in case of n type semiconductor with high impurity


108

concentration? 58. Electron concentration in a semiconductor is 1020m3. Calculate Hall coefficient? 59. What is doping in semiconductors? 60. Evaluate the probability of occupation of an energy level 0.4 eV below the Fermi

energy level in metal at zero Kelvin. 61. Copper has electrical conductivity of 9x107Ω-1m-1 and thermal conductivity of 300

Wm-1K-1 at 305K. Find the Lorentz’s number on the basis of classical free electron theory?

62. If the probability of absence of electron in an energy level of valance band of semiconductor is 0.65 what is the probability of occupation in the same level by a hole?

63. In the band diagram of a p-type semiconductor show the position of the Fermi level when the doping concentration is low?

64. Graphically show the variation of ln(np) with increasing temperature in Kelvin where ne is the electron concentration in an intrinsic semiconductor.

65. A wire of 4 mm radius carries a current of 8A. Find the current density?

Sample Questions 1) Give the postulates of classical free electron theory and explain the failures of classical

free electron theory. 2) Give the success of Quantum free electron theory. 3) Discuss the variation of Fermi factor in metals with temperature. 4) Explain density of states in metals. 5) Explain Fermi Dirac distribution function. Show that at temperatures above 0K

probability of occupancy of Fermi level in metals is 50%. 6) Define Fermi energy at 0K and at above 0K in metals. 7) Derive an expression for the electron concentration in metals at 0K. 8) Derive an expression for the electron concentration in intrinsic semiconductor. 9) Derive an expression for the hole concentration in intrinsic semiconductor. 10) Show that Fermi level of an intrinsic semiconductor lies in the middle of the band gap. 11) With a neat sketch explain how Fermi level changes in n-type semiconductor with the

increase in temperature. 12) With a neat sketch explain how Fermi level changes in p-type semiconductor with the

increase in temperature. 13) Give an account of effect of carrier concentration on Fermi level. 14) What is Hall Effect? Obtain an expression for Hall coefficient. 15) Derive expressions for Hall voltage and Hall coefficient in n-type semiconductors.

UNIT- V: DIELECTRICS AND THERMAL CODUCTIVITY

109

UNIT – V: DIELECTRICS AND THERMAL CONDUCTIVITY Introduction: Materials such as glass, ceramics, polymers and paper are non-conducting materials. They prevent the flow of current through them, therefore they can be used for insulation purposes. When the main function of non-conducting materials is to provide electrical insulation they are called Insulators.

distance

Figure 1: Band diagram of an insulator

When non-conducting materials are placed in an electric field, they undergo appreciable changes as a result of which they act as stores of electric charges. When charge storage is the main function, the insulating materials are called Dielectrics. For a material to be a good dielectric, it must be an insulator. Hence any insulator is a dielectric.

The forbidden energy gap (Eg) between the valence band and conduction band is very large (fig.1) in dielectrics and excitation of electrons from valence band to conduction band is not possible under ordinary conditions. Therefore conduction cannot occur in a dielectric. Even if the dielectric contains impurities, extrinsic conduction cannot occur as observed in case of extrinsic semiconductors. The resistivity of an ideal dielectric is infinity, in practise dielectrics conduct electric current to a negligible extent and their resistivity’s range from 1010 to 1020Ωm. Electric dipole and dipole moment:

Figure 2: Electric dipole


110

A pair of equal and opposite charges separated by a small distance is called an electric dipole and the product of the magnitude of one of the charges and the distance of their separation is called dipole moment (µ). Consider two charges –q and +q with a distance of separation is 2a where a is the distance from the centre of dipole to one of the charge as shown in figure 2. The dipole moment for this arrangement is given by

µ= (2a)q…………(1)

Field due to an electric dipole at any point in a plane

1) Field due to electric dipole at axial point:

Figure 3

A configuration of two equal and opposite charges separated by a distance is a dipole. Let ‘P’ be a point at a distance ‘r’ from the centre ‘O’ on the axial line of the dipole, at which the intensity of the field is to be determined (figure 3). The field E1 at ‘P’ due to the charge ‘-q’ is towards –q and is given by

E1= ( ))1(

41

20

−−−−− arq

πε

Similarly the field E2 at ‘P’ due to ‘+q’ is away from +q and is given by

E2= )2()(4

12

0−−−−−

+ arq

πε

The resultant field intensity at P will be E= E1- E2 Substituting for E1 and E2 from the above equations 1 & 2

E=( ) 2

02

0 )(41

41

arq

arq

+−

− πεπε

E = 220

22

)()(4])()[(

ararararq+−−−+

πε

E = 2220

2222

)(4]22[

arraarraarq

−+−−++

πε

E= 2220 )(44

arqra−πε

If ‘r’ is very much greater that ‘a’, then (r2+a2)2≈r4

E= 30

30

40 24

)2.(24

4rr

aqr

qraπεµ

πεπε==

where µ=q.2a and is the electric dipole moment.


111

2) Field due to an Electric dipole at Equatorial point

Figure 4

Let ‘P’ be a point on the perpendicular bisector (equatorial line) of the dipole at a distance ‘r’ from the centre of O of the dipole. E1 and E2 are the fields due to charges +q and –q respectively (Figure 4). The resultant field at ‘P’ is 21 EEE

+=

The field E1 along AP extended is

20

1 41

APqE

πε= ……….(1)

where AP2=OP2+OA2 = r2+a2 substituting for AP2 in equation (1)

E1= )(4 22

0 arq+πε

The field E2 along PB is

20

1 41

BPqE

πε= ……….. (2)

where BP2=OP2+OA2 = r2+a2 substituting for BP2 in equation (2)

E2= )3.......()(4 22

0 arq+πε

1E = 2E = E=)(4 22

0 arq+πε

……..(4)

when E1 and E2 are resolved parallel to axial line and along the equatorial line, the component along the equatorial line gets cancelled and along the line parallel to axial line gets added up. The vector sum of these two fields has the magnitude E = E1cos θ+ E2cos θ = 2Ecosθ ……(5)


112

From the right angled triangle AOP

cos θ = =ABOA )6......(

)( 22 ar

a

+ Substituting equation 4 & 6 in equation 5 we get

E = )()(4

22222

0 arar

qa

++πε= 2/322

0 )(42

arqa+πε

2/322

02/322

0 )(4)(4)2(

araraq

+=

+=

πεµ

πε If r is very much greater a then, r2+a2≈ r2 and (r2+a2)3/2≈r3

E = 304 rπεµ

where µ is electric dipole moment, 0ε is the permittivity of free space and r is the perpendicular distance from the centre of the dipole to point ‘P’. Polar and Non-polar dielectrics: A dielectric material doesn’t possess any free electrons. All the electrons are bound very strongly to the respective nuclei of the atoms of the parent molecules. Each molecule consists of equal number of positive and negative charges. All the positive charges are concentrated in the nuclei, which are surrounded by electron clouds in which all the negative charges are distributed. If in the molecules of some dielectric materials, the effective centre of the negative charge distribution coincides with the effective centre of the positive charge distribution such materials are called non-polar dielectrics.eg Hydrogen, carbondioxide etc. In some dielectric materials, the effective centres of the negative and positive charges in the molecules do not coincide with each other in the absence of an external electric field. Each molecule behaves like a permanent dipole and the materials comprising of such dipoles are called polar dielectrics. Ex: HCl, H2O etc. Polarization If a dielectric is placed in an electric field of strength ‘E’, the electron cloud will be displaced in the direction opposite to ‘E’ by a distance ‘d’ with respect to the nucleus. The centres of gravity of positive and negative charges in the atom no more coincide.

The atom is equivalent to the system of two charges, q=Ze of equal magnitude but opposite in sign separated by a distance ‘d’ and the atoms behaves like a dipole and it is called induced dipole. The atom is said to be polarized.

E=0 E>0


113

Figure 5 The induced dipole sets up its own electric field which is opposite in direction to the external field. The dipole moment µ is a vector, directed along the axis of the dipole from the negative charge to the positive charge. When the molecule is polarized, restoring forces due to coulomb attraction come into play which tends to pull the displaced charges together. The charges separate until the restoring force balances the force due to the electric field. The induced dipole moment is proportional to the field strength. The larger the field, greater the displacement of charges and hence larger the induced dipole moment. The induced dipole moment is given by µ=αE α is the polarizability of the molecule. It characterizes the capacity of electric charges in a molecule to suffer displacement in an electric field. The unit of polarizability is Fm2. The induced dipole moment vanishes as soon as the electric field is switched off. Dipole in an electric field:

When a polar molecule is placed in a uniform electric field ‘E’ (figure 6), the field exerts a force +qE on the charge +q and –qE on –q. The net force on the dipole is zero since the two forces acting on it are equal and opposite to each other. Therefore, there is no translational force on the dipole in the uniform field. The two forces are anti parallel and separated by perpendicular distance, hence constitutes a couple, which tends to rotate the dipole. The torque on the dipole is given by θτ sin2aqE= = θµ sinE (q2a=µ) Displacement of positive and negative charges in the molecules of a dielectric under the action of applied electric field leading to the development of dipole moment is known as dielectric Polarization.

Figure 7

Figure 6: An electric dipole in a uniform electric field


114

Consider an electrically neutral slab inserted between the plates of a parallel plate capacitor as shown in figure 7. Dielectric is imagined to be divided into large number of identical cells of volume dv. Under the action of external electric field, charges are induced in each cell and each cell acquires a dipole moment dµ. Then intensity of polarization “P” is defined as the total dipole moment per unit volume of the material.

∑= dvdP µ =

vµ

Dielectric constant: For isotropic materials the electric flux density E and the electric induction (or electric displacement) D are related by the equation

D= ε0εrE where ε0=8.854x10-12F/m, is the dielectric constant of vacuum and εr is the dielectric constant or relative permittivity for the material. It has no units. Dielectric susceptibility: The magnitude of polarization is directly proportional to the intensity of the electric field.

Thus, P=χ ε0E (for linear dielectrics) χ (chi) is the proportionality constant and is called the dielectric susceptibility of the material. It characterizes the ease with which the dielectric material can be influenced by an external field. P is a measure of the polarization produced in the material per unit electric field. Relation between εr and χ: In order to describe the combined effects of the applied electric field E and electric polarization P, an auxiliary vector D called Electric displacement vector is introduced. D= ε0E+P Substituting for P=χ ε0E in the above equation

D= ε0E+ χ ε0E D= (1+ χ) ε0E D= ε0 εrE where εr = 1+χ

ε0 is the absolute permittivity of the free space and εr is the relative permittivity or the dielectric constant of the material. Note : Dielectric breakdown or Dielectric strength: When a dielectric is subjected to very high electric fields, a considerable number of co-valent bonds are torn away and electrons get excited to energy levels in the conduction band. These electrons acquire a large kinetic energy and cause localized melting, burning and vapourization of material leading to irreversible degradation and failure of the material. It results in high electrical conductivity and total loss of charge storage property of the dielectric. The formation of such conducting paths in a dielectric under the action of an electric field is termed as dielectric breakdown. Dielectric strength is the limiting electric field intensity above which a breakdown occurs and charge storage property of the dielectric disappears.


115

Types of Polarization 1) Electronic or Atomic Polarization:

Figure 8

This is the polarization that results from the displacement of electron clouds of atoms or molecules with respect to the heavy fixed nuclei to a distance that is less than the dimensions of atoms or molecules (figure 8). This polarization sets in over a very short period of time, of the order of 10-14-10-15s. It is independent of temperature. The polarization is given by Pe=NαeE ……..(1) where N is the number of atoms/unit volume, αe is electronic polarizability.

We have P=χ ε0E or E

P

0εχ = …(2)

Dielectric constant εr =1+χ ….(3) Substituting eqn. (2) in eqn. (3)

εr = 1+E

Pe

0ε …..(4)

Substituting for Pe from eqn (1) in eqn (4)

εr =1+EEN e

0εα

εr = 1+0εαeN ……… (5) or

Nr

e)1(0 −

=εε

α

εr is the dielectric constant of a non-polar gaseous dielectric. The above equation indicates that the dielectric constant depends on the polarizability of a molecule and the number of molecules in a unit volume of the dielectric. 2) Ionic Polarization: Ionic polarization occurs in ionic crystals. It is brought about by the elastic displacement of positive and negative ions from their equilibrium position. Eg. Sodium chloride crystal. A NaCl molecule consists of Na+ ion bound to Cl- ion through ionic bond. If the interatomic distance is ‘d’, the molecule exhibits an intrinsic dipole moment equal to “qd” where q is the charge of the electron and d is the distance of separation.

Figure 9


116

When ionic solids are subjected to an external electric field, the adjacent ions of opposite sign undergo displacement (figure 9). The displacement causes an increase or decrease in the distance of separation between the atoms depending upon the location of the ion pair in the lattice. This polarization takes 10-11-10-14s to build up and is independent of temperature. Ionic polarization is given by Pi= NαiE For most materials, the ionic polarisability is less than electronic polarization. Typically

αi= eα101

3) Orientation or dipole Polarisation: This polarization is a characteristic of polar dielectrics which consists of molecules having permanent dipole moment. In the absence of external electric field, the orientation of dipoles is random resulting in a complete cancellation of each others effect (figure 10). When the electric field is impressed the molecular dipoles rotate about their axis of symmetry and tend to align with the applied field and the dielectric acquires a net dipole moment and it is orientation polarization.

Figure 10 The dipole alignment is counteracted by thermal agitation. Higher the temperature, the greater is the thermal agitation. Hence, orientation polarization is strongly temperature dependent. In case of solids, the rotation of polar molecules may be highly restricted by the lattice forces, leading to a great reduction in their contribution to orientation polarization. Because of this reason, while the dielectric constant of water is about 80, that for solid ice is only 10. As the process of orientation polarization involves rotation of molecules, it takes relatively longer time than other two polarisations. The build up time is of the order of 10-10s or more. The orientation polarizability

α0 = kT3

2µ

and orientation polarization P0 = kT

EN3

2µ

Orientation polarization is inversely proportional to temperature and proportional to the square of the permanent dipole moment. 4) Space charge or Interface polarization: This polarisation occurs in multiphase dielectric materials in which there is a change of resistivity between different phases, when such materials are subjected to an electric field, especially at high temperatures, the charges get accumulated at the interface, because of sudden change in conductivity across the boundary (figure 11). Since the accumulation of


117

charges with opposite faces occurs at opposite parts in the low resistivity phase, in effect it leads to the development of dipole moment within the low resistivity phase domain. Eg. Non homogenous materials such as composites.

Figure 11

Relaxation time: It is the time required for the dipole to reach the equilibrium orientation from the disturbed position in alternating field conditions. The reciprocal of relaxation time is the relaxation frequency. (or) “The average time taken by the dipoles to reorient in the field direction is known as relaxation time (τ)”. Dielectric loss: When a dielectric material is subjected to the influence of an electric field, its molecules come into a state of electrostatic stress due to polarisation. The molecules behave as electric dipoles. If the polarity of the applied voltage is reversed, the stress is also reversed because of which the dipoles switch their orientation. In order to do so, the dipoles are required to overcome a sort of internal friction which involves a loss of energy. This occurs every time the switching or realignment of the dipoles takes place. The energy loss due to friction is always dissipated as heat by the dielectric material to the surroundings. This is termed as dielectric loss and defined as, loss of energy in the form of heat in a dielectric medium due to internal friction that is developed as a consequence of switching action of electric dipoles under ac conditions. This part of energy is lost or wasted as heat and no useful work is done by it. Power loss in dielectrics, when subjected to dc voltages will be very small due to high resistance of dielectric materials. The power loss in an ac field will be quite large. In an ideal dielectric material, loss is zero. Frequency dependence of Dielectric constant: Dielectric constant (εr) remains unchanged when the material is subjected to a dc voltage. But the dielectric constant (εr)` changes when the material is subjected to the influence of an ac voltage, the changes depend on the frequency of the applied voltage.


118

Figure 12: Variation of dielectric constant with frequency

All the four polarization mechanisms that occur in a dielectric material will be effective in static field conditions. But, each of them respond differently at different frequencies under alternating field conditions, since the relaxation frequency of different polarization processes are different as shown in figure 12. If τe, τi and τo are the relaxation times for electronic, ionic and orientation polarisations, then in general τe< τi < τo When the frequency of applied field matches the relaxation frequency of a given polarisation mechanism, the absorption of energy from the field becomes maximum. When the frequency of applied field becomes greater than the relaxation frequency for a particular polarisation mechanism, the switching action of the dipoles cannot keep in step with that of changing field and the corresponding polarization mechanism is halted. Thus as the frequency of applied ac is increased, different polarization mechanisms disappear in the order- interface, orientation, ionic and electronic. f0 <fi<fe In addition, εr becomes a complex quantity and is expressed as εr

*= εr’-j εr

”

where εr’ and εr

” are the real and imaginary parts of the dielectric constant. εr

’ represents part of the dielectric constant that is responsible for the increase of capacitance of a capacitor when the dielectric is placed between the plates of a capacitor. εr

” represents the loss. The loss that occurs in a dielectric material is essentially due to the phase lag of voltage behind current in the capacitor between the plates of which dielectric material lies. Such a loss in a capacitor is expressed by a factor called tan δ. A large value of tan δ signifies higher dielectric loss. It is also referred to as tangent loss

tan δ = r

r

'

"

ε

ε


119

Internal field in a solid for one dimensional infinite array of dipoles:

When a dielectric material, either solid or liquid is subjected to an external electric field, each of the atoms develope a dipole moment and acts as an electric dipole. Hence the resultant field at any given atom will be the sum of applied electric field and the electric field due to the surrounding dipoles. The resultant local field is called the internal field (Ei)and is defined as the electric field that acts at a site of any given atom of a solid or liquid dielectric subjected to an external electric field and is the resultant of the applied field (E) and the field due to all the surrounding dipoles (E') . Ei=E+ E'

Figure 13: Linear array of atoms in an electric field

Consider a dielectric material kept in an external uniform electric field of strength E. In the material let us consider an array of equidistant atomic dipoles arranged parallel to the direction of the field as shown in figure 13. In the linear array let us consider an atom X. The distance from X to the dipole A1 is d, to the dipole B1 is 2d and so on ( d is the distance between two consecutive dipoles). The dipole moment of each of the atomic dipole is µ. The total field at X due to all the other dipoles in the array is evaluated as follows.

Field at X due to 30

30

1 22

41

ddA

πεµµ

πε== ……(1)

Consider the dipole A2, since it is situated symmetrically on the other side of X,

its field at X will also be = 30

30 2

24

1dd πε

µµπε

=

Therefore the field E1 at X due to both the dipoles A1 and A2 gets added up

E1 = 2[ 302 dπεµ ] = 3

0dπεµ ……….(1)

Similarly if we consider the two dipoles B1 and B2 each of which is located at a distance 2d, then the field E2 at X due to both of them is

E2 = 30

30 )2(

])2(2

[2dd πε

µπε

µ= ……(2)

Similarly the field at X due to the dipoles C1 and C2 is


120

E3 = 30 )3( dπεµ …….(3)

Therefore the total field E' induced at X due to all the dipoles in the linear array is E'= E1+ E2+ E3+………………………En

= 30dπεµ + 3

0 )2( dπεµ + 3

0 )3( dπεµ +………………… 3

0 )(ndπεµ

= 30dπεµ [1+ .............

31

21

33 ++ ]

= 30dπεµ ∑

∞

=13

1

n n where n=1,2,3,………….∞

By summation of infinite series

2.11

13 =∑

∞

=n n

E'= 30

2.1dπεµ

Therefore the total field at ‘X’ which is the internal field Ei, is the sum of the applied field (E) and the field due to all the dipoles (E') Ei=E+ E'

=E+ 30

2.1dπεµ

Thus, the combined effect of induced dipoles of neighbouring atoms is to produce a net field at the location of a given atom, which is larger than the applied field. Clausius Mosotti Equation: Let us consider a solid dielectric, which exhibits electronic polarizability. If αe is the electronic polarizability per atom, it is related to the bulk polarization P through the relation P=αeNEi

Therefore αe= iNE

P …. (1)

where N is the number of atoms per m3 and Ei is the local field.

From Lorentz field equation Ei=E+ 03ε

P …. (2)

Substituting equation (2) in equation (1) we get

αe = ]

3[

0εPEN

P

+

We have E=)1(0 −r

Pεε

Substituting for E in the above equation, we obtain


121

αe =]

3)1([

00 εεεPPN

P

r+

−

]

31

11[

1

0 +−

=

r

eN

εεα =

])1(3

2[

1

−+

r

r

εε

0)2(

)1(3εα

εε e

r

r N=

+−

03)2()1(

εα

εε e

r

r N=

+−

The above equation is known as Clausius Mosotti equation. Appendix : (Additional information) Loss Angle and Loss Tangent Let us consider a parallel plate capacitor C, constituted by plates of area A and separated by a distance ‘d’. Let a dielectric having permittivity εr fill the space between the plates. Let a sinusoidal voltage V of angular frequency ω be applied to the capacitor. The current through the capacitor is given by

RVCVjI += ω

I=jIr+Ia The above relations indicate that two kinds of current flow through the dielectric - the conduction current Ia=V/R and the displacement current Ir given by Ir=ωCV The resultant current =I 22

ar II + lags behind the displacement current by an angle δ.

In case of an ideal dielectric, R≈∞ and Ia=0, and it would not absorb electric energy. In such a case, the resultant current ‘I’ would be ahead of the voltage ‘V’ precisely by a phase angle ϕ=900 and the current would have been purely reactive current Ir.

C

φ

δ

Ir

Ia

I


122

d

AVwII rr

εε 0==

However, for a lossy dielectric, the total current is I=Ia+j|Ir| The phase angle ϕ between V and I is now slightly less than 900. The angle δ=(90-ϕ) is called the loss angle. It is given by

r

a

II

=δtan

or CRCV

RVωω

δ 1/tan ==

tan δ is known as the loss tangent. It represents the electrical power lost which is in the form of heat. Hence it is also called dissipation factor. The real power loss in the dielectric is given by PL=VIa=VIr tan δ= ωCV2 tan δ.

= δεωε tan

20

dAVr

Complex relative permittivity: In most of the materials, the dielectric behaviour is more complex indicating the presence of other sources of dielectric loss. To include losses from all sources,

I=Ia+j|Ir|=j|Ir|

−

||1

r

a

IIj = ( )δεεω tan10 j

dAVj r −

The above equation suggest that the lossy dielectric can be described with the aid of a complex relative permittivity εr

* given by εr

*=εr’(1-j tanδ)

or εr*=εr

’-jεr”

where '

"tan

r

r

εε

δ =

Therefore the product (εr’ tan δ) is known as the loss factor. A lossy dielectric is represented

by a resistance parallel to the capacitor.

C

R

φ

δ

εr’

εr*

εr’’


123

Lorentz field

Figure 16 The local field in a three dimensional solid is determined by the structure of the solid. Let us consider a dielectric slab kept in a uniform electric field, E (Figure 15). Let a molecule be at the point O and is surrounded by a spherical cavity of radius r. Let r be arbitrary but sufficiently large compared to molecular dimensions and sufficiently small compared to the dimensions of the dielectric slab. The spherical cavity contains many molecules within it. The molecule at O experiences three electric fields acting on it.

i) The external electric field E. ii) The field E1 due to induced charges on the surface of the spherical cavity. iii) The field E2 due to the molecular dipoles present in the spherical cavity.

Therefore, the total internal field intensity, Ei is given by Ei=E+ E1+E2 To calculate E1, let us imagine that the dielectric is removed from the sphere. For the actual pattern of the electric field not to be distorted, a surface electric charge should be placed on the spherical surface. At each point of the sphere, the surface charge density is given by σ=P cos θ where θ is the angle between radius vector r and the direction E. The charge on the element dS of the surface of the sphere will be dq= σdS =P cos θ dS This charge will produce electric field intensity dE1 at the centre of the sphere

dE1= 204 r

dqπε

= 204 r

Pπε

cosθ dS

This electric field can be resolved into two components: one component dE1 cos θ parallel to the direction of E and the other dE1sinθ perpendicular to the direction of E.

dE1cosθ= 204 r

Pπε

cos2 θ dS

dE1cosθ= 204 r

Pπε

cosθ sinθ dS


124

It is obvious that the perpendicular components of the upper and lower half of the sphere cancel each other and only the parallel components contribute to the total intensity E1. E1 is obtained by integrating dE1 over the whole surface area of the sphere. Thus,

E1 = dSr

PdSdE θε

θππ

∫∫ Π=

0

22

001 cos

4cos

But dS=2πr2 sinθ dθ. Therefore,

E1= θθθε

π

dP sincos2

0

2

0∫

Let cosθ= x and therefore, -sin θ dθ =dx. when θ=0, cos θ=1 and θ=π, cos π=-1

E1= -0000

1

1

3

0

1

1

2

0 362

32

231

31

2322 εεεεεεPPPPxPdxxP

==

−−

=−

−−=

−=

−−

∫

As there exists symmetrical distribution of molecular dipoles around the molecules at O within the cavity, their contribution cancel each other. Therefore E2=0. Hence the total internal field is given by Ei=E+ E1

Ei=E+ 03ε

P

The field given by the above equation is called Lorentz field or local field.


125

THERMAL CONDUCTIVITY Introduction: Matter exists in solid, liquid or gaseous states. The particular state of a substance depends on its temperature. According to kinetic theory, the molecules of a substance in the solid state have less degrees of freedom than a substance in the liquid or gaseous state. The molecules are free to move in the gaseous state. The change of state can be brought about by supplying or withdrawing heat from the substance. Ice is the solid state of water. By supplying heat to ice, it can be changed into water. Similarly by supplying heat to water, it can be converted into the gaseous state (steam). The reverse processes occur when heat is withdrawn from steam. This is true for all substances. Even permanent gases like oxygen, nitrogen, hydrogen etc. can be liquefied at low temperature.

For a given substance, the change of state takes place at a fixed temperature and pressure. When the substance changes from solid to liquid state, the heat supplied at constant temperature (called melting point) is used in overcoming the forces of intermolecular attraction. The mean molecular distance increases and the molecules are more free to move. Similarly, when the substance changes from the liquid to gaseous state at a fixed temperature (called the boiling point) and pressure, the heat supplied is used in increasing the mean molecular distance. The molecules become free to move about in the whole space available to them.

Heat can be transferred from one place to the other in three different ways i.e., Conduction, Convection and Radiation. Conduction is the process in which heat is transmitted from one point to the other through the substance without the actual motion of the particles. When one end of a metal bar is heated, the molecules at the other end vibrate with higher amplitude (kinetic energy) and transmit the heat energy from one particle to the next and so on. However, the particles remain in their mean positions of equilibrium. This process of conduction is prominent in the case of solids. Metals are good conductors of heat. The space between the two walls of a thermo flask is evacuated because vacuum is a poor conductor of heat. The air enclosed in the woollen fabric helps in protecting us from cold because air is a poor conductor of heat. Convection is the process in which heat is transmitted from one place to the other by the actual movement of the particles. It is prominent in the case of liquids and gases. Land and sea breezes are formed due to convection. Radiation is the process in which heat is transferred from one place to other directly without the necessity of the intervening medium. We get heat radiations directly from the sun without affecting the intervening medium. Example is Radiation from sun. Linear flow of heat through a long bar; the steady state Pioneer work on conduction was done by Fourier and followed by Ohm. In his arrangement one end of a uniform rod is heated with a heat source as shown in figure 1.


126

The rod has equally spaced holes along its length. A number of thermometers are inserted in them. It was found that the thermometer nearest to the hot end first registered a rise in temperature, followed by other in succession, as the heat travelled from the hotter to the colder end of the rod. Such a rise in temperature along the rod actually depends upon the quantity of heat conducted, the specific heat and density of the material of the rod. With the temperature thus changing along it, the rod is said to be in variable state of temperature.

The rod has equally spaced holes along its length. A number of thermometers are inserted in them. It was found that the thermometer nearest to the hot end first registered a rise in temperature, followed by other in succession, as the heat travelled from the hotter to the colder end of the rod. Such a rise in temperature along the rod actually depends upon the quantity of heat conducted, the specific heat and density of the material of the rod. With the temperature thus changing along it, the rod is said to be in variable state of temperature. Of the total heat entering any element or section of the rod, a part of it is absorbed or retained by the element to raise its own temperature, another part is conducted to the adjoining element and the rest of the heat is lost to the surroundings by radiation from the surface of the rod. With the passage of time, as the element gets hotter, the proportion of heat absorbed by it increases and attains a constant value, with the result that each segment of the rod attains a constant temperature. The excess of heat entering an element over that absorbed by it and conducted is lost by radiation. This state is called a steady state or stationary state. The temperature of any element of the rod in the steady state is purely a function of its distance from the hotter end and is independent of time. Since no change of temperature is involved, the flow of heat and the distribution of temperature along the rod now cease to be dependent on the density or specific heat of its material.

During the steady state, the quantity of heat flowing across the two opposite faces of an element (figure 2) is directly proportional to (i) the difference of temperature (θ1- θ2) between

Figure 1

Figure 2


127

them (ii) the area of cross section A of each face (iii) the time t, for which heat is allowed to flow through it and (iv) inversely proportional to the thickness (or length) of the element (l)

Q α tl

A )( 21 θθ −

Q = K tl

A )( 21 θθ − ……. (1)

where K is a constant and depends on the nature of the material of the rod, it is called the coefficient of thermal conductivity. The fraction (θ1- θ2)/l, which is the fall of temperature per unit length of the rod (or the rate of fall in temperature with distance) along the direction of heat flow is called the temperature gradient. If dθ is the difference in temperature for an element of length dl, then the temperature gradient is dθ/dt. Hence equation (1) is written as

Q= -KA tdldθ ………….. (2)

The negative sign indicates that there is a decrease in temperature. Coefficient of thermal conductivity of a material is numerically equal to the rate of flow of heat at steady state through the conductor of unit cross-sectional area with a unit temperature gradient between its opposite faces perpendicular to the direction of flow of heat. Units of thermal conductivity is Joules/second meter kelvin or watts/meter kelvin. (Note: It may be noted that the relations (1) and (2) holds good only when the steady state has been reached.) Searle’s method: Searle’s method is used for the determination of Thermal conductivity (K) of a good conductor. In this experiment, two conditions must be satisfied (i) the rate of heat flow through the specimen must be measurable, i.e a sufficient amount of heat must flow through it per second and (ii) the temperature gradient along it must be steep enough to be accurately measured.

In a metal the heat flow is fast enough hence the first condition is satisfied. If length ‘L’ is very much greater than cross sectional area ‘A’ of the bar, the second condition is satisfied. The arrangement of the apparatus is as shown in the figure 3.

KLDJSJIRT A Figure 3


128

‘Y’ is a cylindrical rod of a metal whose coefficient of thermal conductivity is to be determined. One end of the rod is enclosed in a steam chamber and steam is continuously passed through it. A copper tube is wound helically around the rod through which cold water enters at inlet I1 from a constant level tank and hot water exits at exit outlet I2. Thermometers T1 and T2 placed in two mercury filled cups C1 and C2 measure the temperatures θ1 and θ2 at C1 and C2 respectively. Care is taken to see that placing of the thermometers do not alter the rectilinear flow of heat. The bar is lagged with a non conducting material like felt and enclosed in a wooden box to avoid conduction and convection of heat to the atmosphere. This helps to reduce heat loss from the bar, and maintains a constant temperature gradient. This arrangement permits us to measure temperature gradient at one part of the bar and the rate of heat flow in the other part without much error. The temperatures θ3 and θ4 of the incoming and outgoing water in the copper coil are measured by the thermometers T3 and T4

respectively. Heat is conducted along the rod and when steady state is reached, the readings θ1, θ2, θ3 and θ4 will not change.

At steady state, heat flowing through any section of the rod is the same. In this state, heat flowing through the rod is taken up entirely by the circulating water and none by the apparatus.

Heat flowing per second through the rod is

Q= KAx

)( 21 θθ − …………(1)

where A is the cross section of the rod, θ1 and θ2 are the readings of thermometers T1 and T2

respectively, K is thermal conductivity of the rod and ‘x’is the length of the rod between T1

and T2. If ‘m’ is the mass of water flowing for a time ‘t’ second through the coil, then heat

absorbed by the rod is mc(θ3- θ4). Therefore, rate of heat absorbed from the rod is

[mc(θ3- θ4)]/t ………….(2) where ‘c’ is specific heat of water Under steady state conditions, the two quantities of heat are equal, that is heat absorbed by water = heat lost by the rod

Therefore KAx

)( 21 θθ − = [mc(θ3- θ4)]/t

or the themal conductivity of the good conductor is given by tAxmcK

)()(

21

43

θθθθ

−−

=

Lee’s and Charlton method:

This method is used to determine the thermal conductivity of a poor or bad conductor. The specimen disc ‘S’ (poor conductor) whose thermal conductivity is to be determined is sandwiched between a circular brass disc ‘B’ and a metal steam chamber ‘H’. The heating chamber H has a facility for steam-in and steam-out as shown in Figure 4.


129

Figure 4

Thermometers T1 and T2 are inserted into the heat chamber ‘H’ and brass disc ‘B’ respectively as shown in the figure 4. The whole assembly is suspended in air by a string such that the top of the brass plate ‘B’ is horizontal. Steam is passed continuously till the thermometers T1 and T2 attain constant temperatures. If ‘A’ and‘t’ are the surface area and thickness of the poor conductor ‘S’ respectively then, the rate of heat conducted through it is

t

KAQ)( 21 θθ −

= ………… (i)

To determine the rate of loss of heat by the brass disc ‘B’, we remove the poor conductor ‘S’ and heat brass disc ‘B’ directly from heat chamber ‘H’ till its temperature rises about 10°C above the temperature θ2. Then remove the heat chamber ‘H’ and cover the brass disc ‘B’ with thick layer of felt and allow it to cool. The temperature of the brass disc ‘B’ is noted at equal intervals of time and a graph is plotted as shown in Figure 5. A tangent to this cooling curve is drawn at a point corresponding to its previous steady temperature θ2. The rate of cooling at θ2 will be the slope of the curve at θ2

slope= 2

)( θθ

dtd

Rate of heat radiated by the brass disc ‘B’= mcdtdθ …………..(ii)

where m and c are the mass and specific heat of the brass disc ‘B’ respectively. At steady state, rate of heat conducted by the poor conductor ‘S’= rate of heat radiated by the brass disc ‘B’…..(iii)

Therefore dtdmc

tKA θθθ

=− )( 21

or thermal conductivity of the poor conductor ‘S’ is given by

)( 21 θθ

θ

−=

Adtdmct

K

T

Time (t) in minutes

θ2

Figure 5


130

Importance of Dielectrics and Thermal conductivity: Almost any type of electrical equipment employs dielectric materials in some form or another. Wires and cables that carry electric current, for example are always coated or wrapped with some type of insulating (dielectric) material. Sophisticated electronic equipment such as rectifiers, transducers, amplifiers etc. contain or are fabricated from dielectric materials. Transformer oil is a natural or synthetic substance (eg. Mineral oil, silicon oil or organic esthers) that has the ability to insulate the coils of a transformer both electrically and thermally. Thermal Conductivity is important in materials science, research, electronics, building insulation and related fields, especially where high operating temperatures are achieved. High energy generation using turbines require the use of materials with high thermal conductivity such as copper, aluminium or silver. On the other hand, materials with low thermal conductivity such as polysterene and Alumina are used in furnaces in an effort to slow the flow of heat i.e for insulation purposes. Problems: 1. Calculate the dielectric constant of NaCl, if a NaCl crystal is subjected to an electric field

of 1000 V/m and the resulting polarization is 4.3x10-8 C/m2. EP r )1(0 −= εε

0049.01010856.8

103.4)1( 319

8

0===−

−

−

xxx

EP

r εε

0049.1=rε 2. The dielectric constant of Helium at 0°C is 1.000074. The density of atoms is

2.7x1025/m3. Calculate the dipole moment induced in each atom when the gas is placed in an electric field of 3x104 V/m. Dipole moment induced in each atom is Eeα where the electronic polarizability

4125

120 104255.2

107.2000074.010854.8)1( −

−

==−

= xx

xxNr

eεε

α

Dipole moment= Cmxxxx 37441 102767.7103104255.2 −− = 3. An elemental solid dielectric material has a polarizability 7x10-40 Fm2.Assuming the

internal field to be Lorentz field, calculate the dielectric constant for the material if the material has 3x1026 atoms/m3.

7906.0

10854.83107103

3)2()1(

12

4028

0===

+−

−

−

xxxxxN e

r

r

εα

εε

33.122094.05812.2

7906.0)2()1(

==

+=−

r

rr x

ε

εε


131

4. A slab of thermal insulator is 100cm2 in cross section and 2cm thick. Its thermal conductivity is 2x10-4 cal/cm sec C. If the temperature difference between the opposite faces is 100°C, how many calories flow through the slab per day?

calxxxxxx

tKAQ 86400

2360024100100102( 4

)21 ==−

=−θθ

5. One end of a 25cm metal bar is in steam and the other end is in contact with ice. If 12g of

ice melts per minute in steady state, what is the thermal conductivity of the metal? Cross section of the bar is 25cm2 and latent heat of ice is 80 cal/g

Q=mL=12x80=960calories.

( )( ) unitsCGS

xxx

tAQd

dtKAQ 16.0

601002525960

21

21 ==−

=−

=θθ

θθ

6. Two metals made of different metals are similar in shape and dimensions. They are filled

with ice at 0°C. By the heat from outside all the ice in one vessel melts in 25 minutes and in the other vessel in 20 minutes. Compare their thermal conductivities.

K1t1=K2t2

1

21

2 tt

KK

=

2520

21 =

KK =0.8


132

Sl. No Questions Marks COs 1. What is electric dipole moment in dielectrics? 01 1 2. What is dielectric susceptibility? 01 1 3. Write the relationship between εr and χ . 01 2 4. Sketch the frequency dependence of dielectric constant. 01 2 5. What is the formula for Claussius Mossoti equation? 01 1 6. Write the formula for field due to a dipole at an axial and an

equatorial point. 02 1

7. Distinguish between polar and non-polar dielectrics. 04 1 8. What is an induced dipole? 01 1 9. What is relaxation time in context of polarisation? 01 2 10. How will the increase in temperature effect orientation

polarisation? 2

11. Define tangent loss in dielectrics. 01 1 12. If the electronic polarisability of argon is 3x10-40 Fm2. Find the

induced dipole moment when it is subjected to an electric field of 105Vm-1.

02 3

13. Write the mathematical expression for atomic polarisability of a monoatomic gas.

01 1

14. Derive the expression for thermal conductivity at steady state for a linear flow of heat through a long bar.

05 2

15. Derive the expression for a field due to a dipole at its axial and equatorial point.

07 2

16. What are dielectric materials? Explain the effect of electric field on dielectrics.

04 1

17. Explain the four types of polarizations in dielectric materials. 08 2 18. Derive an expression for internal field in case of one-dimensional

array of atoms in dielectric solids. 06 2

19. Explain the variation of dielectric constant with frequency. 06 2 20. Derive an expression for Clausius Mossotti equation. 05 2 21. Determine the thermal conductivity of a good conductor by

Searle’s method. 07 2

22. Explain Lee’s and Charlton method to determine the thermal conductivity of a poor conductor.

07 2

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