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I.B.Tech (CSE/EEE/IT & ECE)
Engineering Physics Syllabus
UNIT-I
- Crystal Structures: Lattice points, Space lattice, Basis, Bravais lattice, unit cell and lattice parameters, Seven Crystal Systems with 14 Bravais lattices , Atomic Radius, Co-ordination Number and Packing Factor of SC, BCC, FCC, Miller Indices, Inter planer spacing of Cubic crystal system.
- Defects in Crystals: Classification of defects, Point Defects: Vacancies, Substitution, Interstitial, Concentration of Vacancies, Frenkel and Schottky Defects, Edge and Screw Dislocations (Qualitative treatment), Burger’s Vector.
- Principles of Quantum Mechanics: Waves and Particles, de Broglie Hypothesis, Matter Waves, Davisson and Germer’s Experiment, Heisenberg’s Uncertainty Principle, Schrodinger’s Time Independent Wave Equation-Physical Significance of the wave Function-Particle in One Dimensional Potential Box. UNIT – II
- Electron Theory of Metals: Classical free electron theory, Derivation of Ohm’s law, Mean free path, Relaxation time and Drift velocity, Failures of Classical free electron theory, Quantum free electron theory, Fermi-Dirac distribution, Fermi energy, Failures of Quantum free electron theory.
- Band Theory of Solids: Electron in a periodic potential, Bloch Theorem, Kronig-Penny Model(Qualitative Treatment), origin of Energy Band Formation in Solids, Classification of Materials into Conductors, Semi Conductors & Insulators, Effective mass of an Electron.
- Semiconductor Physics: Intrinsic Semiconductors and Carrier Concentration, Extrinsic Semiconductors and Carrier Concentration, Fermi Level in Intrinsic and Extrinsic Semiconductors, Hall Effect and Applications. UNIT – III
- Dielectric Properties: Electric Dipole, Dipole Moment, Dielectric Constant, Polarizability, Electric Susceptibility, Displacement Vector, Types of polarization: Electronic, Ionic and Orientation Polarizations and Calculation of Polarizabilities (Electronic & Ionic) -Internal Fields in Solids, Clausius -Mossotti Equation, Piezo-electricity and Ferro- electricity.
- Magnetic Properties: Magnetic Permeability, Magnetic Field Intensity, Magnetic Field Induction, Intensity of Magnetization, Magnetic Susceptibility, Origin of Magnetic Moment, Bohr Magnetron, Classification of Dia, Para and Ferro Magnetic Materials on the basis of Magnetic Moment, Hysteresis Curve on the basis of Domain Theory of Ferro Magnetism, Soft and Hard Magnetic Materials, Ferrites and their Applications. UNIT – IV
- Lasers: Characteristics of Lasers, Spontaneous and Stimulated Emission of Radiation, Meta-stable State, Population Inversion, Einstein’s Coefficients and Relation between them, Ruby Laser, Helium-Neon Laser, Semiconductor Diode Laser, Applications of Lasers.
- Fiber Optics: Structure and Principle of Optical Fiber, Acceptance Angle, Numerical Aperture, Types of Optical Fibers (SMSI, MMSI, MMGI), Attenuation in Optical Fibers, Application of Optical Fibers, Optical fiber Communication Link with block diagram.
UNIT – V
- Nanotechnology: Origin of Nanotechnology, Nano Scale, Surface to Volume Ratio, Bottom-up Fabrication: Sol-gel Process; Top-down Fabrication: Chemical Vapor Deposition, Physical, Chemical and Optical properties of Nano materials, Characterization (SEM, EDAX), Applications.
Unit -1:Crystal Structures,Crystal Defects & Principles of Quantum Mechanics
Part-A (SAQ-2Marks)
1) Define a) Space Lattice b) Basis c) Co-ordination number d) Packing factor e) Miller Indices. a) Space lattice: is defined as an infinite array of points in three dimensions in which every point has surroundings identical to that of every other point in the array. b) Basis: Group of atoms or molecules identical in composition. Lattice + basis = crystal structure c) Co-ordination number : The no of equidistant neighbors that an atom has in the given structure .Greater the co-ordination no, the atoms are said to be closely packed. For Simple Cubic: 6, BCC: 8, FCC: 12 d) Packing factor (PF): It is the ratio of volume occupied by the atoms or molecule in unit cell to the total volume of the unit cell.
Atomic Packing Factor (APF) = Volume of all the atoms in Unit cell 𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑈𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 For Simple Cubic: 52%, BCC: 68%, FCC: 74% e) Miller Indices: are the reciprocals of intercepts made by the planes on the crystallographic axis when reduced to smallest integers.
2) Describe seven crystal systems with lattice parameters and Bravais Lattice points.
S:No Name of the crystal systems
Primitives Interfacial angles Bravais Lattice points 1 2 3 4 5 6 7 Cubic Tetragonal Orthorhombic Monoclinic Triclinic Trigonal Hexagonal
a= b= c a= b≠ c a≠ b≠ c a≠ b≠ c a≠ b≠ c a= b= c a= b≠ c
α=β=γ=90o α=β=γ=90o α=β=γ=90o α=β=90o≠ γ α≠β≠γ≠90o α=β=γ≠90o α=β=90oand γ=120o
3(P,I,F)
2(P,I)
4(P,C,I,F)
2(P,C)
1(P)
1(P)
1(P)
3) Define a) Crystal Structure b) Lattice Parameters c) Unit Cell d) Atomic radius (r).
a) Crystal structure: periodic arrangement of atoms or molecules in 3D space. b) Lattice parameters: the primitives (a,b,c) and interfacial angles (α,β,γ,) are the basic lattice parameters which determine the actual size of unit cell. c) Unit cell: is a minimum volume cell which on repetition gives actual crystal structure.
Co-ordination number =
Nearest neighbor distance = 𝑎√ 2 Lattice constant = a= 4𝑟 √ Number of atoms per unit cell = v= 1 Volume of all atoms in unit cell = v = 2 × 4/3 π r^3
Volume of unit cell =V= a^3 =( 4𝑟 √
)^3
Atomic Packing Factor is
2×^43 𝜋𝑟^3 (4𝑟√3)
Ex: - Li, Na, K, and Cr. Face centered structure (FCC) In FCC there are 8 atoms at 8 corners of the unit cell and 6 atoms at 6 faces. Considering the atoms at the face center as origin, it can be observed that this face is common to 2 unit cells and there are 12 points surrounding it situated at a distance equal to half the face diagonal of the unit cell.
Co- ordination number = N= 12 Number of atoms in unit cell = 8 ×1/8+ 6×1/2=
Lattice constant =a=2r = √2𝑎 2
Volume of the unit cell =V= a^3 =( 4𝑟 √2)
3
Volume of all atoms in unit cell = v= 4 × 4 3 𝜋𝑟
3
Atomic Packing Factor = 𝑣 𝑉 =^
4×^43 𝜋𝑟^3 (4𝑟√2)
Ex:- Cu, Al, Pb, and Ag. By the above values of Atomic packing factors we can say that FCC is the closest packed structure of all the three cubic structures.
2) Explain the significance of Miller indices and derive an expression for interplaner distance in terms of Miller indices for a cubic Structure.
Miller indices: are the reciprocals of intercepts made by the crystal planes on the crystallographic axes when reduced to smallest integers. Important features of Miller indices: Miller indices represent a set of parallel equidistant planes. All the parallel equidistant planes have the same Miller indices. If a plane is parallel to any axis, then the plane intersects that axis at infinity and Miller indices along that direction is zero. If the miller indices of the two planes have the same ratio (844,422,211), then the planes are parallel to each other. If a plane cuts an axis on the – ve side of the origin, then the corresponding index is – ve, and is indicated by placing a minus sign above the index. Ex: if a plane cuts – ve y-axis, then the miller index of the plane is (h 𝑘̅ l) Derivation: Consider a crystal in which the three axes are orthogonal and the intercepts are same. Take ‘o’ as origin, and the reference plane passes through the origin i.e entirely lies on the axis. The next plane ABC is to be compared with the reference plane which makes the intercepts 𝑎 ℎ ,^
𝑏 𝑘 ,^
𝑐 𝑙 on x,y,z axes respectively. Let (h k l) be the miller indices. Let ON=d be a normal drawn to the plane ABC from origin ‘o’ which gives the distance of separation between adjacent planes. Let the normal ON makes an angles α,β,γ with x,y,z axes respectively. Angle α= NOA, angle β=NOB, angle c = NOC.
Then form Δ le NOA Cos α = 𝑂𝑁 𝑂𝐴 =^
𝑑 𝑎/ℎ =^
𝑑ℎ 𝑎 Similarly cos β = 𝑂𝑁 𝑂𝐵 =^
𝑑 𝑏/𝑘 =^
𝑑𝑘 𝑏 Cos γ = 𝑂𝑁 𝑂𝐶 =^
𝑑 𝑐/𝑙 =^
𝑑𝑙 𝑐 According to cosine law of directions, 𝑐𝑜𝑠^2 α + 𝑐𝑜𝑠^2 β +𝑐𝑜𝑠^2 γ = 1 Therefore ( 𝑑ℎ 𝑎 )
2
2
2 = 1
𝑑^2 [
ℎ^2 𝑎^2 +^
𝑘^2 𝑏^2 +^
𝑙^2 𝑐^2 ] = 1 In a cubic crystal a = b = c, Therefore 𝑑^2 [ ℎ^2 𝑎^2 +^
𝑘^2 𝑎^2 +^
𝑙^2 𝑎^2 ] = 1
In case of ionic crystals imperfections appear in crystals while maintaining the electrical neutrality. Two types of defects (point defects) occur in ionic crystals.
- Frenkel defect 2.Schottky defect.
Frenkel defect: When an ion is displaced from a regular lattice site to an interstitial site is called Frenkel defect. Generally cations which are small in size are displaced to an interstitial site as the interstitial space is small .A Frenkel imperfection does not change the overall electrical neutrality of the crystal. Schottky defect: A pair of one cation and one anion missing from the original lattice site on to the surface of the crystal so that charge neutrality is maintained in the crystal is called Schottky defect.
4) Write a short note on line defects. (or) What are edge and screw dislocations?
Line defects (or) dislocations (one dimensional defect) are defined as the disturbed region between the two perfect parts of the crystal and these defects are formed in the process of deformation. Edge dislocation: A perfect crystal is composed of several parallel vertical planes which are extended from top to bottom completely and parallel to side faces. The atoms are in equilibrium positions and the bond lengths are in equilibrium value. If one of the vertical planes does not extend from top to bottom face of the crystal, but ends in midway within the crystal, then crystal suffers with a dislocation called edge dislocation. In imperfect crystal all the atoms above the dislocation plane are squeezed together and compressed there by the bond length decreases. And all the atoms below the dislocation plane are elongated by subjecting to the tension and thereby the bond length increases. There are two types of edge dislocation. They are 1.Positive edge dislocation 2.Negative edge dislocation. Positive edge dislocation: if the vertical plane starts from top of the crystal and never reaches to the bottom. Negative edge dislocation: if the vertical plane starts from bottom of the crystal and never reaches top.
Screw dislocation: Atoms are displaced in two separate planes perpendicular to each other or defects forming a spiral around the dislocation line. A screw dislocation marks the boundary between slipped and unslipped parts of the crystal that can be produced by cutting the crystal partway and then sheering down one part relative to the other by atomic spacing horizontally.
5) What is a Burger’s vector? Explain the significance of Burger’s vector.
Burger’s vector: It gives the magnitude and direction of dislocation line.
𝐹 = 𝑛𝐸𝑣 − 𝐾𝑇(𝑁𝑙𝑜𝑔𝑁 − (𝑁 − 𝑛) log(𝑁 − 𝑛) − 𝑛𝑙𝑜𝑔𝑛)…. (7) At thermal equilibrium, the free energy is minimum and constant .i.e. 𝑑𝐹 𝑑𝑛 = 0^ in (7) 𝑑𝐹 𝑑𝑛 = 𝐸𝑣^ − 𝐾𝑇(0 − (𝑁 − 𝑛)^
1 (𝑁−𝑛) (−1)-log(𝑁 − 𝑛) (−1) − 𝑛^
1 𝑛 − 𝑙𝑜𝑔𝑛)
0 = 𝐸𝑣 − 𝐾𝑇(1 + log(𝑁 − 𝑛) − 1 − 𝑙𝑜𝑔𝑛)
0 = 𝐸𝑣 − 𝐾𝑇 [𝑙𝑜𝑔 (
)]
Taking exponential on both sides
𝑒𝐸𝑣/𝐾𝑇^ =
The number of vacancies in a crystal is very small when compared with the number of atoms.𝑁 ≫ 𝑛 𝑁 − 𝑛 ≅ 𝑁 Therefore 𝑒−𝐸𝑣/𝐾𝑇 𝑛 𝑁 𝑛 = 𝑁𝑒𝑥𝑝−𝐸𝑣/𝐾𝑇
7)Derive an expression for the energy required to create a Frenkel defect.(or) Derive an expression for the no of Frenkel defects created in a crystal a at a given temperature.
Let ‘N’ be the number of atoms, ’Ni’ be the number of interstitial atoms, let ‘ Ei ’s the energy required to create ‘n’ number of vacancies and the total energy required is u = nEi…..(1) The total number of ways in which Frenkel defects can be formed is given by p = Nnc^ × Ninc
p = N! (N−n)!n! ×^
Ni! (Ni−n)!n!........ (2) The increase in entropy (s) due to Frenkel defect is given by s = K logp
S = Klog [ N! n!(N−n)! ×^
Ni! (NI−n)!n!]....... (3) This increase in entropy produces change in Free energy F = u − TS……….. (4) Substitute (1),(3) in (4)
F = nEi − KTlog [
N!
n! (N − n)!
×
Ni! (Ni − n)! n!
]
Using Sterling’s approximation,logx! = xlogx − x
F = nEi − KT [log
N!
n! (N − n)!
Ni! (NI − n)! n!
]
F = nEi − KT[logN! − logn! − log(N − n)! + logNi! − log(Ni − n)! − logn!] F = nEi − KT[(NlogN − N) − (nlogn − n) − [(N − n) log(N − n) − (N − n)] + NilogNi − NI − [(Ni − n) log(Ni − n) − (Ni − n)] − (nlogn − n)] 𝐹 = nEi − KT[NlogN − N − nlogn + n − (N − n) log(N − n) + (N − n) + NilogNI − Ni − (Ni − n) log(Ni − n) + (Ni − n) − nlogn + n] F = nEI − KT[NlogN + NilogNi − (N − n) log(N − n) − (Ni − n) log(Ni − n) − 2nlogn] Differentiating w.r.to ‘n’, and equating to 0, we get
dF
dn
= Ei − KT [0 + 0 − [(N − n)^
(N − n)
(−1)) + log(N − n) (−1)]
− [(Ni − n)
(Ni − n)
(−1) + log(Ni − n) (−1)] − 2 [n ×
n
0 = Ei − KT[1 + log(N − n) + 1 + log(Ni − n) − 2 − 2logn]
0 = Ei − KT [log
(N − n)(Ni − n) n^2
]
Ei = KT [log
(N − n)(Ni − n) n^2
]
As n ≪ N, N − n ≅ N, similarly Ni − n ≅ Ni
Ei = KTlog (
NNi n^2
Thus Ei = KT[logNNi − 2logn] Ei
KT
= log(NNi) − 2logn
2logn = log(NNi) −
Ei KT
logn =
log(NNi) −
Ei 2KT Taking exponentials on both sides
n = (NNi)
1 (^2) exp
−Ei 2KT 8)Derive an expression for the energy required to create a Schottky defect.(or) Derive an expression for the no of Schottky defects created in a crystal a at a given temperature.
Let ‘N’ be the number of atoms, ’Ep’ is the energy required to create a pair of vacancies and ‘n’ be number of vacancies created. The total energy required to create vacancies is U= nEp…………….(1) The number of ways in which ‘n’ vacancies created is p = Nnc^ × Nnc^ = (Nnc^ )^2
p = [
N!
(N − n)! n!
]
2
The relation between the disorder parameter ‘p’ and entropy ‘s’ is given by
s = Klogp = klog [ N! (N−n)!n!]
2 ………….(2)
By applying Sterling’s approximation
log [
N!
(N − n)! n!
]
2 = 2[logN! − log(N − n)! − logn!]
= 2[NlogN − N − ((N − n) log(N − n) − (N − n)) − (nlogn − n)] = 2[NlogN − N − (N − n) log(N − n) + N − n − nlogn + n] = 2[NlogN − (N − n) log(N − n) − nlogn] There fore s = 2K[NlogN − (N − n) log(N − n) − nlogn]…..(3) Free energy of the atoms in the crystal is given by F = U − TS…..(4) Substitute (1),(3) in (4) F = nEp − 2KT[NlogN − (N − n) log(N − n) − nlogn]
Differentiating above equation w.r.to ‘n’ and equating it to zero, we get dF
dn
= Ep − 2KT[log(N − n) + 1 − logn − 1 = 0]
By substituting the values of h=6.625× 10−34𝐽𝑠𝑒𝑐, m 0 = 9.1× 10−31𝐾𝑔 and c= charge of electron=1.6× 10−19C λ =
√V A
(^0) …… (8), Where V= in volt and λ = in A 0
Experimental validity: Davison and Germer Experiment: The first experimental evidence of the wave nature of atomic particles was proved by C.J Davison and L.H Germer in 1927. They were studying scattering of electrons by a metal target and measuring the density of electrons scattered in different directions.
From fig, the electron beam from electron gun which consists of a tungsten filament ‘F’ heated by a low tension battery ‘B 1 ’ are accelerated to a desired velocity by applying suitable potential from a high tension battery ‘B 2 ’. The accelerated electrons are collimated into a fine beam by allowing them to pass thorough a system of pinholes in the cylinder ‘C’. The fast moving electron beam is made to strike the target (nickel crystal) capable of rotating about an axis perpendicular to the plane of diagram. The electrons are scattered in all directions by atomic planes of a crystal and intensity of scattered electron beam in all directions can be measured by the electron collector and can be rotated about the same axis as the target. The collector is connected to a sensitive galvanometer whose deflection is proportional to the intensity of electron beam entering the collector.
When electron beam accelerated by 54 V was directed to strike the given nickel crystal, a sharp max in the electron diffraction occurred at an angle of 50^0 with the incident beam. The incident beam and the diffracted beam make an angle of 65^0 with the family of Bragg’s planes. The whole instrument is kept in an evacuated chamber. The spacing of planes in Nickel crystal as determined by x-ray diffraction is 0.091nm From Bragg’s law 2dsinθ = n λ i.e 2 × 0.091 × 10−9^ × 𝑠𝑖𝑛65°^ = 1 × λ λ =0.615nm Therefore for a 54 V electron beam, the de-Broglie wavelength associated with the electron is given by λ. =
√^ A°= 0.166nm This wavelength agrees well with the experimental value. Thus division experiment provides a direct verification of de-Broglie hypothesis of wave nature of moving particles.
10) Explain the Physical significance of 𝛙 (𝐰𝐚𝐯𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧).
The wave function 𝛙 enables all possible information about the particle. 𝛙 is a complex quantity and has no direct physical meaning. It is only a mathematical tool in order to represent the variable physical quantities in quantum mechanics. Born suggested that, the value of wave function associated with a moving particle at the position co-ordinates (x,y,z) in space, and at the time instant ‘t’ is related in finding the particle at certain location and certain period of time ‘t’. If 𝛙 represents the probability of finding the particle, then it can have two cases. Case 1: certainty of its Presence: +ve probability Case 2: certainty of its absence: - ve probability, but – ve probability is meaningless, Hence the wave function 𝛙 is complex number and is of the form a+ib Even though 𝛙 has no physical meaning, the square of its absolute magnitude |𝛙^2 | gives a definite meaning and is obtained by multiplying the complex number with its complex conjugate then |𝛙^2 | represents the probability density ‘p’ of locating the particle at a place at a given instant of time. And has real and positive solutions. 𝛙 (𝐱, 𝐲, 𝐳, 𝐭) = 𝐚 + 𝐢𝐛 𝛙∗(𝐱, 𝐲, 𝐳, 𝐭) = 𝐚 − 𝐢𝐛 𝐩 = 𝛙𝛙∗^ = |𝛙^2 | = 𝑎^2 + 𝑏^2 𝑎𝑠 𝑖^2 = − Where ‘P’ is called the probability density of the wave function. If the particle is moving in a volume ‘V’, then the probability of finding the particle in a volume element dv, surrounding the point x,y,z and at instant ‘t’ is Pdv
∫|𝛙^2 |𝑑𝑣 = 1 𝑖𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑒𝑙 𝑖𝑠 𝑝𝑟𝑒𝑠𝑒𝑛𝑡
∞
− ∞
. = 0 if particle does not exist
This is called normalization condition.
11 ) Describe Heisenberg’s uncertainty principle?
According to Classical mechanics, a moving particle at any instant has fixed position in space and definite momentum which can be determined simultaneously with any desired accuracy. This assumption is true for objects of appreciable size, but fails in particles of atomic dimensions.
𝑑^2 ψ 𝑑𝑥^2 +^
4𝜋^2 𝑚^2 𝑣^2 ℎ^2 ψ^ = 0…. (4)
For a moving particle, the\ total energy is 𝐸 = 𝑈 + 𝑉 𝑖. 𝑒 𝑈 = 𝐸 − 𝑉 ….(5)
Where E= total energy, V= potential energy, U= kinetic energy = 1 2 𝑚𝑣
2
2𝑚𝑢 = 𝑚^2 𝑣^2 …. (6), substitute (5) in (6) 2𝑚(𝐸 − 𝑉) = 𝑚^2 𝑣^2 …. (7) Substitute (7) in (4) 𝑑^2 ψ
𝑑𝑥^2
4𝜋^2 2𝑚(𝐸 − 𝑉)
ℎ^2
ψ = 0
𝑑^2 ψ 𝑑𝑥^2 +^
8𝜋^2 𝑚(𝐸−𝑉) ℎ^2 ψ^ = 0…. (8) This equation is known as Schrodinger’s time independent wave equation in one dimension. In three dimensions, it can be written as
∇^2 ψ + 8𝜋^2 𝑚(𝐸−𝑉) ℎ^2 ψ^ = 0… (9)
∇^2 ψ +
ℎ^2
ψ = 0 For a free particle, the P.E is equal to zero i.e V=0 in equation (9) Therefore the Schrodinger’s time independent wave equation for a free particle is
∇^2 ψ +
8𝜋^2 𝑚𝐸
ℎ^2
ψ = 0
13) Derive an expression for the energy states of a Particle trapped in 1-Dimensional potential box:
The wave nature of a moving particle leads to some remarkable consequences when the particle is restricted to a certain region of space instead of being able to move freely .i.e when a particle bounces back and forth between the walls of a box. If one – dimensional motion of a particle is assumed to take place with zero potential energy over a fixed distance, and if the potential energy is assumed to become infinite at the extremities of the distance, it is described as a particle in a 1-D box, and this is the simplest example of all motions in a bound state. The Schrodinger wave equation will be applied to study the motion of a particle in 1-D box to show how quantum numbers, discrete values of energy and zero point energy arise. From a wave point of view, a particle trapped in a box is like a standing wave in a string stretched between the box’s walls. Consider a particle of mass ‘m’ moving freely along x- axis and is confined between x=0 and x= a by infinitely two hard walls, so that the particle has no chance of penetrating them and bouncing back and forth between the walls of a 1-D box.
If the particle does not lose energy when it collides with such walls, then the total energy remains constant. This box can be represented by a potential well of width ‘a’, where V is uniform inside the box throughout the length ‘a’ i.e V= 0 inside the box or convenience and with potential walls of infinite height at x=0 and x=a, so that the PE ‘V’ of a particle is infinitely high V=∞ on both sides of the box. The boundary condition are 𝑣(𝑥) = 0 , 𝜓(𝑥) = 1𝑤ℎ𝑒𝑛 0 < 𝑥 < 𝑎…. (1) 𝑣(𝑥) = ∞ , 𝜓(𝑥) = 0𝑤ℎ𝑒𝑛 0 ≥ 𝑥 ≥ 𝑎… (2) Where 𝜓(𝑥) is the probability of finding the particle. The Schrodinger wave equation for the particle in the potential well can be written as 𝑑^2 ψ 𝑑𝑥^2 +^
8𝜋^2 𝑚 ℎ^2 E^ ψ^ = 0, as V = 0 for a free particle… (3) In the simplest form eqn (3) can be written as
𝑑^2 ψ 𝑑𝑥^2 + 𝑘
(^2) ψ = 0…. (4) Where k= propagation constant and is given by 𝑘 = √^8 П^2 𝑚𝐸 ℎ^2 ….(5) The general solution of equation (4) is ψ(x) = Asinkx + Bcoskx… (6) Where A and B are arbitrary constants, and the value of these constant can be obtained by applying the boundary conditions. Substitute eqn(1) in (6) 0 = 𝐴𝑠𝑖𝑛𝑘(0) + 𝐵𝑐𝑜𝑠𝑘(0) → B=0 in eqn (6) ψ(x) = Asinkx… (7) Substituting eqn (2) in (7) 0 = 𝐴𝑠𝑖𝑛𝑘(𝑎) → 𝐴 = 0 𝑜𝑟 𝑠𝑖𝑛𝑘𝑎 = 0, But ‘A’ ≠ 0 as already B=0 & if A= 0, there is no solution at all.
Therefore sinka=0( if sinθ=0,then general solution is θ=nП), i.e Ka=nП
𝑛𝜋 𝑎 …….(8), Where^ n= 1,2,3,4,…and n≠0,because if^ n=0,k=0,E=0 everywhere inside the box and the moving particle cannot have zero energy.
From (8) 𝑘^2 = ( 𝑛𝜋 𝑎 )
2
From (5) 8 П^2 𝑚𝐸 ℎ^2 =^
𝑛^2 𝜋^2 𝑎^2
𝑛^2 ℎ^2
8𝑚𝑎^2
𝑛^2 ℎ^2 8𝑚𝑎^2 = the discrete energy level… (9)
The lowest energy of a particle is given by putting n=1 in the eqn (9), 𝐸1= ℎ^2 8𝑚𝑎^2 = lowest energy, minimum energy, ground state energy or zero point energy of the system.
𝐸𝑛=𝑛^2 𝐸 1
For cu at 20˚ c, = 1.69x 10ˉ 8 ohm-𝑚−1,eˉ concentration n = 8.5 x 1028 𝑚^3.
= 𝟐. 𝟓 × 𝟏𝟎−𝟗𝐦
The experimental value of was obtained nearly 10 times its theoretical value. So classical theory could not explain the large variation in ‘' value. Resistivity: - According to the classical free electron theory, the resistivity is given by the
equation, = √3𝑘𝑇𝑚 𝑛𝑒^2 Which means the resistivity is proportional to the square root of absolute temperature. But according to theory at room temperature it does not change up to 10K and in intermediate range of temperature is proportional to T^5. The conductivity of semiconductors and insulators cannot be explained by the free electron theory.
2) What are the applications of Hall Effect?
Determination of the type of Semi-conductors: The Hall coefficient 𝑅𝐻 is -ve for an n-type semiconductor and +ve for p-type semiconductor. Thus the sign of Hall coefficient can be used to determine whether a given Semi-conductor is n or p-type. Calculation of carrier concentration:
𝑅𝐻 = I =^
I 𝑛𝑒 (for eˉ s) 𝑅𝐻 = I 𝑒 (for holes) => n = I 𝑒𝑅 ᴎ =>^ ^ =^
I 𝑒𝑅 ᴎ Determination of Mobility: σ = neμ μ = σ ne = σ^ 𝑅𝐻 μ = σ 𝑅𝐻 Measurement of Magnetic Flux Density: Hall Voltage is proportional to the magnetic flux density B for a given current I. So, Hall Effect can be used as the basis for the design of a magnetic flux density metal.
3) Define Fermi energy level.
The highest energy level that can be occupied by an electron at 0 K is called Fermi energy level. It is denoted by 𝐸𝐹.
4) Distinguish between conductors, Insulator and Semiconductors.
Solids are classified into three types based on energy gap. Conductors(metals) Insulators Semiconductors In case of conductors, valence band and conduction band almost overlap each other and no significance for energy gap. The two allowed bands are separated by Fermi energy level. Here there is no role in Eg, as a result conduction is high.
Conductors insulators Semiconductors
In case of insulator, valence band and conduction band are separated by large energy gap, hence conductivity is zero. In case of semiconductors, the valence band and conduction band are separated by relatively narrow energy gap; hence the conductivity lies in between conductors and insulators.
5) Define the following terms. i. Collision time ii. Relaxation time iii. Mean free path iv .Drift velocity v. Mobility
i. Collision time: The time taken by the electron to complete one collision with the +ve ion center. ii.Relaxation time: The time taken by the electron to reduce its velocity to 1/e of its initial velocity. iii.Mean free path: The average distance covered by the electron between two successive collisions. iv.Drift velocity: The steady state velocity of the electrons in the presence of Electric field. v.Mobility: The steady state velocity of the electrons per unit electric field.
Part- B (Descriptive- 10marks)
1) What are the salient features of classical free electron theory of metals? What are its drawbacks? Drude and Lorentz proposed free electron theory of on the basis of some assumptions. In conductors (metals), there are large number of free electrons moving freely within the metal i.e. the free electrons or valence electrons are free to move in the metal like gaseous molecules, because nuclei occupy only 15% metal space and the remaining 85% space is available for the electrons to move. Since free electrons behave like gaseous molecules, the laws of kinetic theory of gases can be applied. The mean K.E of a free electron is equal to that of a gas molecule at same temperature. In the absence of any electric field, the electrons move randomly while undergoing scattering at +ve ion centers. The collisions are regarded as elastic (no loss of energy). The electron speeds are distributed according to the Maxwell- Boltzmann distribution law. When an electric filed is applied, the free electrons are accelerated in a direction opposite to that of the field. The free electrons are confined to the metal due to surface potential. The electrostatic force of attraction between the + ve ion cores and the free electrons is assumed to be negligible. Drawbacks:
CB
Eg =5.4 eV
VB
CB
Eg =1.1 eV
VB
CB
VB
