Engineering Physics
( 17PHY12/22)
MODULE WISE
QUESTIONS AND
ANSWERS SEM-I/II VTU ENGINEERING PHYSICS
C.REDDAPPA, M.Sc; B.Ed; Prof. & HOD,CBIT,KOLAR.
IMPORTANT IA TEST & VTU EXAMINATION QUESTIONS
MODULE-1 Modern Physics and Quantum Mechanics
1. State Planck’s quantum hypothesis(postulates) and explain black body radiation spectra based on planck’s radiation law. 2. State Planck’s law of radiation and deduce Wein’s law and Rayleigh-Jeans law from Planck’s law of radiation. 3. Explain Compton effect experimentally. 4. What are matter(de-Broglie) waves ? Mention the properties of matter waves. 5. Derive the relation between Phase velocity and Group velocity. 6. Define Phase velocity and Group velocity and show that Group velocity is equal to Particle velocity. 7. State and explain Heisenberg’s uncertainty Principle and Show that electrons non exist in the nucleus using Heisenberg’s uncertainty Principle . 8. Set up Schrodinger 1D( one dimensional) time independent wave equation. 9. Apply Schrodinger time independent wave equation to a particle in an infinite depth potential well/Box to find eigen energies and eigen functions. 10. Mention the properties of wave function.
MODULE-2 Electrical properties of Materials.
1. Explain the terms : Drift velocity, Mean collision time, Mean free path, Relaxation time. 2. Explain the failures of CFET(Classical free electron theory) 3. State the assumptions of QFET(Quantum free electron theory) and explain the merits of QFET. 4. Discuss the dependence of Fermi factor on temperature and energies. 5. Derive the expression for electrical conductivity based on QFET. 6. State law of mass action and Derive an expression for concentration of electrons and holes ( carrier density) in intrinsic semiconductors. 7. Write a note on Meissner’s effect and Maglev vehicles. 8. Distinguish between Type-I and Type-II superconductors. 9. Explain BCS theory qualitatively. 10. State and explain Matheissen’s rule ( Dependence of resistance on impurities and temperature) 11. Write a note on Fermi-Dirac statistics.
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ENGINEERING PHYSICS (15PHY12/22)
SEMESTER - I / II [ As per CBCS Scheme with effect from the academic year 2015-2016 ]
MODULE-1 : MODERN PHYSICS AND QUANTUM MECHANICS
MODERN PHYSICS: Q:What is meant by Black Body radiation spectra ? explain it briefly.
E
Rayleigh-Jean’s law
푬풎 Experimental B.B.R spectra
Wein’s law
ퟎ 흀풎 흺
1. If we plot intensity of the thermal radiations emitted by a Black Body against the corresponding wave lengths at different temperatures, we get Black body radiation (BBR)spectra as shown in the graph.
2. From the graph it is clear that the wavelength (휆푚) corresponding to the maximum intensity 푬풎varies inversely as the absolute temperature(T ) of the black body. This is ퟏ called Wein’s displacement law , 풊풆: 흀 ∝ or 흀 푻 = 풄풐풏풔풕풂풏풕 According to this law 풎 푻 풎 ′휆푚′ shifts towards shorter wavelength side with the increase of ‘ T ’ of the body.
3. Wein’s law states that the energy density in the wave length interval λ & λ +dλ −5 −퐶2/휆푇 is 푈휆 dλ = 퐶1휆 푒 dλ where 퐶1 & 퐶2 are constants. Wein’s law explained only the shorter wavelength region of BBR spectra below 휆푚 and it failed to explain the longer wavelength region of the BBR spectra beyond 휆푚.
4. Rayleigh-Jean’s law states that the energy density in the wave length interval λ & 8휋푘푇 λ +dλ is 푈 dλ = dλ ,where k = Boltsmann’s constant. Rayleigh-Jean’s law 휆 휆4 explained the longer wavelength region of BBR spectra beyond 휆푚 and it failed explain the shorter wavelength region of BBR spectra below 휆푚.
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5. Planck’s law states that the energy density in the wavelength interval λ & λ +dλ is
8휋ℎ푐 1 푈휆 dλ = ℎ푐 dλ . 휆5 ( ) 푒 휆푘푇 −1 Planck’s law explained the BBR spectra completely ie: both shorter and longer wavelength sides of the spectra.
Q: State Planck’s quantum postulates and explain Planck’s explanation of BBR specta.
The quantum postulates are : 1. The black body is made up of a large number of simple Harmonic oscillating particles, which can oscillate in all possible frequencies. 2. An oscillating particle can have a set of energies ,which are the integral multiples of a
lowest finite quanta of energy(h휈) i.e: 퐸푛 = n.hν ,where n = Quantum number. 3. The oscillating particles absorb or emit energy in discrete units of hν, only when they undergo transitions between any two allowed energy states.
E
Rayleigh-Jean’s law
푬풎 Experimental B.B.R spectra Wein’s law
ퟎ 흀풎 흺
Planck based on quantum postulates, derived an expression for energy density in the wavelength range λ and λ+dλ called Planck’s law given by
8휋ℎ푐 1 dλ = ℎ푐 dλ. 푈휆 휆5 ( ) 푒 휆푘푇 −1
Based on this, Planck explained the BBR spectrum completely ie: both shorter and longer wavelength side of the spectrum.
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Q:Deduce/derive Wein’s law & Rayleigh-Jeans law from planck’s law of radiation or Reduce Planck’s law to Wein’s law and Rayleigh-Jeans law 8휋ℎ푐 1 Planck’s law of radiation is given by 푈휆dλ = ℎ푐 dλ ……………(1) 휆5 ( ) 푒 휆푘푇 −1
a) Wein’s law: ℎ푐 ( ) For shorter wave lengths, ‘λ ‘is very small ,hence 푈휆 & 푒 휆푘푇 are
ℎ푐 ℎ푐 ℎ푐 ( ) ( ) ( ) Large i.e: 푒 휆푘푇 ≫ 1 so that 푒 휆푘푇 − 1 = 푒 휆푘푇 ……….(2)
8휋ℎ푐 1 ∴ from eqn 1 &2,we get 푈휆dλ = ℎ푐 dλ 휆5 ( ) 푒 휆푘푇 푪 − ퟐ ℎ푐 ∴ 푼 dλ = 푪 흀−ퟓ풆 흀푻 dλ 푇ℎ𝑖푠 𝑖푠 푤푒𝑖푛’푠 푙푎푤 .where 퐶 = 8휋ℎ푐 & 퐶 = 흀 ퟏ 1 2 푘
b) Rayleigh-Jeans law: For longer wave lengths, ‘λ ‘is very large, ℎ푐 ℎ푐 ( ) ( ) hence 푒 휆푘푇 𝑖푠 푣푒푟푦 푠푚푎푙푙 ,Expanding 푒 휆푘푇 as power series , ℎ푐 ( ) ℎ푐 ℎ푐 we get , 푒 휆푘푇 = 1+ ( ) + ( )2 +………. 휆푘푇 휆푘푇 ℎ푐 ℎ푐 = 1 + ( ) , 푛푒푔푙푒푐푡𝑖푛푔 ℎ𝑖푔ℎ푒푟 푝표푤푒푟푠 표푓 ( ) 휆푘푇 휆푘푇 ℎ푐 ( ) ℎ푐 𝑖푒: 푒 휆푘푇 − 1 = ………(3) 휆푘푇
8휋ℎ푐 1 ∴ from eqn 1 &3,we get 푈휆dλ = ℎ푐 dλ 휆5 휆푘푇 ퟖ흅풌푻 풊풆: 푼 dλ = dλ , 푇ℎ𝑖푠 is Rayleigh-Jeans law 흀 흀ퟒ
Q: What is meant by “ultraviolet catastrophe” or failures of Rayleigh-Jeans law ?
1. As per Rayleigh-Jeans law, 푈휆 → ∞ 푎푠 휆 → 0
2. 퐵푢푡 experimentally observed that 푈휆 → 0 푎푠 휆 → 0.
3. This failure of R-J’s law beyond ultra-violet region is called “ultraviolet catastrophe”.
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Q:Explain Compton effect experimentally & mention it’s significance.
Bragg’s spectrometer
Slits X - rays Collimated X - rays 휃
푆1 푆2 ∅
Graphite target Recoil electron
1. Compton experimental arrangement is as shown in the diagram. 2. X-ray photons of wave length λ from X-ray tube are collimated by passing through two slits S1 &S2. 3. The collimated X-rays are made to fall on graphite target. 4. X-rays collide with the electrons of the target at rest and transfer part of their energy resulting in the recoil of electrons in a direction making an angle ∅ 푎푛푑 푆푐푎푡푡푒푟푒푑 푋 − 푟푎푦푠 푚푎푘푒푠 푎푛 푎푛푔푙푒 휃 푟푒푠푝푒푐푡𝑖푣푒푙푦 푤𝑖푡ℎ the incident X-rays direction. 5. The intensity of the scattered X-rays in different directions are measured using Bragg’s spectrometer. 6. Compton calculated the wavelengths of scattered X-rays at different scattering angles and found that scattered X-rays consists of two components namely ‘unmodified’ component having the same wavelength(λ) as incident X-rays & ’modified’ component having slightly higher wavelength 휆′ 7. This scattering of X-rays due to recoil of electrons is called “Compton effect”. 8. The change in wavelength (휆′ − 휆)is called ‘Compton wavelength’ ,which depends only 휃 푎푛푑 𝑖푛푑푒푝푒푛푑푒푛푡 표푓 ힴ. 9. Based on conservation of energy & momentum laws, considering the collision between X- rays & electrons as ‘particle-particle’ collision , Compton wavelength is given by ℎ (휆′ − 휆) = ∆휆 = (1−퐶표푠휃) 푚0퐶 ℎ where = 휆0 called “Compton wavelength of electron”= 0.02426 Å(constant) 푚0퐶 10. ∆흀 푣푎푟ies from ‘0’ for 휃 = 0° 푡표 2휆0 for 휃 = ퟏퟖퟎ° 11. 퐶표푚푝푡표푛 푒푓푓푒푐푡 푠𝑖푔푛𝑖푓𝑖푒푠 푡ℎ푒 푝푎푟푡𝑖푐푙푒 푛푎푡푢푟푒 표푓 푋 − 푟푎푦푠(푙𝑖푔ℎ푡)
Q: What is meant by Dual nature of matter ? Matter is made up of particles like electrons, protons, neutrons, atoms etc ,also waves are associated with these matter particles under suitable conditions. Thus matter exhibiting both particle & wave nature is called dual nature of matter.
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Q: What are de-Broglie (matter) waves ? de-Broglie(matter) waves are the waves associated with material particles in motion.
Q: Derive an expression for de-broglie wavelength(λ). According to Einstein’s mass-energy relation E=m퐶2……(1) where m=mass, C=speed of light. ℎ퐶 also, from Planck’s quantum theory, E = …………(2) where h=planck’s contant. 휆 ℎ퐶 ν=frequency. From eqns 1&2, m퐶2 = 휆 풉 ∴ 흀 = 풎푪
Similarly, for a particle of mass ‘m’ moving with a velocity ‘풗 ′,
풉 풉 De-broglie wavelength, λ = or λ = 풎퓥 푷
Q: De-broglie’s wavelength of an electron accelerated in a potential difference ’V’ volt/ ퟏퟐ.ퟐퟖ Show that λ = Å for an electron accelerated in a potential difference ’V’ volt √푽
The K.E of an electron of mass ‘m’,charge ‘e’ accelerated in a potential ‘V’ volt is given 1 by 푚푣2 = 푒푉 , multiplying both Nr. and Dr. of LHS by ‘m’ 2 푚2푣2 we get, = 푒푉 2푚 𝑖푒; 푚2푣2 = 2푚푒푉 푃2 = 2푚푒푉 ∵ 푚푣 = 푃 푷 = √ퟐ풎풆푽 풉 풉 But λ = ∴ 흀 = 푷 √ퟐ풎풆푽 ퟔ.ퟔퟐퟓ풙10−34 = √ퟐ풙 ퟗ.ퟏ풙ퟏퟎ−ퟑퟏ풙ퟏ.ퟔ풙ퟏퟎ−ퟏퟗ푽 ퟏ.ퟐퟐퟖ풙ퟏퟎ−ퟗ = m √푽 1.228 ퟏퟐ.ퟐퟖ = 푛푚 = Å √푽 √푽 Q:Mention the characteristics of matter waves or Write a note of matter waves. 1. MW are non mechanical waves. 2. MW are associated with moving particles . 3. MW are non electromagnetic waves. 4. MW of microscopic particles can be measured . 5. MW are charge independent .
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6. MW of macroscopic particles cannot be measured . 7. Different MW have different phase velocity. 8. Phase velocity of MW is greater than that of light. 9. Velocity of MW depends on the velocity of material particles.
Q:Define Phase velocity/What is meant by Phase velocity ?
“Phase Velocity ” is defined as the velocity with which the uni-phase particles on 휔 the wave travels and is given by 푣 = 푃 푘 where 휔 = 푎푛푔푢푙푎푟 푣푒푙표푐𝑖푡푦 푎푛푑 푘 = 푤푎푣푒 푐표푛푠푡푎푛푡.
Q: Define Group velocity/What is meant by Group velocity ?
“Group Velocity ” is defined as the velocity with which the resultant wave packet of 푑휔 the group of waves travels and is given by 푣 = 𝑔 푑푘 where d휔 = 푐ℎ푎푛푔푒 𝑖푛 푎푛푔푢푙푎푟 푣푒푙표푐𝑖푡푦 푎푛푑 푑푘 = 푐ℎ푎푛푔푒 𝑖푛 푤푎푣푒 푐표푛푠푡푎푛푡.
Q: Derive the relation between group velocity (풗품) & phase velocity(풗푷)/ 풅푉푃 Show that 풗 = 풗 − 흀 품 푷 풅흀
Consider a particle of mass ‘m’ having phase velocity ‘ 풗푷 ‘ and group velocity ‘′풗품′. Let the wave have the wavelength ’λ’ , frequency ‘ν, wave number k & angular velocity ω
휔 푑휔 We know that 푣 = …(1) and 푣 = ….(2) 푉 푃 푘 𝑔 푑푘 푃
푑(푉 .푘) From eqns 1 & 2 , we get 푣 = 푃 𝑔 푑푘 푑푉 = 푣 + 푘 푃 푃 푑푘 푑푉 푑휆 = 푣 + 푘. 푃 ……(3) 푃 푑휆 푑푘
2휋 푑휆 푑 2휋 Also, λ = ∴ = ( ) 푘 푑푘 푑푘 푘
−2휋 = ……..(4) 푘2
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푑푉 −2휋 From eqns 3 & 4 ,we get 푣 = 푣 + 푘. 푃 ( ) 𝑔 푃 푑휆 푘2
2휋 푑푉 = 푣 − 푋 푃 푃 푘 푑휆
풅푉 2휋 ie: 풗 = 풗 − 흀 푃 ∵ λ = 품 푷 풅흀 푘
Note: 풗품 < 풗푷 for a dispersive medium
and 푣𝑔 = 푣푃 = 퐶 푓표푟 푎 푛표푛 푑𝑖푠푝푒푟푠𝑖푣푒 푣푎푐푐푢푚 푚푒푑𝑖푢푚
Q: Deduce the relation between group velocity(풗품) & particle velocity(풗푷풂풓풕풊풄풍풆)/
Show that 풗품 = 풗푷풂풓풕풊풄풍풆 or 풗품 = 풗
′ Consider a particle of mass ‘m’ having particle velocity ‘풗푷풂풓풕풊풄풍풆 & 푔푟표푢푝 푣푒푙표푐𝑖푡푦 ′풗품′. 푑휔 We know that, 푣 = … … . . (1) 𝑔 푑푘 퐸 but 휔 = 2휋휈 & 퐸 = ℎ휈 ,we get 휔 = 2휋 ℎ 2휋 ∴ d휔 = ( ) 푑퐸 … (2) ℎ 2휋 ℎ 2휋 Also from k = & 휆 = we get k = ( )P 휆 푃 ℎ
2휋 ∴ dk = ( )dP ……(3) ℎ 2휋 ( )푑퐸 푑휔 ℎ 푑퐸 From eqns 1,2 &3,we get 푣𝑔 = = 2휋 = …….(4) 푑푘 ( )푑푃 푑푃 ℎ
푚2푣2 Also , from E = ½m푣2 = & P = m퓥 , 2푚 푃2 푤푒 푔푒푡 퐸 = 2푚 2푃 ∴ dE = 푑푃 2푚 푑퐸 푃 푚 푣 𝑖푒; = = 푃푎푟푡𝑖푐푙푒 푑푃 푚 푚 = 푣푃푎푟푡𝑖푐푙푒 ….(5)
From eqns 4 & 5,we get 풗품 = 풗푷풂풓풕풊풄풍풆 Thus group velocity =Particle velocity
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QUANTUM MECHANICS
Q: State & explain Heisenberg’s Uncertainty Principle (HUP). Mention its significance.
HUP states that “the product of the uncertainty ‘Δ푥’ in the position and the uncertainty
‘ΔP푥’ in the momentum of a particle at any instant is equal to or greater than ( h/4훑) “ ℎ i.e.: Δ푥 ΔP푥 ≥ ,Where h is Planck’s constant. 4휋
The significance of HUP is that, it is impossible to determine simultaneously both the position and momentum of the particle accurately at the same instant. ℎ NOTE : Other HUP relations are ∆퐸.∆푡 ≥ , 푤ℎ푒푟푒 ∆퐸 = 푒푛푒푟푔푦, ∆푡= time, 4휋 ℎ & ∆퐿.∆휃 ≥ 4휋 where , ∆퐿 = 푎푛푔푢푙푎푟 푚표푚푒푛푡푢푚 & ∆휃 = 푑𝑖푠푝푙푎푐푒푚푒푛푡.
Q: Show that electrons do not present in the nucleus using Heisenberg’s Uncertainty Principle.
Electron to be present in the nucleus, maximum uncertainty in position Δ푥=10-14 m (diameter) According to HUP, ℎ The minimum uncertainty in momentum ΔP푥 ≥ 4휋 훥푥 6.625 푥 10−34 ≥ 4 푥 3.14 푥 10−14 ΔP푥 ≥ 5.275 x 10-21 kg m/s = P(say) 푚 Using E = m퐶2 , 푃 = 푚푉 푎푛푑 푚 = 0 , 푤푒 푐푎푛 푠ℎ표푤 푡ℎ푎푡 the minimum energy of the 푉2 √1− 퐶2 2 2 2 2 4 2 4 electron in the nucleus is given by 퐸 = P c +푚표 푐 , but the rest mass energy 푚표 푐 is very very small compared to P2c2,neglecting 푚2푐4 , 표 NOTE: 2 4 2 6 2 2 4 푚표퐶 푚표퐶 퐸 = 푚 퐶 = 푣2 = 2 2 −21 8 (1− ) (퐶 −푣 ) we have E ≈ PC = 5.275 x 10 x 3 x 10 J 퐶2 2 2 2 5.275 푥 10−21 푥 3 푥 108 2 2 2 2 2 푚표푣 퐶 푃 퐶 = 푚 푣 퐶 = 푣2 = = MeV (1− ) 1.6 푥 10−13 퐶2 푚2퐶4푣2 표 = 9.89 MeV (퐶2−푣2) 2 4 2 2 = 10 MeV 푚표퐶 (퐶 −푣 ) 퐸2 − 푃2퐶2 = (퐶2−푣2) But the maximum energy of the electrons(훃-particle) emitted ퟐ ퟐ ퟐ ퟐ ퟒ ∴ 푬 = 푷 푪 + 풎풐푪 from the nucleus does not exceed 4MeV,hence electrons do not present in the nucleus.
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Q: Set up 1D time independent Schrodinger wave equation for a free particle.
One dimensional wave function 횿 describing the de-broglie wave for a particle moving freely in the positive direction of 푥-direction is given by 횿 = A푒𝑖(푘푥−휔푡) = A푒𝑖푘푥 푒−휔푡, Where A푒𝑖푘푥 represent the time independent part of the wave function and is represented by 휓 = A푒𝑖푘푥 ……(1) differentiating eqn (1) w.r.t ‘푥 ‘ twice we get 푑휓 = 퐴(𝑖푘) 푒𝑖푘푥 and 푑푥 푑2휓 =A(𝑖푘) (𝑖휔) 푒𝑖푘푥 푑푥2 = 𝑖2 푘2 A푒𝑖푘푥 …….(2) From eqns 1 & 2 we get 2 2 푑 휓 4휋 ퟐ흅 = − 휓 ……..(3) ∵ 풊ퟐ = −1 & 풌 = 푑푥2 휆2 흀 ℎ But, de-broglie wave length λ = 푚푣 1 푚2푣2 ∴ = 휆2 ℎ2 2푚(½푚푣2) = ℎ2 2푚(퐸 ) = 퐾 ∵ 퐸 = ½푚푣2 ℎ2 퐾 Also , the kinetic energy (퐸퐾 )in terms of the total energy (E) & the potential energy( V) is given by 퐸퐾=(E−푽) 1 2푚(퐸−푽) ∴ = ……(4) 휆2 ℎ2 푑2휓 4휋22푚(퐸−푽) From eqns 3 & 4 ,we get = − 휓 푑푥2 ℎ2
푑2휓 8휋2푚(퐸−푽) + 휓 = 0 ……(5) 푑푥2 ℎ2
This is Schrodinger time independent equation for a particle
For a free particle ,V=0
푑2휓 8휋2푚퐸 ∴ + 휓 = 0 푑푥2 ℎ2 This is Schrodinger time independent equation for free particle.
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Q: Obtain normalized wave function for a free particle in a infinite walled potential Box/Well using Schrodinger 1D time independent equation.
a
∞ ∞ V=∞ V=0 Box/Well ●
particle 푥=0 푥=a
Consider a particle of mass ‘m’ moving by reflection at infinitely high walls of a Box/well of width ‘a’ moving between 푥=0 & 푥=a Potential V=0 inside the Box/well and V=∞ outside the Box/well .one dimensional Schrodinger equation for particle is given by 푑2휓 8휋2푚(퐸−푽) + 휓 = 0 ……(1). 푑푥2 ℎ2 For a particle inside the Box/well V=0 푑2휓 8휋2푚퐸 ∴ + 휓 = 0 ….(2) 푑푥2 ℎ2 8휋2푚퐸 Putting = 퐾2 …..(3) in equation (2) , ℎ2 푑2휓 we get + 퐾2휓 = 0 ……(4) 푑푥2 The general solution of the quadratic equation (4) is of the form ψ (푥)=A sin(K푥) +B Cos(K푥) ……(5) where A & B are constants determined from boundary conditions as follows : ψ (푥)=0 at 푥=0 from eqn (5), 0=A x 0 +B x 1 ∴ B=0 also, ψ (푥)=0 at 푥=a ∴ from eqn(4) 0=C Sin(Ka)+0xCos(Ka) 0= C Sin(Ka) Sin(Ka) = 0 as C≠ 0 ∴ Sin(Ka) =0= Sin(n휋) 푛휋 ie: Ka=n휋 or K= …..(6) 푎 8휋2푚퐸 푛2휋2 From eqns 3 & 6 ,we get = ℎ2 푎2 푛2ℎ2 ∴ E = or 8푎2푚 푛2ℎ2 In general En= …(7) 8푎2푚
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where n=1,2,3..called quantum number.
The values of En are called Eigen energy values which satisfy Schrodinger wave equation. ℎ2 n=1 gives E1 = = Eo called end point /ground state energy . 8푎2푚 4ℎ2 This can be omitted n=2 gives E2 = = 4 Eo called 1st excited state energy 8푎2푚 9ℎ2 n=3 gives E3 = =9 Eo ,called 2nd excited state energy & so on 8푎2푚
푛휋 Substituting the values of B=0 & K= in eqn 5, we get 푎 휓 (푥) = A Sin( 푛휋푥 ) …….(8) this represents the permitted solutions 푛 푎 To find ’A’ by normalization: +∞ Applying the normalization condition 퐼훹퐼2푑푥 = 1 to eqn 8 for x=0 & x=a, we get ∫−∞ 푎 퐴2 푆𝑖푛2 ( 푛휋푥 )푑푥 = 1 ∫0 푎 푎 퐴2 ½[1 − 퐶표푠 ( 2 푛휋푥 )]푑푥 = 1 ∵ 푆𝑖푛2 (휃)=½[1-Cos(2휃)] ∫0 푎 푎 푎 ½퐴2 [ 1푑푥 − 퐶표푠( 2 푛휋푥 )푑푥 ] = 1 ∫0 ∫0 푎 푎 2푛휋푥 푆𝑖푛(푚푥) ½퐴2 { [푥]푎 − [ Sin( )]푎 } = 1 ∵ ∫ 퐶표푠(푚푥) 푑푥 = 0 2푛휋 푎 0 푚 푎 푎 ½퐴2 { [푎 − 0] − [ Sin(2n휋) − Sin(0)] }=1 2푛휋 2푛휋 ½퐴2 { [푎 − 0] − [0 − 0] }=1 퐴2 푎 = 2 ⇨ 퐴2 = 2/푎 or A =√ퟐ/풂 …(9) From eqns 8 &9, the normalized Eigen functions are 푔𝑖푣푒푛 푏푦 흍 (푥)= ퟐ/풂 Sin( 풏흅풙 ) …..(10) 풏 √ 풂
2 푛 = 3 흍3 퐼흍3퐼 . 2 푛 = 2 흍2 퐼흍2퐼
2 푛 = 1 흍1 퐼흍1퐼 푥=0 푥=a 푥=0 푥=a
2 2 2 흍2 ,흍3 , … . 푎푛푑 푡ℎ푒𝑖푟 푝푟표푏푎푏𝑖푙𝑖푡푦 푑푒푛푠𝑖푡𝑖푒푠 퐼흍1퐼 , 퐼흍2퐼 , 퐼흍3퐼 … 퐸𝑖푔푒푛 푓푢푛푐푡𝑖표푛푠 흍1 , are represented as shown in the diagrams.The probability of finding the particle at the anti-nodes is maximum and at nodes is zero.(The particle never found at nodes)
NOTE: Eigen Functions are the acceptable wave functions [휓푛(푥)] Eigen Energy values are energy values for which Schrodinger equation can be solved.
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Q: What is a wave function and mention it’s properties/limitations The variable quantity that characterizes the de-broglie wave of the particle is called a ‘Wave function’ denoted mathematically by the symbol ‘횿’(Psi)
Properties of the wave functions are:- 1. Wave function (횿) is single valued everywhere , 2. 횿 is finite everywhere 3. 횿 is continuous every where 4. First derivatives of 횿 are continuous every where 5. 퐼훹퐼2 or 훹훹 * is called probability density. +∞ 6.푃푟표푏푎푏𝑖푙𝑖푡푦 표푓 푓𝑖푛푑𝑖푛푔 푝푎푟푡𝑖푐푙푒 𝑖푛 푠푝푎푐푒 𝑖푠 푔𝑖푣푒푛 푏푦 퐼훹퐼2푑푉 = 1 ∫−∞
PROBLEMS SECTION
MODULE-1 : MODERN PHYSICS AND QUANTUM MECHANICS
ℎ Formulae needed : λ = where P = mv = √2푚퐸 =√2푚푒푉 , 푃 1 푃2 ℎ퐶 ℎ E = 푚푉2 = ;Photon energy E’ = hν = ; ∆푥. ∆푝 ≥ and 2 2푚 휆 4휋 ℎ ℎ퐶∆휆 ∆푝 = 푚. ∆푣 ; ∆퐸. ∆푡 ≥ 푎푛푑 ∆퐸 = ; also ∆퐸 = ℎ ∆휈 ; 4휋 휆2 2 2 푥2 2 푛 ℎ P = ∫ 퐼훹퐼 dx ; 퐸푛 = 푥1 8푚푎2
1. Compare the energy of a photon with that of an electron when both are associated with wavelength 0.2 nm.(Dec 2014/Jan2015)
−9 −34 −31 8 Given: 휆푃 = 휆푒 = 0.2 푛푚 = 0.2푥10 푚 , h=6.63x10 Js, m=9.1x10 kg ; C=3x10 m/s, 퐸 푃 = ? 퐸푒 ℎ퐶 ℎ2 Using 퐸푃 = and 퐸푒 = 2 휆푃 2푚휆푒 퐸 퐶2푚휆 2 We get , 푃 = 푒 퐸푒 ℎ휆푃 3푥108푥2푥9.1푥10−31( 0.2푥10−9)2 = 6.63푥10−34 푥 0.2푥10−9 = 1.647x 102
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2. Calculate the kinetic energy of an electron of wavelength 18 nm.(Jun/Jul 14)
Given: 휆 = 18 푛푚 = 18푥10−9 푚 , h=6.63x10−34 Js, m=9.1x10−31kg ; E= ? ℎ Using λ = √2푚퐸 ℎ2 We get ,퐸 = 2푚휆2 ( 6.63푥10−34 )2 = 2푥9.1푥10−31푥( 18푥10−9)2 = 7.454x10−22 J 7.454푥10−22 = 1.6푥10−19 = 4.66x10−3 eV
3. An excited atom has an average life time of 10−8 seconds. During this period. it emits photon and returns to the ground state. What is the minimum uncertainty in the frequency of this photon ? (Jun/Jul 14)
Given: ∆푡 = 10−8 S, ℎ = 6.63푥10−34 퐽푠 ; ∆휈 = ?
ℎ Using ∆퐸 = ℎ. ∆휈 and ∆퐸. ∆푡 ≥ 4휋 1 We get , ∆휈 ≥ 4휋.∆푡 1 ≥ 4푥휋푥10−8 = 7.96 x 106 Hz
4. Calculate the wavelength associated with electrons whose speed is 0.01 part of the speed of light. (Dec 2013/Jan2014)
Given: v =0.01 C = 0.01x 3x108m/s ; h=6.63x10−34 Js; 푚 = 9.1푥10−31푘푔 ; 휆 = ?
ℎ Using λ = 푚푣 6.63푥10−34 = 9.1푥10−31푥0.01푥 3푥108
= 2.43 x10−10 m
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5. An electron is bound in one dimensional infinite well of width 0.12nm.Find the energy value and de-Broglie wavelength in the first excited state. (Dec 2013/Jan2014)
Given: a = 0.12nm = 0.12x10−9 m ; n = 2 ; h=6.63x10−34 Js; 푚 = 9.1푥10−31푘푔 ;
퐸2 = ? & λ =? 푛2ℎ2 Using 퐸 = 푛 8푚푎2 22(6.63푥10−34 )2 퐸 = 2 8푥9.1푥10−31푥(0.12푥10−9)2 = 1.68x10−17 J
ℎ Also , λ = √2푚퐸2 6.63푥10−34 = 2푥9.1푥10−31푥1.68푥10−17
= 1.19 x10−10 m
6. Calculate the de-Broglie wavelength associated with an electron of energy 1.5eV. (Jun13)
Given: E=1.5 eV=1.5x1.6x10−19 퐽 , h=6.63x10−34 Js; m=9.1x10−31kg; e=1.6x10−19 C; λ =
ℎ Using λ = √2푚퐸 6.63푥10−34 = √(2푥9.1푥10−31푥1.5푥1.6푥10−19 ) = 1.003 x 10−9 m
7. A spectral line of wavelength 5461Å has a width of 10−4Å. Evaluate the minimum time spent by the electrons in the upper energy state. (Jun/Jul 2013)
Given:, ℎ = 6.63푥10−34 퐽푠 ; 휆 = 5461Å = 5461푥10−10 푚 ; ∆푡 = ?
ℎ퐶 ℎ Using ∆퐸 = and ∆퐸. ∆푡 ≥ 휆 4휋 휆 We get , ∆푡 ≥ 4휋.퐶 5461푥10−10 ≥ 4푥휋푥3푥108 = 1.45 x 10−16 S
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8. Find the de-Broglie wavelength of an electron accelerated through a potential difference of 182 volts and object of mass 1 kg moving with a speed of 1 m/s. Compare the results and comment. (Jun 2012)
−34 −31 −19 Given:V=182V ; 푚0 =1 kg, 푉0 = 1 m/s , h=6.63x10 Js; m=9.1x10 kg; e=1.6x10 C , 휆푒 = ? 휆0 = ? ℎ Using 휆 = 푒 √2푚푒푉 6.63푥10−34 = √2푥9.1푥10−31푥1.6푥10−19 푥182 = 9.11 x10−11 m ℎ Also,using 휆0 = 푚0푉0 6.63푥10−34 = 1푥1 = 6.63x10−34 푚
Comment : Waverlength is inversely proportional to the mass ,as 휆푒 ≫ 휆0
9. A quantum particle confined to one-dimensional box of width ‘a’ is in its first excited 푎 state. What is the probability of finding the particle over an interval of ( ) marked 2 퐽푢푛 symmetrically at the centre of box. ( 2011) 퐽푢푙
푎 3푎 Given: 1st excited state , n=2 ;the interval ( , ) 4 4 푥 3푎/4 2 Using P = ∫ 2 퐼훹퐼2dx = ∫ (√ 푆𝑖푛 ( 2휋푥 ))2푑푥 푎/2 푥1 푎/4 푎 푎 3푎 4 ퟐ 2 2휋푥 = ∫푎 푆𝑖푛 ( )푑푥 0 a/4 a/2 3a/4 a 풂 푎 4 3푎 ퟐ 4 2 휋푥 2 = ∫푎 ½[1 − 퐶표푠 ( )]푑푥 ∵ 푆𝑖푛 (휃)=½[1-Cos(2휃)] 풂 푎 4 3푎 3푎 ퟐ 4 4 2 휋푥 = ½ ∫푎 1푑푥 − ∫푎 퐶표푠( )푑푥 풂 푎 4 4 3푎 3푎 4 푎 2휋푥 4 = 푎 { [푥]푎 − [ Sin( )]푎 } 2휋 푎 4 4 3푎 푎 3푎 푎 푎 2휋 푎 2휋 = 푎 { [ − ] − [ Sin( 4 )− Sin( 4 )] } 4 4 2휋 푎 2휋 푎 푎 = 푎 { [ ] − [0 − 0 } = ½ = 0.5 or 50% 2
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10. A particle moving in one dimension box is described by the wave function 훹 = 푥[√3] 푓표푟 0<푥<1 and Ψ =0 elsewhere. Find the probability of finding the particle within the interval (0,½). (퐽푢푛 2012)
1 Given : 훹 = 푥[√3] ; (푥 , 푥 )( 0, ) ; P = ? 1 2 2
푥 Using Probability, P = ∫ 2 퐼훹퐼2dx 푥1 1 1 3 푥 1 1 = ∫2 3푥2dx = [3. ]2 = ( )3 – (0)3 = = 0.125 or 12.5 % 0 3 0 2 8 11. Calculate the energy of electron that produces Bragg’s diffraction of first order at glancing angle of 22° ,when incident on crystal with inter-planar spacing of 1.8Å. (퐽푢푛 2012)
Given : n =1 ; 휃 = 22° ; 푑 = 1.8 Å = 1.8푥 10−10 m ; h=6.63x10−34 Js; m=9.1x10−31kg; λ = ? & E = ?
2푑 푠𝑖푛 휃 Using 1 ) λ = 푛 2푥1.8푥 10−10푥 푠𝑖푛 22 = 1 =1.348x 10−10 m ℎ Also from, λ = √2푚퐸 ℎ2 ℎ2 We get, E = J OR E = eV 2푚휆2 2푚휆2 푒 (6.63푥10−34)2 (6.63푥10−34)2 = = 2푥9.1푥10−31푥(1.348푥 10−10 )2 2푥9.1푥10−31푥(1.348푥 10−10 )2푥1.6 푥 10−19 = 1.329 x 10−17 J = 83.07 eV = 83.07 eV
12. Find the energy of the neutron in eV whose de-Broglie wavelength is 1Å. (퐷푒푐 2011)
Given; λ = 1Å = 1x 10−10 푚 ;h=6.63x10−34 Js, m=1.678x10−27kg ; e=1.6x10−19 C ;E = ?
ℎ Using , λ = √2푚퐸 ℎ2 We get , E = joules 2푚휆2 ℎ2 = eV 2푚푒휆2 (6.63푥10−34 )2 = 2푥1.678푥10−27푥1.6푥10−19 푥(1푥 10−10 )2 = 0.082 eV
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13. An electron is confined to a box of length 10−9 m, calculate the minimum uncertainty in it’s velocity. (퐷푒푐 2011)
Given: ∆푥 = 10−9 m ; h=6.63x10−34 Js, m=9.1x10−31kg ;∆푣 = ?
ℎ Using ∆푥. ∆푝 ≥ and ∆푝 = 푚. ∆푣 4휋 ℎ We get , ∆푣 ≥ 4휋푚∆푥
6.63푥10−34 ≥ 4휋 푥9.1푥10−31푥10−9
= 58 x103m/S
14. Compute the de-Broglie wavelength for a neutron moving with one tenth part of the 퐽푢푛 velocity of light. ( 2011) 퐽푢푙
퐶 3 푋 108 Given; V = = = 0.3 푥 108 푚/푠;h=6.63x10−34 Js, m=1.678x10−27kg ; 10 10 C = 3 X 108 m/s ; λ = ? ℎ Using λ = 푚푉 6.63푥10−34 = = 1.317x 10−14 m. 1.678푥10−27푥0.3 푥 108
15. An electron has a de-Broglie wavelength 3 nm and rest mass 511keV. Determine its group velocity and kinetic energy.
−9 −34 Given: λ = 3nm = 3x 10 m ; h=6.63x10 Js ; 푉𝑔 = V = ? and 퐸퐾 = ? m =E/ 퐶2 =511x103x1.6x 10−19/(3푥108)2 kg = 9.08x10−31 kg
ℎ ℎ Using λ = ,we get 푉 = V = 푚푣 𝑔 푚휆 6.63푥10−34 = 9.08푥10−31푥3푥 10−9 = 2.43x 106 m/s
1 1 푎푙푠표, 퐸 = 푚푉2 = 푥 9.08푥10−31 푥 (2.43푥 106 )2 퐾 2 2 = 2.681x 10−18 J
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16. A spectral line of wavelength 546.1 nm has a width 10−5 nm. Estimate the minimum time spent by electrons in the excited state during transitions. (퐷푒푐 2010)
Given:, ℎ = 6.63푥10−34 퐽푠 ; 휆 = 546.1 푛푚 = 546.1푥10−9 푚 ; ∆푡 = ?
ℎ퐶 ℎ Using ∆퐸 = and ∆퐸. ∆푡 ≥ 휆 4휋 휆 푤푒 푔푒푡 , ∆푡 ≥ 4휋.퐶 546.1푥10−9 ≥ = 1.45x 10−16 S 4휋푥3푥108
17. Calculate the momentum of the particle and de-Broglie wavelength associated with an electron with a kinetic energy of 1.5KeV. (퐽푢푛 2010)
Given: E =1.5 Kev =1.5 x 103 x 1.6x10−19 J ; h = 6.625푥10−34 퐽푆 ;e = 1.6x10−19 J ; m = 9.1푥10−31 푘푔 P =? & λ = ?
Ans: Using P = √2푚퐸 = √2푥9.1푥10−31푥1.5 푥 103 푥 1.6푥10−19 = 2.09 x 10−23 kg.m/s ℎ Also, λ = 푃 6.625푥10−34 = = 3.17x10−11 푚 2.09 푥 10−23
18. An electron is bound in one dimensional potential well of width 0.18 nm. Find the energy value in eV of the second excited state. (퐽푢푛 2010 & 퐶퐵퐶푆 − 퐽푢푛/퐽푢푙16)
Given: n=3(For 2nd excited state), a=0.18 nm =0.18x10−9m,h=6.63x10−34 Js; −31 −19 m=9.11x10 kg & e=1.6x10 C; 퐸2 = ?
푛2ℎ2 푛2ℎ2 Using 퐸 = J OR 퐸 = eV 푛 8푚푎2 푛 8푚푎2푒 32(6.63푥10−34 )2 32(6.63푥10−34 )2 퐸 = 퐸 = 2 8푥9.11푥10−31푥(0.18 푥10−9)2 2 8푥9.11푥10−31푥(0.18 푥10−9)2 푥1.6푥10−19 = 1.675 x10−17 J = 104.7 eV 1.675 푥10−17 = =104.7 eV 1.6푥10−19
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19. Calculate the energy in eV, for the first excited state of an electron in an infinite potential well of width 2 Å (4 marks)
Given: n=2(For 1st excited state), a=2Å=2x10−10m,h=6.63x10−34 Js; m=9.1x10−31kg & e=1.6x10−19 C;
푛2ℎ2 푛2ℎ2 Using 퐸 = J OR 퐸 = eV 푛 8푚푎2 푛 8푚푎2푒 22(6.63푥10−34 )2 22(6.63푥10−34 )2 = = 8푥9.1푥10−31푥(2푥10−10)2 8푥9.1푥10−31푥(2푥10−10)2 푥1.6푥10−19 = 6.038x10−18 J = 37.74 eV 6.038푥10−18 = =37.74 eV 1.6푥10−19
20. The wavelength of a fast neutron of mass 1.675x10−27kg is 0.02 nm.Calculate the group velocity and phase velocity of its de-Broglie waves.( 4 marks)
−27 −9 8 −34 Given: m=1.675x10 kg, λ = 0.02 nm=0.02x10 m, C=3x10 m/s, ℎ = 6.63푥10 퐽푠, 푣𝑔 & 푣푃 =?
ℎ 6.63푥10−34 푈푠𝑖푛푔, 푣 = v = = = 1.979x104 m/s , 𝑔 푚휆 1.675푥10−27푥0.02푥10−9
2 8 2 퐶 (3푥10 ) 12 Also 푣푃 = = 4 = 4.55X10 m/s 푣𝑔 1.979푥10
21. Calculate the de-Broglie wavelength associated with neutron of mass 1.674 X 10−27 kg with one tenth part of the velocity of light (4 marks - CBCS Jun/July16)
1 1 Given : v = C = x 3x108m/s = 3x107m/s; h=6.63x10−34 Js; 푚 = 9.11푥10−31푘푔 ; 휆 = ? 10 10
ℎ Using λ = 푚푣 6.63푥10−34 = 9.11푥10−31푥 3푥107
= 2.43 x10−11 m
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22. Calculate the deBroglie wavelength of an electron moving with K.E of 50KeV.
Given: E= 50keV= 5x103푥 1.6x10−19 퐽 , h=6.63x10−34 Js; m=9.1x10−31kg; e=1.6x10−19 C; λ =? ℎ Using λ = √2푚퐸 6.63푥10−34 = √(2푥9.1푥10−31푥 50푥103푥 1.6푥10−19 )
= 5.495 x 10−12 m
22. X-rays of wavelength 0.75Å are scattered from a target at an angle of 45°. Calculate the wavelength of scattered X-rays.
Given: 휆 = 0.75Å ; 휃 = 45° ; h=6.63x10−34 Js; m=9.1x10−31kg ; C=3x108m/s ;흺λ’ = ?
ℎ Using , λ’−휆 = (1−푐표푠휃) 푚퐶 6.63푥10−34 λ’ −0.75Å = (1−푐표푠45) 9.1푥10−31푥 3푥108
λ’ = 0.75Å + 7.1x10−13m = 0.75Å + 0.0071Å = 0.7571Å
223. A particle of mass 940 MeV/퐶2 has kinetic energy 0.5 KeV. Find its de- Broglie wavelength , C is velocity of light.
940 푥 106 푒푉 940 푥 106 푥1.6푥10−19 퐽 Given: m = 940 MeV/퐶2 = = = 1.671x10−27 푘푔 (3푥108 )2 (3푥108 )2 E = 0.5 keV = 0.5 푥103eV = 0.5 푥103푥1.6푥10−19 퐽 ; 휆 = ?
ℎ Using; 흺 = √(2푚퐸) 6.63푥10−34 = √(2푥1.671푥10−27푥0.5 푥103푥1.6푥10−19)
= 1.282x10−12 m
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24. The first excited state energy of an electron in an infinite well is 240 eV. what will be its ground state energy when the width of the potential Well is doubled.
Given: 퐸2 = 240 eV ; 퐸1 = ? 푛2ℎ2 Using 퐸 = 푛 8푚푎2 22ℎ2 퐸 = 2 8푚푎2 4ℎ2 240 = 8푚푎2 4ℎ2 푎2 = 8푚푥240 ℎ2 ∴ a = √( ) 480푚 12ℎ2 480푚푥ℎ2 푎푙푠표 , 퐸1 = = = 15 eV ℎ2 8푚푥4푥ℎ2 8푚(2√( ) )2 480푚 푂푅
푛2ℎ2 퐸 푎2 퐸 푎2 ℎ2 Using ; 퐸 = ,we can show that 2 2 = 1 1 = 푛 8푚푎2 22 12 8푚
Given; 퐸2 = 240푒푉 , 푎1 = 2푎2
2 2 퐸2 푎2 240푎2 ∴ 퐸1 = 2 2 = 2 = 15 eV 푎12 4푎2푥4
25. Calculate the deBroglie wavelength associated with neutron of mass 1.674 x10−27kg with one tenth part of the velocity of light .
퐶 3 푋 108 Given; V = = = 0.3 푥 108 푚/푠;h=6.63x10−34 Js, 10 10 m=1.674x10−27kg ; C = 3 X 108 m/s ; λ = ?
ℎ 6.63푥10−34 Using λ = = 푚푉 1.674푥10−27푥0.3 푥 108
= 1.320x 10−14 m.
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26. An electron is bound in one dimensional potential well of width 0.18 nm. Find the energy value in eV of the second excited state.
Given: n=3(For 2rd excited state), a=0.18 nm =0.18 x10−9m,h=6.63x10−34 Js; m=9.1x10−31kg & e=1.6x10−19 2 2 푛 ℎ 푛2ℎ2 Using 퐸 = J OR 퐸 = eV 푛 8푚푎2 푛 8푚푎2푒 32(6.63푥10−34 )2 = = 8푥9.1푥10−31푥(0.18 푥10−9)2 32(6.63푥10−34 )2
8푥9.1푥10−31푥1.6푥10−19 푥(0.18 푥10−9)2
= 1.677x10−17 J = 104.83 eV
6.038푥10−18 = 1.6푥10−19
= 104.81 eV
27. Find deBroglie wavelength of a particle of mass 0.58 MeV/푪ퟐ has a kinetic energy 90 eV , where C is the velocity of light. ( 04 Marks) DEC2016 0.58 푥 106 푒푉 Given: m = 0.58 MeV/퐶2 = (3푥108 )2 0.58 푥 106 푥1.6푥10−19 퐽 = = 1.0324x10−30 푘푔 (3푥108 )2 E = 90eV = 90푥1.6푥10−19 퐽 ; 휆 = ? ℎ Using; 흺 = √(2푚퐸) 6.63푥10−34 = √(2푥1.0324푥10−30 푥 90푥1.6푥10−19) = 1.2142푥10−10 m
28. The inherent uncertainty in the measurement of time spent by Iridium- 19 nuclei in the excited state is found to be 1.4 x ퟏퟎ−ퟏퟎ S. Estimate the uncertainty that results in its energy in eV in the excited State. ( 04 Marks) Given:, ℎ = 6.625푥10−34 퐽푠 ; ∆푡 = 1.4 x ퟏퟎ−ퟏퟎ S: ∆퐸 =? ℎ ℎ Using ∆퐸. ∆푡 ≥ 푤푒 푔푒푡 , ∆퐸 ≥ 4휋 4휋.∆푡 6.625푥10−34 ≥ J 4휋푥ퟏ.ퟒ 풙 ퟏퟎ−ퟏퟎ 6.625푥10−34 ≥ eV 4휋푥ퟏ.ퟒ 풙 ퟏퟎ−ퟏퟎ풙ퟏ.ퟔퟎퟐ풙ퟏퟎ−ퟏퟗ ≥ 2.351푥 10−6 eV
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29. A spectral line of wavelength 5896 Å has a width of ퟏퟎ−ퟓ Å. Evaluate the minimum time spent by the electrons in the upper energy state between the excitation and de-excitation processes. ( 04 Marks)
Given: 흀=5896 Å = 5896 xퟏퟎ−ퟏퟎ m ,∆흀 = ퟏퟎ−ퟓÅ = ퟏퟎ−ퟓ xퟏퟎ−ퟏퟎm , h = 6.625 x ퟏퟎ−ퟑퟒ JS , C = 3 x ퟏퟎퟖ풎풔−ퟏ , ∆풕 = ?
풉 풉풗 ∆흀 풉풄 Using ∆푬. ∆풕 ≥ and ∆푬 = ∵ E = ퟒ흅 흀ퟐ 흀 흀ퟐ We get , ∆풕 = ퟒ흅풄∆흀 (ퟓퟖퟗퟔ 풙ퟏퟎ−ퟏퟎ)ퟐ ∴ ∆풕 = ퟒ흅풙ퟑ 풙 ퟏퟎퟖ풙ퟏퟎ−ퟓ 풙ퟏퟎ−ퟏퟎ = 9.221 x ퟏퟎ−ퟖ S 29. Compare the energy of a photon with that of a neutron when both are associate with a wavelength 0.25 nm,mass of neutron is 1.675 X ퟏퟎ−ퟐퟕ kg. ( 04 Marks) −ퟗ −ퟐퟕ Given : 흀풑 = 흀풏 = ퟎ. ퟐퟓ 풏풎 = 0.25 x ퟏퟎ m , 풎풏 = 1.675 X ퟏퟎ kg 흀 h = 6.625 x ퟏퟎ−ퟑퟒ JS , C = 3 x ퟏퟎퟖ풎풔−ퟏ , 풑 = ? 흀풏 풉풄 ퟔ.ퟔퟐퟓ 풙 ퟏퟎ−ퟑퟒ풙ퟑ 풙 ퟏퟎퟖ Using , 푬 = = = 7.95 x ퟏퟎ−ퟏퟔ J 풑 흀 ퟎ.ퟐퟓ 풙 ퟏퟎ−ퟗ
ퟐ −ퟑퟒ ퟐ 풉 (ퟔ.ퟔퟐퟓ 풙 ퟏퟎ ) −ퟐퟏ Also , 푬풏 = ퟐ = ퟐ = 2.096 xퟏퟎ J ퟐ풎흀 ퟐ풙ퟏ.ퟔퟕퟓ 푿 ퟏퟎ−ퟐퟕ풙 (ퟎ.ퟐퟓ 풙 ퟏퟎ−ퟗ) −ퟏퟔ 푬풑 ퟕ.ퟗퟓ 풙 ퟏퟎ ∴ = = 3.793 x ퟏퟎퟓ 푬풏 ퟐ.ퟎퟗퟔ 풙ퟏퟎ−ퟐퟏ
**end**
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM MODULE-2: ELECTRICAL PROPERTIES OF MATERIALS:
Q:Explain the terms: Drift velocity, Mean collision time, Mean free path & relaxation time.
a) Drift velocity(풗풅) is the average velocity with which the free electrons drift in a direction opposite to that of the applied electric field b) Mean collision time(흉) is the average time elapsed between two consecutive collisions. c) Mean free path(λ)is the average distance travelled by the electron between two consecutive collisions.
1 d) Relaxation time(흉 ) is time during which the drift velocity is reduced to of its 풓 푒 velocity when the field is cut off.
Q:Obtain an expression for Drift velocity.
Let an electron of mass ‘m’ ,charge ‘e’ moves with drift velocity ’풗풅 ‘ when an electric field ‘E ‘ is applied to it. Then the driving force acting on the electron F = −푒퐸 ……(1) 푚푣 Also the retarding force acting on the electron F’= 푑 …..(2) 휏 where 휏 = 푚푒푎푛 푐표푙푙𝑖푠𝑖표푛 푡𝑖푚푒 For a system in steady state F’ = −퐹 푚푣 ie: 푑 = 푒퐸 휏 풆흉푬 ∴ 풗 = 풅 풎 Q:Derive an expression for electrical conductivity and hence electrical resistivity. 풆흉푬 The drift velocity of electron 풗 = ……(1) 풅 풎 풋 Also from Ohm’s law electrical conductivity , 휎 = ..(2)where j=current density & 푬 E=electric field But j = ne풗풅 … (ퟑ) 풆흉푬 ∴ eqns 1&3 ,we get j = ne. 풎 풋 풏풆ퟐ흉 ∴ = ….(4) 푬 풎 풏풆ퟐ흉 From eqns 2 &4,we get 흈 = …..(5) 풎 1 By definition electrical resistivity, 휌 = …(6) 휎 풎 ∴ from eqns 5&6, we get 흆 = 풏풆ퟐ흉
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM Q:Write the assumptions of classical free electron theory(CFET) or Drude-Lorentz theory. The assumptions of CFET are:- 1. All metals contains a large number of free electrons called conduction electrons . 2. The free electrons are called electron gas which have 3 degrees of freedom as they are treated as gas molecules. 3. Free electrons obey laws of kinetic theory of gases and hence they have mean free path, mean collision time, drift velocity. 4. Free electrons move in a constant electric or potential field due to positive ion cores. 5. Electro-static forces between electron-ions and electron-electrons are negligible. 6. Free electrons move with drift velocity in a direction opposite to the direction of the applied field. 3 7. Average kinetic energy of free electrons at temperature T is 푘푇, 2 where k = Boltzmann constant.
Q:Explain the failures of classical free electron theory(CFET). Failures of CFET are:- 1.Specific heat :According to CFET the molar specific heat of electron gas at constant 3 volume is given by 퐶 = 푅 ,where R=universal gas constant. But experimentally 푣 2 −4 determined 퐶푣 𝑖푠 given by 퐶푣 = 10 푅푇 ,where T=absolute temperature. Thus CFET fail to explain the the dependence of 퐶푣 표푛 T and numerical constant.
2.Dependance of electrical conductivity (흈) on T: 1 3 According to CFET kinetic energy 푚푣2 = 푘푇 2 2 ∴ 푣 ∝ √푇 …(1) 1 But mean collision time 휏 ∝ 푣 1 ∴ 휏 ∝ … . (2) ∵ 푣 ∝ √푇 √푇 푛푒2휏 Also from 휎 = ,we get 휎 ∝ 휏 … . (3) 푚 ퟏ ퟏ From eqns 2 & 3 we get 흈 ∝ ,but experimentally it has been observed that 흈 ∝ , √푇 푻 thus CFET fail to explain the dependence of 휎 표푛 푇. 3.Dependance of ′흈′풐풏 풆풍풆풄풕풓풐풏 풄풐풏풄풆풏풕풓풂풕풊풐풏 ‘n’ : 푛푒2휏 퐹푟표푚 휎 = , we get 휎 ∝ 푛 , according to CFET 흈 표푓 tri-atomic metals must be more 푚 than that of di-atomic and mono-atomic metals but from experimental results, it has been found that 흈 of mono-atomic metals of low n are more than that of di and tri-atomic metals of large n .Thus CFET fail to explain dependence of 흈 on concentration n.
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM 4. Mean free path’ λ’: Mean free path of Cu from CFET is given by λ= 푣휏=2.85nm. But the experimentally determined value of λ=28.5 nm, which is 10 times more than that λ of CFET.
Q:State/Mention the assumptions of (Sommerfeld’s)quantum free electron theory(QFET)
The assumptions of QFET are: 1.The energy of free electrons are quantized. 2. Free electrons obey Pauli’s exclusion principle. 3. The distribution of free electrons in energy levels is governed by Fermi-Dirac statistics. 4. Free electrons move in uniform potential field due to ionic cores in a metal . 5. The electrostatic electron-ion attractions and electron-electron repulsions are negligible. 6. Electrons are considered as wave like particles.
Q:Explain the success of QFET.
QFET explained the following experimental facts which were not explained by CFET. 2푘 1.Specific heat: According to QFET the molar specific heat is given by 퐶푣 = 푅푇 퐸퐹 2푘 −4 −4 where k = Boltzmann constant and 퐸퐹=Fermi energy. But = 10 ∴ 퐶푣 = 10 푅푇 퐸퐹 which agrees with experimental value.
2.Dependance of electrical resistivity ′흈′ on ‘T’(temperature):
풏풆ퟐ흉 휆 We know that(WKT) 흈 = ∗ and 휏 = where 푣퐹=Fermi velocity, λ=wavelength 풎 푣퐹 풏풆ퟐ 흀 ∴ 흈 = ∗ ∴ 흈 ∝ 흀 … . . (1) 풎 풗푭 1 1 1 1 Also it has been shown that λ ∝ ∝ ∝ ⇨ λ ∝ …(2) 퐴 푟2 푇 푇
where A=area, r=radius & T=temperature. ퟏ ∴ 푓푟표푚 푒푞푛푠 1&2,we get 흈 ∝ which is the experimentally determined relation. 푻 3.Dependance of ′흈′풐풏′풏′:
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM ퟐ 풏풆 흀 흀 WKT conductivity 흈 = ∗ , thus 흈 푑푒푝푒푛푑푠 푏표푡ℎ 표푛 풏 푎푛푑 .With the decreases of 풎 풗푭 풗푭 흀 .흀 atomicity, n decreases and increases so that n . increases ,so that 흈 is more for 풗푭 풗푭 monoatomic metals than that of di&tri-atomic metals.This explains the conductivity of mono-atomic metals is more than that of di & tri-atomic metals.
4.Mean free path obtained from QFET is about 28.5 nm which is experimentally determined value.
Q:Distinguish /(Write the differences )between CFET and QFET. CFET QFET
1.Free electron energy levels are 1.Free electron energy levels are dis- continuous . continuous 2.Free electrons may possess 2.No two electrons can possess same energy, as same energy they obey Pauli’s exclusion principle. 3.Distribution of electrons in 3.Distribution of electrons in energy levels energy levels obeys Maxwell- obeys Fermi-Dirac statistics Boltzmann statistics
Q: Derive an expression for conductivity and hence resistivity based on QFET. ℎ 2휋 According to de-broglie hypothesis λ = ,also λ = where symbols have usual 푚푣 푘 ℎ푘 significance. ∴ 푚푣 = , where v & k are vectors 2휋 differentiating this w.r.t ‘t’, 풅풗 풉 풅풌 We get m = ….(1) 풅풕 ퟐ흅 풅풕 풅풗 If an electric field E is applied ,the electron experience a force m = −푒퐸 …(2) 풅풕 풉 풅풌 ∴ 푓푟표푚 푒푞푢푛푠1&2, 푤푒푔푒푡 , = −푒퐸 ퟐ흅 풅풕 2휋 ∴ 푑푘 = − 푒퐸 푑푡 ℎ 푂푛 Integration between the limits 0 to t ,we get 푡 2휋 푡 푑푘 = − 푒퐸 ∫ 푑푡 ∫0 ℎ 0 2휋 k(푡) − 푘(0) = − 푒퐸 (푡 − 0) ℎ 2휋 휕푘 = − 푒퐸 푡 ℎ The Fermi sphere displace through 휕푘 in a direction opposite to the direction of applied electric field E in a time ‘t’ as shown in the diagram.
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If t = 휏 , 푡ℎ푒푛 푑푢푒 푡표 푐표푙푙𝑖푠𝑖표푛푠 푡ℎ푒 푚푒푎푛 푑𝑖푠푝푙푎푐푒푚푒푛푡 휕푘푎푣 표푓 푓푒푟푚𝑖 푠푝ℎ푒푟푒 is given 2휋 by 휕푘 = − 푒퐸 휏 ….(3), 푎푣 ℎ ℎ Using mv = 푘 2휋 ℎ we get 푚 휕푣 = 휕푘 푎푣 2휋 푎푣 ℎ ∴ 휕푣 = 휕푘 ….(4) 푎푣 2휋푚 푎푣 푒퐸 From eqns 3&4,we get 휕푣 = − 휏 푎푣 푚 but 휕푣푎푣 = 푣푑 drift velocity as initial average velocity of electrons is zero. 푒퐸 ie: 푣 = − 휏 … (5) 푑 푚 E
Also current density J = −푛푒푣푑 ….(6) 푛푒2휏 ∴ 푓푟표푚 푒푞푛푠 5&6, we get J = E 푚 0 퐽 푛푒2휏 ∴ = ….(7) 휕푘 퐸 푚 퐽 But from Ohm’s law conductivity 휎 = …..(8) 퐸 풏풆ퟐ흉 ∴ 푓푟표푚 푒푞푛푠 7&8 ,we get 흈 = …(7), 풎 where m is called effective electron mass usually denoted by 풎∗ 1 Also resistivity 휌 = …(8) 휎 풎 ∴ 푓푟표푚 푒푞푛푠 7&8 ,we get 흆 = 풏풆ퟐ흉
Q:Explain the terms: Mobility, Fermi velocity, Fermi temperature, Fermi Level &Fermienergy. a) Mobility(μ) of electrons is defined as the drift velocity (풗풅)acquired by the electrons per unit 풗 흈 풆흉 electric field(E). ie: μ = 풅 ⇨ μ = ⇨ μ = 푬 풏풆 풎 b) Fermi velocity(풗푭) of an electron is defined as it’s velocity when it’s energy is equal to the 1 Fermi energy. ie: 푚푣2 = 퐸 ⇨푣 = (√2퐸퐹) 2 퐹 퐹 퐹 푚 c) Fermi temperature(푻푭) is defined as the temperature at which the average thermal 푬 energy of the electron is equal to the Fermi energy at 0K. ie: 푻 = 푭 where k=Boltzmann 푭 풌 constant. d) Fermi level is defined as the highest filled energy level in a metal at 0K.
e) Fermi energy(푬푭)is defined as the energy of the highest occupied level in a metal at 0K.
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM Q:What is Fermi-Dirac statistics ? Explain. Fermi –Dirac statistics is the statistical rule applied to the distribution of identical, 1 indistinguishable particles of spin called fermions like electrons, which obey Pauli’s 2 exclusion principle. The probability of occupation of a state by an electron is given by Fermi factor/Fermi- Dirac distribution function given by 1 f(E)= 1+푒(퐸−퐸퐹)/푘푇 where 퐸퐹 = 퐹푒푟푚𝑖 푒푛푒푟푔푦, E=energy of State, k=Boltzmann constant and T= temperature.
Case1. At T=0K & E < 퐸퐹 ,f(E)= 1 Case2. At T=0k & E > 퐸퐹 ,f(E)= 0 and Case3. At T> 0퐾 & E = 퐸퐹 ,f(E)= 0.5 Variation of f(E) with E is as shown in the graph.
f(E) 1 T = 0K
0.5 T> 표퐾 0 E
퐸 퐹 Q:What is meant by Fermi factor ?Explain the variation of Fermi factor with temperature & energy.
Fermi factor is the probability of occupation of a given energy state by an electron in a metal ퟏ at thermal equilibrium. It is given by the relation f(E) = ,where 퐸 퐹푒푟푚𝑖 푒푛푒푟푔푦, ퟏ+풆(푬−푬푭)/풌푻 퐹 E = energy of State, k = Boltzmann constant and T = temperature The variation of f(E) with temperature and energy is discussed below: When T = 0K (푬−푬푭)/풌푻 −∞ Case-1: If E < 퐸퐹 ⇨ (푬 − 푬푭)𝑖푠 − 푣푒 ,then 풆 = 풆 = 0 ퟏ ퟏ ∴ f(E) = = =1, ퟏ+풆(푬−푬푭)/풌푻 ퟏ+ퟎ Thus the probability of occupation up to Fermi level is 100%.
(푬−푬푭)/풌푻 ∞ Case-2: If E > 퐸퐹 ⇨ (푬 − 푬푭)𝑖푠 + 푣푒 ,then 풆 = 풆 = ∞ ퟏ ퟏ 1 ∴ f(E) = = = =0, ퟏ+풆(푬−푬푭)/풌푻 ퟏ+∞ ∞ Thus the probability of occupation above Fermi level is 0%.
ퟎ (푬−푬푭)/풌푻 Case-3: If E = 퐸퐹 ⇨ (푬 − 푬푭) = 0 ,then 풆 = 풆ퟎ indeterminate
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM When T > 0K ퟏ ∴ f(E)= ퟏ+풆(푬−푬푭)/풌푻
(푬−푬푭)/풌푻 ퟎ f(E) Note: 푇 = 500푘 < 푇 =1000k 푰풇 푬 = 푬푭 ⇨ (푬 − 푬푭) = ퟎ 푡ℎ푒푛 풆 = 풆 = 1 1 2 ퟏ ퟏ ퟏ ∴ f(E) = = = Thus the probability ퟏ+풆(푬−푬푭)/풌푻 ퟏ+ퟏ ퟐ of occupation of Fermi level is 50% above 0K. 1 T = 0K The variation of f(E) with temperature(T) and 0.5 푇 < 푇 energy(E)is shown in the graph . 1 2 0 E
퐸 퐹
Q:What is meant by Density of states ? Explain.
Density of states g(E) is defined as the number of electronic states present in a unit 8√2휋푚3/2 energy range. Mathematically g(E) 푑퐸 = 퐸1/2 푑퐸 is a continuous function and ℎ3 the product g(E)dE=dN gives the number of states per unit volume in an energy range (dE)between E and E+dE The number of electrons per unit volume ,n=∫ 푔(퐸)푓(퐸)푑퐸 and theVariation of g(E) with E is shown in the graph.
g(E)
0 E
Q:What is meant by effective mass? Explain.
When an electric field is applied to a metal, electrons in the K-shell are not at all accelerated as they are tightly bound to the nucleus. These electrons possess infinite mass called effective mass denoted by 풎∗ Effective mass is equal to the true mass if the electron is in vacuum.
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM Q:Explain the dependence of resistivity on temperature and impurity Or State and explain Matthiessen’s rule.
휌
휌𝑖
0 T
1. Variation of resistivity 흆 of metals with temperature T is given by
Matthiessen’s rule 흆 = 흆풊 + 흆푷풉 (T) Where 휌𝑖 = 푡푒푚푝푒푟푎푡푢푟푒 𝑖푛푑푒푝푒푛푑푒푛푡 푟푒푠𝑖푠푡𝑖푣𝑖푡푦 푑푢푒 푡표 𝑖푚푝푢푟𝑖푡𝑖푒푠 & 𝑖푚푝푒푟푓푒푐푡𝑖표푛푠 푙푎푡푡𝑖푐푒 휌 (T)= 푡푒푚푝푒푟푎푡푢푟푒 푑푒푝푒푛푑푒푛푡 푟푒푠𝑖푠푡𝑖푣𝑖푡푦 푑푢푒 푡표 푣𝑖푏푟푎푡𝑖표푛푠. 푃ℎ 푝ℎ표푛표푛 2. Variation of resistivity 흆 with temperature T is shown graphically.
At 0K, 흆푷풉 (T) = 0 ⇨ = 흆풊 , 3. For pure metals 휌𝑖 = 0 ,but 푝푟푎푐푡𝑖푐푎푙푙푦 푛표푡 푝표푠푠𝑖푏푙푒 푡표 푝푟표푑푢푐푒 푝푢푟푒 푚푒푡푎푙푠, ∴ 휌𝑖 ≠ 0
4. and at large T , 휌𝑖 is negligible compared to 푙푎푟푔푒 휌푃ℎ (T) ∴ 휌 = 휌푃ℎ (T)
Q:Derive an expression for electrical conductivity of a semi-conductor. In a semiconductor the net current is due to both electrons and holes .
The current due to electrons is 퐼푒 = 푛푒푒푎푣푒 ,where 푛푒 = 푛푢푚푏푒푟 푑푒푛푠𝑖푡푦, 푎 = 푎푟푒푎 , 푣푒 = 푑푟𝑖푓푡 푣푒푙표푐𝑖푡푦 & 푒 = 푐ℎ푎푟푔푒 Also current due to holes is 퐼ℎ = 푛ℎ푒푎푣ℎ But the total current I= 퐼푒 + 퐼ℎ = ea(푛푒푣푒 + 푛ℎ푣ℎ) 퐼 = e (푛 푣 + 푛 푣 ) 푎 푒 푒 ℎ ℎ 퐼 푊퐾푇, 푐푢푟푟푒푛푡 푑푒푛푠𝑖푡푦 퐽 = 푎 ∴ J= e (푛푒푣푒 + 푛ℎ푣ℎ) but from Ohm’s law J=휎퐸
ie: 휎퐸 = 푒 (푛푒푣푒 + 푛ℎ푣ℎ) 푣 푣 휎 = 푒 (푛 푒 + 푛 ℎ) 푒 퐸 ℎ 퐸 흈 = e(풏풆흁풆+풏풉흁풉) where 휇푒&휇ℎ mobilities of electrons and holes respectively. For an intrinsic semiconductors 푛푒 = 푛ℎ = 푛𝑖 called the density of intrinsic charge carriers.
For n-type semiconductor 푛푒 ≫ 푛ℎ ⇨ 흈 = e풏풆흁풆
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For p-type semiconductor, 푛푒 ≪ 푛ℎ ⇨ 흈 = e풏풉흁풉 Q: State law of mass action. Obtain an expression for the intrinsic carrier density.
Law of mass action states that the product of the concentration of charge carriers
electrons ( 푛푒)and holes( 푛ℎ) in an intrinsic semiconductor is equal to the 2 square of the intrinsic charge carriers( 푛𝑖 ) at any temperature. ퟐ ie: 풏풆. 풏풉 = 풏풊
Q: Derive an expression for electron concentration in an intrinsic semiconductor. or Derive an expression for electron charge carrier density in an intrinsic semiconductor. We know that the number of energy levels between the energy interval E and E+dE 8√2휋(푚∗)3/2 is given by g(E)dE = ( 푒 ) 퐸1/2푑퐸 ,where 푚∗ = effective mass of ℎ3 푒 electron. Also, the probability of occupation of an energy level is given by Fermi factor 1 f(E) = 퐸−퐸 where 퐸퐹 = 퐹푒푟푚𝑖 푒푛푒푟푔푦 & 푘 = 퐵표푙푡푧푚푎푛푛 푐표푛푠푡푎푛푡 ( 퐹 ) 푒 푘푇 +1
Then ,the concentration of electron charge carriers is given by ∞ 푛푒 = ∫ 푓(퐸) g(E)dE 퐸푔 ∗ 3/2 ∞ 1 8√2휋(푚푒) 1/2 = ∫ 퐸−퐸 ( ) 퐸 푑퐸 퐸푔 ( 퐹 ) ℎ3 푒 푘푇 +1 ∗ where 푚푒 = effective mass of electron.
On integration we can show that 퐸 −퐸푔 휋푚∗ 푘푇 3/2 ( 퐹 ) 푛 = 4√2 ( 푒 ) 푒 푘푇 푒 ℎ2
Similarly, the concentration of hole charge carriers is given by
0 푛ℎ = ∫−∞ 푓(퐸) g(E)dE ∗ 3/2 0 1 8√2휋(푚ℎ) 1/2 = ∫ 퐸−퐸 ( ) 퐸 푑퐸 −∞ ( 퐹 ) ℎ3 푒 푘푇 +1 ∗ where 푚ℎ = effective mass of electron.
On integration we can show that −퐸 휋푚∗ 푘푇 3/2 ( 퐹) 푛 = 4√2 ( ℎ ) 푒 푘푇 ℎ ℎ2
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Q: Based on law of mass action derive an expression for intrinsic charge density (concentration) in an intrinsic semiconductor.
Law of mass action states that the product of the concentration of charge carriers
electrons ( 푛푒)and holes( 푛ℎ) in an intrinsic semiconductor is equal to the square 2 of the intrinsic charge carriers( 푛𝑖 ) at any temperature. ퟐ ie: 풏풊 = 풏풆. 풏풉 퐸 −퐸푔 휋푚∗푘푇 3/2 ( 퐹 ) But , WKT, 푐표푛푐푒푛푡푟푎푡𝑖표푛 표푓 푒푙푒푐푡푟표푛푠, 푛 = 4√2 ( 푒 ) 푒 푘푇 and 푒 ℎ2 −퐸 휋푚∗ 푘푇 3/2 ( 퐹) Concentration of holes, 푛 = 4√2 ( ℎ ) 푒 푘푇 ℎ ℎ2 From law of mass action, 2 푛𝑖 = 푛푒. 푛ℎ ∗ 퐸퐹−퐸푔 ∗ −퐸퐹 2 휋푚푒푘푇 ( ) 휋푚ℎ푘푇 3/2 ( ) ∴ 푛 = 4√2 ( )3/2 푒 푘푇 푋 4√2 ( ) 푒 푘푇 𝑖 ℎ2 ℎ2
퐸 −퐸푔 −퐸 휋푚∗푘푇 ( 퐹 ) 휋푚∗ 푘푇 ( 퐹) 푛 = √4√2 ( 푒 )3/2 푒 푘푇 푋 4√2 ( ℎ )3/2 푒 푘푇 𝑖 ℎ2 ℎ2 on simplification we can we can show that, −푬품 흅풎∗풌푻 흅풎∗ 풌푻 ( ) 풏 = 4√ퟐ ( 풆 )ퟑ/ퟒ( 풉 )ퟑ/ퟒ 풆 ퟐ풌푻 풊 풉ퟐ 풉ퟐ Q:Explain the terms: Fermi velocity & Fermi temperature.
Fermi velocity (푣퐹)is defined as the velocity of the electron having the energy 1 2퐸 equal to Fermi energy(퐸 ) ,mathematically , 푚푣2 = 퐸 ∴ 푣 = √ 퐹 퐹 2 퐹 퐹 퐹 푚
Fermi temperature (푇퐹)is defined as the ratio of Fermi energy(퐸퐹) to 퐸 the Boltzmann constant(k). Mathematically,푇 = 퐹 퐹 푘
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SUPERCONDUCTIVITY:
Q: Explain the terms :- a. Superconductivity is the phenomenon in which the resistivity/resistance of a substance becomes zero. b. Superconducting state is the zero resistivity/resistance state of the substance. c. Critical/Transition temperature is the minimum temperature at which the substance changes from normal state to superconducting state. d. Critical magnetic field is the minimum magnetic field at which the substance changes to normal state from superconducting state. e. Flux quantization is the quantization of the magnetic flux through a superconducting ring 풏풉 and is given by ∅ = ,where n =1,2,3…..h = Planck’s constant and e = electron charge. ퟐ풆 f. Isotopic effect states that the transition temperature 푇퐶varies inversely 1 as the square root of the atomic mass M ie: 푇 ∝ 퐶 √푀
Q: temperature dependence of resistivity in metals and superconducting materials.
휌 Normal conductor
휌𝑖 Super conductor
0 푇 퐶 T
1. The variation of the resistivity of the metals and superconductor is as shown in the graph. 2. When metals are cooled their resistivity decreases and reaches the minimum value 휌𝑖 called the residual resistivity at 0K , 휌𝑖 is due to impurities and imperfections in 푡ℎ푒 metals. 3. The resistivity of the metals increases with temperature due to ion vibrations .
4. The resistivity of the metals is given by Mathessian’s rule 휌 = 휌𝑖 + 휌푃ℎ(T). 5. But the resistivity of superconductors decreases with the decrease of temperature and
suddenly becomes zero at the temperature 푇퐶 called critical/transition temperature. 6. The transition temperature depends on the impurities present in the metals.
Magnetic materials reduce 푇퐶 to large extent.
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Q:Explain Meissner’s effect.
H H
T> 푻푪 푆푢푝푒푟 푐표푛푑푢푐푡표푟 T< 푻푪
1) Meissner effect is the phenomenon in which the magnetic flux in the material Is completely expelled out of the material at temperature called Critical temperature. 2) Below Critical temperature the material behave as perfect dia-magnet. 3) Consider a superconducting material placed in a magnetic field H, the magnetic lines penetrate through the conductor. 4) When the temperature of the conductor is cooled below the critical temperature
푇퐶,the magnetic flux inside the conductor is completely expelled out of the conductor.
5) Above 푇퐶 ,B≠ 0 and below 푇퐶 ,B= 0. 6) Meissner effect is the principle of Maglev vehicles, Squid etc.
Q:Write a note on Type-I and Type-II superconductors. Type-I
−푀 Type-I
Super Conducting Normal State State O 퐻퐶 H
1) Type-I Superconductors(SC’s) are also called soft SCs. 2) The magnetization of Type-I SC’s increases with the increase of the applied magnetic field
and suddenly drops to zero at the magnetic field 퐻퐶 called the critical magnetic field as shown graphically.
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3) Below 퐻퐶 SC becomes perfectly diamagnetic obeying Meissner effect and beyond 퐻퐶 SC changes to normal state. 4) The Type-I SC’s exists only in two states namely superconducting and normal
Type-II −푀
Super Intermediate Normal conducting state state state
O 퐻퐶1 퐻퐶 퐻퐶2 H
1) Type-II Superconductors(SC’s) are also called hard SC’s.
2) The magnetization of Type-II SC’s increases with the increase of the applied magnetic
field and begin to decrease at the magnetic field 퐻퐶1 called the lower critical field and
finally becomes zero at the upper critical magnetic field 퐻퐶2.
3) Up to 퐻퐶1 Type-II SC’s exhibit Meissner effect with perfect diamagnetism.
4) Between 퐻퐶1 & 퐻퐶2 they exhibit meissner effect partially.ie: Magnetic flux begins to penetrate at 퐻퐶1 푎푛푑 푐표푚푝푙푒푡푙푦 penetrates at 퐻퐶2.
5) Thus Type-II SC’s exists in three states namely SC state up to 퐻퐶1, intermediate state between 퐻퐶1 & 퐻퐶2 and normal state beyond 퐻퐶2.
6) Type-II SC’s exhibit superconductivity electrically even in the intermediate state .They can carry large currents.
7) These SC’s are used as coils for producing very large magnetic fields.
8) Type-I can be changed in Type-II , Type-I SC Lead can be changed to Type-II by adding indium as impurity to it.
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Q; Explain the temperature dependence of critical field.
H
퐻푂
Superconducting normal State state
O 푇퐶 T
1) Critical magnetic field(퐻퐶) is the minimum magnetic field applied to the superconductor to change from superconducting state to normal state. 2) The variation of the critical field with temperature is given by the relation 푇2 퐻퐶 = 퐻0 [1 − 2], 푇퐶
where 퐻0 is called the critical field at 0K , T = temperature and 푇퐶 =Critical temperature. 3) Thus critical field decreases with the increase of temperature.
Q:Explain BCS theory of superconductivity qualitatively.
( 푘1 − 푞 ) ( 푘2 +q )
q
e e
푘1 푘2
1) BCS(Bardeen-Cooper-Schrieffer) theory successfully explained superconductivity quantum mechanically. 2) BCS theory is based on electrons interactions through phonons/lattice as mediators. 3) An electron distorts the lattice due to electrostatic attraction when it approach a positive ion lattice, Smaller the mass of the ion distortion is more. 4) The oscillatory distorted lattice is quantized in terms of phonons . 5) This distorted lattice attract another electron by reducing its energy. 6) Thus the second electron attracts the first electron through lattice. This results in the pairing of two electrons called “Cooper Pair”. The attraction between the electrons is maximum when the two electrons have equal and opposite spin and momentum.
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM 7) The formation of cooper pair is considered as emission and absorption of the phonon by the two electrons as shown in the diagram.
8) The cooper pair formation begins at critical temperature 푇퐶 and is completed at 0K. 9) Thus at 0K all the electrons are paired in to ordered cooper pairs, which can pass through lattice without collisions when a potential difference is applied. This explains the zero resistivity of superconductors. 10) When the temperature increases the breaking of the cooper pairs begin beyond
0K and completed 푎푡 푇퐶 as a result the superconductor attains normal state. 11) BCS theory explained the existence of energy gap in superconductors, Meissner effect ,Coherence length and Flux quantization, etc.
Q: Write a note on High temperature superconductors(HTSC). 1. Generally Superconductors of transition temperature more than 10K are called High temperature superconductors(HTSCs). 2. First HTSC found was ceramic material (oxide of lanthanium, barium &
copper)whose 푇퐶 =30~35K, by Bednorz and Muller 3. Later many HTSCs are found of 푇퐶 more than 150K using liquid Nitrogen. In 2015 HTSC found was Hydrogen Sulphide of 푇퐶 = 203K at very high pressure. 4. Also 1-2-3 compound HTSCs are found which are alloys of 1 rare Earth element
atom, 2 barium atoms,3 Copper atoms and 7 Oxygen atoms. Ex:Y퐵푎2퐶푢3푂7 etc 5. HTSCs are all Type-II superconductors having complicated structures called Perovskite , Tetragonal and Orthorhombic. 6. In all the HTSCs 1 to 4 Cu-O layers are placed between other layers. 7. HTSCs are used in Squids, MagLev vehicles ,MRI, NMR, LHC, and other applications in various fields like Industry, Medicine, Physics, Chemistry, Engineering ,Scientific research etc 8. HTSCs are not perfectly diamagnetic since they exhibit partial Meissner’s effect.
Q;Explain the Principle,Construction and working of Maglev vehicles.
RW V ehi cle RW
Superconducting magnet
Aluminium guide way
1) Maglev vehicles works on the principle of Meissner effect & magnetic levitation.
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM 2) Floating of a magnet placed over a superconducting magnet is called magnetic levitation. 3) The schematic diagram of Maglev vehicle is as shown in the diagram. Superconducting magnet is fitted in to the base of the vehicle, which is placed over aluminium guide way. 4) Magnetic field produced due to current in the guide way lifts/levitate the vehicle and also propel it. 5) The vehicle is provided with retractable wheels which can be pulled out when the vehicle is brought to rest and can be pulled in when levitated and propelled. 6) Maglev vehicle trains have attained very high speeds of 603 km/hr.
PROBLEMS SECTION
MODULE-2 : ELECTRICAL PROPERTIES OF MATERIALS
1 푚 1 푛푒2휏 Formulae needed : f(E) = ; 휌 = ; 휎 = ; 휎 = ; 퐸−퐸퐹 푛푒2휏 휌 푚 1+푒 푘푇 푥퐷푁 1 푒퐸휏 휎 1 n = 퐴 ; 휇 = ; ½ m푉 2 = 퐸 ; 푉 = ; λ = 푉. 휏 ; n = = ; 퐻 = 푀 푛푒휌 퐹 퐹 푑 푚 휇푒 휌휇푒 퐶 푇2 퐻0 [1 − 2] 푇퐶
1. Find the relaxation time of conduction electrons in a metal of resistivity 1.54 x10−8 Ω-m, if the metal has 5.8 x1028 Conduction electrons per 푚3.( JAN2015)
Given: 휌 = 1.54 x10−8 Ω-m ; n = 5.8 x1028 /푚3 ; 푚 = 9.1푥10−31푘푔 ; −19 푒 = 1.6푥10 퐶 ; 휏푟= 휏 =? 푚 푚 Using 휌 = ,we get 휏 = 푛푒2휏 휌푛푒2 9.1푥10−31 = 1.54 푥10−8푥5.8 푥1028푥(1.6푥10−19)2 = 3.98 푥10 −14 S
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM 2. Calculate the mobility and relaxation time of an electron in copper assuming that each 3 3 atom contributes one free electron ,whose density is 8.92 x 10 kg/푚 and 푁퐴=6.02 x 1026/kg mole. (Jun/Jul 14) −8 3 3 26 Given: 휌 = 1.73 x 10 Ω-m ; M = 63.5 ; D = 8.92 x 10 kg/푚 ; 푁퐴=6.02 x 10 /kg mole ; 푚 = −19 9.1푥10−31푘푔 ; 푒 = 1.6푥10 퐶 ; 푥 = 1 푒푙푒푐푡푟표푛/푎푡표푚, 푛 = ? ; 휇 = ? & 휏 = ? 푥퐷.푁퐴 푈푠𝑖푛푔 n = 푀 1푥8.92 푥 103푥.6.02 푥 1026 = 63.5 = 8.456x1028 atoms. 1 휇 = 푛푒휌 1 휇 = = 4.27x10−3 8.456푥1028푥1.6푥10−19푥1.73 푥 10−8 푒휏 휇푚 also 휇 = , we get 휏 = 푚 푒 4.27푥10−3푥9.1푥10−31 = 1.6푥10−19 = 2.43 x10−14 S
3. Find the temperature at which there is 1.0% probability that a state with an energy 0.5eV above Fermi energy will be occupied. (Dec 2013/Jan2014)
−19 −23 Given: (퐸 − 퐸퐹) = 0.5 푒푉 = 0.5푥1.6푥10 J , k=1.38x10 J/K ; f(E) = 1 % = 1/100 =0.01 ; T= ? 1 (퐸−퐸퐹) Using f(E)= ,we get T = 1 1+푒(퐸−퐸퐹)/푘푇 푘.퐼푛( –1) 푓(퐸) 0.5푥1.6푥10−19 = 1 1.38푥10−23.퐼푛( –1) 0.01 0.5푥1.6푥10−19 = 1.38푥10−23.퐼푛(100 –1) = 1262 K =1.262x103K −14 3 4. Calculate the conductivity of sodium given 휏푚 =2 x 10 S. Density of sodium is 971 kg/푚 ,its atomic weight is 23 and has one conduction electron/atom. (퐽푢푛 2012) 3 26 −31 −14 Given: M = 23 ; D = 971 kg/푚 ; 푁퐴=6.02 x 10 /kg mole ; 푚 = 9.1푥10 푘푔 ; 휏 = 휏푚 =2 x 10 S 푒 = 1.6푥10−19퐶 ; 푥 = 1푒푙푒푐푡푟표푛/푎푡표푚, 푛 = ? ; 휎 = ? 푥퐷.푁퐴 푈푠𝑖푛푔 n = 푀 1푥971푥.6.02 푥 1026 = 23 = 2.54x1028 atoms.
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM 푛푒2휏 Also, 휎 = 푚 2.54푥1028푥( 1.6푥10−19)2푥2 푥 10−14 = 9.1푥10−31 = 1.42x107 Ωm.
5. Calculate the Fermi velocity and the mean free path for the conduction electrons in silver, given that its Fermi energy is 5.5 eV and relaxation time for electrons is 3.97 X10−14S. (퐷푒푐 2011)
−19 −14 −31 Given: 퐸퐹 = 5.5 푒푉 = 5.5 푥 1.6푥10 퐽 ; 휏 = 휏푟 = 3.97 X10 S ; 푚 = 9.1푥10 푘푔 푉퐹 = ? : λ = ? 2퐸 Using ½ m푉 2 = 퐸 ,we get 푉 = √ 퐹 퐹 퐹 퐹 푚
2푥5.5 푥 1.6푥10−19 = √ 9.1푥10−31 = 1.391x 106 m/s and 5 −14 −9 Also, λ =푉퐹. 휏푟 = 1.391x 10 x 3.97 X10 = 5.522푥 10 m
6. The Fermi level of potassium is 2.1eV. What are the energies for which the probabilities of occupancy at 300 K are 0.99 and 0.5 ? (퐽푢푛/퐽푢푙2011)
−19 퐽 Given: 퐸퐹 =2.1 eV = 2.1x1.6x10 ;T = 300K ; f(퐸1) =0.99 ; f(퐸2) =0.5 ; −23 K =1.38x10 J/K ; 퐸1 =? ; 퐸2 =? 1 Using f(E)= , 1+푒(퐸−퐸퐹)/푘푇 1 we get E = 퐸 + 푘푇 퐼푛 [ – 1] 퐹 푓(퐸) 1 ∴ 퐸 = 2.1푥1.6푥10−19 + 1.38x10−23푥300 퐼푛 [ – 1] 1 0.99 = 3.36 푥10−19 − 0.1902 푥10−19 = 3.17x10−19 퐽 3.17푥10−19 = eV 1.6푥10−19 = 1.98eV
1 Also, 퐸 = 2.1푥1.6푥10−19 + 1.38x10−23푥300 퐼푛 [ – 1] 2 0.5 = 3.36 푥10−19 − 0 = 3.36 푥10−19 퐽 3.36 푥10−19 = eV 1.6 푥10−19 = 2.1 eV
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM 7. The Fermi level for a metal is 2.1eV, calculate the energies for which the probability of occupancy at 300 K are 98% and 50%.(퐷푒푐 2010)
−19 퐽 Given: 퐸퐹 =2.1 eV = 2.1x1.6x10 ;T = 300K ; f(퐸1) =98%=0.98 ; f(퐸2)=50%=0.5 −23 k =1.38x10 J/K ; 퐸1? , 퐸2 =?
1 Using f(E)= , 1+푒(퐸−퐸퐹)/푘푇 1 we get E = 퐸 + 푘푇 퐼푛 [ – 1] 퐹 푓(퐸) 1 ∴ 퐸 = 2.1푥1.6푥10−19 + 1.38x10−23푥300 퐼푛 [ – 1] 1 0.98 = 3.36 푥10−19 − 0.16 푥10−19 = 3.2x10−19 퐽 3.2푥10−19 = eV 1.6푥10−19 =2eV 1 Also , 퐸 = 2.1푥1.6푥10−19 + 1.38x10−23푥300 퐼푛 [ – 1] 2 0.5 = 2.1푥1.6푥10−19 − 0 2.1푥1.6푥10−19 = eV 1.6푥10−19 = 2.1 eV
8. A uniform silver wire has resistivity 1.54 x10−8훺푚 at room temperature for an electric field 2 V/m. Calculate relaxation time and drift velocity of the electrons, assuming that there are 5.8 X 1022 conduction electrons per 푐푚3 of the material. (퐽푢푛 2010)
Given: 휌 = 1.54 x 10−8 Ω-m ; E = 2 V/m ; 푛 = 5.8 푋 1022/푐푚3 = 5.8 푋 1022푥10−6/ 3 −31 −19 푚 ; 푚 = 9.1푥10 푘푔 ; 푒 = 1.6푥10 퐶 ; 휏 = ? & 푉푑 = ?
휌푚 푈푠𝑖푛푔: 휏 = 푛푒2 1.54 푥 10−8 푥9.1푥10−31 = 5.8 푋 1022푥10−6( 1.6푥10−19)2 = 9.44 x10−18 S
푒퐸휏 푉 = 푑 푚 1.6푥10−19푥2푥9.44 푥10−18 = 9.1푥10−31 = 3.32 x 10−6 m/s
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM 9. Calculate the probability of an electron occupying an energy level 0.02 eV above the Fermi level at 300K.( 4 marks)
−19 −23 Given: (퐸 − 퐸퐹) = 0.02 푒푉 = 0.02푥1.6푥10 J , k=1.38x10 J/K and T=300K
1 Using f(E) = 1+푒(퐸−퐸퐹)/푘푇 1 = 1+푒(0.02푥1.6푥10−19)/1.38푥10−23 푥 300 1 = 1+푒0.773 1 = 1+2.166 1 = 3.166 =0.32 or 32 %
10. Calculate the concentration at which the acceptor atoms must be added to a germanium sample to get a P-type semiconductor with conductivity 0.15 per Ohm-metre. Given the mobility of holes=0.17 푚2/Vs. ( 4 marks)
Given: 휎 =0.15 /Ωm ,휇 = 0.17 푚2/푉푠 , e=1.6x10−19 C, n= ?
휎 Using, n = 휇푒 0.15 = 0.17푥1.6푥10−19 = 5.515 x1018 electrons
13 . A superconducting tin has a critical field of 306 gauss at 0K and 217 gauss at 2K. Find the critical temperature of superconducting tin.
Given : 퐻표 = 306 푔푎푢푠푠, 퐻퐶 = 217 푔푎푢푠푠, 푇 = 2퐾 , 푇퐶 =?
푇2 Using: 퐻퐶 = 퐻표 [1 − 2] 푇퐶
푇2 We get , 푇퐶 = 퐻 √[1− 퐶] 퐻표
22 = √ 217 [1− ] 306
=3.71K
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM 14. Calculate the mobility of electrons in copper assuming that each atom Contribute one free electron for conduction. Resistivity of copper=1.7 X 10−8 Ωm, atomic weight =63.54,density =8.96 X 103kg/푚3( CBCS-Jun/Jul16)
−8 3 3 26 Given: 휌 = 1.7 x 10 Ω-m ; M = 63.54 ; D = 8.96 x 10 kg/푚 ; 푁퐴=6.025 x 10 /kg mole ; 푚 = 9.11푥10−31푘푔 ; 푒 = 1.6푥10−19퐶 ; 푥 = 1푒푙푒푐푡푟표푛/푎푡표푚, 푛 = ? ; 휇 = ?
푥퐷.푁 푈푠𝑖푛푔 n = 퐴 푀 1푥8.96 푥 103푥.6.025 푥 1026 = 63.54 = 8.5x1028 atoms. 1 휇 = 푛푒휌 1 = = 4.325x10−3 푚2/vs 8.5푥1028푥1.6푥10−19푥1.7 푥 10−8 15. For intrinsic gallium arsenide, the room temperature electrical conductivity is 10−6 /Ωm.The electron and hole mobilities are respectively 0.85 푚2/푉푠 and 0.04 푚2/푉푠. 퐶alculate the intrinsic carrier concentration at room temperature.
−6 2 2 Given: 휎𝑖 = 10 /Ωm ; 휇푒 = 0.85 푚 /푉푠 휇푒 = 0.04 푚 /푉푠; 푛𝑖 =? 푈푠𝑖푛푔, 휎𝑖 = 푛𝑖e (휇푒 + 휇ℎ), 휎𝑖 푤푒 푔푒푡 , 푛𝑖 = 푒 (휇푒 +휇ℎ) 10−6 푛 = 𝑖 1.6푥10−19 (0.85+0.04) = 7.023푥1012/푚3
16. The effective mass for the electron in germanium is 0.55 푚표,where 푚표𝑖푠 푡ℎ푒 free electron mass. Find the electron concentration in Germanium at 300 K, assuming that the Fermi level lies exactly in the middle of the energy gap, given that the energy gap for Germaium is 0.66 eV.
−31 ∗ −31 Given: 푚표 = 9.1 x 10 kg ; 푚푒 = 0.55x 9.1 x 10 kg ; T= 300K ; 퐸푔 = 0.66푒푉; 퐸 0.66 퐸 = 푔 = = 0.33 푒푉 퐹 2 2
3/2 퐸 −퐸푔 휋푚∗ ( 퐹 ) 푒 푘푇 푘푇 Using ; 푛푒 = 4√2 ( ) 푒 ℎ2 (0.33−0.66)푥1.6푥10−19) 휋푥0.55푥 9.1 푥 10−31푥1.38푥10−23푥300 = 4√2 ( )3/2 x 푒 1.38푥10−23푥300 (6.625푥10−34)2 = 5.657x1.807x1024 x2.892x10−6 19 3 푛푒 = 2.956x10 /푚
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM 17. For intrinsic Gallium Arsenide , the electric conductivity at room temperature is ퟏퟎ−ퟔ풐풉풎−ퟏ풎−ퟏ. The electron and hole mobilities are respectively 0.85 풎ퟐ/푽푺 풂풏풅 ퟎ. ퟎퟒ 풎ퟐ/푽푺. Calculate the intrinsic carrier concentration at room temperature. (04 Marks)
−6 2 2 Given: 휎𝑖 = 10 /Ωm ; 휇푒 = 0.85 푚 /푉푠 휇푒 = 0.04 푚 /푉푠; 푛𝑖 =? 푈푠𝑖푛푔, 휎𝑖 = 푛𝑖e (휇푒 + 휇ℎ), 휎𝑖 푤푒 푔푒푡 , 푛𝑖 = 푒 (휇푒 +휇ℎ) 10−6 푛 = 𝑖 1.6푥10−19 (0.85+0.04) = 7.023푥1012/푚3 18. Calculate the probability of finding an electron at an energy level 0.02 eV above Fermi level at 200 K . ( 04 Marks) −19 −23 Given: (퐸 − 퐸퐹) = 0.02 푒푉 = 0.02푥1.602푥10 J , k=1.38x10 J/K and T=200K
1 Using f(E) = 1+푒(퐸−퐸퐹)/푘푇 1 = 1+푒(0.02푥1.602푥10−19)/1.38푥10−23 푥 200 1 = 1+푒1.161 1 = 1+3.193 1 = 4.193 = 0.2385 or 23.85 % 19. Gold has one free electron/atom.Its density,atomic weight and resistivity are 19300 kg/푚3 ,197 and 2.21 x 10−8 Ω푚 . Calculate the free electron concentration and mobility of conduction electron. ( 04 Marks) Given : D=19300 kg/풎ퟑ , M =197 , 흆 = 2.21 x 10−8 Ω푚 , 풙 = 1 electron/atom, ퟐퟖ −ퟏퟗ 푵푨 = 6.02 x ퟏퟎ /Kmol , e=1.6 x ퟏퟎ C , n =? , 흁 = ? 풙 푫 푵 Using, n = 푨 푴 ퟏ풙 ퟏퟗퟑퟎퟎ풙 ퟔ.ퟎퟐ풙 ퟏퟎퟐퟖ ∴ 풏 = = 5.898 x ퟏퟎퟑퟎ electrons/풎ퟑ ퟏퟗퟕ ퟏ 푨풍풔풐 , using 흁 = 흆풏풆 1 ∴ 흁 = = 4.795x10−5 푚2/Vs 2.21 푥 10−8푥5.898 푥 1030푥1.6 푥 10−19
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REDDAPPA C MODULE-2 INTERLINE PUBLISHING.COM 20. Calculate the drift velocity and thermal velocity of conduction electrons in copper at a temperature of 300K,when a copper wire of length 2 mand resistance 0.02 Ω carries a current of 15 A. Given the mobility of free electrons in copper is 4.3xퟏퟎ−ퟑ풎ퟐ/Vs. ( 04 Marks) Given: T = ퟑퟎퟎ푲 ,L = ퟐ 풎 , 푹 = 0.02 Ω , I = 15 A , 흁 = 4.3xퟏퟎ−ퟑ 풎ퟐ/Vs, −ퟑퟒ −ퟐퟑ m= 풎풆 = 9.1 x ퟏퟎ kg , k = 1.38 x ퟏퟎ J/K , 풗풅 =? , 풗푭 =? 푽 Using 풗 = 흁E but E = and V= IR 풅 푳 흁푰푹 ퟒ.ퟑ풙ퟏퟎ−ퟑ풙 ퟏퟓ 풙 ퟎ.ퟎퟐ we get 풗 = = = 6.45 x ퟏퟎ−ퟒ m/s 풅 푳 ퟐ ퟑ ퟏ Also using 풌푻 = 풎풗ퟐ ퟐ ퟐ 푭 ퟑ풌푻 ퟑ풙ퟏ.ퟑퟖ 풙 ퟏퟎ−ퟐퟑ풙ퟑퟎퟎ We get , 푽 = √ = √ =1.168xퟏퟎퟓ m/s 푭 풎 ퟗ.ퟏ 풙 ퟏퟎ−ퟑퟒ
@@ end@@
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM MODULE-3 : LASERS AND OPTICAL FIBRES LASERS Q: Explain Einstein’s explanation of interaction of radiation with matter (or) Explain Induced absorption , Spontaneous emission &Stimulated emission.
Let N1 & N2 be the population of the energy states E1 & E2 respectively so that (E2 −E1) = hν & E2 >E1 According to Einstein radiation interacts with matter in 3 ways namely:
1) Induced absorption:
∗ 퐸 푁 푨 • 2 2 hν ⇝ ⇒
퐸 • 푨 푁 1 2
Induced absorption is the phenomenon in which an atom(A) in the lower energy state E1 absorb the incident photon of energy ‘hν’ & excite to the higher energy state E2 if (E2 −E1) = hν . Mathematically it(induced absorption) is represented as hν +A → 푨∗ or photon +atom→ 푎푡표푚∗
Also, Rate of induced absorption = 퐵12 푁1푈휈 ,where 푈휈 = 푒푛푒푟푔푦 푑푒푛푠𝑖푡푦 & ′ 퐵12 = 퐸𝑖푛푠푡푒𝑖푛 푠 𝑖푛푑푢푐푒푑 푎푏푠표푟푝푡𝑖표푛 푐표푒푓푓𝑖푐𝑖푒푛푡.
2.Spontaneous emission: ∗ 퐸2 푨 • 푁2
⇒ hν ⇝
퐸1 푁1 푨 •
Spontaneous emission is the phenomenon in which an atom (A) in the excited state of energy E2 de-excite to the lower energy state E1 without any external influence by emitting a photon of energy hν =(E2 −E1). Mathematically, it is represented as 푨∗ → A + hν
Also, Rate of Spontaneous emission = 퐴21 푁2 ′ 푤ℎ푒푟푒 퐴21 = 퐸𝑖푛푠푡푒𝑖푛 푠 푠푝표푛푡푎푛푒표푢푠 푒푚𝑖푠푠𝑖표푛 푐표푒푓푓𝑖푐𝑖푒푛푡.
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3.Stimulated emission: ∗ 퐸2 푨 • 푁2
hν ⇝ ⇨ hν ⇝+ hν ⇝
퐸1 푁1 •A
퐸2 푁2 • Stimulated emission is the phenomenon in which an atom (푨∗) in the excited state of
energy E2 de-excite to the lower energy state E1 under the influence of an external photon (hν) by emitting an identical photon of energy hν =(E2 −E1). Mathematically,𝑖푡 𝑖푠 푟푒푝푟푒푠푒푛푡푒푑 푎푠 풉흂 + 푨∗ → A +2 hν
Also, Rate of Stimulated emission= 퐵21 푁2푈휈,where 푈휈 = 푒푛푒푟푔푦 푑푒푛푠𝑖푡푦 & ′ 퐵21 = 퐸𝑖푛푠푡푒𝑖푛 푠 푆푝표푛푡푎푛푒표푢푠 푒푚𝑖푠푠𝑖표푛 푐표푒푓푓𝑖푐𝑖푒푛푡.
Q: Explain the terms : Active medium ; Population Inversion ; Pumping ; Meta stable state and Laser Cavity( Optical resonator) Or Explain the requisites of a laser system.
The requisites of a laser system are :
1. Active medium is a Solid/Liquid/Gas medium in which stimulated emission and amplification of the radiations can be achieved.
2. Pumping is the supply of energy to the atoms in the lower states in order to excite them to higher states. The methods of pumping are Optical pumping, Electrical pumping, Forward bias pumping ,Chemical pumping, Elastic one-one collisions.
3. Population Inversion is condition of system in which the population of higher energy states exceed the population of lower states.
4. Meta stable state is an intermediate state in which the average life of the atoms is of the order of 10−3s ie: their life is 105 times more than that of normal states.
5. Laser Cavity( Optical resonator) is a pair of parallel/con-focal/concentric mirrors between which active medium is placed so that stimulated emitting photons are used to cause further Stimulated emissions and to amplify the beam. One mirror is highly silvered and the other partially silvered. The distance between the 푛휆 mirrors is given by L = ,where λ = wavelength and n = number of stationary waves 2 produced.
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 푸:Derive an expression for energy density in terms of Einstein’s coefficients (or) Derive the relation between Einstein’s coefficients Induced absorption Stimulated emission
퐸2 푁2 • •
. hν ⇝ ⇨ ⇨ hν ⇝ hν ⇝ ⇨ 2hν ⇝ ⇝
퐸1 • 푁1 • •
Spontaneous emission
Consider a system of atoms in thermal equilibrium with radiation of energy density
푈휈 & frequency ‘ν’. Let N1 & N2 be the population of the energy states E1 & E2 respectively, where (E2 >E1 ) We know that,
Rate of induced absorption = 퐵12 푁1푈휈 , ′ where 퐵12 = 퐸𝑖푛푠푡푒𝑖푛 푠 푐표푒푓푓𝑖푐𝑖푒푛푡 표푓 𝑖푛푑푢푐푒푑 푎푏푠표푟푝푡𝑖표푛 .
Rate of Spontaneous emission = 퐴21 푁2 푤ℎ푒푟푒 퐴21 = 퐸𝑖푛푠푡푒𝑖푛′푠 푐표푒푓푓𝑖푐𝑖푒푛푡 표푓 푠푝표푛푡푎푛푒표푢푠 푒푚𝑖푠푠𝑖표푛 푎푛푑
Rate of Stimulated emission= 퐵21 푁2푈휈 , ′ where 퐵21 = 퐸𝑖푛푠푡푒𝑖푛 푠 푐표푒푓푓𝑖푐𝑖푒푛푡 표푓 푆푡𝑖푚푢푙푎푡푒푑 푒푚𝑖푠푠𝑖표푛 . At thermal equilibrium, 푅푎푡푒 표푓 푆푝표푛푡푎푛푒표푢푠 푅푎푡푒 표푓 푆푡𝑖푚푢푙푎푡푒푑 Rate of induced absorption = { } + { } 푒푚𝑖푠푠𝑖표푛 푒푚𝑖푠푠𝑖표푛
ie: 퐵12 푁1푈휈 = 퐴21 푁2 + 퐵21 푁2푈휈
∴ 푈휈( 퐵12 푁1 − 퐵21 푁2) = 퐴21 푁2 퐴21 푁2 𝑖푒: 푈휈 = dividing both Nr & Dr by 퐵21 푁2 ,we get ( 퐵12 푁1− 퐵21 푁2) 퐴21 1 = 퐵 푁 ……….(1) 퐵21 [ 12 1 −1] 퐵21 푁2 푁 According to Boltzmann’s law, 1 = 푒(퐸2 −퐸1)/퐾푇 = 푒 ℎ휈/퐾푇 …….(2) 푁2 퐴21 1 From eqns 1 &2 , we get. 푈휈 = 퐵 ……..(3) 퐵21 [ 12 푒ℎ휈/퐾푇 −1] 퐵21 8휋ℎ휈3 1 But .the energy density given by Planck’s law is 푈휈 = ……(4) 퐶3 ( 푒ℎ휈/퐾푇 −1) ퟑ 푨ퟐퟏ ퟖ흅풉흂 퐴 ℎ휈 21 퐾푇 Comparing eqns 3 & 4 ,we get = ퟑ or = 푈휈 ( 푒 − 1) and 푩ퟏퟐ = 푩ퟐퟏ 푩ퟐퟏ 푪 퐵21 Thus coefficient of stimulated absorption = coefficient of stimulated emission.
푨ퟐퟏ ퟏ Thus energy density 푼흂 = 풉흂/푲푻 푩ퟐퟏ [풆 –ퟏ]
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Q:Explain the fundamental mode of vibration in CO2 molecule.
In CO2 molecule there are three fundamental modes of vibrations, namely
1. Symmetric mode :
Molecular axis
O C O
Symmetric mode is the mode in which both the oxygen atoms oscillate simultaneously to & fro about the stationary carbon atom along the molecular axis.
2. Asymmetric mode:
Molecular axis
O C O
Asymmetric mode is the mode in which both the oxygen atoms move in one direction and the carbon atom move in the opposite direction along the molecular axis.
3. Bending mode:
Molecular axis
C
O O
Bending mode is the mode in which both oxygen atoms and carbon atom move in opposite directions perpendicular to the molecular axis. The internal vibrations of CO2 molecule are the combination of the above three modes.
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Q:Explain the principal, construction and working of C푶ퟐ laser.
푪푶 laser diagram ퟐ
C푂2 푁2 퐻푒 gases outlet
Water outlet
푀1 -optical cavity(oc) 푀2 (oc)
5 m
A electrodes C 2.5 cm Laser
BW BW=Brewster window
Q:Explain the Waterconstruction inlet & working of Ba CO 2 Laser.
Energy level diagram
푁2 Resonant transfer of energy C푂2
휈 = 1 퐶5 (meta stable state)
퐶4 10.6 휇푚 laser
Excitations by collission 퐶3 9.6 휇푚 laser
퐶2
휈 = 0 Ground state 퐶1
Principle : CO2 laser works on the principle of stimulated emission. Construction: 1) The schematic diagram CO2 Laser is as shown in the diagram invented by CKN Patel an Indian engineer 2) It consists of a (glass)discharge tube of length 5 m & diameter 2.5 cm filled with a mixture of gases CO2 ,N2 ,He in the ratio 1:2:3 3) High DC voltage can be applied to the gas between the electrodes A&C . 4) Ends of the tube is fitted with ( NaCl ) Brewster windows to get polarized laser beam 5) Two con-focal silicon mirrors coated with aluminum are provided at the ends of the tube which act as optical resonators. 6) Cold water is circulated through a tube surrounding the discharge tube
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM Working: 1) CO2 Laser is a four level molecular gas laser which produce continuous or pulsed laser beam. 2) It works on the principle of stimulated emission between the rotational sublevels of an upper & lower vibrational levels of CO2 molecules. 3) Ionisation takes place due to electric discharge when high DC voltage is applied between electrodes producing electrons. 4) The accelerated electrons excite both N2& CO2 atoms to their higher energy levels ‘ 퓥 =1 & C5 from their ground states 0 & C1 due to collision as follows: ∗ ∗ e+푵ퟐ → 푵ퟐ+e’ and e+푪푶ퟐ → 푪푶ퟐ+e’ where e& e ’are the energies of electron before and after collision. ∗ 5) 푵ퟐ molecule in excited level collide with CO2 molecules in their ground state C1 & excite it to metastable state C5 by resonant energy transfer as level C5 of CO2 is same as ∗ ∗ level 퓥=1 of 푁2 given by 푵ퟐ+푪푶ퟐ → 푪푶ퟐ+푵ퟐ 6) As this process continues due to electric discharge pumping , population inversion takes place betweenC5 &C4 and C5 & C3. 7) The transitions/de-excitations takes place as follows: C5 → C4 producing laser 10.6휇푚 (IR region) C5 → C3 producing laser 9.6휇푚 (IR region)
C4 → C2
C3 → C2 Radiation less transitions C2 → C1 8) Due to high thermal conductivity of He, it removes heat from mixture and de-populate the lower states C3 &C2 quickly . 9) Laser beam is amplified by using optical resonators. 10) The laser output is 100kW for continuous mode and 10 kW in pulsed mode. Q: Explain the construction and working of Semi-conductor laser.
Semiconductor laser Energy level giagram
Metalic coated(MC) surface P-type pn-Jn n-type
RS
퐸퐹푒 PS CB P-type Ba p P Laser 퐸 h+e Pn-jn 퐹푝 n-type VB
n-type Polished surfac(PS) Laser
Roughened surface(RS) MC
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM Principle: SC laser works on the principle of stimulated emission. Construction: 1. The schematic diagram of GaAs semi-conductor device is as shown in the diagram. 2. It consists of heavily doped n-region of GaAs doped with tellurium and p-region of GaAs doped with zinc. 3. The upper and lower surfaces are metalized so that pn-junction is forward biased . 4. Two surfaces perpendicular to the Jn are polished so that they act as optical resonators and the other two surfaces roughened to prevent lasing in that direction.
Working: 1. Semi-conductor laser are made up of highly de-generate semi-conductors having direct band gap like Gallium Arsenide (GaAs). 2. When GaAS diode is forward biased with voltage nearly equal to the energy gap voltage, electrons from n-region & holes from p-region flow across the junction creating population inversion in the active jn region. 3. As the voltage is gradually increased due to forward biasing population inversion is achieved between the valence band and conduction band which in turn result in stimulated emission. 4. Photons produced are amplified between polished optical resonator surfaces producing laser beam. 5. GaAs laser produce laser beam of wavelength 8870Å in IR region , GaAsP produce laser beam of 6500Å in visible region etc.
Q:Mention the characteristics of laser beam/light. The laser beam characteristics are: 1. They are highly monochromatic.
2. They are highly coherent.
3. They are highly directional.
4. They are highly focusable.
5. They are least divergent.
Q:Mention the uses of laser beam.
Laser beam is used in Holography, Laser welding, Laser cutting laser drilling and measurement of atmospheric pollutants.
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM Q: What is holography ? Holography is the process in which the details of an object in 3-dimensions can be recorded on a 2-dimensional aid based on the principle of interference of light.
Q: Construction/Recording of Hologram.
Object
Reflected beam Object spherical beam
Laser beam ( λ ) Hologram
Beam splitter reference beam
1. The schematic diagram for the construction of a Hologram is as shown in the diagram. 2. A laser beam of wavelength (λ) is made to fall on beam splitter, which split the beam in to two beams. One beam which passes through splitter is called ‘reference beam’. The other reflected beam is made to fall on the object whose hologram is to be produced. 3. The beam reflected by the object is called ‘spherical object beam’. 4. The reference and object beams produce concentric circular rings called “Gaber zones “where both intensity and phase are recorded due to interference on the photographic plate. 5. On developing photographic plate we obtain ‘hologram’. 6. At each and every point on the hologram, complete information/details of the object are recorded. 7. If the hologram is cut in to any number of pieces, each piece produce the complete image of the object with less resolution.
Q: Re-construction of Hologram.
Hologram Observation direction
Laser beam (λ)
Object Virtual image Object Real image
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 1. The schematic diagram for the reconstruction of the image from Hologram is as shown in the diagram. 2. The hologram is illuminated with the same laser beam of wavelength(λ) which used for construction of hologram. 3. The hologram act as diffraction grating. Due to diffraction and interference two images of the object are produced. 4. One which is formed on the side of incident laser is virtual image and the other formed on the other side is the real image of the object. 5. By changing the direction of the observation ,we can see all the details of the object originally hidden from view.
Q:Explain laser welding
Laser beam
Condenser lens
Materials to be weld
1. The schematic arrangement of laser welding is as shown in the diagram.
2. High intensity and high focussability of lasers is used for laser welding. 3. Laser beam is focused on to the spot to be welded. Due to generation of high heat the material melts in a short period & the impurities float to the surface.
4. On cooling the joint, it becomes homogeneous stronger joint. 5. As laser welding is contac tless welding ,no foreign materials enter in to the joint. 6. Laser welding can be done more precisely by using pre-programmed computer assisted
welding. 7. CO2 laser can be used to weld both metallic and non-metallic substances. Q:Explain laser cutting:
푂2/푁2 inlet
Condenser lens
material
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 1. The schematic arrangement of laser cutting is as shown in the diagram. 2. High intensity and high focussability of lasers is used for laser cutting. 3. Laser cutting involve melting and gas assisted blowing out the material. 4. Laser beam focused on to the surface to be cut ,melts it and the high speed gas O2 /N2 passed through the nozzle blows away the molten material. O2 gas need low power laser than for N2. 5. This process continues till the material is cut. 6. Laser cutting can be done more precisely by using pre-programmed computer assisted cutting. 7. CO2 laser can be used to cut both metallic and non-metallic substances
Q:Explain laser drilling.
Laser beam
Condenser lens
material
1..The schematic arrangement of laser cutting is as shown in the diagram. 2.High intensity and high focussability of lasers is used for laser cutting. 3.The pulsed laser beam is focused to the spot where the hole to be drilled. High heat generated melts the material. 4.Melted material absorb the heat further and vaporizes. 5.with continues heating by the laser vapours ionize into plasma state. 6. The plasma state material further absorb laser and emit black body radiations, which in turn generate detonation waves. These waves and high power pressure expel the material out of hole. 7.. Laser drilling can be done more precisely by using pre-programmed computer assisted drilling. 8. CO2 laser can be used to drill both metallic and non-metallic substances.
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM Q:Measurement of atmospheric pollutants using LIDAR(Light Detection And Ranging).
Atmosphere
• • ⋅ ͦ • • • ⋅ ͦ •
pollutants
• • ⋅ ͦ • • • ⋅ ͦ • laser Detector
Analyser
1. The schematic experimental arrangement to measure atmospheric pollutants is as shown in the diagram. 2. LIDAR (LIght Detection And Ranging)method is used to find the concentration of atmospheric pollutants using laser. 3. Laser beam is sent in to the atmosphere in a particular direction and the scattered laser beam by the pollutants are detected using detector. 4. The received scattered laser beam is analysed using analyser. 5. The measured intensity gives the concentration of the pollutants in a particular distance and direction. 6. This method does not give information regarding the nature of the pollutants. 7. In order to know the nature of pollutants we have to adopt Raman spectroscopy method .n this comparing the spectra of atmospheric pollutants and comparing them with standard spectra of pollutants, we can identify pollutants.
OPTICAL FIBRES:
Q:What is an Optical Fibre? Optical Fibre is a transparent di-electric material (like glass/plastic) which guides/ carry) light along it based on the principle of total reflection of light.
Optical fibre consists of a cylindrical transparent di-electric material of high refractive index called core. It is surrounded by another di-electric transparent material of low refractive index called cladding Cladding in turn is surrounded by cylindrical insulator called Sheath, which gives mechanical strength & protect the fibre from absorption, scattering etc
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM Q:What is an acceptance angle of an optical fibre ? Find the expression for acceptance angle or Numerical aperture.
B Cladding ( 풏 ) ퟐ
훷 Core (푛1) Launch medium (푛표 ) Fibre axis
휃 A 푟 D 휃𝑖
O
Acceptance angle is the maximum angle submitted by the ray with the axis of the fibre so that light can be accepted and guided along the fibre.
Let 푛1 , 푛2 & 푛표 be the RI of core, cladding and launch medium respectively. Also OA incident ray,AB refracted ray,BC totally reflected ray, 휃𝑖, 휃푟 & ∅ be the angles of incidence, refraction at A & angle of incidence at B respectively. 푛1 By snell’s law at A, 푛표 푠𝑖푛 휃𝑖 = 푛1 푠𝑖푛 휃푟 ⇨ 푠𝑖푛 휃𝑖 = 푠𝑖푛 휃푟 ……(1) 푛표 푙푒 But, from ∆ 퐴퐷퐵, 휃푟 =(90 ͦ−∅)⇨ 푠𝑖푛 휃푟 = 푠𝑖푛(90 ͦ − ∅)=cos(∅) … . (2)
when 휃𝑖 𝑖푠 푚푎푥𝑖푚푢푚 = 휃표 , 푎푐푐푒푝푡푎푛푐푒 푎푛푔푙푒, 푡ℎ푒푛 ∅ = ∅퐶 Critical angle ……..(3) 푛1 From eqns 1,2 &3 ,we get 푠𝑖푛 휃표 = 푐표푠(∅퐶) …….(4) 푛표
2 2 √푛1−푛2 푛2 2 Also, 푠𝑖푛(∅퐶) = and 푐표푠(∅퐶) = √(1 − 푠𝑖푛 ∅퐶) = …(5) 푛1 푛1
2 2 2 2 √푛1−푛2 √푛 −푛 −1 1 2 From eqns 4 &5,we get 푠𝑖푛 휃표 = or 휃표 = 푠𝑖푛 푛표 푛표
for air 푛표 = 1
Q: What is meant by numerical aperture(NA) and mention the expression for it. Numerical aperture is the ability of the optical fibre to accept the light and guide along the fibre and is numerically equal to sine of the acceptance angle.
2 2 √푛1−푛2 ie: NA = 푠𝑖푛 휃표 = 푛표
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM Q:What is meant by fractional refractive index change(∆). The ratio of the difference between refractive indices of core and cladding to that (푛 −푛 ) of core is called fractional refractive index change. ie: ∆ = 1 2 푛1 Q:What is attenuation and explain types of attenuation in optical fibres. Loss of power of light signal as it is guided along the fibre is called attenuation. Attenuation is measured in terms of dB/km. There are three types of attenuations in the fibre namely: 1. Absorption losses are the losses due to impurities & material itself and they are two types namely a) Impurity losses are the losses due to the impurities(Cu, Fe, etc) present in the fibre, which can be minimised by taking care during manufacture of the fibre. b) Intrinsic losses are the losses due to the material itself, these losses decreases with the increase of wavelength. 2. Scattering losses are the losses due to imperfections of the fibre called Rayleigh scattering losses which varies inversely as the 휆4. 3. Radiation losses are the losses are two types namely:- a) Microscopic losses are the losses due to non-linearity of the fibre axis ,which can be minimized by providing compressible jacket & taking care during manufacture of the fibre. a) Macroscopic losses are the losses due to large curvature/bending of the fibre when it is wound over a spool/bent at corners. These losses increase exponentially up to threshold radius and there afterwards losses becomes large. Q: Obtain expression for attenuation coefficient in an optical fiber of length L 푑푃 Lambert’s law states that the rate of decrease of intensity of light with distance(− ) 푑퐿 in a medium is directly proportional to the original intensity(P) 푑푃 ie: − ∝ 푃 푑퐿 푑푃 = −훼푃 …(1) ,where 훼 is called the attenuation coefficient. 푑퐿 푑푃 Rewriting(1 ) as = −훼. 푑퐿 , 푃 Integrating this between the limits (푃𝑖, 푃표) 푓표푟 푃 푎푛푑 (0, 퐿)푓표푟 퐿, ,푃 푑푃 퐿 ∫ 표 = −훼 ∫ 푑퐿 푃𝑖 푃 0 [푙표푔 P],푃표 = −훼 [L]퐿 10 푃𝑖 0 ,푃표 푙표푔10 ( ) = −훼L……(2) where 푃𝑖, 푃표 are 𝑖푛 & out put powers. 푃𝑖 ퟏퟎ ,푷풐 Equation (2),can be written as 휶 = − 풍풐품ퟏퟎ ( ) dB/km. 푳 푷풊 −휶푳 푷풐 or = 풆 ퟏퟎ 푷풊
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM Q:What is meant by modes of propagation.
The paths along which the light is guided in the fibre are called modes of 푉2 propagation and the number of modes of the fibre is given by N = 2
Q: What is V-Parameter/number ?
It is the quantity which represent the number of modes of the fibre given by 2 2 휋푑 휋푑 √푛1 −푛2 V = . NA = . 휆 휆 푛0 where d = diameter , λ = wavelength,
푛1, 푛2 &푛0 RI’s of Core, Cladding & medium
Q: Explain the construction and working of Types of optical Fibres.
There are three types of optical fibres namely:-
1) Single mode step index fibre(SMF
Cladding Core
Input output Index profile Cross section profile Modes profile Pulse profile
1. SMF has a core diameter 8 − 10휇푚 of uniform RI and cladding diameter 60-70휇푚 has 푢푛𝑖푓표푟푚 푅퐼 2. The step index, cross section, modes and pulse profiles of SMF are as shown in the diagram. 3. SMF guides light in a single mode as shown. 4. The V-number is < 2.4. 5. Numerical aperture is < 0.12 . 6. Attenuation is in the range 0.25-0.5 dB/km. 7. Information carrying capacity is very large. They are long haul carriers. 8. The output and input pulses are almost same. 9. Laser source is used. Connectors are costly
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 2) Multimode step index fibre(MMF)
Cladding Core
Input output Input output Index profile Cross section profile Modes profile Pulse profile
1) MMF has a core diameter 50 − 200휇푚 of uniform RI and cladding diameter 100-250휇푚 has 푢푛𝑖푓표푟푚 푅퐼. 2) The step index, cross section,modes and pulse profiles are as shown in the diagram. 3) MMF guides light in multi-modes as shown . 4) The V-number is > 2.4. 5) Numerical aperture is 0.2 푡표 0.3 . 6) Attenuation is in the range 0.5- 4 dB/km. 7) Information carrying capacity is small to medium and short haul carriers. 8) The output pulse is widened. 9) LED source is used & Connectors are cheap.
3) Graded index multimode fibre(GRIN
Cladding
Core Input output
Index profile Cross section profile Modes profile Pulse profile
1) GRIN has a core diameter 50 − 200휇푚 of variable RI and cladding diameter 100-250휇푚 has 푢푛𝑖푓표푟푚 푅퐼. 2) The graded index, cross section ,modes and pulse profiles are as shown in the diagram. 3) MMF guides light in multi-modes as shown in the diagram. 4) The V-number is > 2.4. 5) Numerical aperture is 0.2 푡표 0.3 .
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 6) Attenuation is in the range 0.5- 4 dB/km. 7) Information carrying capacity is large and efficient and are short haul carriers. 8) The output pulse and input pulse are as shown. 9) Laser/LED source is used. Connectors are cheap. 10) Easy to splice and interconnect but expensive.
Q:Explain point to point communication system using optical fibre. The schematic block diagram of point to point communication system using optical fibre is as shown in the diagram.
Input Information Coder Optical Optical fibre REPEATER Audio-video receiver AES BES transmitter carrier Signal AVS AES BES OS Modulator [ Receiver
Cum
Transmitter ] Output Information De-Coder Optical Optical fibre Audio-video transmitter BES AES receiver carrier Signal AES AVS De- modulator OS BES
Note: AVS =audio/video signal. AES=analog electrical signal, BES=binary electrical signal & OS = optical signal. 1. Information receiver-receives , convert input AVS in to AES & fed to coder. 2. Coder- receives, convert AES in to BES and fed in to optical transmitter after modulating it with carrier signal. 3. Optical transmitter-receives, convert BES in to OS and fed in to carrier optical fibre. 4. Carrier optical fibre-receive OS and guide it along the fibre. Weakened OS is fed in to repeater. 4. Re-peater( Receiver cum transmitter)-receives the Weakened OS, restore to original strength and fed back in to carrier optical fibre again, which in turn guide OS and fed in to optical receiver. 6. Optical receiver- receive ,convert OS in to BES & fed in to de-coder. 7. De-coder-receive, de-modulate & convert BES in to AES & fed in to information transmitter. 8. Information transmitter-finally receive, convert AES in to AVS as output
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM PROBLEMS SECTION
MODULE-3 : LASERS AND OPTICAL FIBRES
푛휆 10 푃표 푁2 −(ℎ퐶/휆푘푇) ℎ퐶 Formulae needed: L = ; 훼 = − log[ ] ; = 푒 ; E= P.t and E =n. 2 퐿 푃𝑖 푁1 휆 ; 2 2 2 √푛1 −푛2 휋푑 푉 (푛1−푛2) NA = ; Sin휃표 = NA ; V = X NA ; N = ; ∆ = ; 푛표 휆 2 푛1 1. Find the number of modes of standing waves in the resonator cavity of length 1 m in He- Ne temperature of laser operating at wavelength 632.8nm.( JAN2015) Given: L= 1 m ; λ = 632.8 nm = 632.8푥10−9 m. n = ? 푛휆 2퐿 Using L = ,we get n = 2 휆 2푥1 = 632.8푥10−9 = 3.161 x 106 푚표푑푒푠
2. A fiber with an input power of 9μW has a loss of 1.5 dB/km. If the fiber is 3000 m long, calculate the output power.(JAN2015)
−6 Given: 푃𝑖 = 9휇푤 = 9푥10 푤 ,훼 = 1.5 푑퐵/푘푚 ,L = 3000m = 3 km , 푃표 = ? −훼퐿 10 푃표 Using 훼=− log ⟦ ⟧ we get 푃표 = 푃𝑖 푒 10 퐿 푃𝑖 −1.5푥3 −6 = 9푥10 푒 10 −6 = 5.74 푥10 푤
3. Find the ratio of populations of two energy levels in a laser if the transition between them produces light of wavelength 6493Å,assuming the ambient temperature as 27℃.(Jun/Jul 14) Given: λ = 6493Å = 6493x10−10 m, t=27℃⇨ T = 273+27 =300K; h=6.63x10−34 Js; 푁 k=1.38x10−23 J/K ;C=3x108m/s ; 2 = ? 푁1 ℎ퐶 푁2 −( ) Using = 푒 휆푘푇 푁1 6.63푥10−34푥3푥108 −( ) = 푒 6493푥10−10푥1.38푥10−23푥300 = 푒−79.99 = 7.33x10−33
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 4. The numerical aperture of an optical fibre is 0.2 when surrounded by air. Determine the R.I of its core, given the R.I of the cladding is 1.59.Also find the acceptance angle when the fibre is in water of R.I 1.33(Jun/Jul 14)
Given: 푛표푥 NA = 0.2 , 푛2 = 1.59 , 푛′표 = 1.33 , 푛1 =? a푛푑 휃′표 = ? 2 2 √푛1 −푛2 2 2 For air , Using NA = , we get 푛1 = √(푛표푥푁퐴) + 푛2 푛표 = √( 0.22 + 1.592 ) = 1.603
√푛 2−푛 2 √ 2 2 ′ −1 1 2 −1 1.603 −1,59 Also 휃 표 = 푆𝑖푛 ( ′ ) = 푆𝑖푛 ( ) = 8.65° 푛 표 1.33
5. A pulse laser has an average power output 1.5mW per pulse and pulse duration is 20ns.The number of photons emitted per pulse is estimated to be 1.047 x108.Find the wavelength of the emitted laser. (Dec 2013/Jan2014) Given: P=1.5mW = 1.5x10−3 푤 ,t=20 ns =20x10−9 s , C=3x108m/s ; h=6.63x10−34 Js, n = 1.047 x108 ,λ = ? ℎ퐶 푛ℎ퐶 Using E= Pxt and E = n. , 푤푒 푔푒푡 , λ = 휆 푃푡 1.047 푥108푥6.63푥10−34푥3푥108 = 1.5푥10−3 푥20푥10−9 = 6.942x10−7 m. 6. The angle of acceptance of an optical fibre is 30° 푤ℎ푒푛 kept in air. Find the angle of acceptance when it is in a medium of refractive index 1.33. (Dec 2013/Jan2014)&(Jun/Jul 퐽푢푛 2013) ( 2011) 퐽푢푙
Given: 휃표 =30° , 푛′표=4/3=1.333 , 푛표 = 1 푓표푟 푎𝑖푟 , 휃′표 =? 2 2 Using 푛′표푆𝑖푛 휃′표 = 푛표Sin휃표 = √푛1 − 푛2 = constsnt −1 We get , 휃′표 = 푆𝑖푛 (푛표푆𝑖푛휃표⁄푛′표) = 푆𝑖푛−1 (1 푆𝑖푛30°⁄1.333) =22.03° 7. Calculate the NA,V-number and number of modes in an optical fibre of core diameter 50μm,core and cladding refractive indices 1.41 and 1.4 at wavelength 820 nm. (퐽푢푛 2012) −9 −6 Given: 푛1 = 1.41 ; 푛2 = 1.4 ; λ = 820 nm= 820 x 10 m ; d = 50μm =50x 10 m ; 푛표 =1 for air NA = ? ; V = ? & N = ? 2 2 2 2 √푛1 −푛2 √(1.41) −(1.4) a. NA = = = 0 .168 푛표 1
휋푑 휋푥50푥 10−6푥0 .168 b. V = X NA = = 32.18 휆 820 푥 10−9
푉2 (32.17)2 c. N = = = 517.5 = 518 2 2
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 8. The refractive indices of the core and cladding of a step index fibre are 1.45 and 1.40 respectively and its core diameter is 45μm.Calculate its relative refractive index difference, V- number at wavelength 1000 nm and number of modes. (퐷푒푐 2011) −9 −6 Given: 푛1 = 1.45 ; 푛2 = 1.4 ; λ = 1000 nm= 1000 x 10 m ; d = 45μm =45x 10 m ; 푛표 = 1; ∆ = ? ; V = ? & N = ? (푛1−푛2) (1.45−1.4) 1) ∆ = = = 0.0345 푛1 1.45 ퟐ ퟐ −6 2 2 휋푑 √풏ퟏ −풏ퟐ 3.14푥45푥 10 푥√1.45 −1.4 2) V = 푋 = −9 = 53.34 휆 풏풐 1000 푥 10 푥1 푉2 (53.34)2 3) N = = = 1422.6 =1423 2 2
9. A He-Ne gas laser is emitting a laser beam with an average power of 4.5 mW. Find the number of photons emitted per second by the laser. The wavelength of the emitted radiation is 6328Å. (퐷푒푐 11) Given: P=4.5mW = 4.5x10−3 푤 ,t= 1 S , C=3x108m/s ; h=6.63x10−34 Js, λ = 6328Å = 6328x10−10 m ; n = ? ℎ퐶 휆푃푡 Using E= Pxt and E =n. ⇨ 푛 = 휆 ℎ퐶 −10 6328푥10 푥4.5푥10−3푥1 = 6.63푥10−34 푥3푥108 = 1.432x1016 photons. 10. Find the number of modes of standing waves and their frequency separation in the resonant cavity of 1 m length of He-Ne operating at wavelength of 632.8 nm. (퐽푢푛/ 퐽푢푙2011) Given: L= 1 m ; λ = 632.8 nm = 632.8푥10−9 m. n = ? and 푛휆 2퐿 Using L = , we get , n = 2 휆 2푥1 = 632.8푥10−9 = 3.161 푥 106 푚표푑푒푠
11. The ratio of population of two energy states in a laser 1.059 x10−30.If the temperature of the system is 57°퐶, 푤ℎ푎푡 𝑖푠 푡ℎ푒 푤푎푣푒푙푒푛푔푡ℎ 표푓 푡ℎ푒 푙푎푠푒푟 ? 푁 Given: 2 = 1.059x10−30, t = 57°퐶, ⇨ T =273+57=330K; h=6.63x10−34 Js; k=1.38x10−23 푁1 J/K ;C=3x108m/s ; λ = ?
푁2 −(ℎ퐶/휆푘푇) 푁2 푁2 푁2 Using = 푒 ,we get , In( ) = −(ℎ퐶/휆푘푇) 푆𝑖푛푐푒: 푙표푔푒 ( ) = In( ) 푁1 푁1 푁1 푁1 푁 λ = −(ℎ퐶/퐼푛 ( 2) 푘푇) 푁1 −6.63푥10−34 푥3푥108 = 퐼푛(1.059푥10−30) 푥1.38푥10−23푥330 = 6.328x10−7 푚
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 12. A signal with input power 200mW loses 10% of its power after travelling a distance of 3000 m. Find the attenuation coefficient of the fiber. (퐷푒푐 2010) −3 −3 Given: 푃𝑖 = 200푚푊 = 200푥10 푤 , 푃표 = 90% 푃𝑖 = 0.9푥200푥10 푤 ; L = 3000 m = 3 km ; 훼 =? , 10 푃표 10 0.9푥200푥10−3 Using 훼= − log[ ] = − log [ −3 ] = 0.1525 푑퐵/푘푚 퐿 푃𝑖 3 200푥10
13. Calculate on the basis of Einstein’s theory the number of photons emitted per second by He-Ne laser source emitting light of wavelength 6328Å with an optical power 10mW. (퐽푢푛 2010) Given: P=10mW = 10x10−3 푤 ,t=1 s , C=3x108m/s ; h=6.63x10−34 Js, , λ = 6328 Å = 6328 x 10−10 m ; n = ? ℎ퐶 휆푃푡 Using E= Pxt and E =n. , 푤푒 푔푒푡 푛 = 휆 ℎ퐶 6328 푥 10−10 푥10푥10−3 푥1 = 6.63푥10−34 푥3푥108 = 3.18 x1016 /푚3 14. An optical fiber has core R.I 1.5 and R.I of cladding is 3% less than the core index. Calculate the numerical aperture, angle of acceptance and internal critical acceptance angle. (퐽푢푛 2010)
Given: 푛1 = 1.5 ; 푛2 =97% 푛1 = 0.97 x1.5 =1.455 ; 푛0 = 1 ; 푁퐴 = ? ; 휃표 = ? & 휃퐶 = ? √푛 2−푛 2 Using 1) NA = 1 2 푛표 √1.52−1.4552 = = 0.365 1 −1 2) 푠𝑖푛 휃0= 푁퐴 ⇨ 휃표 = 푆𝑖푛 (푁퐴) = 푆𝑖푛−1(0.365) = 21.41° 1.455 푛2 −1 푛2 −1( ) 3) 푠𝑖푛 휃퐶 = ⇨ 휃퐶 = 푆𝑖푛 ( ) = 푆𝑖푛 1.5 = 75.93° 푛1 푛1
A 5W pulsed laser emits light of wavelength 694 nm. If the duration of each pulse is 20 ns, calculate the number of photons emitted per pulse.( 4 marks
Given: P=5W , λ =694 nm=694x10−9 m ,t=20 ns =20x10−9 s , C=3x108m/s ; h=6.63x10−34 Js, n =? ℎ퐶 Using E=P.t and E =n. , 휆 푃푡휆 푤푒 푔푒푡 n = ℎ퐶 5푥20푥10−9 푥694푥10−9 = 6.63푥10−34 푥3푥108 =3.489x1011 photons/pulse.
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 15. The angle of acceptance of an optical fiber kept in air is 30°.Find the angle of acceptance when the fiber is in a medium of refractive index 4/3.( 4 marks)
Given: 휃표 =30° , 푛′표=4/3=1.333 , 푛표 = 1 푓표푟 푎𝑖푟 , 휃′표 =?
ퟐ ퟐ Using 푛′표푆𝑖푛 휃′표 = 푛표Sin휃표 =constant (√풏ퟏ − 풏ퟐ ) for a fibre ′ −1 푛표푆𝑖푛휃표 We get 휃 표 = 푆𝑖푛 ( ′ ) 푛 표 −1 1 푆𝑖푛30° = 푆𝑖푛 ( ) 1.333 = 22.03°
16. The ratio of population of two energy levels out of which one corresponds to metastable state is 1.059 x10−30.Find 푡ℎ푒 푤푎푣푒푙푒푛푔푡ℎ 표푓 푙𝑖푔ℎ푡 푒푚𝑖푡푡푒푑 푎푡 330퐾.(CBCS-Jun/Jul16)
푁 Given: 2 = 1.059x10−30, T =330K; h=6.63x10−34 Js; k=1.38x10−23 J/K ; 푁1 C=3x108m/s ; λ = ?
푁2 −(ℎ퐶/휆푘푇) Using = 푒 푁1
푁2 푁2 푁2 we get , In( ) = −(ℎ퐶/휆푘푇) 푆𝑖푛푐푒: 푙표푔푒 ( ) = In( ) 푁1 푁1 푁1 푁 λ = −(ℎ퐶/퐼푛 ( 2) 푘푇) 푁1 −6.63푥10−34 푥3푥108 = 퐼푛(1.059푥10−30) 푥1.38푥10−23푥330 = 6.328x10−7 푚
17. The refractive indices of the core and cladding of a step index fibre are 1.45 and 1.40 respectively and its core diameter is 45μm.Calculate the refractive index change and numerical Aperture.(CBCS-Jun/Jul16)
−6 Given: 푛1 = 1.45 ; 푛2 = 1.4 ; ; d = 45μm =45x 10 m ; 푛표 = 1, ∆ = ? ; & NA = ?
(푛 −푛 ) 1) ∆ = 1 2 푛1 (1.45−1.4) = = 0.0345 1.45 ퟐ ퟐ √풏ퟏ −풏ퟐ 2) NA = 풏풐 √1.452−1.42 = = 0.3775 1
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 19.The average power output of a laser beam of wavelength 6500Å is 10 mW.Find the number of photon emitted per second by the laser source. Given: P =10mW = 10x10−3 푤 ; t=1 s ; C=3x108m/s ; h=6.63x10−34 Js ; λ = 6500 Å = 6500 x 10−10 m ; n = ? ℎ퐶 Using E= Pxt and E =n. 휆 휆푃푡 We get, 푛 = ℎ퐶 6500 푥 10−10 푥10푥10−3 푥1 = 6.63푥10−34 푥3푥108 =3.268 x 1016 푝ℎ표푡표푛푠/푚3
20. An optical signal propagating in a fiber retains 85% of input power after travelling a distance of 500 m in the fiber. Calculate the attenuation coefficient.
Given: 푃표 = 85% 푃𝑖 = 0.85 푃𝑖 ; L = 500m = 0.5 km; 훼 =? , 10 푃표 Using 훼 = − log[ ] 퐿 푃𝑖 10 0.85 푃 10 = − log [ 𝑖 ] = − log [0.85] 0.5 푃𝑖 0.5 = 1.412 dB/km 21. The attenuation of light in an optical fiber is 2 dB/km. What fraction of its intensity remains after (i) 2 km, (ii) 5 km ?
푑퐵 푃 푃 Given: 훼 = 2 , ( 표) =?, ( 표) =? 푘푚 푃𝑖 퐿=2푘푚 푃𝑖 퐿=5푘푚 −훼퐿 푃표 Using ( ) = 10 10 푃𝑖 퐿 −훼퐿 −2푥2 푃표 ( ) = 10 10 = 10 10 = 0.3981 푃𝑖 퐿=2푘푚 −훼퐿 −2푥5 푃표 ( ) = 10 10 = 10 10 = 0.1 푃𝑖 퐿=5푘푚 22. The refractive indices of the core and cladding of a step-index optical fibre are 1.45 and 1.40 respectively and its core diameter is 45μm. Calculate its fractional refractive index change and numerical aperture. −6 Given: 푛1 = 1.45 ; 푛2 = 1.4 ; ; d = 45μm =45x 10 m ; ∆ = ? ; & NA = ?
(푛1−푛2) (1.45−1.4) 1) ∆ = = = 0.0345 푛1 1.45 √풏 ퟐ−풏 ퟐ 2) NA = ퟏ ퟐ 풏풐 √1.452−1.42 = 1 = 0.3775
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 23. Find the ratio of population of two energy levels in a medium at thermal equilibrium , if the wavelength of light emitted at 291 K is 6928 Å . ( 04 Marks)
Given: λ = 6928Å = 6928x10−10 m, t=27℃⇨ T =291K; h=6.625x10−34 Js; 푁 k=1.38x10−23 J/K ;C=3x108m/s ; 2 = ? 푁1 ℎ퐶 푁2 −( ) Using = 푒 휆푘푇 푁1 6.625푥10−34푥3푥108 −( ) = 푒 6928푥10−10푥1.38푥10−23푥291 = 푒−71.44 = 9.419x10−32 or = 9.441x10−32 24. Calculate the numerical aperture and angle of acceptance for an optical fibe having refractive indices 1.563 and 1.498 for core and cladding respectively. ( 04 Marks)
Given: 푛1 = 1.563; 푛2 = 1.498 ,푛표 = 1 ; 푁퐴 =? a푛푑 휃표 = ? 2 2 √푛1 −푛2 Using NA = 푛표 √1.5632−1.4982 = 1 = 0.4461
−1 −1 Also 휃표 = 푆𝑖푛 (푁퐴 ) = 푆𝑖푛 (0.4461) = 26.49°
25. A pulsed laser emits photons of wavelength 820 nm with 22 nW average power/pulse. Calculate the number of photons contained in each pulse,if the pulse duration is 12 ns. ( 04 Marks)
Given: 흀 = 820 nm = 820 x 10−9 m , P = 22 mW = 22 x 10−3W , t = 12 ns = 12 x 10−9s , h = 6.625 x 10−34 JS , C = 3 x 108푚푠−1 , n = ? ℎ퐶 Using E = Pt and E = 푛 푥 휆 휆푃푡 820 푥 10−9푥 22 푥 10−3푥 12 푥 10−9 We get , n = = = 1.089 x 109 electrons/푚3 ℎ퐶 6.625 푥 10−34푥 3 푥 108
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C.REDDAPPA MODULE-3 INTERLINE PUBLISHING.COM 26. A glass clad fiber is made with core glass of refractive index 1.5 and cladding is doped to give a fractional index difference of 0.0005.Determine the cladding index and numerical aperture. ( 04 Marks)
Given: 푛1 = 1.5 ,∆ =0.0005 ,푛표 = 1 , 푛2 = ? , 푁퐴 =? (푛 −푛 ) Using ∆ = 1 2 푛1
We get 푛2 = 푛1 − ∆ 푛1 = 1.5 – 0.0005푥 1.5 = 1.49925 (1.499)
√ 2 2 푛1−푛2 √1.52−1.499252 Also using NA = = = 0.0474 (0.054) 푛0 1
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM MODULE-4 CRYSTAL STRUCTURE : • • • • •
• • • • •
• • • • •
Lattice sites/points
Space lattice :The 3 dimensional periodic array(arrangement) of geometrical points in Space is called space lattice/crystal lattice. Each point is called lattice point/site which has identical surroundings. Basis/Primitive/Translational vectors: A co-ordinate system is used to represent latticepoints . Three basic vectors 푎⃗, 푏⃗⃗ & 푐⃗ used to represent lattice points along x,y & z- axis are called Basis/Primitive Vectors.
Translational vectors are position vector of lattice point in space given by ⃗⃗ 푅⃗⃗= 푛1푎⃗+ 푛2푏+ 푛3푐⃗ , where 푛1, 푛2 & 푛3 푎푟푒 𝑖푛푡푒푔푒푟푠.
Q: What is Bravais lattice & non-Bravais lattice ? Bravais lattice –is the lattice in which all the lattice points are equivalent , each lattice point represent an identical set of one or more atoms/molecules. Non-Bravais lattice- is the lattice in which all the lattice points are not equivalent . Superposition/overlapping of two or more bravais lattice forms non-bravais lattice.
Q: What is Basis /Pattern? Unit assembly of identical composition of atoms/molecules is called Basis/pattern.
Q: What is crystal structure? Crystal structure is obtained if basis is added to each and every lattice point.
• • • • • • + = • • • • • • Lattice + Basis = Crystal structure
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM Q: What is unit cell ? and explain types of cells.
Unit cell-is the smallest block/geometrical figure from which the crystal can be built by repetition of it in three dimensions.
Primitive cell-is the unit cell having lattice points only at the corners /vertices of it or is the unit cell containing only one lattice point in it.
Non-Primitive cell-is the unit cell which have lattice point within it in addition to lattice points at the corners /vertices.
Note: All primitive cells are unit cells but all unit cells are not primitive cells.
Q: What are lattice parameters ?
Z
c
훽 훼 0 b y a 훾
x
The Six quantities that describe the unit cell completely are called lattice parameters.
Three basis vectors are 푎 , b & c and three interfacial angles 훼, 훽& 훾 푎푟푒 푐푎푙푙푒푑 푡ℎ푒 푙푎푡푡𝑖푐푒 푝푎푟푎푚푒푡푒푟푠.
Q:What are Miller Indices of a plane ?
Miller Indices are the three smallest integers which represent the position & orientation of the crystal planes having the same ratio as the reciprocals of the intercepts of the plane on x,y and z axes.
Q:What are Crystallographic direction and Crystallographic plane ?
Crystallographic direction is the line joining origin to a lattice point(s) and Crystallographic plane is the plane passing through all lattice points in a particular crystallographic direction.
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM Q: Explain the seven crystal systems.
Based on six lattice parameters all the crystals are classified in to seven crystal systems as follows:- 1) Cubic 2) Tetragonal
푎=푏 ≠푐 c 푎=푏 =푐 a 훼 = 훽 = 훾=90 ͦ 훼 = 훽 = 훾=90 ͦ Types: SC/BCC Types: SC/BCC/FCC 훽 훼 a Ex: SnO2,TiO2 훽 훼 Ex: Copper/Gold a 0 훾 0 a
a 훾
3 ) Orthorhombic 4) Triogonal/Rhombohedral
푎≠푏 ≠푐 c 푎=푏 =푐 a 훼 = 훽 = 훾=90 ͦ 훼 = 훽 = 훾 ≠90 ͦ Types:SC/BaC/BCC/F 훽 0 훼 b Types:SC 훽 훼 Ex:KNO3, MgSO3 a 훾 Ex: As, Sb, 0 a a 훾
5 ) Hexagonal 6) Monoclinic
푎 = 푏 ≠ 푐 푎≠푏 ≠푐 훼 = 훽 =90 ͦ, 훾=120 ͦ c 훼 = 훾 =90 ͦ ≠ 훽ͦ c Types: SC Types:SC/BaC Ex:Zn,Mg, 훽 0 훼 a Ex: CaSO4,2H2O 훽 훼 b a 훾 a 0 훾
7) Triclinic
푎≠푏 ≠푐 훼 ≠ 훽 ≠ 훾 ≠90 ͦ c Types:SC Ex: 5H2O CuSO4 훽 0 훼 a 훾 b
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM Q:Explain the procedure for finding miller indices.
1, Find the intercepts of the plane along X,Y & Z axes in terms of basis vectors 푎,푏 &푐.
Ex: Let X,Y &Z axes intercepts are 3푎,2푏 &1푐 respectively.
2. Find the coefficients of 푎,b &푐.
ie: 3,2 & 1.
3. Find the reciprocals of the coefficients of basis vectors.
ퟏ ퟏ ퟏ ie:The reciprocals are , , ퟑ ퟐ ퟏ
4. Find the LCM of the denominators .
ie: The LCM of 3,2 &1 is 6.
5. Multiply each reciprocal term by LCM & write the results within the brackets like(h k l),which gives the miller indices of the plane.
ퟏ ퟏ ퟏ ie:The miller indices are 풙 ퟔ , 풙 ퟔ, 풙 ퟔ = ퟐ, ퟑ, ퟏ = (ퟐ ퟑ ퟏ) ퟑ ퟐ ퟏ
CHARACTERISTICS OF THREE CUBIC LATTICES
Note: 1. For an intercept at infinity, the miller indices is zero (0)
2. For negative intercepts, the corresponding miller index is negative. ex: h or 2 (Read as ‘bar’ h or 2
3. The miller indices ( h k l ) do not represent a single plane but set of parallel planes.
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM Q:Derive an expression for inter-planar distance/spacing in terms of miller indices.
Z
C
N B Y
휃 휃 푧 푦
0 휃푥
A X
1. Consider the plane ABC of a crystal which intersect the X,Y & Z axes at A,B & C respectively. 2. Let ( h k l ) be the miller indices of the family of ABC planes. 3. Draw ‘ON ’ ⊥푙푟 to the plane ABC ,then ON=풅 represent inter-planar distance/spacing. 푎 푏 푐 Then OA= , OB= &OC = ,where 푎,푏& 푐 are basis vectors ℎ 푘 푙 푂푁 푑 ℎ푑 From right 푎푛푔푙푒 ∆푙푒 OAN, Cos 휃 = = = …….(1) 푋 푂퐴 푎/ℎ 푎 푂푁 푑 푘푑 From ∆푙푒 OBN, Cos 휃 = = = …….(2) and 푌 푂퐵 푏/푘 푏 푂푁 푑 푙푑 also, from ∆푙푒 OCN, Cos 휃 = = = ……(3) 푍 푂퐶 푐/푙 푐
2 2 2 For orthogonal co-ordinates 푐표푠 휃푋+푐표푠 휃푌 + 푐표푠 휃푍 = 1 ……(4)
ℎ2푑2 푘2푑2 푙2 푑2 From eqns 1,2,3 &4 we get, + + = 1 푎2 푏2 푐2 ℎ2 푘2 푙2 ie: 푑2 [ + + ] = 1 푎2 푏2 푐2 ퟏ ∴ d = 풉ퟐ 풌ퟐ 풍ퟐ √ + + 풂ퟐ 풃ퟐ 풄ퟐ
For a cubic crystal 푎=푏 =푐
풂 ∴ d= √ 풉ퟐ +풌ퟐ +풍ퟐ
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM Q:Derive an expression for space lattice constant ‘푎’ for a cubic lattice. Let 휌 푏푒 푡ℎ푒 푑푒푛푠𝑖푡푦 & M the molecular weight of a cubic crystal of lattice constant’푎’. Then, volume of the unit cell = 푎3 and mass of unit cell =휌 푎3 …….(1)
If n be the number of molecules per unit cell and 푁퐴 be the Avagadro’s number. Then, mass of each molecule = (M/푁퐴) and mass of molecules in the unit cell=n (M/푁퐴)….(2) 3 ퟑ 풏푴 From eqns 1& 2,we get 휌 푎 = n (M/푁퐴)⇨ 풂 = 흆푵푨 Q:Find the relation between atomic radius (R) and lattice constant(푎) for SC ,BCC & FCC crystals:- a) Simple cubic(SC): a
2 R
If ‘ r ’ be the rdius and ‘ 푎 ‘ be the lattice constant/side of the unit cell, then from the diagram 푎=2R ⇨ 푎/2 = R
a) Body centered cubic (BCC) :
D
4R a
C A
a B AA A In a bcc cube AB=BC=CD=푎 and AC=√푎2 + 푎2=√2푎
Also from ∆푙푒 ACD , 퐴퐷2 = 퐴퐶2 + 퐶퐷2 ie: (4푅)2 = (2푎)2 +푎2 = 3푎2 4R = √3푎 3푎 푂푟 푅 = √ 4
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM c) Face centered cubic (FCC)
C
4R a
A B a
In a fcc cube AB=BC=푎 and AC=4R
Also from ∆푙푒 ABC, 퐴퐶2 = 퐴퐵2 + 퐵퐶2 ie: (4푅)2 = (푎2 +푎2 = 2푎2 Or 4R =√2푎 √2푎 푎 Or R = = 4 2√2 Q:Find the number of atoms in unit SC.BCC & FCC cell.
SC : In SC cell each atom is shared by 8 neighbour cells & hence no of atoms per unit cell 1 due to Corner atoms = x 8 =1 8
1 BCC: no of atoms per unit bcc cell= x 8 corner atoms + one atom at the body = 1 8
1 1 FCC: no of atoms per unit fcc cell = 푥 8 푐표푟푛푒푟 푎푡표푚푠 + x 6 face atoms=1+3=4 8 2
Q: What is meant by Co-ordination number ? & mention the co-ordination number of SC, BCC & FCC cells.
Co-ordination number of any atom in a crystal is the number of nearest equidistant neighbouring atoms Surrounding that atom.
SC: Co-ordination number of SC =6 BCC: Co-ordination number of bcc = 8 FCC: Co-ordination number of fcc =12
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM SC BCC FCC
• • •
Q:What is meant by Atomic Packing Factor(APF) ? Atomic packing factor is defined as the fraction of the volume of the unit cell occupied by the atoms present in the unit cell. (푛표 표푓 푎푡표푚푠 𝑖푛 푡ℎ푒 푢푛𝑖푡 푐푒푙푙)푥푉표푙푢푚푒 표푓 푒푎푐ℎ 푎푡표푚 ie: Atomic packing factor = (푉표푙푢푚푒 표푓 푡ℎ푒 푢푛𝑖푡 푐푒푙푙) 푛 4 ie: APF = 푥 휋푅3 where a = Lattice constant & R = Radius of the atom 푎3 3
Q: Calculate the atomic packing factors of SC,BCC & FCC cells. 풂 For SC, n =1, R = ퟐ 푛 4 Using, APF = 푥 휋푅3 푎3 3 1 4 푎 = 푥 휋( )3 푎3 3 2 1 4 푎3 = 푥 휋 푎3 3 8 휋 = = 0.52 or 52% 6 √ퟑ풂 For BCC , n = 2, R = ퟒ 푛 4 Using, APF = 푥 휋푅3 푎3 3 2 4 √3푎 = 푥 휋( )3 푎3 3 4 2 4 3√3푎3 = 푥 휋 푎3 3 64 √3휋 = = 0.68 or 68% 8
푎 For FCC, n = 4, R = 2√2 푛 4 Using, APF = 푥 휋푅3 푎3 3 4 4 푎 = 푥 휋( )3 푎3 3 2√2 4 4 푎3 = 푥 휋 푎3 3 16√2 휋 = = 0.74 or 74% 3√2
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM NOTE: Important data related to SC , BCC & FCC cells. Sl. Particulars Simple Body Face No Centere Centered 1 Volume of a unit cell 푎3 푎3 푎3 2 No. of atoms per cell 1 2 4 3 Co-ordination number 6 8 12 (No. of nearest neighbours) 4 Nearest neighbour distance 푎 (√3/2)푎 푎/√2 (2R) 5 Packing Faction 0.52 0.68 0.74
Q:Explain the structure of diamond crystal. Note: = 8 corner atoms, = 6 face atoms & = 4 body diagonal atoms
O 1/2 O
3/4 1/4
1/2 O 1/2 OR
B 1/4 3/4
2R O 1/2 O A 푎
4
푎 1. Diamond is formed by two intervening FCC sub-lattices one of them is moved by 4 along the body diagonal as shown in the diagram. 2. The origin of one FCC is ( 0,0,0) and the origin of the other FCC is (a/4,a/4,a/4) 3. There are 8 atoms at the corners, 6 atoms at the faces and 4 atoms along the diagonals. 4. Each diagonal atom form bond with One nearest corner atom and Three nearest face atoms. 5. The coordination number of diamond is 4. 1 1 6. Total atoms in unit diamond cell = x 8+ x 6+4 =8 8 2 7. If R be the radius and a the lattice constant ,then we can show that
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푎2 푎2 푎2 AB=2R=√ + + 16 16 16
3푎2 = √ 16 3푎 = √ or 4 ퟑ풂 R= √ ퟖ ퟑ풂 For Diamond n = 8 & R = √ ퟖ 푛 4 ∴ Using APF = 푥 휋푅3 푎3 3 8 4 3푎 = 푥 휋( √ )3 푎3 3 8 8 4 3 3푎3 = 푥 휋 √ 푎3 3 8푥8푥8 휋√3 = 16 = 0.34 or 34% Q: What is perovskite ? explain.
Perovskie is the common name for all the oxides of type AB푶ퟑ,where A & B are different metals.
Ex: 퐵푎푇𝑖푂3 ( 퐵푎푟𝑖푢푚 푡𝑖푡푎푛푎푡푒) , 퐶푎푇𝑖푂3( calcium titanate) etc The common structure of perovskite is as follows:
= 8 Ba/Ca atoms form are at the edges of the cell.,
= 6 푂3atoms are at the face of the cell and = 1 푇𝑖 atom at the body centre of the cell. Perovskites exhibit superconductivity. Perovskies exhibit both piezo-electric & ferro-electric properties . Perovskies are used as dielectric in capacitors. Perovskies are used as piezo-electric in microphone. Type II superconductors have Perovskite structure.
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM Q: Explain Allotrophy and Polymorphism.
Allotrophy is the property of the material by the virtue of which it can have more than one type of crystal structures. All the structures will have the same chemical properties and different physical properties Carbon,Sulphur ,Phosphorous exhibit allotrophy. Ex: 1.Diamond and graphite are the allotrophic forms of carbon. 2. Diamond is hard and good insulator . 3. Graphite is soft and good conductor.
Polymorphism is the ability of the materials to crystallize in several solid phases that posses different lattice structures at different temperatures. Ex: Iron ( Alpha Iron)exhibit BCC structure from room temperature to 910℃. Iron ( Gamma Iron)exhibit FCC structure from 910℃ to upto 400℃ Iron ( Delta Iron)exhibit BCC structure from 1400℃ to 1540℃( melting point. All these phases are reversible. Q:Derive Bragg’s law of diffraction.
Incident rays -AD A C reflected rays-CF Glancing angle D 휃 휃 F ₀ ₀ B ₀ ₀ ₀
G 휃 휃 H d = Interplanar spacing
₀ ₀ ₀ E ₀ ₀
Let the incident parallel x-rays of wavelength λ, incident on two atomic planes of Separation d at glancing angles 휃 and reflect along BC and EF respectively. Draw BG & BH ⊥푙푟 to DE & EF respectively. From the ∆푙푒 BGE & BHE we have GE=BE.Sin휃 & 퐻퐸 = 퐵퐸. 푆𝑖푛휃 where BE = d ,interplanar distance. ∴ 푝푎푡ℎ 푑𝑖푓푓푒푟푒푛푐푒, 푝푑 = 퐺퐸 + 퐻퐸 = 푑 푆𝑖푛휃 + 푑 푆𝑖푛휃 = 2푑 푆𝑖푛휃 … . (1) Also for constructive interference pd= nλ ……(2) ∴ 푓푟표푚 푒푞푛푠 1 &2 , 푤푒 푔푒푡 nλ = ퟐ풅 푺풊풏휽 ,
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM This equation is called the Bragg’s law
Q:Describe the construction & working of Bragg’s X-ray spectrometer.
Vernier- 푉1 Circular scale
Slits glancing angle-휃 Crystal target
푆1 푆2
X-rays Vernier-푉2
Intensity Collimated X-rays
푆3
푆4 E -electrometer
Ionisation chamber (D) Ba
0 휃
휃 1 휃 2 휃 3
The labeled schematic diagram of the Bragg’s X-ray Spectrometer is as shown in the diagram. To verify Bragg’s law: 1. Collimate the X-rays from X-ray tube by passing them through slits S1 & S2. 2. Allow the collimated X-rays to incident on the surface of a given crystal mounted vertically at the centre of a turn table at a glancing angle 휃 3. The position of the crystal is noted using the vernier V1 and horizontal circular scale. 4. Adjust the ionization chamber to receive the scattered X-rays and measure the ionization current (I)which is a measure of intensity of the X-rays using electrometer E. 5. Repeat the experiment as before by increasing the glancing angle 휽 gradually and note the corresponding ionization current each time. ′ 6. Plot a graph of ‘I‘v/s ‘휃 푎푛푑 푓푟표푚 푡ℎ푒 푔푟푎푝ℎ note down the glancing angles 휃1, 휃2& 휃3 for the 1st,2nd &3rd order diffraction respectively.
7 Then substituting for 휃1, 휃2& 휃3 in the Bragg’s equation 2d Sin휃 = 푛휆 , It will be found that ,Sin휽ퟏ: 푺풊풏휽ퟐ: 푺풊풏휽ퟑ = ퟏ: ퟐ: ퟑ,which verifies Bragg’s law. To find the nature of the crystal: 1. Collimate the X-rays from X-ray tube by passing them through slits S1 & S2. 2. Allow the collimated X-rays to incident on the surface of a given crystal mounted vertically at the centre of a turn table at a glancing angle 휃. The position of the
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crystal is noted using the vernier V1 and horizontal circular scale 3. Adjust the ionization chamber to receive the scattered X-rays and measure the ionization current (I) produced by the X-rays using electrometer E.
4. Adjust 푑100 plane of the crystal to the incident X-rays and find the glancing angle st 휃1 for 1 order Spectrum. 5. Repeat the experiment as before for the crystal surfaces 푑110 & 푑111 and find st the corresponding 1 order glancing angles 휃2 & 휃3 . 6. Then using bragg’s equation, 2d Sin휃 = 푛휆 , ퟏ ퟏ ퟏ find 풅ퟏퟎퟎ ∶ 풅ퟏퟏퟎ ∶ 풅ퟏퟏퟏ = : : 푺풊풏휽ퟏ 푺풊풏휽ퟐ 푺풊풏휽ퟑ 1 1 1 7. The nature of the crystal is identified from the ratios of : : 푆𝑖푛휃1 푆𝑖푛휃2 푆𝑖푛휃3 as follows: ퟏ ퟏ ퟏ ퟏ ퟏ For SC; : : = 1: : 푺풊풏휽ퟏ 푺풊풏휽ퟐ 푺풊풏휽ퟑ √ퟐ √ퟑ ퟏ ퟏ ퟏ ퟐ ퟏ For BCC; : : = 1: : 푺풊풏휽ퟏ 푺풊풏휽ퟐ 푺풊풏휽ퟑ √ퟐ √ퟑ ퟏ ퟏ ퟏ ퟏ ퟐ For FCC; : : = 1: : 푺풊풏휽ퟏ 푺풊풏휽ퟐ 푺풊풏휽ퟑ √ퟐ √ퟑ
PROBLEMS SECTION
MODULE-4 : CRYSTAL STRUCTURE
푎 Formulae needed : nλ = 2dSin휃 ; d= ; √ℎ2 +푘2 +푙2 4 For SC ,n=1& a = 2R , BCC, n=2 & a = R , FCC , n=4 & a = 2√2 푅 √3 푛 4휋푟3 Atomic Packing Factor (APF) = x 푎3 3 1. Copper has fcc structure with atomic radius 0.127nm.Calculate the inter-planar spacing for (3 2 1) plane.( JAN2015)
Given: r = 0.127 nm = 0.127x10−9푚 , (ℎ 푘 푙 ) = ( 3 2 1), 푓표푟 푓푐푐 푎 = 2√2 푟 = 2√2 x 0.127x10−9푚 , 푑 = ?
푎 Using d = √ℎ2 +푘2 +푙2 2√2 푥 0.127푥10−9 = √32 +22 +12 = 9.60 x 10−11 푚
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2. 퐷푟푎푤 the following planes in a cubic unit cell: i) (2 0 0) ii) (2 1 0)iii)(1 3 2)
ퟏ ퟏ ퟏ ퟏ (2 0 0) ⇨ 퐱 = , 퐲 = ∞ & 푧 = ∞ (2 1 0)⇨ 퐱 = − , , 퐲 = ퟏ & 푧 = ∞ (1 3 2 )⇨ 퐱 = ퟏ, 퐲 = − & 푧 = ퟐ ퟐ ퟑ ퟐ Z Z Z
½ 1 − ½ 3
½ 0 y 0 1 Y 0 Y 1
X X X
3. The minimum order of Bragg’s reflection occurs at an angle of 20° in the plane (2 1 2 ).Find the wavelength of X-rays if lattice constant is 3.614Å. (Dec 2013/Jan2014)
Given: n = 1 ; 휃 = 20° ; ( h k l ) = ( 2 1 2 ) ; a = 3.614Å = 3.614x10−10 m ; λ = ?
푎 Using nλ = 2dSin휃 & d= √ℎ2 +푘2 +푙2 2푎 푆𝑖푛휃 We get , λ = 푛√ℎ2 +푘2 +푙2 2푥3.614푥10−10 푆𝑖푛20° = = 8.240 x 10−11 m 1푥√22 +12+22
4. Monochromatic X-rays of wavelength 0.82 Å undergo first order reflection from a crystal of cubic lattice with lattice constant 3 Å at a glancing angle of 7.855°. Identify the possible planes which give rise to this reflection in terms of Miller indices. (퐷푒푐 11 & 퐽푢푛 − 퐽푢푙16)
Given: n = 1 ; 휃 = 7.855° ; a = 3 Å = 3x10−10 m ; λ = 0.82 Å =0.82 x 10−10 m ; ( h k l ) = ?
푎 Using nλ = 2dSin휃 & d= √ℎ2 +푘2 +푙2 2푎 푆𝑖푛휃 We get , √ℎ2 + 푘2 + 푙2 = 푛휆 2푥3푥10−10 푆𝑖푛(7.855) = = 1 1푥0.82 푥 10−10
Thus the possible planes ( h k l ) =( 1 0 0 ) ,(0 1 0 ) ,( 0 0 1)
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5. Calculate the glancing angle of the (1 1 0) plane of a simple cubic crystal (a=2.814 Å ) corresponding to second order diffraction maximum for the x-rays of wavelength 0.710 Å.
Given: n = 2 ; ( h k l ) = (1 1 0 ) ; a = 2.814Å = 2.814x10−10 m ; λ = 0.710 Å =0.710 x 10−10 m ; 휃 =?
푎 Using nλ = 2dSin휃 & d= √ℎ2 +푘2 +푙2 푛휆√ℎ2 +푘2 +푙2 We get , 휃 = 푠𝑖푛−1( ) 2푎 2푥0.710 푥 10−10푥√12 +12 +0 = 푠𝑖푛−1( ) = 20.91° 2푥2.814푥10−10
6. Draw the following planes in the unit cube: i) ( 1 0 2 ) ii) ( 1 1 2 ). (퐷푒푐 2010) 1 1 Ans: i) ( 1 0 2 ) ⇨ 푥 = −1, 푦 = ∞ & 푧 = & ii) (1 1 2 ) ⇨ 푥 = −1, 푦 = 1 & 푧 = − 2 2
ퟏ ퟏ ( 1 0 2 ) ⇨ 풙 = −ퟏ, 풚 = ∞ & 푧 = (1 1 2 ) ⇨ 풙 = −ퟏ, 풚 = ퟏ & 푧 = − ퟐ ퟐ Z −1 Z 0 1 Y
1 ½ −1 X(− ) 2 0 Y
x
7. A monochromatic X-ray beam of wavelength 1.5Å undergoes second order Bragg reflection from the plane (2 1 1) of a cubic crystal, at a glancing angle of 54.38°,calculate the lattice constant. (퐽푢푛 2010)
Given: n = 2 ; ( h k l ) = (2 1 1 ) ; λ = 1.5 Å =1.5 x 10−10 m ; 휃 = 54.38° ; 푎 = ?
푎 Using nλ = 2dSin휃 & d= √ℎ2 +푘2 +푙2 푛휆√ℎ2 +푘2 +푙2 We get , 푎 = 2푆𝑖푛휃 2푥1.5 푥 10−10 푥√22 +12 +12 = 2푥푆𝑖푛 54.38 = 4.52 x 10−10 푚
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8. In a calcite crystal, second order Bragg’s reflections occur from the planes with d-spacing 3 Å,at a glancing angle of 24° .Calculate the path difference between X-rays reflected from the two adjacent planes. Also calculate the wavelength of x-rays.( 4 marks)
Given: d = 3 Å =3x10−10 푚 , 휃 = 24° , n = 2 ,Pd = ? and λ = ? Path difference,Pd = 2d Sin휃 =2x3x10−10 xSin24° =2.44 x10−10 m
푃푑 Also, λ = 푛 2푑푆𝑖푛휃 = 푛 2.44 푥10−10 = 2 = 1.22 x10−10 m
9. The atomic radius of gold is 0.144nm.Determine the interplanar distance for ( 1 1 0) planes assuming that gold belongs to FCC system.( 4 marks)
Given: r = 0.144 nm =0.144x10−9 m , (h k l)=(1 1 0), For FCC a =2√2 푟 =2√2 푥0.144푥10−9 푚 , d = ?
푎 Using d= √(ℎ2+푘2+푙2 ) 2√2 푥0.144푥10−9 = √(12+12+02 ) = 2.88 x 10−10 푚
10. Calculate the glancing angle for incidence of of X-rays of wavelength 0.058 nm on the plane o (1 3 2)of NaCl which results in 2nd order diffraction maxima taking the lattice spacing as 3.81 Å.
Given: n = 2 ; ( h k l ) = (1 3 2 ) ; a = 3.81Å = 3.81x10−10 m ; λ = 0.058nm=0.058 x 10−9 m ; 휃 =? 푎 Using nλ = 2dSin휃 & d= √ℎ2 +푘2 +푙2 푛휆√ℎ2 +푘2 +푙2 We get , 휃 = 푠𝑖푛−1( ) 2푎 2푥0.058 푥 ퟏퟎ−ퟗ푥√12 +32 +22 −1 ) = 푠𝑖푛 ( 2푥3.81푥10−10
= 34.72° ( also d = 1.018 X 10−10 m)
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11. Draw the following planes in a cubic unit cell.
i) (2 0 0) ii) (Ι Ī 0) iii) (Ι 0 2) iV) (Ι Ι 2)
(200)⇒X = ½ ,Y = ∞ & 푍 = ∞ (Ι Ī 0)⇒X = 1 ,Y = −1 & 푍 = ∞ Z z
0 Y 0 y 1/2 1 X x
(102 )⇒X = 1,Y = ∞ & 푍 = ½ (112)⇒X = 1,Y = 1 & 푍 = ½ Z
Z
½ ½ −1 0 Y
0 1 Y 0 Y
X 1 X 1 1
12. An X-ray beam of wavwlength 0.7 Å undergoes first order X Bragg’s reflection from the plane ( 302) of a cubic crystal at glancing angle 35° , 풄풂풍풄풖풍풂풕풆 풕풉풆 lattice constant. ( 04 Marks)
Given: n = 1 ; ( h k l ) = ( 302 ) ; λ = 0.7 Å =0.7 x 10−10 m ; 휃 = 35° ; 푎 = ?
푎 Using nλ = 2dSin휃 & d= √ℎ2 +푘2 +푙2
푛휆√ℎ2 +푘2 +푙2 We get , 푎 = 2푆𝑖푛휃
1푥0.7 푥 10−10 푥√32 +02 +22 = 2푥푆𝑖푛 35
= 2.200 x 10−10 푚 [ Note: d = 6.102x10−11m]
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13.Draw the crystal planes ( 102 ) ( 111 ) ( 011 ) and ( 002 ) in a cubic crystal. ( 04 Marks)
(102) ⇨ X =1/1 =1, Y =1/0 = ∞ and Z =1/2 ⇨ (111) X =1/1 =1, Y =1/1 =1, and Z 1/1 =1, Z Z 1
1/2
0 Y 1
1 0 Y
X X 1
(011) ⇨ X =1/0 = ∞, Y =1/1 =1and Z =1/1 = 1 (002) ⇨ X =1/0 = ∞, Y =1/0 = ∞ and Z =1/2
Z 1 Z
½
0 1 Y 0 Y
X
X
14. Find the Miller indices of a set of parallel planes which make intercepts in th ratio 3a : 4b and parallel to z-axis and also calculate the interplanar distance of the planes taking thr lattice to be cubic with a = b = c = 2 Å. ( 04 Marks) Given: Intercepts 3a:4b: ∞푐 , a= 푏 = 푐 =2Å=2 x 10−10m , (h k l ) =? ,d = ? Coefficients of a,b and c are 3 ,4, ∞ 1 1 Reciprocals of the coefficients are , ,0 3 4 LCM 0f the denominators 3 & 4 is 12 1 1 Miller indices are 푥12 = 4 , 푥12 = 3 , 0 x 12= 0 3 4 ∴ ( h k l ) = ( 4 3 0 ) 퐴푙푠표 푤ℎ푒푛 푎 = 푏 = 푐 푎 2 푥 10−10 We have d = ∴ d = = 4 x 10−11m √ℎ2+푘2+푙2 √42+32+02
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REDDAPA C MODULE-4 INTERLINE PUBLISHING.COM 14. Draw the following planes in a cubic unit cell, i) ( 0 0 I ) ii) ( I ī 0 ) iii) ( I I 2) iV) ( 0 2 0 ) ( 04 Marks)
(001) ⇨ x =1/0= ∞, y=1/0=∞ & z =1/1=1 ( I ī 0 ) ⇨ x =1/1= 1, y =1/-1=−1 & z =1/0 =∞
Z 1 z
0 y 0 1 y X X 1
(112) ⇨ x =1/1= 1, y=1/1=1 & z =1/2 (020) ⇨ x =1/0= ∞, y =1/2 & z = 1/0 = ∞
z Z
½
0 ½ y 0 1 y X 1 x
@@end@@
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REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM MODULE-5
SHOCK WAVES AND SCIENCE OF NANO MATERIALS:
SHOCK WAVES
Q:What are shock waves ? Explain.
Shock waves are the waves produced due to the sudden release/dissipation of energy. Shock waves are the waves in which the pressure, density and temperature changes are large Ex: Shock waves are produced during the burst of crackers, Explosion of dynamite/bombs, Volcanic eruptions, etc.
Q:Mention different types of shock waves. There are FOUR types of shock waves namely :- Stationary shock waves, Moving shock waves, Normal shock waves and Oblique shock waves,
Q:Mention methods of producing shock waves.
Shock waves can be produced by the following methods ,namely By detonation of crackers/explosives, by volcanic eruptions, supersonic objects/waves, by Reddy shock tube in the laboratory.
Q:What are Acoustic waves? Mention the types of acoustic waves.
Acoustic waves are the longitudinal waves which travel with the speed of sound in a medium (Solid/liquid/gas)
Acoustic waves are classified in to THREE types namely: 1. Infrasonic waves(Infrasonics) are the Acoustic waves of frequency less than 20 Hz. 2. Audible waves are the Acoustic waves of frequency between 20 Hz and 20 kHz. 3. Ultrasonic waves(Ultrasonics) are the Acoustic waves of frequency greater than 20 kHz. (Elephants detect Infrasonics, Human ear detect Audible waves & Bats/Dogs detec Ultrasonics)
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REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM Q: Define Mach number, subsonic waves supersonic waves and Mach angle.
1. Mach Number(M) is the ratio of the speed of an object (V) through a fluid to the speed of 푽 sound(a) in the fluid at that point. Mathematically, M = , M is dimensionless quantity. 풂
2. Subsonic waves are the mechanical waves whose speed is less than that of sound in the same medium. mach number of Subsonic waves is less than 1. Ex: Motor cycle, Bus, Train , aeroplanes etc produce subsonic waves.
3. Supersonic waves are the mechanical waves whose speed is greater than that of sound in the same medium. for which the mach number of Supersonic waves is greater than 1. Ex: Fighter planes, Rockets, Missiles,tornedo etc produce supersonic waves
4. Mach angle(흁) is the half the angle of cone of sound waves formed and is given by ퟏ 흁 = 푺풊풏−ퟏ ( ) 푴
Q: Explain the basic conservation laws and Rankine-Hugonite/Normal shock wave Relations. There are three basic conservation laws namely Conservation of mass, Conservation of momentum and Conservation of energy.
1. Conservation of mass states that the total mass of the system always remains constant as the mass can neither be created nor destroyed. Mathematically, 휌푣 = 푐표푛푠푡푎푛푡 or
∴ 휌1푣1 = 휌2푣2 , Where 휌1, 휌2densities & 푣1 , 푣2 velocities.
2. Conservation of momentum states that the sum total momentum of the system always remain constant. 2 2 2 Mathematically,푃 + 휌푣 = 푐표푛푠푡푎푛푡 표푟 푃1 + 휌1푣1 = 푃2 + 휌2푣2 Where 푃1, 푃2 pressures , 푣1, 푣2 velocities and 휌1, 휌2densities.
3. Conservation of energy states that the sum total energy of a system is always remains constant. 푣2 푣2 푣2 Mathematically, ℎ + = 푐표푛푠푡푎푛푡 표푟 ℎ + 1 = ℎ + 2 2 1 2 2 2 Where ℎ1, ℎ2 enthalpies and 푣1, 푣2 velocities.
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REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM Q: RH equations or Normal shock wave equations.
RH equation /Normal shock wave equations are derived from basic conservation laws ,which relate the temperature ratio and density ratio in terms of pressure ratio for normal shock waves. The pressure ratio in terms of upstream Mach number M is given by 푷ퟐ ퟐ휸 2 휸−ퟏ = 푀 − …….(1) 푷ퟏ 휸+ퟏ 휸+ퟏ 푻ퟐ ퟏ 휸−ퟏ ퟐ휸 2 ퟐ(휸−ퟏ) But temperature ratio, = ( 2 + ) ( 푀 − 1) ( ퟐ) ……..(2) 푻ퟏ 푀 ퟐ 휸−ퟏ (휸+ퟏ)
Substituting for 푀2 푓푟표푚 푒푞푛 1, 𝑖푛 푒푞푛 2, 푤푒 푔푒푡
휸−ퟏ 푷 푷 [ퟏ+ ퟐ] ퟐ 푻 휸+ퟏ 푷 푷 ퟐ ퟏ ퟏ 2 2 = 푷 휸−ퟏ …..(3) ∵ 4ab+(푎 − 푏) = (푎 + 푏) 푻ퟏ ퟐ+ 푷ퟏ 휸+ퟏ
푷ퟐ 흆ퟐ 푷ퟏ Also , density ratio, = 푻 …….(4) 흆ퟏ ퟐ 푻ퟏ
From eqns 1,3 &4,we get,
푷ퟐ 흆 푷 ퟐ = ퟏ 휸 − ퟏ 푷 푷 흆ퟏ [ퟏ + ퟐ] ퟐ 휸 + ퟏ 푷ퟏ 푷ퟏ 푷 휸 − ퟏ ퟐ + 푷ퟏ 휸 + ퟏ 휸+ퟏ 푷 ퟐ+ퟏ 휸−ퟏ 푷ퟏ = ( 휸+ퟏ 푷 ) ….(5) + ퟐ 휸−ퟏ 푷ퟏ
eqns 3 & 5 are called RH relations/NS relations.
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REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM Q:What is a (Reddy)shock tube ? Describe the construction and working of simple Reddy shock tube .
1. Reddy Shock tube is a device used to produce and study shock waves in the laboratory. 2. Schematic labeled diagram of the original Reddy shock tube is as shown in the diagram.
Piston Diaphragm S 푆1( 푃4) 푠푒푛푠표푟푠 푆2(푃5)
푃2 푃1 x =7 cm
Driver section/gas Driven section/gas 2.9cm Plunger
49 cm 51 cm
Construction : 1. RST consists of a steel tube of length 100 cm and diameter 2.9 cm. 2. A diaphragm of thickness 0.1cm divides the tube in to two compartments of length 49 cm fitted with piston called Driver section filled with driver gas. The other compartment of length 51 cm is called Driven section filled with driven gas.
3. Sensor S fitted to driver section measures the rupture pressure 푃2,temparature푇2. 4. Two sensors 푆1 & 푆2 separated by a distance ∆푋 fitted to driven section measures the pressures 푃4,푃5 and temperatures 푇4, 푇5 respectively.
Working : 1. Driver section is filled with gas at high pressure and Driven section is filled with gas of
low pressure 푃1 & 푡푒푚푝푒푟푎푡푢푟푒 푇1 . 2. Diaphram is ruptured to produce shock waves by pushing the piston and the rupture
pressure 푃2 & temperature is measured using sensor S. 3. The time’풕′ taken by the shock wave to travel the distance ‘x’ is measured using CRO. also
the pressures 푃4,푃5 and temperatures 푇4, 푇5 are measured using the sensors 푆1 & 푆2 respectively. 푥 4. The speed of the shock waves is calculated using V = . 푡 푉 5. Then the match number of the shock waves is calculated using M = ,where a is the 푎 speed of sound at temperature 푇1. 6. Also the mach number M can be calculated using the RH relations 푷ퟐ ퟐ휸 2 휸−ퟏ = 푀 − , by finding 푃2, 푃1 for the gas of known 훾. 푷ퟏ 휸+ퟏ 휸+ퟏ
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 4
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM Q: Uses of Shock waves. 1. Shock waves (SW)are used in the treatment of kidney stones. 2. SW are used in the pencil industry for softening of pencil wood and painting. 3. Sw are used in the extraction of sandal wood. 4. Sw sre used to rejunevate/activate dried bore wells. 5. Sw are used for needleless drug delivery. 6. Sw are used to push DNA in to the cell. 7. SW are used for the treatment of orthopedic diseases.
SCIENCE OF NANOMATERIALS:
Q:What are nano-particles? Nano-particles are the material particles whose size is in the range of 1nm to 100 nm and their properties are size dependant.
Q:What is mesoscopic state?
Mesoscopic state is the size/state of the matter at which its physical properties changes and becomes size dependent.
Q:Explain different structures(3D,2D,1D & 0D) based on their energy density graphically.
3D structure/Bulk metal : g(E)
0 E
1. The 3D structure is as shown in the diagram, which have all the 3 dimensions. ퟖ√ퟐ흅풎ퟑ/ퟐ푬ퟏ/ퟐ 2. The density of states for 3D structure is given by g(E) dE = 푑퐸 풉ퟑ Where m = mass, E = energy, h = Planck’s constant 3. The density of states g(E) increases with energy E is as shown in the graph 4. The electrons are not confined to any direction.
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 5
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM 2D structure/Quantum Film/Well:
g(E)
.
0 퐸 퐸 퐸 퐸 1 2 3
1. Quantum film/well is obtained when 3D structure reduced to nano scale in one dimension as shown in the diagram. 4휋푚 2. The density of states for quantum well is given by g(E) 푑퐸 = dE where m = mass, ℎ2 h =Planck’s constant. 3. For each quantum state the density of states is constant. 4. The variation of g(E) with E is as shown in the graph. 5. The electrons are confined to 1 direction.
1D structure/Quantum Wire :
g(E)
0 퐸1 퐸2 퐸3 퐸 0
1. Quantum wire is obtained when 3D structure reduced to nano Scale in two dimensions as shown in the diagram. 2√2푚1/2퐸−1/2 2. The density of states for quantum well is given by g(E)dE = dE ℎ where m = mass, E = energy, h = Planck const. 3. The variation of g(E) with E is as shown in the graph 4. The electrons are confined to 2 directions.
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 6
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM 0D structure /Quantum Dot :
g(E)
0 0 퐸1 퐸2 퐸4 퐸
1. Quantum Dot is obtained when 3D structure reduced to nano scale all the 3 dimensions 2. The variation of g(E) with E is as shown in the graph. 3. The electrons are confined to all the 3 directions 4. and g(E) has discrete structure as shown.
Q: Explain the TWO approaches/methods adopted to obtain nanoparticles.
Top Down method Bottom Up method Bulk material
Nanoparticals Powder
Clusters /Groups
Nanopartical Nanoparticalss Atoms/molecules
TOP DOWN approach
1. In the TOP DOWN approach ,bulk material is crushed in to small pieces. 2. Crushed pieces are further grinded in to smaller and smallers pieces. 3. Crushing and grinding process is continued till nanoparticles are obtained.
BOTTAM UP approach
1. Basic atoms/molecules are grouped to form clusters/globules. 2. The clusters /globules are further grouped in to nanoparticles.
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 7
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM Q: Explain Ball Mill method of synthesis/producing nanoparticles.
Steel container Steel Balls
↺
Nanopowder
Stand
1. Ball mill method is a mechanical method based on “ Top Down ” approach to synthesis nanoparticles of metals and alloys on large scale. 2. It consists of a steel container filled with the material whose nanoparticles to be produced along with large number of heavy steel balls. 3. The container is capable of rotation about an axis inclined to the horizontal. 4. The steel balls rotate circularly about the axis and spin about their own axis. Due to these motions of steel balls ,the material is continuously crushed and powdered. 5. As this process is continued finally we get powdered nanoparticles. 6. Merits : By ball mill method we can produce nanoparticles on large scale economically. 7. Demerits: Nanoparticles produced are irregular in shape and impure.
Q: Explain Sol-Gel method of synthesis/producing nanoparticles.
Precursor + Solution = Sol + Dehydration = Gel
Spray/dip adding Surfactants Slow heating
Gelled spheres Zero gel
Substrate On Calcination
On calcination On calcination Dense ceramic
Nano film Nano powder
1. Sol-gel method is a wet chemical process in which bottom-up approach is adopted. 2. The precursor is dissolved in suitable liquid to form Sol.
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 8
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM 3. Sol is then dehydrated to get Gel. 4. On slow drying of gel zerogel is obtained. 5. On calcinations zerogel change in to nano dense ceramic. 6. If surfactants are added to Sol, gelled spheres are formed,which on calcinations form nano powder 7. If Sol sprayed on to substrate by spinning/dipping, gel is formed on the substrate, which on calcinations form thin nano film. 8. By sol-gel method we can obtain pure nanoparticles.
Q: what are Carbon nano tubes? Mention and explain structures of different types of CNT’s..
1. Carbon nanotube(CNT) structure is imagined to be the cylinder formed by rolling a hexagonal graphene sheet of carbon atoms and then closing the ends with fullerene hemispheres.
2. There are three types of CNT structures namely Aramchair CNT,Zigzag CNT and Chiral CNT.
Arm chair CNT Zig-Zag CNT Chiral CNT
Axis
Types of CNTs: 1.Sigle walled nanotubes(SWNT) consists of single graphene sheet.
2.Multiwalled nanotubes(MWNT) consists of nanotube with in nanotubes.
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 9
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM Q:Expain the production of CNTS by Arc method.
퐻 푔푎푠 𝑖푛푙푒푡 퐻 gas outlet 푒 푒
Chamber
Graphite anode
500 torr pressure 5 Ba
Graphite cathode
1. The schematic diagram of Arc discharge method is as shown in the diagram. 2. It consists of a chamber in to which two graphite electrodes separated by 1 mm and diameter 5-20 휇푚 are sealed. 3. Helium gas is circulated through the chamber at a pressure of 500 torrand a voltage of 20-25V can be applied between the electrodes. 4. When a voltage of 20-25 V is applied between the electrodes, graphite evaporates and is deposited on the cathode. 5. If the anode is coated with catalytic agents like Iron, cobalt or nickel ,SWCNT’s are produced. 6. MWCNT’s are produced if the anode is not coated with catalytic agents. 7. Pure CNT’s can be obtained by using pure graphite rods.
Q:Expain the production of CNTS by Pyrolysis method. Furnace
Chamber ⊙ ⊙ ⊙ ⊙ ⊙
C Catalyst
Pressure gauge
푵ퟐ 푪ퟐ푯ퟐ ⊙ ⊙ ⊙ ⊙ ⊙ ⊙ 푵ퟐ 푪ퟐ푯ퟐ Substrate on graphite sheet
1. The schematic diagram of Pyrolisis method ⊙ is as shown in the diagram. ⊙
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 10
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM 2. It consists of a chamber in an electric furnace through which a mixture of nitrogen and acetylene is passed and let out through the out let. 3. A substrate is placed on the quartz sheet inside the chamber. 4. The temperature in the chamber is maintained at about 700-800℃. 5. Due to high temperature acetylene breaks down in to carbon atoms. 6. When these atoms come near substrate they get attracted and deposited as carbon nanotubes in the presence of catalysts which are MWCNT’s. 7. SWCNT’ s are obtained when acetylene is replaced with methane or carbon monoxide at 1200℃
Q: Mention the properties of CNTS’.
1. CNT’s are highly elastic. 2. Young’s modulus of CNTs is about 9 times sronger than that of steel. 3. CNT’s exhibit large strength in tension. 4. CNT’s can be bent without breaking. 5. Electrical properties of CNTs ranges from semiconductor to good conductor. 6. CNT’s have low resistivity and low heat dissipation. 7. Electrical Conductivity of CNTs is maximum along the axis and very less along the perpendicular direction. 8. CNT’s exhibit magneto-resistance ie:their resistance decreses with increasing mag.field. 9. Thermal conductivity of CNTs is maximum along the axis and very less along the perpendicular direction 10. CNTs have very high strength to weight ratio and have low density. 11. CNTs are chemically more inert compared to other forms of carbon.
Q: Mention the uses of CNTS’.
1. CNTs can store lithium hence they are used in the manufacture of batteries 2. CNTs can store hydrogen hence they are used in fuel cells. 3. CNTs are used as atomic force microscope probe tips. 4. CNTs are used to produce flat panel display of television and computer monitors . 5. CNTs are used as light weight shield for electromagnetic radiation. 6. Semiconducting CNTs are used to produce field effect transistors used in computers whose processing capacity is 104 faster than present processors. 7. CNTs are used to produce light weight high strength materials for aircrafts, rockets automobiles etc 8. CNTs are used as chemical sensors to detect gases.
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 11
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM Q: Explain the Principle, Construction and Working of SEM (Scanning Electron Microscope). Electron gun
To Vacuum pump
Magnetic condenser coil -1
Magnetic condenser coil-2
Scanning coil
Objective magnetic coil
퐷1 back scattered electron detector
퐷2 Secondary electron detector
퐷3 X-rays detector
TV monitor Evacuated chamber Specimen Base
Principle: 1. SEM is a device used to produce very high resolving images based on the principle of wave nature of electrons. 5 2. The resolving power of SEM is 10 times more than that of a best optical microscope. 3. The schematic diagram of SEM is as shown in the labeled diagram. Working: 1. The electrons produced from the electron gun are passed through two magnetic condensing coils 1&2 in a highly evacuated chamber to condense the beam 2. The condensed beam is passed through scanning coil which will scan the entire specimen. 3. The scanning beam is focused on to the specimen by the magnetic objective coil. 4. When the electron beam incident on the specimen a few electrons are reflected back called
back scattered electrons which are detected using the detector 퐷1. 5. Secondary electrons produced due to interaction with valence electrons are detected by the detector 퐷2
6. and X-rays produced due to deep penetration are detected using the detector 퐷3 7. High resolution 3D image can be seen on the TV monitor to which 퐷1, 퐷2& 퐷3 connected. 8. Biological specimens must be dehydrated and non conducting samples must be coated with a thin conducting layer.
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 12
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM Q:Mention the uses of SEM. 1. SEM is used to study reflectivity, roughness of the surfaces . 2. SEM is used to study the composition of a compound and the abundance of the constituents. 3. SEM is used to study biological specimens like pollen grains, blood cells tissues, bacteria etc 4. SEM is used to study the structures, corroded layers . 5. SEM is used in forensic science for examining gunshot residues etc 6. SEM is used in textile industry for the evaluation of fabric.
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PROBLEMS SECTION MODULE-5 : SHOCK WAVES AND SCIENCE OF NANO MATERIALS 푋 푉 ℎ 1.228 푋10−9 Formulae needed:V = ; M = ; λ = or ( For an electron) 푡 푎 √2푚푒푉 √푉
1. The distance between the two pressure sensors in a shock tube is 100 mm.The time taken by a shock wave to travel this distance is 200 microsecond.If the velocity of sound under the same conditions is 340 m/s. find the Mach number of the shock wave. ( 4 marks) Given: x =100 mm=100x10−3 m,t =200μS=200푥10−6 S,a=340 m/s ,M=? 푋 푉 Using, ,V= and M = 푡 푎 푋 100푥10−3 We get, M = = 푎푡 340푥200푥10−6 =1.471
2. In a scanning electron microscope ,electrons are accelerated by an anode potential difference of 60 kilo volt. Estimate the wavelength of the electrons in the scanning beam. ( 4 marks) Given: V=60KV=60x103 V , h=6.63x10−34 Js; m=9.1x10−31kg; e=1.6x10−19 C; λ =? ℎ Using λ = √(2푚푒푉) 6.63푥10−34 = = 5.016 x 10−12 m √(2푥9.1푥10−31푥1.6푥10−19 푥60푥103)
1.228 푋10−9 or λ = √푉 1.228 푋10−9 = = 5.013 x 10−12 m √(60푥103)
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 13
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM 3. In a Reddy shock tube,it was found that, the time taken to travel between the two sensors is 195 μs.If the distance between the two sensors is 100mm,find the Match number.(CBCS- Jun/Jul16)
Given: x=100 mm=100x10−3 m,t =195μS=195푥10−6 S,a=?(Not given) ,M=? 푋 Using, ,푉 = 푆 푡 100푥1 0−3 = 195푥10−6 = 512.8 m/s and
푉 Also, M = 푆 푎 512.8 = 푎 4. Calculate the wavelength of an electron accelerated under a potential difference of 100 V in scanning electron microscope.
Given: V=100V , h=6.63x10−34 Js; m=9.11x10−31kg; e=1.6x10−19 C; λ =? ℎ Using λ = √( 2푚푒푉) 6.63푥10−34 = √( 2푥9.11푥10−31푥1.6푥10−19 푥100) = 1.228 x 10−10 m
1.228 푋10−9 or λ = √푉 1.228 푋10−9 = √(100) = 1.228 x 10−10 m
5. The distance between two pressure sensors in a shock tube is 150 mm. The Time taken by a shock wave to travel this distance is 0.3 ms.If the velocity of sound under the same condition is 340m/s.Find the Mach number of the shock wave.
Given: x =150 mm=150x10−3 m,t = 0.3mS= 0.3푥10−3 S, a=340 m/s ,M=? 푋 푉 Using, ,V = and M = 푡 푎 푋 We get, M = 푎푡 150푥10−3 = = 1.471 340푥0.3푥10−3
VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 14
REDDAPPA C MODULE-5 INTERLINE PUBLISHING.COM 6. Calculate the wavelength of an electron accelerated under a potential difference of 100 V in scanning electron microscope.
Given: V=100V , h=6.63x10−34 Js; m=9.11x10−31kg; e=1.6x10−19 C; λ =? ℎ Using λ = 1.228 푋10−9 √(2푚푒푉) OR λ = √푉 6.63푥10−34 = 1.228 푋10−9 √(2푥9.11푥10−31푥1.6푥10−19 푥100) = √(100) = 1.228 x 10−10 m −10 = 1.228 x 10 m
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VTU ENGINEERING PHYSICS(17PHY 12/22) SEM-I/II Page 15
IMPORTANT IA TEST & VTU EXAMINATION QUESTIONS
MODULE-3 Lasers and Optical Fibres.
1. Explain : Induced absorption, Spontaneous emission , Stimulated emission and Derive an expression for energy de sit i ter s of Ei stei ’s coefficie ts. 2. Explain the principle,construction and working of laser and Semiconductor laser. 3. Explain Recording and Reconstruction of images in Holography. CO2 4. Explain types of Optical fibres with diagrams. 5. Explain point to point communication using optical fibres with block diagram. 6. Define attenuation and explain attenuation mechanisms/losses in optical fibres. 7. Derive an expression for acceptance angle (Theory of optical fibre) and Numerical aperture. 8. Explain the applications of lasers: Measurement of atmospheric pollutants, Welding, Cutting & Drilling. 9. Explain the Requisites of Laser system.
MODULE-4 Crystal Structure. 1. Explain seven crystal systems with diagrams. 2. What are Miller Indices and explain the steps/method of finding the Miller Indices of a plane. 3. Derive an expression for inter-planar spacing/distance. 4. Define Co-ordination number and (APF)Atomic Packing factor).Also find the APF of SC,BCC&FCC. 5. Write a note on structures of Diamond and Perovskites with diagrams. 6. Deri e Bragg’s la a d e plai ho to erif Bragg’s la a d Ide tif cr stals usi g Bragg’s X-ray spectrometer(Diffractometer) 7. Explain Polymorphism and Allotrophy.
MODULE-5 Shock Waves and Science of Nano Materials.
1. Explain the terms : Acoustic waves, Infrasonic waves, Audible waves, Ultrasonic waves, Mach waves, subsonic waves , supersonic waves and Shock waves. 2. State and explain basics of conservation of mass, momentum and energy. 3. Derive/explain Rankine-Hugonite( Normal shock wave) equations. 4. Describe the principle, construction and working of operated Reddy shock tube. 5. Explain density of states with diagrams. 6. Explain the synthesis of nano particles by Ball mill method and Sol-Gel method. 7. Explain the synthesis of CNTs by Arc discharge method and Pyrolysis method. 8. Explain the principle, construction , working and uses of SEM(Scanning Electron Microscope) 9. Methods of producing shock waves and properties and uses of shock waves. 10. Explain different types and structures of CNTs and mention properties & uses of CNTs
HINTS FOR SURE SUCCESS……..
“ SLOW AND STEADY WINS THE RACE “
1. Adopt Easy to Difficult approach. First thorough with the easy questions of the topic and then attempt difficult questions, which you will feel easy.
2. Practice writing diagrams again and again, because no marks are awarded for derivations without relevant diagrams wherever needed.
3. Memorize all the formulae so that you can solve problems and score maximum marks with less effort.
4. Always express the given data in terms of simple SI units ,then write the relevant formula and substitute the data.
5. While answering in the tests/exams, First choose the questions which you can answer correctly and completely.
6. Attempt all the questions by writing whatever you know about it even if you are not sure about the answers.
7. Practice the use of calculator by solving the solved problems again and again, so that you will not find any difficulty in the tests/exams.
8. Take all the tests without fail before the main examination so that you will know your strengths and weakness .This will help you to rectify the mistakes and score more marks.
9. Pay attention to the units of the physical quantities so that you can improve your marks.
10.Write the diagrams neatly and answers legibally.
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