IA TEST -3 SCHEME & SOLUTION TEST ON : 10-05-2016 Marks: 40 Subject & Code: ENGG.PHYSICS & 15PHY22 Sec : A, B, C, D & E Name of faculty: Dr.Gajanan v Honnavar, Dr. M. Faisal & Mohana Lakshmi Time : 11.30 AM to 1.00 PM Q. no. Scheme Marks P Definition of A a i) Mean free path ii) Relaxation time iii) Drift velocity 3 x 01 R 1 b Definition of Fermi factor. 02 T Variation of Fermi factor with temperature 3 x 01 a Success of QFET over CFET 3 x 02 1 2 b Equation , Substitution & Answer 02 P a Statement and explanation of law of mass action 02 + 02 A 3 b Equation , Substitution & Answer 04 R Use of equations of ohms law , current density, electrical T a current , mobility of charge carriers and arriving at the final 04 expression electrical conductivity of an ISC 2 4 b Equation , Substitution & Answer 04 P Temperature dependence of resistivity in metals 02 A 5 a Temperature dependence of resistivity in superconductors 02 R Definition of Superconductivity 01 T b Explanation of BCS theory of superconductivity 03 6 a Description of Type I and Type II superconductors 2 x 03 3 b Explanation of maglevs 02 P a Construction and working of shock tube 02 A 7 Diagram, Principle, Working 03 R b Any three applications of shock wave 03 T 8 a Explanation of ultrasonic and supersonic waves 01 +01 4 B Three conservation laws, Normal shock relationships 03 + 03 P 9 Principle, construction, working and applications of SEM 08 A R a Synthesis of carbon nanotubes by the Arc-discharge method 04 T b Graphical representation of the variation of density of states 10 in 0D, 1D, 2D and 3D quantum structures 04 5 SOLUTION PART 1 1. a) Define the terms i ) Mean free path ii) Relaxation time iii) Drift velocity (3M) Mean free path (λ): It is the average distance travelled by the free electrons between successive collisions with the lattice ions. Relaxation time (흉r): When the electric field applied across a conductor is switched off, the drift velocity decreases exponentially to zero. The time during which the drift velocity reduces to 1/e times the initial value is called relaxation time. Drift velocity (vd): When an electric field is applied, some of the free electrons acquire a constant, terminal velocity in a direction opposite to that of the field. This steady and very small velocity is called drift velocity. 1. b) Explain Fermi factor. Discuss the probability of occupation of various energy states by electrons at T=0oK, and T>0oK, on the basis of Fermi factor. (5M) Fermi factor: Under thermal equilibrium the free electrons occupy various energy levels in accordance with a statistical rule known as Fermi – Dirac statistics. Fermi – Dirac statistics enables the evaluation of probability of finding electron in energy levels over a certain range of energy values. The evaluation is done with the help of a quantity called Fermi factor f (E) given by: 1 f E EEF kT e 1 Effect of temperature on Fermi function Case 1: Probability of occupation for E< EF at T= 0k. Substituting this condition in F-D equation, we get 1 푓(퐸) = 1 + 푒−∞ 1 푓(퐸) = 1 + 0 푓(퐸) = 1 The distribution 푓(퐸) = 1 means that at T = 0k, all the energy levels below the Fermi level are fully occupied by electrons leaving the upper levels vacant. I.e. there is 100 % probability of finding an electron below the Fermi energy level. Case 2: Probability of occupation for E >EF at T= 0k. Substituting this condition in F-D equation, we get 1 푓(퐸) = 1 + 푒∞ 1 푓(퐸) = 1 + ∞ 1 푓(퐸) = = 0 ∞ This indicates that at T= 0k, the energy levels above the Fermi level is not occupied by electrons, they are vacant. I.e. there is 0% probability of finding the electron above the Fermi level at absolute zero K Case 3: Probability of occupation at ordinary temperatures. At temperatures above 0 K and E = EF . Substituting this condition in F-D equation, we get 1 푓(퐸) = 1 + 푒0 1 푓(퐸) = 1 + 1 1 푓(퐸) = = 0.5 2 At temperatures above 0K, there is only 50 % probability for an electron to occupy Fermi level. 2. a) Describe how quantum free electron theory was successful in overcoming the failures of classical free electron theory. (6M) Merits of Quantum Free Electron Theory: 1. Specific heat of free electrons: From quantum theory of free electrons, the specific heat of free electrons is given by Cv = (2k/EF) RT For a typical value of EF = 5 eV, we get Cv = 10−4RT. Hence gives the correct molar specific heat at constant volume for free electrons in metals. 2. Temperature dependence of conductivity of metals: The quantum free electron theory gives the correct temperature dependence of conductivity. From the equation for conductivity according to Q.F.E.T, 흈 ∝ 흀 ퟏ But, 흀 ∝ 흅풓ퟐ Where ‘r’ is the amplitude of lattice vibrations which is directly proportional to temperature i.e. 푟2 ∝ 푇 1 ∴ 휆 ∝ 푇 Since, 흈 ∝ 흀 ퟏ ∴ 흈 ∝ 푻 3. Dependence of electrical conductivity on electron concentration: ne2 The electrical conductivity in metals is given by * , m vF It is clear from this equation that the electrical conductivity depends on the electron *. concentration, the ratio of λ/vF and m Now by taking above expression, if we calculate σ, we get the observed conductivities of the metals. - 2. b) Find the relaxation time of conduction electrons in a metal of resistivity 1.54 x 10 8 28 3 ohm-m, if the metal has 5.8 x 10 conduction electrons per m . (2M) σ = n e2휏/m 흉 = ퟑ. ퟗퟕퟒ × ퟏퟎ−ퟏퟒ 풔 PART 2 3. a) State and explain the law of mass action for a semiconductor. (4M) Law of Mass Action: The law of mass action states that the product of electron concentration (ne) in the conduction band and the hole concentration (nh) in the valence band is equal to the square of the intrinsic carrier concentration (ni). We have the equation for electron concentration and hole concentration as Concentration of Free Electrons in the Conduction Band ne: E E 4 2 3 F g n m kT 2 e kT e h3 e Concentration of Holes in the Valence Band nh: 3 E 4 2(m kT) 2 F n h e kT h h3 In both the equations, we see the dependence of both ne and nh on EF Let us consider the product of ne and nh 3 E 2 g 32(kT) * * 3 2 kT n n m m e e h h6 e h The above equation shows that the product of nenh does not depend on EF, but remains a constant at a given temperature ne nh = a constant This condition is applicable for both extrinsic and intrinsic semiconductors. The above equation is called the law of mass action for semiconductors. 2 In case of intrinsic semiconductor , ne nh = ni 3. b) The Fermi level for a metal is 3.1eV. Calculate the energies for which the probabilities of occupancy at 300 K are 98% and 50 %. (4M) 1 f E EEF e kT 1 1 E EF kT ln 1 f E EF = 3.1 eV i) For f (E) = 98% = 0.98 E = 2.999 eV ii) For f (E) = 50% = 0.50 E = 3.1 eV 4. a) Derive the expression for electrical conductivity of an intrinsic semi-conductor The current in a semiconductor is contributed by both electrons and holes, moving in opposite directions. Consider a semiconductor with area of cross-section, A, free electron concentration ne and number density of holes, nh. The current due to free electrons is: Ie = ne A e ve, Where e is electronic charge and ve is the drift velocity of electrons. And, the current due to holes is: Ih = nh A e vh, Where vh is the drift velocity of holes. Hence the total current, I = Ie + Ih = ne A e ve + nh A e vh = A e (ne ve + nh vh) Current density J = I/A = e (ne ve + nh vh) But, current density, J = 휎 E, where 휎 is conductivity and E, the field across the semiconductor. 휎 E = e (ne ve + nh vh) 휎 = e (ne ve/E + nh vh/E) Or, 흈 = e (ne 흁 e + nh 흁 h) Where, ve/E = e is the mobility of electrons and vh/E = h is the mobility of holes. For intrinsic semiconductors ne = nh = ni, where ni is the density of intrinsic charge carriers. Hence for intrinsic semiconductors, = e ni (e + h) For n-type semiconductors, ne >> nh and therefore, = e ne e For p-type semiconductors, np >> ne and therefore, = e nh h 4. b) The effective mass of the electron in silicon is 0.31 m0, where m0 is the free electron mass. Find the electron concentration for silicon at 300 K assuming that the Fermi level lies exactly in the middle of the energy gap, given that the energy gap of silicon is 1.11 eV. (4M) Sol: T= 300K ; Eg= 1.11 eV; EF= Eg/2 15 3 ne= 2.07 x 10 /m PART 3 5. a. Explain the temperature dependence of resistance in metals and superconductors. (4M) In case of metals the resistance decreases witht temperature and reaches a minimum value at T=0 K. Even at ‘0K’, metals offer finite resistance, called residual resistance which is due to the scattering of electrons by impurities and crystal defects present in the material.