Problem Solving Section 1

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Clinton Arnold
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1 Problem Solving Section 1 Problem 1: Copper has a mass density ρ m 8.95 gmcm 3 and an electrical resistivity ρ ohm m at room temperature. Calculate, (a). The concentration of the conduction electrons. (b). The mean free time τ. (c). The Fermi energy E F. (d). The Fermi velocity v F and the mean free path at Fermi level. Solution: (a) We have to find the concentration of conduction electrons n N V. Given that, mass density of copper Cu mass volume 8.95 gcm 3. n N mass/mass density (1) Cu has valency 1 and mass number is 64 amu. By plugging N 1 and mass density in Eq(1), we get 1(8.95) n 64( ) cm 3. (b) The relation for electrical conductivity is given by, σ ne2 τ m, (2) σ 1 ρ, (3) Feb. 24,

2 where, τ is relaxation time and ρ is resistivity. For copper, given that ρ cu ohm m. Using Eq(2) and Eq(3), relaxation time is τ m ne 2 ρ ( ) s. (c) The Fermi energy for 3-D is given by, ( ) 2/3 E F h2 3N (4) 8m e πv ( ) 0.66 h2 3n. 8m e π (6.625 ( ) ) 2 3( ). 8( ) ev 7 ev. (d) The Fermi momentum is given by, p F v F k F mv F k F m e From Eq(4), k F is k F (3π 2 n) 1/3 ( 3(3.14) 2 ( ) m 1. ) 1/3 Feb. 24,

3 Thus, Fermi velocity is ( ) v F 2(3.14) ms 1, and mean free path of a conduction electron at Fermi level is l F v F τ ( )( ) m Problem 2: Find the Hall coefficient for germanium if for a given sample (length 1 cm, breadth 5 mm. thickness 1 mm) a current of 5 milliamperes flown from a 1.35 volts supply develops a Hall voltage of 20 millivolts across the specimen in a magnetic field of 0.45 Wbm 2. Solution: We have to find Hall coefficient R H, E y 1 nec (Hj x) R H Hj x, where E y is Hall s field and R H is R H E y Hj x Feb. 24,

4 Therefore, firstly we have to find E y E y V y d voltm 1. (d thickness) Area of cross section (Breadth)(thickness) ( )( ) m 2. j x current density current ( ) area ampm 2. The Hall coefficient is, R H 2 voltm 1 (0.45 wbm 2 )( ampm 2 ) voltm 3 amp 1 wb 1. Problem 3: Use the equation, ( dv v m dt + τ ) e E, Feb. 24,

5 for the electron drift velocity v to show that the conductivity at frequency ω is ( ) 1 + iωτ σ(ω) σ(0), 1 + (ωτ) 2 where σ(0) ne 2 τ/m. Solution: Given that, the magnitude form of equation of motion is For v v 0 e iωt, we have ( dv m dt + v ) ee. (5) τ dv dt iωv 0e iωt iωv. By plugging v and dv dt in Eq(5), m ( iωv 0 e iωt + v ) 0e iωt ee τ ( 1) ee iω + v τ m v ee/m ( 1 iω) τ eeτ m ( 1 + iωτ 1 + ω 2 τ 2 ), and the current density is Feb. 24,

6 Using the relation for electrical conductivity, J nqv ( ) eeτ 1 + iωτ n( e) m 1 + ω 2 τ ( ) 2 ne2 Eτ 1 + iωτ m 1 + ω 2 τ 2 ( ne 2 τ ) ( ) 1 + iωτ E (6) m 1 + ω 2 τ 2 By comparing Eq(6) and Eq(7), σ 0 ne2 τ m J σe. (7) ( ) σ(ω) ne2 τ 1 + iωτ m 1 + ω 2 τ ( 2 ) 1 + iωτ σ(ω 0), 1 + ω 2 τ 2 which is the required condition. Problem 4: a. Calculate the mean free energy for Mg at 0 K. The density of Mg is 1.74 gcm 3. b. How does E F compare to kt for Mg at room temperature? What is the value of the Fermi temperature? Solution: (a) The mean free energy at T 0 K is < E > E N 3 5 k BT F 3 5 E F, (8) Feb. 24,

7 where, Fermi energy is E F Given that, density of Mg 1.74 gcm 3. h2 8m e ( ) 2/3 3N. (9) πv n N V 2( ) 24( ) m 3. Plug n, m e, h in Eq(9) E F (6.625 ( ) 2 3( ) 8( ) J, ) 2/3 and mean free energy is < E > 3 5 ( ) J. (b) In order to compare E F to KT at T room 300 K, take their ratio E F k B T ( ) 275 it means that E F 275k B T. The Fermi temperature is the temperature at which E F k B T F, Feb. 24,

8 T F E F k B , 608 K. Problem 5: Estimate the electronic contribution of specific heat kmol of copper at 4 K and 300 K. The Fermi energy of copper is 7.05 ev and is assumed to be temperature independent. Solution: Given that, Fermi energy of Cu is E F 7.05 ev. Using the relation for electronic specific heat C v, C v π2 (k B T ) nkb 2 E F π2 2 k 2 B T E F n. Therefore, electronic contribution of specific heat kmol at T 4 K is C v (3.14)2 2 ( ) ( ) 2.00 Jkmol 1 K 1. Similarly, C v at T 300 K is C v 150 Jkmol 1 K 1. Problem 6: A uniform copper wire of length 0.5 m and diameter 0.3 mm has a resistance of Feb. 24,

9 0.12 Ω at 293 K. If the thermal conductivity of the specimen at the same temperature is 390 Wm 1 K 1, calculate the Lorentz number. Compare this value with the theoretical value. Solution Given that, Length of Cu wire 0.5 m. diameter m. radius m. resistance 0.12 Ω. thermal conductivity 390 Wm 1 K 1. We have to calculate the Lorentz number L as, where σ is electrical conductivity. L K e σt, (10) σ 1 ρ 1 RA/L ( ) Ω 1 m 1. A πr m 2 By plugging the values of σ, K e and T 293 K in Eq(10) 390 L exp (293) WΩK 2. Feb. 24,

10 On the other hand, the theoretical value of Lorentz number can be evaluated by using the relation L theo π2 3 ( ) 2 kb e (3.14)2 3 ( ) 2 ( ) WΩK 2, L theo > L exp. Comparing the above two values of Lorentz numbers, we observe that the theoretical value is about 1.26 times higher than the experimental one. Problem 7: Calculate the Hall coefficient of sodium based on free electron model. structure and the side of the cube is 4.28 A 0. Sodium has bcc Solution Given that, Sodium has bcc structure, therefore number of atoms per unit cell is 2 and the side of the cube is a 4.28 A 0 n Number of electrons volume 2 a 3 2 ( ) m 3. Hall coefficient is given by, Feb. 24,

11 R H 1 nec 1 ( )( )( ) m 2 s 1. Feb. 24,

Nasser S. Alzayed.

Nasser S. Alzayed. Lecture #4 Nasser S. Alzayed nalzayed@ksu.edu.sa ELECTRICAL CONDUCTIVITY AND OHM'S LAW The momentum of a free electron is related to the wavevector by mv = ћk. In an electric field E and magnetic field