Problem Solving Section 1 Problem 1: Copper has a mass density ρ m 8.95 gmcm 3 and an electrical resistivity ρ 1.55 10 8 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(1.67 10 27 ) 8.47 10 22 cm 3. (b) The relation for electrical conductivity is given by, σ ne2 τ m, (2) σ 1 ρ, (3) Feb. 24, 2011 1
where, τ is relaxation time and ρ is resistivity. For copper, given that ρ cu 1.55 10 8 ohm m. Using Eq(2) and Eq(3), relaxation time is τ m ne 2 ρ 9.11 10 31 8.47 10 28 (1.602 10 19 ) 2 1.55 10 8 2.677 10 14 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 ( ) 10 34 ) 2 3(8.47 10 28 0.66 ). 8(9.11 10 31 ) 3.14 6.7 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 (8.47 10 28 ) 1.326 10 10 m 1. ) 1/3 Feb. 24, 2011 2
Thus, Fermi velocity is ( ) 6.625 10 34 1.326 10 10 v F 2(3.14) 9.11 10 31 1.536 10 6 ms 1, and mean free path of a conduction electron at Fermi level is l F v F τ (1.536 10 6 )(2.704 10 14 ) 4.153 10 8 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, 2011 3
Therefore, firstly we have to find E y E y V y d 20 10 3 1 10 3 2 voltm 1. (d thickness) Area of cross section (Breadth)(thickness) (5 10 3 )(1 10 3 ) 5 10 6 m 2. j x current density current ( ) area 5 10 3 5 10 6 1 10 3 ampm 2. The Hall coefficient is, R H 2 voltm 1 (0.45 wbm 2 )(1 10 3 ampm 2 ) 4.44 10 3 voltm 3 amp 1 wb 1. Problem 3: Use the equation, ( dv v m dt + τ ) e E, Feb. 24, 2011 4
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, 2011 5
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, 2011 6
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(1.74 10 3 ) 24(1.67 10 27 ) 8.68 10 28 m 3. Plug n, m e, h in Eq(9) E F (6.625 ( 10 34 ) 2 3(8.68 10 28 ) 8(9.11 10 31 ) 3.14 1.14 10 18 J, ) 2/3 and mean free energy is < E > 3 5 (1.14 10 18 ) 6.84 10 19 J. (b) In order to compare E F to KT at T room 300 K, take their ratio E F k B T 1.14 10 18 300(1.38 10 23 ) 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, 2011 7
T F E F k B 1.14 10 18 1.38 10 23 82, 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 (1.38 10 23 ) 2 4 7.05(1.602 10 19 ) 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, 2011 8
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 0.3 10 3 m. radius 1.5 10 4 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 0.5 0.12(7.065 10 8 ) 5.89 10 7 Ω 1 m 1. A πr 2 7.065 10 8 m 2 By plugging the values of σ, K e and T 293 K in Eq(10) 390 L exp 5.89 10 7 (293) 2.26 10 8 WΩK 2. Feb. 24, 2011 9
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 (1.38 10 23 ) 2 (1.602 10 19 ) 2 2.84 10 8 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 (4.28 10 10 ) 3 2.55 10 28 m 3. Hall coefficient is given by, Feb. 24, 2011 10
R H 1 nec 1 (1.602 10 19 )(2.55 10 28 )(3 10 8 ) 8.163 10 19 m 2 s 1. Feb. 24, 2011 11