Chapter 27

# Chapter 27 - CHAPTER 27 The Microscopic Theory of...

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CHAPTER 27 The Microscopic Theory of Electrical Conduction 1* In the classical model of conduction, the electron loses energy on average in a collision because it loses the drift velocity it had picked up since the last collision. Where does this energy appear? The energy lost by the electrons in collision with the ions of the crystal lattice appears as Joule heat ( I 2 R ). 2 A measure of the density of the free-electron gas in a metal is the distance r s , which is defined as the radius of the sphere whose volume equals the volume per conduction electron. ( a ) Show that r s = (3/4 π n ) 1/3 , where n is the free-electron number density. ( b ) Calculate r s for copper in nanometers. ( a ) The volume occupied by one electron is 1/ n = (4/3) r s 3 . Thus, r s = (3/4 n ) 1/3 . ( b ) n = 8.47 × 10 28 m –3 (See Table 27-1). Evaluate r s r s = 1.41 × 10 –10 m = 0.141 nm 3 ( a ) Given a mean free path λ = 0.4 nm and a mean speed v av = 1.17 × 10 5 m/s for the current flow in copper at a temperature of 300 K, calculate the classical value for the resistivity ρ of copper. ( b ) The classical model suggests that the mean free path is temperature independent and that v av depends on temperature. From this model, what would be at 100 K? ( a ) Use Equ. 27-7 ( b ) v av T 1/2 = 10 4 ) 10 (1.6 10 8.47 10 1.17 10 9 10 2 19 28 5 31 × × × × × × × × . m = 0.123 µ . m 100 = (0.123 . m)(100/300) 1/2 = 0.071 . m 4 Calculate the number density of free electrons in ( a ) Ag ( = 10.5 g/cm 3 ) and ( b ) Au ( = 19.3 g/cm 3 ), assuming one free electron per atom, and compare your results with the values listed in Table 27-1. ( a ), ( b ) n = N A / M ; = density, M = molar mass ( a ) n Ag = 5.86 × 10 22 el./cm 3 ; ( b ) n Au = 5.90 × 10 22 el./cm 3 . The results agree with Table 27-1 5* The density of aluminum is 2.7 g/cm 3 . How many free electrons are present per aluminum atom? n e = electrons/atom = nM / N A n e = 10 6.02 2.7 26.98 10 18.1 23 22 × × × × = 3.00 electrons/atom 6 The density of tin is 7.3 g/cm 3 . How many free electrons are present per tin atom?

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Chapter 27 The Microscopic Theory of Electrical Conduction See Problem 5 n e = 10 6.02 7.3 118.7 10 14.8 23 22 × × × × = 4.00 electrons/atom 7 Calculate the Fermi temperature for ( a ) Al, ( b ) K, and ( c ) Sn. ( a ), ( b ), ( c ) T F = E F / k ; k = 8.625 × 10 –5 eV/K ( a ) T F = (11.7 eV)/ k = 1.36 × 10 5 K; ( b ) T F = 2.45 × 10 4 K ( c ) T F = 1.18 × 10 5 K 8 What is the speed of a conduction electron whose energy is equal to the Fermi energy E F for ( a ) Na, ( b ) Au, and ( c ) Sn? ( a ), ( b ), ( c ) m / E 2 = e F F u ; use Table 27-1 for E F ( a ) u F = 1.07 × 10 6 m/s; ( b ) u F = 1.39 × 10 6 m/s; ( c ) u F = 1.89 × 10 6 m/s 9* Calculate the Fermi energy for ( a ) Al, ( b ) K, and ( c ) Sn using the number densities given in Table 27-1. ( a ), ( b ), ( c ) Use Equ. 27-15 b ( a ) E F = 0.365(181) 2/3 eV = 11.7 eV; ( b ) E F = 2.12 eV; ( c ) E F = 10.2 eV 10 Find the average energy of the conduction electrons at T = 0 in ( a ) copper and ( b ) lithium. ( a ), ( b ) E av = 0.6 E F ( a ) E av = 0.6 × 7.04 eV = 4.22 eV; ( b ) E av = 2.85 eV 11 Calculate ( a ) the Fermi temperature and ( b ) the Fermi energy at T = 0 for iron.
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Chapter 27 - CHAPTER 27 The Microscopic Theory of...

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