ch41 - 1 The number of atoms per unit volume is given by n...

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1. The number of atoms per unit volume is given by n d M = / , where d is the mass density of copper and M is the mass of a single copper atom. Since each atom contributes one conduction electron, n is also the number of conduction electrons per unit volume. Since the molar mass of copper is 63.54g / mol, A = 23 1 22 / (63.54g / mol)/(6.022 10 mol ) 1.055 10 g A M A N = = × = × . Thus, n = × = × = × 8 96 1055 10 8 49 10 8 49 10 22 22 3 28 . . . . . g / cm g cm m 3 3
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2. We note that n = 8.43 × 10 28 m – 3 = 84.3 nm – 3 . From Eq. 41-9, E hc m c n F e = = × = 0121 0121 1240 511 10 84 3 7 0 2 2 2 3 3 3 2 3 . ( ) . ( ( . ) . / / eV nm) eV nm eV 2 where the result of problem 83 in Chapter 38 is used.
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3. (a) Eq. 41-5 gives 3/ 2 1/ 2 3 8 2 ( ) m N E E h π = for the density of states associated with the conduction electrons of a metal. This can be written 1/ 2 ( ) n E CE = where 3/ 2 31 3/2 56 3/2 3 3 3 34 3 8 2 8 2 (9.109 10 kg) 1.062 10 kg / J s . (6.626 10 J s) m C h π π × = = = × × (b) Now, 2 2 1J 1kg m /s = (think of the equation for kinetic energy K mv = 1 2 2 ), so 1 kg = 1 J·s 2 ·m – 2 . Thus, the units of C can be written ( ) ( ) / / J s m J s J m 2 2 3 3/2 = 3 2 3 2 3 3 . This means C = × × = × ( . )( . . . / 1062 10 1602 10 681 10 56 3 19 27 3 3 2 J m J / eV) m eV 3/2 3/2 (c) If E = 5.00 eV, then n E ( ) ( . )( . ) . . / = × = × 681 10 500 152 10 27 3 1 2 28 1 3 m eV eV eV m 3/2
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4. We note that there is one conduction electron per atom and that the molar mass of gold is 197g mol / . Therefore, combining Eqs. 41-2, 41-3 and 41-4 leads to n = × = × ( . / )( / ) ( / ) . . 19 3 10 197 590 10 3 6 3 3 28 g cm cm m g mol) / (6.02 10 mol m 23 1 3
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5. (a) At absolute temperature T = 0, the probability is zero that any state with energy above the Fermi energy is occupied. (b) The probability that a state with energy E is occupied at temperature T is given by P E e E E kT F ( ) ( )/ = + 1 1 where k is the Boltzmann constant and E F is the Fermi energy. Now, E – E F = 0.0620 eV and 5 ( ) / (0.0620eV) /(8.62 10 eV / K)(320K) 2.248 F E E kT = × = , so 2.248 1 ( ) 0.0955. 1 P E e = = + See Appendix B or Sample Problem 41-1 for the value of k .
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6. We use the result of problem 3: n E CE ( ) . ( ) ( . . . / / = = × = × 1 2 27 3 2 3 28 3 681 10 8 0 19 10 m eV eV) m eV 1/2 1 This is consistent with Fig. 41-5.
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7. According to Eq. 41-9, the Fermi energy is given by E h m n F = F H G I K J 3 16 2 2 3 2 2 3 π / / where n is the number of conduction electrons per unit volume, m is the mass of an electron, and h is the Planck constant. This can be written E F = An 2/3 , where A h m = F H G I K J = F H G I K J × × = × 3 16 2 3 16 2 6 626 10 9109 10 5842 10 2 3 2 2 3 34 31 38 π π / / ( . . . / J s) kg J s kg . 2 2 2 Since 1 1 2 2 J kg m s = / , the units of A can be taken to be m 2 ·J. Dividing by 1602 10 19 . × J / eV , we obtain A = × 365 10 19 . m eV 2 .
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8. Let E 1 = 63 meV + E F and E 2 = – 63 meV + E F . Then according to Eq. 41-6, P e e E E kT x F 1 1 1 1 1 1 = + = + ( )/ where x E E kT F = ( ) / 1 . We solve for e x : e P x = = = 1 1 1 0 090 1 91 9 1 .
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