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 nd 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 MAN −− == × = × . Thus, n = × 896 1055 10 8 4 91 0 8 4 0 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 mc n F e == × = 0121 0121 1240 511 10 84 3 7 0 2 2 23 3 323 .( ) (. ) . // 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 82 () m NE E h π = for the density of states associated with the conduction electrons of a metal. This can be written nE CE = where 31 3/2 56 3/2 3 3 33 4 3 8 2 8 2 (9.109 10 kg) 1.062 10 kg / J s . (6.626 10 J s) m C h ππ × == = × ×⋅ (b) Now, 22 1J 1kg m /s =⋅ (think of the equation for kinetic energy Km v = 1 2 2 ), so 1 kg = 1 J·s 2 ·m – 2 . Thus, the units of C can be written ( ) ( ) // Js m J s J m / 2 ⋅⋅ = −− 32 3 3 . This means C × = × (. ) . . / 1062 10 1602 10 681 10 56 3 19 27 3 3 2 J m J / eV) m eV (c) If E = 5.00 eV, then nE () ( . ) ( . ) . . / = × 500 152 10 27 3 1 2 28 1 3 me V e V e Vm
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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 36 3 3 28 gcm cm m g mol) / (6.02 10 mol m 2 31 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 PE e EE k T 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 T −= × = , so 2.248 1 ( ) 0.0955. 1 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: nE CE () . ( ) ( . . . // == × = × −− 12 27 3 23 28 3 681 10 80 19 10 me V e V ) m e V 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 23 2 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 34 31 38 ππ // (. . ./ Js ) kg Jsk g . 2 22 Since 1 1 Jk g 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 e V 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 ee EE k T x F 1 1 1 1 1 1 = + = + ( )/ where xE E k T F =− ( 1 . We solve for e x : e P x = = 1 1 1 0 090 1 91 9 1 . . Thus, 21 2 ( ( 1 11 1 1 0.91, 1 ( 9 1 / 9 ) 1 FF E E kT E E kT x P e −− == = = = ++ + + where we use E 2 E F = – 63 meV = E F E 1 = – ( E 1 E F ).
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9. The Fermi-Dirac occupation probability is given by Pe EkT FD =+ 11 / / c h , and the Boltzmann occupation probability is given by B = −∆ / . Let f be the fractional difference. Then f PP P e e e = = + BF D B / / / .
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This note was uploaded on 10/01/2007 for the course PHYS 2213 taught by Professor Perelstein,m during the Fall '07 term at Cornell University (Engineering School).

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

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