Chapter41 - 1. (a) At absolute temperature T = 0, the...

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1. (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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2. 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 gm o l ) / ( 6 . 0 21 0m o l m 23 1 3
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() // Js m J s J m 22 33 / 2 ⋅⋅ = −− 32 3 3 . This means C × = × (. ) . . / 1062 10 1602 10 681 10 56 3 19 27 3 3 2 Jm J / e V ) m e V 3/2 (c) If E = 5.00 eV, then 27 3 3/2 1/2 28 1 3 ( ) (6.81 10 m eV )(5.00eV) 1.52 10 eV m . NE = × 3. (a) Eq. 41-5 gives 3 82 m 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 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, 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 as
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4. 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 we have used 1240eV nm. hc =⋅
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5. 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 49 10 8 49 10 22 22 3 28 . . .. . g/cm g cm m 3 3
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6. 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 T T x P e −− == = = = ++ + + where we use E 2 E F = – 63 meV = E F E 1 = – ( E 1 E F ).
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E h m n F = F H G I K J 3 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 . 7. According to Eq. 41-9, the Fermi energy is given by 16
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8. We use the result of Problem 41-3: 1/2 27 3 2/3 1/2 28 3 1 ( ) 6.81 10 m (eV) (8.0eV) 1.9 10 m eV . NE CE −− ⎡⎤ == × = × ⎣⎦ This is consistent with Fig. 41-6.
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9. (a) Eq. 41-6 leads to 15 1 ln ( 1) 7.00eV (8.62 10 eV / K)(1000K)ln 1 0.900 6.81eV.
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter41 - 1. (a) At absolute temperature T = 0, the...

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