# CH-41 - Chapter 41 1 The number of atoms per unit volume is...

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Unformatted text preview: Chapter 41 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-- ﾴ ﾴ . Thus, n ---8 96 1055 10 8 49 10 8 49 10 22 22 3 28 . . . . . g / cm g cm m 3 3 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. 3. (a) Eq. 41-5 gives 3/ 2 1/ 2 3 8 2 ( ) m N E E h p 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 p--ﾴ ﾴ ﾴ ﾴ ﾴ (b) Now, 2 2 1J 1kg m /s (think of the equation for kinetic energy K mv 1 2 2 ), so 1 kg = 1609 CHAPTER 41 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 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 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-- ﾴ , so 2.248 1 ( ) 0.0955. 1 P E e See Appendix B or Sample Problem 41-1 for the value of k . 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. 7. According to Eq. 41-9, the Fermi energy is given by 74 E h m n F F H G I K J 3 16 2 2 3 2 2 3 p / / 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 p p / / ( ....
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## This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.

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CH-41 - Chapter 41 1 The number of atoms per unit volume is...

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