# ch41 - 1543 Chapter 41 1 According to Eq 41-9 the Fermi...

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Unformatted text preview: 1543 Chapter 41 1. 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 / / 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 2/3 2/3 2 34 2 38 2 2 31 3 3 (6.626 10 J s) 5.842 10 J s / kg . 9.109 10 kg 16 2 16 2 h A m Since 2 2 1 J 1 kg m / s , the units of A can be taken to be m 2 ·J. Dividing by 19 1.602 10 J/eV, we obtain 19 2 3.65 10 m eV. A 2. Equation 41-5 gives 3/ 2 1/ 2 3 8 2 ( ) m N E E h 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 27 3 2/3 8 2 8 2 (9.109 10 kg) 1.062 10 kg / J s (6.626 10 J s) 6.81 10 m (eV) . m C h Thus, 1/ 2 27 3 2/3 1/2 28 3 1 ( ) 6.81 10 m (eV) (8.0eV) 1.9 10 m eV . N E CE This is consistent with that shown in Fig. 41-6. 3. 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 . CHAPTER 41 1544 Thus, n 8 96 1055 10 8 49 10 8 49 10 22 22 3 28 . . . . . g / cm g cm m 3 3 4. Let E 1 = 68 meV + E F and E 2 = – 68 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 . . Thus, P 2 1 e ( E 2 E F )/ kT 1 1 e ( E 1 E F )/ kT 1 1 e x 1 1 (91/ 9) 1 1 0.91, where we use E 2 – E F = – 68 meV = E F – E 1 = – ( E 1 – E F ). 5. (a) Equation 41-5 gives 3/ 2 1/ 2 3 8 2 ( ) m N E E h 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 (b) Now, 2 2 1 J 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 as ( ) ( ) / / J s m J s J m 2 2 3 3/2 3 2 3 2 3 3 ....
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ch41 - 1543 Chapter 41 1 According to Eq 41-9 the Fermi...

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