20101ee2_1_HW2 solution

20101ee2_1_HW2 solution - Solutions to HW#2 Problem 1(a n=...

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Solutions to HW #2 Problem 1 (a) 3 28 3 22 3 23 / 10 9 . 5 / 10 5.9 tom electron/a free 1 mole / 108 / 5 . 10 atom/mole 10 02 . 6 n m cm g cm g V N × = × = × × × = = eV joule n m h 5 . 5 10 8 . 8 π 10 9 . 5 3 10 1 . 9 8 ) 10 6 . 6 ( π 3 8 E 19 3 / 2 28 31 2 34 3 / 2 2 F = × = × × × × × × = = (b) Average speed = dE E D E f dE E D E vf ) ( ) ( ) ( ) ( Density of State per unit energy, E c E m E D = = 2 / 1 2 / 3 2 2 ) ( h π where c is a constant Velocity m E v 2 = Probability that a state is occupied by an electron + = kT E E E f F exp 1 1 ) ( For T = 0 K, f(E) = 1, E < E F = 0, E > E F Therefore, average speed = = F F F F E E E E dE E c cEdE m dE E c dE E c m E 0 0 0 0 2 2 m E E E m F E F 2 4 3 2 / 3 2 2 0 2 / 3 2 = =
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s m / 10 04 . 1 10 1 . 9 10 8 . 8 2 4 3 6 31 19 × = × × × = Average Energy = = F F F F E E E E dE E dE E dE E c dE E Ec 0 0 2 / 3 0 0 eV eV E E E F E F 3 . 3 5 . 5 5 3 5 3 2 / 3 2 / 5 0 2 / 3 2 / 5 =
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