p42_032 - F √ E e E − E F/kT 1 dE R ∞ √ E e E − E...

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32. (a) At T = 300 K f = 3 kT 2 E F = 3(8 . 62 × 10 5 eV / K)(300 K) 2(7 . 0eV) =5 . 5 × 10 3 . (b) At T = 1000 K, f = 3 kT 2 E F = 3(8 . 62 × 10 5 eV / K)(1000 K) 2(7 . 0eV) =1 . 8 × 10 2 . (c) Many calculators and most math software packages (here we use MAPLE) have built-in numerical integration routines. Setting up ratios of integrals of Eq. 42-7 and canceling common factors, we obtain frac = R E
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Unformatted text preview: F √ E/ ( e ( E − E F ) /kT + 1) dE R ∞ √ E/ ( e ( E − E F ) /kT + 1) dE where k = 8 . 62 × 10 − 5 eV / K. We use the Fermi energy value for copper ( E F = 7 . 0 eV) and evaluate this for T = 300 K and T = 1000 K; we ±nd frac = 0 . 00385 and frac = 0 . 0129, respectively....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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