p42_016 - E = 63 meV and P ( E F + 63 meV) + P ( E F 63...

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16. According to Eq. 42-6, P ( E F +∆ E )= 1 e ( E F +∆ E E F ) /kT +1 = 1 e E/kT +1 = 1 e x +1 where x =∆ E/kT . Also, P ( E F E )= 1 e ( E F E E F ) /kT +1 = 1 e E/kT +1 = 1 e x +1 . Thus, P ( E F +∆ E )+ P ( E F E )= 1 e x +1 + 1 e x +1 = e x +1+ e x +1 ( e x +1)( e x +1) =1 . A special case of this general result can be found in problem 13, where ∆
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Unformatted text preview: E = 63 meV and P ( E F + 63 meV) + P ( E F 63 meV) = 0 . 090 + 0 . 91 = 1 . 0....
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