p42_024 - E = kT ln 1 P 1 = (1 . 38 10 23 J / K)(300 K) ln...

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24. Let the energy of the state in question be an amount ∆ E above the Fermi energy E F . Then, Eq. 42-6 gives the occupancy probability of the state as P = 1 e ( E F +∆ E E F ) /kT +1 = 1 e E/kT +1 . We solve for ∆ E to obtain
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Unformatted text preview: E = kT ln 1 P 1 = (1 . 38 10 23 J / K)(300 K) ln 1 . 10 1 = 9 . 1 10 21 J , which is equivalent to 5 . 7 10 2 eV = 57 meV....
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