p42_021

# p42_021 - e ∆ E/kT = 1 P − 1 Now we take the natural...

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21. (a) We evaluate P ( E )=1 / ( e ( E E F ) /kT +1 ) for the given value of E , using kT = (1 . 381 × 10 23 J / K)(273 K) 1 . 602 × 10 19 J / eV =0 . 02353 eV . For E =4 . 4eV, ( E E F ) /kT =(4 . 4eV 5 . 5 eV) / (0 . 02353 eV) = 46 . 25 and P ( E )= 1 e 46 . 25 +1 =1 . 00 . Similarly, for E =5 . 4eV, P ( E )=0 . 986, for E =5 . 5 eV, P ( E )=0 . 500, for E =5 . 6eV, P ( E )= 0 . 0141, and for E =6 . 4eV, P ( E )=2 . 57 × 10 17 . (b) Solving P =1 / ( e E/kT +1 ) for e E/kT ,weget
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Unformatted text preview: e ∆ E/kT = 1 P − 1 . Now, we take the natural logarithm of both sides and solve for T . The result is T = ∆ E k ln ³ 1 P − 1 ´ = (5 . 6 eV − 5 . 5 eV)(1 . 602 × 10 − 19 J / eV) (1 . 381 × 10 − 23 J / K) ln ³ 1 . 16 − 1 ´ = 699 K ....
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