ch41-p013 - 13(a We evaluate P(E = 1(e E EF kT 1 for the...

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13. (a) We evaluate P ( E ) = ( ) 1/( 1) F EE k T e + for the given value of E , using kT = × × = (. . .. 1381 10 1602 10 0 02353 23 19 J / K)(273K) J/eV eV For E = 4.4 eV, ( E – E F )/ kT = (4.4 eV – 5.5 eV)/(0.02353 eV) = – 46.25 and 46.25 1 () 1 . 0 . 1 PE e == + (b) Similarly, for E = 5.4 eV, P ( E ) = 0.986 0.99 . (c) For E = 5.5 eV, P ( E ) = 0.50. (d) For E = 5.6 eV, P ( E ) = 0.014. (e) For E = 6.4 eV, P ( E ) = 2.447 × 10 – 17
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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