F
21. (a) We evaluate P (E) =1/ e(E-E )/kT +1 for the given value of E, using
(1.381 × 10-23 J/K)(273 K)
kT = =0.02353 eV .
1.602 × 10-19 J/eV
For E =4.4eV, (E - EF )/kT =(4.4eV- 5.5eV)/(0.02353 eV) = -46.25 and
1
P (E) = =1.00 .
e-46.25 +1
Similarly, for E =5.4eV, P (E) =0.986, for E =5.5eV, 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 , we get
1
e"E/kT = - 1 .
P
Now, we take the natural logarithm of both sides and solve for T . The result is
"E (5.6eV- 5.5eV)(1.602 × 10-19 J/eV)

T = = = 699 K .
1 1
k ln - 1 (1.381 × 10-23 J/K) ln - 1
P 0.16

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