p42_018

# p42_018 - n ( E ) are 9 . × 10 27 m − 3 · eV − 1 , 9...

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18. We use N 0 = N ( E ) P ( E )= CE 1 / 2 £ e ( E E F ) /kT +1 ¤ 1 ,wh e r e C is given in problem 7(a). At E = 4 . 00 eV, n 0 = ³ 6 . 8 × 10 27 m 3 · (eV) 3 / 2 ´ (4 . 00 eV) 1 / 2 e (4 . 00 eV 7 . 00 eV) / [(8 . 62 × 10 5 eV / K)(1000 K)] +1 =1 . 36 × 10 28 m 3 · eV 1 , and at E =6 . 75 eV, n 0 = ³ 6 . 8 × 10 27 m 3 · (eV) 3 / 2 ´ (6 . 75 eV) 1 / 2 e (6 . 75 eV 7 . 00 eV) / [(8 . 62 × 10 5 eV / K)(1000 K)] +1 =1 . 67 × 10 28 m 3 · eV 1 . Similarly at E =7 . 00 , 7 . 25 and 9 .
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Unformatted text preview: n ( E ) are 9 . × 10 27 m − 3 · eV − 1 , 9 . 5 × 10 26 m − 3 · eV − 1 and 1 . 7 × 10 18 m − 3 · eV − 1 , respectively. We note that the latter value is eFectively zero (relative to the other results)....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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