p42_012(1)

p42_012(1) - E F = 121 hc 2 m e c 2 n 2 3 = 121(1240 eV ...

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12. We reproduce the calculation of Exercise 6: Combining Eqs. 42-2, 42-3 and 42-4, the number density of conduction electrons in gold is n = (19 . 3g / cm 3 )(6 . 02 × 10 23 / mol) (197 g / mol) =5 . 90 × 10 22 cm 3 =59 . 0nm 3 . Now, using the result of Exercise 3 in Chapter 39, Eq. 42-9 leads to
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Unformatted text preview: E F = . 121( hc ) 2 ( m e c 2 ) n 2 / 3 = . 121(1240 eV · nm) 2 511 × 10 3 eV (59 . 0 nm − 3 ) 2 / 3 = 5 . 52 eV ....
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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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