p42_027

# p42_027 - 27. (a) Setting E = EF (see Eq. 42-9), Eq. 42-5...

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27. (a) Setting E = E F (see Eq. 42-9), Eq. 42-5 becomes N ( E F )= 8 πm 2 m h 3 µ 3 16 π 2 1 / 3 h m n 1 / 3 . Noting that 16 2=2 4 2 1 / 2 =2 9 / 2 so that the cube root of this is 2 3 / 2 =2 2, we are able to simplify the above expression and obtain N ( E F )= 4 m h 2 3 3 π 2 n which is equivalent to the result shown in the problem statement. Since the desired numerical answer uses eV units, we multiply numerator and denominator of our result by c 2 and make use of the mc 2 value for an electron in Table 38-3 as well as the hc value found in problem 3 of Chapter 39: N ( E F )= µ 4 mc 2 ( hc ) 2 3 3 π 2 n 1 / 3 = µ 4(511 × 10 3 eV) (1240 eV · nm) 2 3 3 π 2 n 1 / 3 = ( 4 . 11 nm 2 · eV 1 ) n 1 / 3 which is equivalent to the value indicated in the problem statement.
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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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