ch41-p047 - 47. (a) Setting E = EF (see Eq. 41-9), Eq. 41-5...

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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 37-3 as well as the value 1240eV nm hc =⋅ : NE mc hc nn n F () ( ( (. ) // / = F H G I K J = × F H G I K J −− 4 3 4 511 10 1240 34 1 1 2 2 2 3 13 3 2 3 2 1 ππ eV) eV nm) nm eV 2 which is equivalent to the value indicated in the problem statement. (b) Since there are 10 27 cubic nanometers in a cubic meter, then the result of Problem 41- 5 may be written as n = 849 10 849 28 3 3 .. . mn m The cube root of this is n 1/3 4.4/nm. Hence, the expression in part (a) leads to
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