p42_023

# p42_023 - Now M =(27 0 g mol(6 022 × 10 23 mol − 1 = 4...

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23. Let N be the number of atoms per unit volume and n be the number of free electrons per unit volume. Then, the number of free electrons per atom is n/N . We use the result of Exercise 11 to Fnd n : E F = An 2 / 3 ,where A =3 . 65 × 10 19 m 2 · eV. Thus, n = µ E F A 3 / 2 = µ 11 . 6eV 3 . 65 × 10 19 m 2 · eV 3 / 2 =1 . 79 × 10 29 m 3 . If M is the mass of a single aluminum atom and d is the mass density of aluminum, then N = d/M
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Unformatted text preview: . Now, M = (27 . 0 g / mol) / (6 . 022 × 10 23 mol − 1 ) = 4 . 48 × 10 − 23 g, so N = (2 . 70 g / cm 3 ) / (4 . 48 × 10 − 23 g) = 6 . 03 × 10 22 cm − 3 = 6 . 03 × 10 28 m − 3 . Thus, the number of free electrons per atom is n N = 1 . 79 × 10 29 m − 3 6 . 03 × 10 28 m − 3 = 2 . 97 ....
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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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