ch41-p019 - 19. Let N be the number of atoms per unit...

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3/2 29 3 19 2 11.6eV 1.79 10 m . 3.65 10 m eV F E n A ⎛⎞ == = × ⎜⎟ ×⋅ ⎝⎠ If M is the mass of a single aluminum atom and d is the mass density of aluminum, then N = d / M . 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 29 3 28 3 1.79 10 m
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