Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 18 - Electrical Properties - Questions and Problems - Page 779: 18.11b


$\frac{n}{N_{Al}} = 3.29$

Work Step by Step

Given: electrical conductivity of Aluminum= 3.8 x $10^{7}$ $(Ωm)^{-1}$ electron mobility of Aluminum = 0.0012 $m^{2}$/ Vs Required: number of free electrons per aluminum atom assuming density of 2.7 $\frac{g}{cm^{3}}$ Solution: Determine the number of aluminum atoms per cubic meter, $N_{Al}$ using Equation 4.2 and using the atomic weight of Aluminum of 26.98 g/mol: $N_{Al} = \frac{N_{A}p^{'}}{A_{Al}} = \frac{(6.022 x 10^{23})(2.7 \frac{g}{cm^{3}})(10^{6} \frac{cm^{3}}{m^{3}})}{(26.98 g/mol)} = 6.026 \times 10^{28} m^{-3}$ Computing for the number of free electrons per aluminum atom: $\frac{n}{N_{Al}} = \frac{1.98 \times 10^{29} m^{-3}}{6.026 \times 10^{28} m^{-3}} = 3.29$
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