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: 18.11a


n = 1.98 x $10^{29}$ $m^{-3}$

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 $m^{3}$ for aluminum at room temperature Solution: Using Equation 18.8, n = $\frac{σ}{(|e|µ_{e})}$ = $\frac{3.8 \times 10^{7} (Ωm)^{-1}}{(1.602 \times 10^{-19} C)(0.0012 m^{2}/Vs)}$ n = 1.98 x $10^{29}$ $m^{-3}$
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