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 780: 18.21

Answer

$n_{i}$ = 1.95 x $10^{21}$ $m^{-3}$

Work Step by Step

Given: PbS electrical conductivity = 25 $(Ω-m)^{-1}$ electron mobility = 0.06 $\frac{m^{2}}{V-s}$ hole mobility = 0.02 $\frac{m^{2}}{V-s}$ Required: intrinsic carrier concentration at room temperature Solution; Using Equation 18.15 and substituting the given values; $n_{i}$ = $\frac{σ}{|e|(µ_{e} + µ_{h})}$ $n_{i}$ = $\frac{25 (Ω-m)^{-1}}{(1.602 \times 10^{-19} C)(0.06+ 0.02) \frac{m^{2}}{V-s}}$ $n_{i}$ = 1.95 x $10^{21}$ $m^{-3}$
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