Answer
$\frac{n_{doped}}{n_{pure}} = 5.0 \times 10^5$
Work Step by Step
The ratio
$\frac{n_{doped}}{n_{pure}} = \frac{5.0 \times 10^{21}m^{-3}}{(2)(5.0\times 10^{15} m^{-3})}$
$\frac{n_{doped}}{n_{pure}} = 5.0 \times 10^5$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 41 - Conduction of Electricity in Solids - Problems - Page 1274: 38c
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