Answer
$n = 1.46 \times 10^{22} m^{-3}$
Work Step by Step
Given:
electrical conductivity (σ) of Silicon specimen = 500 $(Ω*m)^{-1}$
hole concentration (p) = $2.0 \times 10^{22} m^{-3}$
Required:
electron concentration (n)
Solution:
From Table 18.3,
electron mobility ($µ_{e}$) = $ 0.145 \frac{m^{2}}{V-s}$
hole mobility ($µ_{h}$) = $ 0.050 \frac{m^{2}}{V-s}$
Using Equation 18.13,
$σ = n|e|µ_{e} + p|e|µ_{h}$
$n = \frac{σ - p|e|µ_{h}}{|e|µ_{e}}$
$n = \frac{500(Ω−m)^{-1} - (2.0 \times 10^{22} m^{-3}) (1.602 \times 10^{-19} C)(0.05 \frac{m^{2}}{V-s})}{(1.602 \times 10^{-19} C) (0.145 \frac{m^{2}}{V-s})}$
$n = 1.46 \times 10^{22} m^{-3}$
