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 $10^{21}$ $m^{-3}$