A semiconductor has an electron concentration of  $0.45\times {10}^{12} {\text{m}}^{-3}$ and a hole concentration of  $5\times {10}^{20}{m}^{-3}$. Calculate its conductivity. Given electron mobility is $0.135m^2{v}^{-1}{s}^{-1}$ . Hole mobility is $0.048m^2{v}^{-1}{s}^{-1}$ 

Conductivity of semiconductor is the sum of the conductivities due to electrons and holes

$\begin{array}{l}Given , n_e=0.45\times {10}^{12}{m}^{-3}\\ n_h=5\times {10}^{20}{m}^{-3}\\ {\mu }_e=0.135m^2{v}^{-1}{s}^{-1}\end{array}$

$\begin{array}{l}{\mu }_h=0.048m^2{v}^{-1}{s}^{-1}\\ Conductivity , \sigma ={\sigma }_e+{\sigma }_h\\ \end{array}$

$\Rightarrow \sigma =n_{e}e{\mu }_e+n_{h}e{\mu }_h$

${\text{n}}_e{\text{<<<<n}}_h\text{ So }{n}_e\text{ is negligible}$

$\begin{array}{l}\Rightarrow \sigma =n_{h}e{\mu }_h\\ \sigma =5\times {10}^{20}\times 1.60\times {10}^{-19}\times 0.048\\ \sigma =5\times {10}^1\times 1.6\times 0.048\\ \sigma =5\times 16\times 0.048\\ \sigma =3.84{\text{sm}}^{-1}\end{array}$