Question

Calculate the resistivity of an n-type semiconductor from the following data: density of conduction electrons $=8\times {10}^{13}{cm}^{-3}$, density of holes $=5\times {10}^{12}{cm}^{-3}$, mobility of conduction electron $=2.3\times {10}^4{cm}^2{V}^{-1}{s}^{-1}$ and mobility of holes $=100{cm}^2{V}^{-1}{s}^{-1}$.

Answer 1

Given,

Density of conductor electron$\left(n_e\right)=8\times {10}^{13}c{m}^{-3}$

Density of holes$\left(n_h\right)=5\times {10}^{12}c{m}^{-3}$

Mobility of conduction electrons $\left({\mu }_e\right)=2.3\times {10}^{4}c{m}^2{v}^{-1}{s}^{-1}$

Mobility of holes$\left({\mu }_n\right)=100c{m}^2{v}^{-1}{s}^{-1}$

And we know,

Resistivity$\left(\rho \right)=\frac{1}{\sigma }$

$\Rightarrow =q(n_e{\mu }_e+n_r{\mu }_r)$

Where,

$q=$charges of electrons$=1.602176\times {10}^{-19}C$

So,

$\sigma =1.602\times {10}^{-19}\times ((8\times {10}^{13}\times 2.3\times {10}^4)+(5\times {10}^{12}\times 100))$

$=0.294$

$\sigma =\frac{1}{\sigma }=\frac{1}{0.294}=3.401{\Omega }^{-3}$

$\rho =0.03401{\Omega }^{-m}$