Question

In a sample of semi conductor mobilities of electrons and holes are $24\times {10}^{3}c{m}^2{V}^{-1}{S}^{-2}$ and $0.2\times {10}^{3}c{m}^2{V}^{-1}{S}^{-1}$ respectively. If the density electrons is $0.8\times {10}^{14}c{m}^{-3}$ and that of holes is $0.4\times {10}^{14}c{m}^{-3}$. Find the nature of semi conductor and its conductivity.

Answer 1

Given : ${\mu }_e=24\times {10}^3\times {10}^{-4}=24\times {10}^{-1}$ $m^2{V}^{-1}{s}^{-1}$

${\mu }_h=0.2\times {10}^3\times {10}^{-4}=0.2\times {10}^{-1}{m}^2{V}^{-1}{s}^{-1}$

$n_e=0.8\times {10}^{14}\times {10}^6=0.8\times {10}^{20}{m}^{-3}$

$n_h=0.4\times {10}^{14}\times {10}^6=0.4\times {10}^{20}{m}^{-3}$

Since, $n_e>n_h$

So, the given semiconductor is n-type.

Conductivity $\sigma =e({\mu }_e{n}_e+{\mu }_h{n}_h)$

Or $\sigma =1.6\times {10}^{-19}[24\times {10}^{-1}\times 0.8\times {10}^{20}+0.2\times {10}^{-1}\times 0.4\times {10}^{20}]=30.85mho-m$