Question

A pure semiconductor has equal electron and hole concentration of ${10}^{16}{m}^{-3}$. Doping by indium increases $n_h$ to $4.5\times {10}^{22}{m}^{-3}.Whatisthevalueof$n_e$ in the doped semiconductor?

• A. ${10}^6{m}^{-3}$
• B. ${10}^{22}{m}^{-3}$
• C. $\frac{{10}^{32}}{4.5\times {10}^{22}}{m}^{-3}$
• D. $4\times {10}^{22}{m}^{-3}$

Answer 1

The correct option is C $\frac{{10}^{32}}{4.5\times {10}^{22}}{m}^{-3}$

We know that,

$n_e\times n_h\approx n_i^2$

$\Rightarrow n_e=\frac{n_i^2}{n_h}$
$n_e=\frac{{10}^{16}\times {10}^{16}}{4.5\times {10}^{22}}=\frac{{10}^{32}}{4.5\times {10}^{22}}{m}^{-3}$.