Question

Pure $Se$ at 300K has equal electron $\left(n_e\right)$ and hole $\left(n_h\right)$ concentration of $1.5\times {10}^6{m}^{-3}$. Doping by indium increases $n_h=4.5\times {10}^{22}{m}^{-3}$. Calculate the doped silicon.

Answer 1

Here $n_1=1.5\times {10}^{16}{m}^{-3},n_h=4.5\times {10}^{22}{m}^{-3}$

Now $n_h{n}_e={n_i}^2$

or $n_e=\frac{{n_i}^2}{n_h}=\frac{{(1.5\times {10}^{16})}^2}{4.5\times {10}^{22}}=5\times {10}^9{m}^{-3}$