Suppose a pure Si crystal has $5\times {10}^{28}atoms{\text{ m}}^{-3.}$. It is doped by 1 ppm concentration of pentavalent  As. calculate the number of electrons and holes. Given that  $n_i=1.5\times {10}^{16}{m}^{-3}.$

Here,  $n_i=1.5\times {10}^{16}{m}^{-3}$
Doping concentration of pentavalent as atoms 
=1ppm=1part per million
 $\therefore \text{ number of pentavalent As atoms,}$
 ${\text{N}}_D=\frac{5\times {10}^{28}}{{10}^6}=5\times {10}^{22}{\text{ atom m}}^{-3}$
Now the thermally generated electrons ($n_i\alpha {10}^{16}{m}^{-3)}$ are negligibly small as compared to those produced by doping ,so  $n_e\simeq N_D=5\times {10}^{22}{m}^{-3}$
$$\begin{gathered} also{\text{ n}}_e{n}_h={n_i}^{\begin{array}{l}2\\ \end{array}} \\ {n}_h=\frac{{n_i}^2}{n_e}=\frac{1.5\times {10}^{16}1.5\times {10}^{16}}{5\times {10}^{22}} \\ =4.5\times {10}^9{m}^{-3} \end{gathered}$$