A semiconductor has equal electron and hole concentration of 2 × 10⁸m^–3. On doping with a certain impurity, the electron concentration increases to $4\times {10}^{10}$m^–3, then the new hole concentration of the semiconductor is

Intrinsic carrier concentration, $n_i=2\times {10}^8{m}^{-3}$

electron concentration $n_e=4\times {10}^{10}{m}^{-3}$

Thus the hole concentration,

We know that,

$$\begin{gathered} n_e{n}_h={n_i}^2 \\ {n}_h=\frac{{n_i}^2}{n_e} \end{gathered}$$

Substituting the values,

$$\begin{gathered} n_h=\frac{{(2\times {10}^8)}^2}{4\times {10}^{10}} \\ {n}_h={10}^6{m}^{-3} \end{gathered}$$

Hence the correct answer is ${10}^6{m}^{-3}.$