• # question_answer A semiconductor has equal electron and hole concentration of $5\times {{10}^{8}}\text{ }{{m}^{-3}}.$ On doping with certain impurity, electron concentration increases to$~8\times {{10}^{12}}\text{ }{{m}^{-3}}.$ (i) Identify the new semiconductor obtained after doping. (ii) Calculate the new hole concentration.

 Here, ${{n}_{i}}=5\times {{10}^{8}}{{m}^{-3}}$ and on doping ${{n}_{e}}=8\times {{10}^{12}}{{m}^{3}}$ (i) As on doping electron concentration has increased, hence, the doped semiconductor should behave as n-type semiconductor. (ii) We know that for a doped semiconductor, ${{n}_{e}}.{{n}_{h}}=n_{i}^{2}$
$\Rightarrow {{n}_{h}}=\frac{n_{i}^{2}}{{{n}_{e}}}=\frac{{{(5\times {{10}^{8}})}^{2}}}{(8\times {{10}^{12}})}=\frac{25\times {{10}^{16}}}{8\times {{10}^{12}}}=3.125\times {{10}^{4}}{{m}^{-3}}$