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

Since, we know, ${{n}_{e}}{{n}_{h}}=n_{i}^{2}$ The number of holes, ${{n}_{h}}=n_{i}^{2}/{{n}_{e}}$             $=\frac{{{(1.5\times {{10}^{16}})}^{2}}}{5\times {{10}^{22}}}\cong 4.5\times {{10}^{9}}{{m}^{-3}}$ The number of electrons thermally generated $({{n}_{i}}\tilde{\ }{{10}^{6}}{{m}^{-3}})$ are negligibly small as compared to those produced by doping. $\therefore$      ${{n}_{e}}\simeq {{N}_{D}}$ 1 ppm = 1 part per million ${{N}_{D}}=\frac{5\times {{10}^{28}}}{{{10}^{6}}}=5\times {{10}^{22}}{{m}^{-3}}$ $\therefore$      ${{n}_{e}}\cong 5\times {{10}^{22}}{{m}^{-3}}$