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.

Answer:

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, \[{{n}_{e}}.{{n}_{h}}=n_{i}^{2}\]

\[\Rightarrow {{n}_{h}}=\frac{n_{i}^{2}}{{{n}_{e}}}=\frac{{{(5\times {{10}^{8}})}^{2}}}{(8\times {{10}^{12}})}=\frac{25\times {{10}^{16}}}{8\times {{10}^{12}}}=3.125\times {{10}^{4}}{{m}^{-3}}\]