12th Class Physics Sample Paper Physics Sample Paper-10

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

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