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NEET Sample Paper NEET Sample Test Paper-26

  • question_answer
    In a pure silicon \[({{n}_{1}}={{10}^{16}}/{{m}^{3}})\] crystal at 300K, \[{{10}^{21}}\] atoms of phosphorus are added per cubic metre. The new hole concentration will be:

    A)  \[{{10}^{21}}\,per\,\,{{m}^{3}}\]              

    B)  \[{{10}^{19}}\,per\,\,{{m}^{3}}\]

    C)  \[{{10}^{11}}\,per\,\,{{m}^{3}}\]                             

    D)  \[{{10}^{5}}\,per\,\,{{m}^{3}}\]

    Correct Answer: C

    Solution :

    By using mass-action law \[n_{1}^{2}={{n}_{e}}{{n}_{h}}\] \[\Rightarrow \]               \[{{n}_{h}}=\frac{n_{1}^{2}}{{{n}_{e}}}=\frac{{{\left( {{10}^{16}} \right)}^{2}}}{{{10}^{21}}}={{10}^{11}}per\,\,{{m}^{3}}.\]


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