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NEET Sample Paper NEET Sample Test Paper-75

  • question_answer
    In a pure silicon \[({{n}_{i}}=1{{0}^{16}}/{{m}^{3}})\] crystals at 300 K, \[{{10}^{21}}\] atoms of phosphorous are added per cubic metre. The new hole concentration will be

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

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

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

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

    Correct Answer: B

    Solution :

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


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