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Uttarakhand PMT Uttarakhand PMT Solved Paper-2008

  • question_answer
    Pure silicon at 300 K has equal electron \[({{n}_{e}})\] and hole \[({{n}_{h}})\] concentration of\[1.5\times {{10}^{16}}{{m}^{-3}}\]. Doping   by   indium   increases   \[{{n}_{h}}\]  to \[4.5\times {{10}^{22}}{{m}^{-3}}\] . The \[{{n}_{e}}\] in the doped silicon is

    A)  \[9\times {{10}^{5}}\]       

    B)  \[5\times {{10}^{9}}\]

    C)  \[2.25\times {{10}^{11}}\]    

    D)  \[3\times {{10}^{19}}\]

    Correct Answer: B

    Solution :

     \[{{n}_{e}}{{n}_{h}}={{({{n}_{i}})}^{2}}\] \[{{n}_{e}}\times 4.5\times {{10}^{22}}={{(1.5\times {{10}^{16}})}^{2}}\] \[{{n}_{e}}=\frac{2.25\times {{10}^{32}}}{4.5\times {{10}^{22}}}\] \[{{n}_{e}}=5\times {{10}^{9}}\]


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