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VMMC VMMC Medical 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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