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CMC Medical CMC-Medical VELLORE Solved Paper-2012

  • question_answer
    In a semiconductor the concentrations of electrons and holes are \[8\times {{10}^{18}}{{m}^{-3}}\]and  \[5\times {{10}^{18}}{{m}^{-3}}\]respectively. If the mobilities of electrons and holes are \[2.3\,\,{{m}^{2}}{{V}^{-1}}{{s}^{-1}}\]and \[0.01\]\[0.01\,{{m}^{2}}{{V}^{-1}}{{s}^{-1}}\]respectivey, then semiconductor is

    A)  n-type and its resistivity is 0.034 \[\Omega \text{-}m\]

    B)  n-type and its resistivity is 0.34 \[\Omega \text{-}m\]

    C)  p-type and its resistivity is 0.034\[\Omega \text{-}m\]

    D)  p-type and its resistivity is 0.34\[\Omega \text{-}m\]

    Correct Answer: B

    Solution :

                    Given,   \[{{n}_{e}}=8\times {{10}^{18}}{{m}^{-3}}\] and        \[{{n}_{h}}=5\times {{10}^{18}}{{m}^{-3}}\] and        \[{{\mu }_{e}}=2.3\,{{m}^{2}}\text{/}V\text{-}s\] and        \[{{\mu }_{h}}=0.01\,\,{{m}^{2}}{{m}^{2}}\text{/}V\text{-}s\] \[\because \,{{n}_{e}}>{{n}_{h}},\] so semiconductor is n-type. Now, conductivity\[\sigma =\frac{\text{1}}{\text{Resistivity}\,\text{( }\!\!\rho\!\!\text{ )}}\] \[=e\,({{n}_{e}}{{\mu }_{e}}+{{n}_{h}}{{\mu }_{h}})\] \[\Rightarrow \frac{1}{\rho }=1.6\times {{10}^{-19}}[8\times {{10}^{18}}\times 2.3+5\times {{10}^{18}}\times 0.01]\] \[\Rightarrow \]               \[\rho =0.34\,\,\Omega \text{-}m\]


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