JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
    Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is \[{{10}^{19}}{{m}^{-3}}\] and their mobility is \[1.6\text{ }{{m}^{2}}/\left( V.s \right)\] then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to: [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) \[0.4\text{ }\Omega \,m\]           

    B) \[2\text{ }\Omega \,m\]

    C) \[4\,\,\Omega \,m\]  

    D) \[0.2\,\,\Omega \,m\]

    Correct Answer: A

    Solution :

    \[\mu \,\,=\,\,\frac{{{V}_{d}}}{E}\] \[I=n\,e\,A\,V{{\,}_{d}}\] \[\Rightarrow \,\,\,\,\mu \,\,\,=\,\,\frac{I}{n\,e\,A\,E}\] \[\mu \,\,\,=\,\,\frac{I\,.\,l}{n\,e\,A\,(El)}\,\,=\,\,\frac{Il}{n\,e\,A\,V}\] \[\mu \,\,\,=\,\,\frac{I\,l}{n\,e\,AI\,.\,R}\] \[\mu \,\,\,=\,\,\frac{I\,A}{n\,e\,A\rho \,l}\,\,=\,\,\frac{l}{n\,e\,\rho }\] \[\Rightarrow \,\,\rho \,\,=\,\frac{1}{ne\mu }\,\,\,=\,\,\,\frac{1}{{{10}^{19}}\times 1.6\,\,\times \,{{10}^{-19}}\,\times \,1.6}\,\] \[\Rightarrow \,\,\rho \,\,=\,\frac{1}{1.6\,\,\times \,\,1.6}\,\,\,\simeq \,\,\,0.4\,\,\Omega m\,\]


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