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JEE Main & Advanced Physics Semiconducting Devices Question Bank Self Evaluation Test - Semiconductor Electronics: Materials, Devices and Simple Circuits

  • question_answer
    The ratio of work function and temperature of two emitters are \[1:2,\] then the ratio of current densities obtained by them will be

    A)  \[4:1\]

    B)         \[2:1\] 

    C)  \[1:2\] 

    D)         \[1:4\]

    Correct Answer: D

    Solution :

    [d] \[\frac{{{J}_{1}}}{{{J}_{2}}}=\frac{AT_{1}^{2}{{e}^{-{{W}_{1}}/k{{T}_{1}}}}}{AT_{2}^{2}{{e}^{-{{W}_{2}}/k{{T}_{2}}}}}\] \[={{\left(\frac{{{T}_{1}}}{{{T}_{2}}}\right)}^{2}}{{e}^{-\frac{{{W}_{1}}}{k{{T}_{1}}}+\frac{{{W}_{2}}}{k{{T}_{2}}}}}={{\left(\frac{1}{2} \right)}^{2}}{{e}^{0}}=\frac{1}{4}\]


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