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JEE Main & Advanced Physics Semiconducting Devices Question Bank Graphical Questions

  • question_answer
    If the thermionic current density is J and emitter temperature is T then the curve between \[\frac{J}{{{T}^{2}}}\] and \[\frac{1}{T}\] will be

    A)                          

    B)                   

    C)                          

    D)                   

    Correct Answer: C

    Solution :

                       \[J=A{{T}^{2}}{{e}^{-b/T}}\] Þ \[\frac{J}{{{T}^{2}}}\propto {{e}^{-b/T}}\] i.e. \[\frac{J}{{{T}^{2}}}\] will vary exponentially with \[\frac{1}{T}\], having negative slope.


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