J & K CET Medical J & K - CET Medical Solved Paper-2010

  • question_answer
    The mobility of free electrons (charge = e, mass = m and relaxation time = x) in a metal is proportional to

    A) \[\frac{e}{m}\tau \]                                       

    B) \[\frac{m}{e}\tau \]

    C) \[\frac{e}{m\tau }\]                                       

    D) \[\frac{m}{e\tau }\]

    Correct Answer: A

    Solution :

                     Drift velocity per unit electric field is called mobility of electron, ie, \[\mu =\frac{{{v}_{d}}}{E}=\frac{E}{\rho neE}\] Now,                     \[\rho =\frac{m}{n{{e}^{2}}\tau }\] \[\therefore \]  \[\mu =\frac{n{{e}^{2}}\tau }{mne}=\frac{e\tau }{m}\]


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