MGIMS WARDHA MGIMS WARDHA Solved Paper-2003

  • question_answer
    Light waves of wavelength \[\lambda \] is incident on a metal of work function \[\theta \] Maximum velocity of the electron is :

    A) \[{{\left[ \frac{2(h\lambda -\theta )}{m} \right]}^{1/2}}\]                            

    B) \[{{\left[ \frac{2(h\lambda -\lambda \theta )}{m\lambda } \right]}^{1/2}}\]

    C) \[\frac{2(hc-\lambda \theta )}{m}\]                       

    D) \[\left[ \frac{hc+\lambda \theta }{m\lambda } \right]\]

    Correct Answer: A

    Solution :

    For a photon of maximum velocity, \[\frac{1}{2}2{{\upsilon }^{2}}+\phi =\frac{hc}{\lambda }\] where \[\phi =\]work function of metal \[\frac{1}{2}m{{\upsilon }^{2}}=\frac{hc-\lambda \phi }{\lambda }\] Or           \[{{\upsilon }^{2}}=\frac{2(hc-\lambda \phi )}{m\lambda }\] Hence,   \[\upsilon ={{\left[ \frac{2(hc-\lambda \phi )}{m\lambda } \right]}^{1/2}}\]       


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