BHU PMT BHU PMT (Mains) Solved Paper-2008

  • question_answer
    The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency v of the incident radiation is doubled. The work function of this metal is

    A)  \[\frac{hv}{4}\]                                               

    B)  \[\frac{hv}{3}\]

    C)  \[\frac{hv}{2}\]                                               

    D)  \[\frac{2hv}{3}\]

    Correct Answer: D

    Solution :

                     \[\frac{\frac{1}{2}m{{v}^{2}}}{\frac{1}{2}m{{(2v)}^{2}}}=\frac{hv-\phi }{h(2v)-\phi }\] Or           \[\frac{1}{4}=\frac{hv-\phi }{2hv-\phi }\] Or           \[\phi =\frac{2hv}{3}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner