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MGIMS WARDHA MGIMS WARDHA Solved Paper-2008

  • question_answer
    When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, light of wavelength 600 nm is sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

    A) 1 : 2                                       

    B) 2 : 1

    C) 4 : 1                                       

    D) 1 : 4

    Correct Answer: B

    Solution :

    Work function is given by \[\phi =\frac{hc}{\lambda }\] Or           \[\phi \propto \frac{1}{\lambda }\] \[\because \]     \[\frac{{{\phi }_{1}}}{{{\phi }_{2}}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{600}{300}=\frac{2}{1}\]


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