AFMC AFMC 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 \[\text{o }\!\!|\!\!\text{ }\,\text{=}\frac{hc}{\lambda }\]or     \[\text{o }\!\!|\!\!\text{ }\,\propto \frac{1}{\lambda }\] \[\because \]      \[\frac{\text{o }\!\!|\!\!\text{ }{{\,}_{1}}}{\text{o }\!\!|\!\!\text{ }{{\,}_{2}}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{600}{300}=\frac{2}{1}\]


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