JCECE Medical JCECE Medical Solved Paper-2009

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


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