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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2006

  • 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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