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CET Karnataka Engineering CET - Karnataka Engineering 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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