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JCECE Engineering JCECE Engineering 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                 \[\phi =\frac{hc}{\lambda }\] or            \[\phi \propto \frac{1}{\lambda }\] \[\because \]     \[\frac{{{\phi }_{1}}}{{{\phi }_{2}}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{600}{300}=\frac{1}{2}\]


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