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NEET NEET SOLVED PAPER 2016 Phase-I

  • question_answer
    When a metallic surface is illuminated with radiation of wavelength \[\lambda ,\] the stopping potential is V. If the same surface is illuminated with radiation of wavelength \[2\lambda ,\] the stopping potential is \[\frac{V}{4}.\] The threshold wavelength for the metallic surface is :-                                                                                                           

    A)  \[4\lambda \]                                  

    B)  \[5\lambda \]

    C)   \[\frac{5}{2}\lambda \]                                               

    D)   \[3\lambda \]

    Correct Answer: D

    Solution :

                     \[eV=\frac{hc}{\lambda }-\frac{hc}{{{\lambda }_{0}}}\]                                                ?(i)                 \[eV/4=\frac{hc}{2\lambda }-\frac{hc}{{{\lambda }_{0}}}\]                                           ?(ii) From equation (i) and (ii) \[\Rightarrow \]               \[4=\frac{\frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}}}{\frac{1}{2\lambda }-\frac{1}{{{\lambda }_{0}}}}\]       On solving \[{{\lambda }_{0}}=3\lambda \]


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