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JEE Main & Advanced Sample Paper JEE Main Sample Paper-34

  • question_answer
    Light of wavelength \[\lambda \] strikes a photosensitive surface and electrons are ejected with kinetic energy E. If the KE is to be increased to 2E, the wavelength must be changed to \[\lambda '\] where

    A)  \[\lambda '=\lambda /2\]          

    B)  \[\lambda '=2\lambda \]

    C)  \[\lambda '>\lambda \]              

    D)  \[\frac{\lambda }{2}<\lambda '<\lambda \]

    Correct Answer: D

    Solution :

    \[E=\frac{hc}{\lambda }-\phi \]             \[2E=\frac{hc}{\lambda '}-\phi \]             \[2\left( \frac{hc}{\lambda }-\phi  \right)\,=\frac{hc}{\lambda '}-\phi \]             \[2\frac{hc}{\lambda }>\frac{hc}{\lambda '}\]             \[\lambda '\,>\frac{\lambda }{2}\] \[\because \] Energy of photoelectron is increased hence \[\lambda \] must be decreased.             i.e., \[\lambda '\,<\lambda \]


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