BCECE Medical BCECE Medical Solved Papers-2010

  • question_answer
    Light of wavelength \[\lambda \], strikes a photoelectric surface and electrons are ejected with an energy E. If E is to be increased to exactly twice its original value, the wavelength changes to\[\lambda \], where

    A)  \[\lambda \] is less than \[\frac{\lambda }{2}\]

    B)  \[\lambda \] is greater than \[\frac{\lambda }{2}\]

    C)  \[\lambda \] is greater than \[\frac{\lambda }{2}\] but less than A-

    D)  \[\lambda \] is exactly equal to \[\frac{\lambda }{2}\]

    Correct Answer: C

    Solution :

    Energy of photoelectron                 \[E=\frac{hc}{\lambda }-W\] \[\Rightarrow \] \[\frac{hc}{\lambda }=E+W\] ... (i) where W is the work function for the metal surface (constant).                 \[2E=\frac{hc}{\lambda }-W\] \[\Rightarrow \] \[\frac{hc}{\lambda }=2E+W\] Dividing Eq. (i) by Eq. (ii), we get                 \[\frac{\lambda }{\lambda }=\frac{E+W}{2E+W}\]                 \[\frac{\lambda }{\lambda }=\frac{(E+W)}{2\left( E+\frac{W}{2} \right)}\] \[\therefore \] \[\frac{\lambda }{\lambda }>\frac{1}{2}\] or            \[\lambda >\frac{\lambda }{2}\] \[\therefore \] \[\lambda >\lambda >\frac{\lambda }{2}\]


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