MCQ
Light of wavelength $\lambda $ strikes a photo-sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$, the wavelength must be changed to $\lambda '$ where
  • A
    $\lambda ' = \frac{\lambda }{2}$
  • B
    $\lambda ' = 2\lambda $
  • $\frac{\lambda }{2} < \lambda ' < \lambda $
  • D
    $\lambda ' > \lambda $

Answer

Correct option: C.
$\frac{\lambda }{2} < \lambda ' < \lambda $
c
(c)$E = \frac{{hc}}{\lambda } - {W_0}$ and $2E = \frac{{hc}}{{\lambda '}} - {W_0}$

$ \ Rightarrow \frac{{\lambda '}}{\lambda } = \frac{{E + {W_0}}}{{2E + {W_0}}}$ 

$\Rightarrow \lambda ' = \lambda \left( {\frac{{1 + {W_0}/E}}{{2 + {W_0}/E}}} \right)$

Since $\frac{{(1 + {W_0}/E)}}{{(2 + {W_0}/E)}} > \frac{1}{2}$ so $\lambda ' > \frac{\lambda }{2}$

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