Question
In a photoemissive cell with executing wavelength $\lambda $, the fastest electron has speed $v.$ If the exciting wavelength is changed to $\frac{{3\lambda }}{4}$, the speed of the fastest emitted electron will be
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Answer
(d) $h\nu - {W_0} = \frac{1}{2}mv_{\max }^2$
$\Rightarrow \frac{{hc}}{\lambda } - \frac{{hc}}{{{\lambda _0}}} = \frac{1}{2}mv_{\max }^2$
$ \Rightarrow hc\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} \right) = \frac{1}{2}mv_{\max }^2$
$ \Rightarrow {v_{\max }} = \sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} \right)} $
When wavelength is $\lambda $ and velocity is $v$, then
$v = \sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} \right)} $…. $(i)$
When wavelength is $\frac{{3\lambda }}{4}$ and velocity is $v$’ then
$v' = \sqrt {\frac{{2hc}}{m}\left[ {\frac{{{\lambda _0} - (3\lambda /4)}}{{(3\lambda /4) \times {\lambda _0}}}} \right]} $….$(ii)$
Divide equation $(ii)$ by $(i)$, we get
$\frac{{v'}}{v} = \sqrt {\frac{{[{\lambda _0} - (3\lambda /4)]}}{{\frac{3}{4}\lambda {\lambda _0}}} \times \frac{{\lambda {\lambda _0}}}{{{\lambda _0} - \lambda }}} $
$v' = v{\left( {\frac{4}{3}} \right)^{1/2}}\sqrt {\frac{{[{\lambda _0} - (3\lambda /4)]}}{{{\lambda _0} - \lambda }}} $ i.e. $v' > v{\left( {\frac{4}{3}} \right)^{1/2}}$
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