If _0 and be threshold wavelength and wavelength of incident ligh…

MCQ

If $\lambda_0$ and $\lambda $ be threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is

A
$\sqrt {\frac{{2h}}{m}\left( {{\lambda _0} - \lambda } \right)} $
B
$\sqrt {\frac{{2hc}}{m}\left( {{\lambda _0} - \lambda } \right)} $
✓
$\sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} \right)} $
D
$\sqrt {\frac{{2h}}{m}\left( {\frac{1}{{{\lambda _0}}} - \frac{1}{\lambda }} \right)} $

✓

Answer

Correct option: C.

$\sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} \right)} $

c
The kinetic enegry of the ejected electron is given by the equation

$hv = h{v_0} + \frac{1}{2}m{v^2}\,\,\,\because \,v = \frac{c}{\lambda }$

or $\frac{{hc}}{\lambda } = \frac{{hc}}{{{\lambda _0}}} + \frac{1}{2}m{v^2}$

$\frac{1}{2}m{v^2} = \frac{{hc}}{\lambda } - \frac{{hc}}{{{\lambda _0}}}$

$ = hc\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} \right)$

$\therefore \,\,{v^2} = \frac{{2hc}}{m}\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} \right)$

or $v = \sqrt {\frac{{2hc}}{m}\left( {\frac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}}} \right)} $

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