If $λ_{0}$ and $λ$ are the threshold wavelength and the wavelength of the incident light respectively on a metal surface, the velocity of the photoelectron ejected from the metal surface is ( $m^{e}$ = mass of electron, h = Planck's constant, c = speed of light) :

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$\sqrt{\frac{2 h}{m_{e}} (\frac{1}{λ_{0}} - \frac{1}{λ})}$

$\sqrt{\frac{2 h (λ_{0} - λ)}{m_{e}}}$

$\sqrt{\frac{2 h c (λ_{0} - λ)}{m_{e}}}$

$\sqrt{\frac{2 h c}{m_{e}} (\frac{λ_{0} - λ}{λ λ_{0}})}$

answer is C.

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Detailed Solution

It is known that;

$\frac{h c}{λ} = \frac{h c}{λ_{0}} + \frac{1}{2} m_{e} v^{2}$

On solving $v = \sqrt{\frac{2 h c}{m_{e}} (\frac{λ_{0} - λ}{λ λ_{0}})}$ .

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