Red light, however bright it is, cannot produce the emission of electrons from a clean zinc surface. But even weak ultraviolet radiation can do so. Why?
X-ray of wavelength ‘λ’ fall on a photosensitive surface, emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-Broglie wavelength of the electrons emitted will be $\sqrt{hλ / 2 mc} .$

Solution

For any material to emit photoelectrons, the threshold frequency has to be greater than or comparable to the frequency of incident radiation. Therefore, the frequency of incident light affects the emission of photoelectrons whereas, intensity does not.
Here, the threshold frequency of zinc surface is not greater than that of the frequency of red light. Hence, no emission of electrons take place.

The kinetic energy of photoelectron is given by,
   $\frac{1}{2} {mv}^{2}_{\max} = hv - ϕ_{0}$
$\Rightarrow$ $\frac{1}{2} {mv}^{2}_{\max} \approx hv = \frac{hc}{λ} (Neglecting the term for work function)$

$\Rightarrow$    $m^{2} {v^{2}}_{\max} = \frac{2 mhc}{λ}$
$\Rightarrow$    $p = {mv}_{\max} = \sqrt{\frac{2 mhc}{λ}}$
But de-Broglie wavelength is given by,

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 mhc / λ}} = \sqrt{\frac{hλ}{2 mc}} .$
Hence, proved.

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