X-rays of wavelength ' ' fall on a photosensitive surface, emitti… - Vidyadip

Question

$X-$rays of wavelength $\ '\lambda\ '$ 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 electrons emitted will be $\sqrt{\frac{\text{h}\gamma}{2\text{mc}}}.$

✓

Answer

$K.E.$ of electrons$, E_k =$ energy of $X-$ray photon $=\frac{\text{hc}}{\lambda}$
$\therefore$ de Broglie wavelength $,\lambda_\text{B}=\frac{\text{h}}{\sqrt{2\text{mE}_\text{k}}}.$
But $\text{E}_\text{k}=\frac{\text{hc}}{\lambda}$
$\therefore\ \lambda_\text{B}=\frac{\text{h}}{\sqrt{2\text{m}\big(\frac{\text{hc}}{\lambda}\big)}}$
$=\sqrt{\frac{\text{h}\lambda}{2\text{mc}}}$

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