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Dual Nature of Radiation and Matter — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsDual Nature of Radiation and Matter2 Marks
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, Ek = 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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