X-rays of wavelength `lambda` fall on photosenstive 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((hlambda)/(2mc))`.

Explanation

To prove that the de Broglie wavelength of electrons emitted from a photosensitive surface when X-rays of wavelength \( \lambda \) fall on it is given by

\[ \lambda_{dB} = \sqrt{\frac{h \lambda}{2mc}}, \]

we will follow these steps:

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