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Question

X- rays of wavelength λ falls 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 hλ2mc

Solution
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Energy of the X-rays of wavelength=hcλ
As work function is neglected,
12mev2e=hcλ
v2e=2hcmeλ
ve=2hcmeλ
De Brogile wavelength =hmeve
De Brogile wavelength =hmλme2hc
=hλ2mec

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