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

Explanation

`E=(hc)/lambda=(P^2)/(2m) therefore P=sqrt((2mnc)/(lambda)),lambda_e=h/Psqrt((hlambda)/(2mc))`