X-rays of wavelength lambda fall onphotosensitive 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)).

X-rays of wavelength lambda fall onphotosensitive surface, emitting el

1.	X-rays of wavelength lambda fall onphotosensitive 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)).
Answer» Solution :`E=(HC)/lambda=(P^2)/(2m) THEREFORE P=sqrt((2mnc)/(lambda)),lambda_e=h/Psqrt((hlambda)/(2MC))`

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X-rays of wavelength lambda fall onphotosensitive 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)).

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