If an em wave of wavelength `lambda` is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength `lambda_1`, prove that
`lambda=((2mc)/h)lambda_1^2`

Options

A. `sqrt(((2mc)/(h)))lamda_(1)`
B. `sqrt((h)/(2mc))xxlamda_(1)`
C. `((2mc)/(h))xxlamda_((2)/(1))`
D. None

Answer

C

Explanation

As K.E. of emitted `e^(-)` = energy of x - Ray
`(1)/(2)mV^(2)=hv`
`(P^(2))/(2m)=(hc)/(lamda)`
`thereforeP=sqrt((2mhc)/(lamda))`
`thereforelamda_(1)=(h)/(P)= (h)/(sqrt((2mhc)/(lamda)))=sqrt((hlamda)/(2mc))`
`or lamda_(1)^(2)=(hlamda)/(2mc)`
or `lamda=((2mc)/(h))lamda_(1)^(2)`