X-rays fall on a photosensitive surface to cause photoelectric emission. Assuming that the work function of the surface can be neglected, find the relation between the de-Broglie wavelength (`lambda`) of the electrons emitted to the energy (`E_(v)`) of the incident photons. Draw the nature of the graph for `lambda` as a function of `E_(v)`.

Explanation

`E_(v)=phi_(0)+K_("max")`
As `phi_(0)=0`
`implies E_(v)=K_("max")`
`implies K_("max")=(p^(2))/(2m)=E_(v)`
`implies p=sqrt(2mE_(v))`
`:.` Wavelength `(lambda)` of emitted electrons ,
`lambda=(h)/(p)=(h)/(sqrt(2mE_(v)))`