An e-m wave of wavelength `lambda` is incident on a photo sensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength `lambda_(1)`. Find relation between `'lambda'` and `'lambda_(1)`'-

Explanation

The frequency of ultraviolet radiation is greater than the threshold frequency for zinc while the frequency of red light is less than this threshold frequency, so ultraviolet radiation can cause emission of electrons from zinc surface while red light cannot.
As the work function of the metal can be neglected, so K.E. of emitted electron = Energy of X-ray photon.
`(1)/(2) mv^(2) = hv`
`(p^(2))/(2m)=(hc)/(lambda) or p=sqrt((2mhc)/(lambda))`
de-Broglie wavelength of emitted electrons,
`lambda_(1)=(h)/(p)=(h)/(sqrt((2mhc)/(lambda))) or lambda_(1) = sqrt((h lambda)/(2mc)) or lambda_(1)^(2)=(h lambda)/(2mc) rArr lambda=((2mc)/(h))lambda_(1)^(2)`.