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. `lamda = ((2mc)/(h))lamda_1^2`
B. `lamda=(2mc)/(h) lamda_1`
C. `lamda=(h)/(2mc) lamda_1^2`
D. `lamda=sqrt((2mc)/(h)) lamda_1`

Answer

D

Explanation

de broglie wavelength of the photoelectron
`lamda_1 = (h)/(mv) = (h)/(sqrt(2mE)) = (h)/(sqrt((2mhc)/(lamda)))`
`because E = h (v - v_0) = hv = (hc)/(lamda)`
`therefore lamda_1^2 = (h^2lamda)/(2hmc) = (hlamda)/(2mc)`
`therefore lamda = ((2mc)/(h)) lamda_^1^2`