If an em wave of wavelength `lambda` is incident on a photosenstive 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`

Explanation

From solution of above problem, we have
`lambda_1=sqrt((hlambda)/(2mc)) or lambda_1^2=(hlambda)/(2mc) of lambda=((2mc)/h) lambda_1^2` (Proved)