A photo-emissive substance is illuminated with radiation of wavelength $\lambda_i$ so that it releases electrons with de Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectrons is $\lambda_0$. The expression for the de Broglie wavelength is: ($m$: mass of the electron, $h$: Planck's constant and $c$: speed of light)
Answer: (A) $\lambda_e = \sqrt{\dfrac{h}{2mc\left(\frac1{\lambda_i} - \frac1{\lambda_0}\right)}}$
The fastest electrons have $K = \dfrac{hc}{\lambda_i} - \dfrac{hc}{\lambda_0}$ (the work function is $hc/\lambda_0$). Their de Broglie wavelength is
$$\lambda_e = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mhc\left(\frac1{\lambda_i} - \frac1{\lambda_0}\right)}} = \sqrt{\frac{h}{2mc\left(\frac1{\lambda_i} - \frac1{\lambda_0}\right)}}$$
Solution by Sreeraj P, M.Sc Physics