Q 12-11-012NEETNEET 2021Top questionMedium
An electromagnetic wave of wavelength '$\lambda$' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength $\lambda_d$, then :
Answer: (D) $\lambda = \left(\dfrac{2mc}{h}\right)\lambda_d^2$
With negligible work function, all the photon energy becomes kinetic energy:
$$\frac{hc}{\lambda} = \frac{p^2}{2m} = \frac{h^2}{2m\lambda_d^2}$$
$$\lambda = \frac{2mc\lambda_d^2}{h} = \left(\frac{2mc}{h}\right)\lambda_d^2$$
Solution by Sreeraj P, M.Sc Physics