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) $\lambda = \left(\dfrac{2mc}{h}\right)\lambda_d^2$
Concept. With zero work function, the photon energy becomes the electron's KE $= \dfrac{h^2}{2m\lambda_d^2}$.
Step 1. Equation. $\dfrac{hc}{\lambda} = \dfrac{h^2}{2m\lambda_d^2}$.
Step 2. Solve. $\lambda = \left(\dfrac{2mc}{h}\right)\lambda_d^2$.
Answer. $\lambda = \left(\dfrac{2mc}{h}\right)\lambda_d^2$ (Option D).
Solution by Sreeraj P, M.Sc Physics