Q. Light of wavelength λ falls on a metal having work function $\frac{hc}{\lambda_{0}}.$ Photoelectric effect will take place only if
A) $\lambda \geq \lambda_{0}$
B) $\lambda \geq 2\lambda_{0}$
C) $\lambda \leq \lambda_{0}$
D) $\lambda = 4\lambda_{0}$
Explanation
(c) The energy of photon
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \frac{hc}{\lambda}$
and work function of metal
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ W = \frac{hc}{\lambda_{0}}$
For photoelectric effect
$E > W$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{hc}{\lambda} > \frac{hc}{\lambda_{0}}$
$\because\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda \leq \ \lambda_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.