Q. When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, light of wavelength 600 nm is sufficient for liberating photoelectrons. The ratio of the work function of the two emitter is
A) 1:2
B) 2:1
C) 4:1
D) 1:4
Explanation
(b) Work function is given by
$\phi = \frac{hc}{\lambda}$
or $\phi \propto \frac{1}{\lambda}$
$\because\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\phi_{1}}{\phi_{2}} = \frac{\lambda_{2}}{\lambda_{1}} = \frac{600}{300} = \frac{2}{1}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.