Q. The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is
A) Double the earlier value
B) Unchanged
C) More than doubled
D) Less than doubled
Explanation
(c) Let $E_{1}$ and $E_{2}$ be the KE of photoelectrons for incident light of frequency $v$ and $2v$ respectively.
Then $hv = E_{1} + \phi$ and $h2v = E_{2} + \phi_{0}$
So, $2\left( E_{1} + \phi_{0} \right) = E_{2} + \phi_{0}$ or $E_{2} = 2E_{1} + \phi_{0}$
It means the KE of photoelectron becomes more than double
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.