Q. The photoelectric threshold frequency of a metal is v. When light of frequency 4v is incident on the metal. The maximum kinetic energy of the emitted photoelectrons is
A) 4 hv
B) 3 hv
C) 5 hv
D) $\frac{5}{2}\ hv$
Explanation
(b) From Einstein’s photoelectric equation the maximum kinetic energy of photoelectrons emitted from metal surface is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{k} = hv_{1} - W$
Where W is work function of metal.
Given, $\ \ \ W = hv$ and $v_{1} = 4v$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{k} = 4hv - hv = 3hv$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.