Q. Photons of energy of 6 eV are incident on a metal surface whose work function is 4 eV. The minimum kinetic energy of the emitted photoelectrons will be
A) Zero
B) 1 eV
C) 2 eV
D) 10 eV
Explanation
(a) $\frac{1}{2}mv^{2} = hv - \phi_{0} = hv - hv_{0}$
For minimum kinetic energy of emitted photoelectron,
$v = v_{0}$
$\therefore\ \frac{1}{2}mv^{2} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.