Q. Light of wavelength 4000 $\mathring{\mathrm{A}}$ is incident on a metal plate whose work function is 2 eV. The maximum KE of the emitted photoelectron would be
A) 0.5 eV
B) 1.1 eV
C) 1.5 eV
D) 2.0 eV
Explanation
(b) Maximum KE$= \frac{hc}{\lambda} - \phi_{0}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{400 \times 10^{- 10}} \times \frac{1}{1.6 \times 10^{- 19}} - 2 = 1.1\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.