Q. Light of wavelength 5000 $\mathring{\mathrm{A}}$ falls on a sensitive plate with photoelectric work functional of 1.9 eV. The kinetic energy of the photoelectron emitted will be
A) 0.58 eV
B) 2.48 eV
C) 1.24 eV
D) 1.16 eV
Explanation
(a) $E_{k} = \frac{hc}{\lambda} - \phi_{0}(in\ eV)$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5000 \times 10^{- 10} \times 1.6 \times 10^{- 19}} - 1.9$
$= 2.48 - 1.9 = 0.58\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.