Q. A light having wavelength 300 nm falls on a metal surface. Work function of metal is 2.54 eV. What is stopping potential?
A) 2.3 V
B) 2.59 V
C) 1.60 V
D) 1.29 V
Explanation
(c) KE of photoelectron =$\ hv - W_{0}$
$= \ \frac{hc}{\gimel} - W_{0}$
$= \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}} - 2.54 \times 1.6 \times 10^{- 19}$
$= 6.63 \times 10^{- 19} - 4.064 \times 10^{- 19}$
$= \ 2.566 \times 10^{- 19}\ eV$
If $V_{0}$ is the stopping potential, then KE of photoelectron =$\ eV_{0}$
or $V_{0} = \frac{KE\ of\ photo\ electron\ }{e}$
= $\frac{2.566 \times 10^{- 19}}{1.6 \times 10^{- 19}} = 1.60V\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.