Q. Maximum velocity of photoelectron emitted is $4.8\ ms^{- 1}$. The $e/m$ ratio of electron is $1.76 \times 10^{11}\ Ckg^{- 1}$, then stopping potential is given by
A) $5 \times 10^{- 10}\ JC^{- 1}$
B) $3 \times 10^{- 7}\ JC^{- 1}$
C) $7 \times 10^{- 11}\ JC^{- 1}$
D) $2.5 \times 10^{- 2}\ JC^{- 1}$
Explanation
(c) $eV_{s} = \frac{1}{2}mv_{m}^{2}\ \ or\ V_{s} = \frac{mv_{m}^{2}}{2e} = \frac{v_{m}^{2}}{2(e/m)}$
$= \frac{(4.8)^{2}}{2 \times 17.6 \times 10^{11}} = 7 \times 10^{11}\ JC^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.