Q. When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 V. I$\ $e/m for the electron is $1.8 \times 10^{11}\ C\ {Kg}^{- 1},$the maximum velocity the ejected electron is
A) $6 \times 10^{5\ }{ms}^{- 1}$
B) $8 \times 10^{5\ }{ms}^{- 1}$
C) $1.8 \times 10^{6}\ {ms}^{- 1}$
D) $1.8 \times 10^{5}\ \ {ms}^{- 1}$
Explanation
(c) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}mv_{\max}^{2} = eV_{0}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{\max} = \sqrt{2\left( \frac{e}{m} \right)V_{0}}$
$= \sqrt{2 \times 1.8 \times 10^{11} \times 9}$
$= \ 18 \times 10^{5}\ {ms}^{- 1}$
$\ = \ 1.8 \times 10^{6}\ {ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.