Q. When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 V. If $e/m$ for the electron is $1.8 \times 10^{11}\ Ckg^{- 1}$, the maximum velocity of ejected electrons is
A) $6 \times 10^{5}ms^{- 1}$
B) $8 \times 10^{5}ms^{- 1}$
C) $10^{6}ms^{- 1}$
D) $1.8 \times 10^{6}ms^{- 1}$
Explanation
(d) $eV = \frac{1}{2}mv^{2}$
or $v = \sqrt{\frac{2eV}{m}} = \sqrt{2 \times (1.8 \times 10^{11}) \times 9} = 1.8 \times 10^{6}ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.