Q. Maximum velocity of the photoelectrons emitted by a metal surface is $1.2 \times 10^{6}\ {ms}^{- 1}$. Assuming the specific charge of the electron to be $1.8{\times 10}^{11}\ C\ {kg}^{- 1},\ $the value of the stopping potential in volt will be
A) 2
B) 3
C) 4
D) 6
Explanation
(c) Specific charge on electron,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{e}{m} = 1.8 \times 10^{11}\ C\ {Kg}^{- 1}$
Maximum kinetic energy of photoelectron
$\frac{1}{2}mv_{\max}^{2} = {eV}_{s}\ $
Where $V_{s}$ is the stopping potential.
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{s} = \frac{mv_{\max}^{2}}{2e} = \frac{v_{\max}^{2}}{2(e/m)}$
= $\frac{\left( 1.2 \times 10^{6} \right)^{2}}{2 \times 1.8 \times 10^{11}}\ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.4 \times 10 = 4V\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.