Q. If maximum velocity with which an electron can be emitted from a photo cell is $4 \times 10^{8}cm/s,$ the stopping potential is (mass of electron $= 9 \times 10^{- 31}kg$)
A) $30\ volt$
B) $45\ volt$
C) $59\ volt$
D) Information is insufficient
Explanation
(b) $v_{\max} = 4 \times 10^{8}\ cm/s = 4 \times 10^{6}\ m/sec$
$\therefore K_{\max} = \frac{1}{2}mv_{\max}^{2} = \frac{1}{2} \times 9 \times 10^{- 31} \times \left( 4 \times 10^{6} \right)^{2}$
$= 7.2 \times 10^{- 18}J = 45\ eV$
Hence, stopping potential $\left| V_{0} \right| = \frac{K_{\max}}{e} = \frac{45eV}{e} = 45volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.