Q. When a monochromatic point source of light is at a distance 0.2 m from a photoelectric cell, the saturation current and cut-off voltage are 12.0 mA and 0.5 V. If the same source is placed 0.4 m away from the photoelectric cell, then the saturation current and the stopping potential respectively are
A) 4 mA and 1V
B) 12 mA and 1V
C) 3 mA and 1V
D) 3 mA and 0.5 V
Explanation
(d) The cut-off voltage or stopping potential measure maximum kinetic energy of the electron. It depends on the frequency of incident light whereas the current depends on the number of photons incident. Hence, cut-off voltage will be 0.5 V. Now by inverse square law,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 12 \propto \frac{1}{(0.2)^{2}}\ \ or\ I \propto \frac{1}{(0.4)^{2}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{I}{12} = \frac{(0.2)^{2}}{(0.4)^{2}} = \frac{1}{4}$
or $I = \frac{12}{4} = 3mA\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.