Question

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 $0.5\, V$, respectively. If the same source is placed $0.4 m$ away from the photoelectric cell, then the saturation current is $x$ milliampere and the stopping potential is $y$ volt. Find $\frac{x}{y}$.

Answer 1

The cut-off voltage or stopping potential measures 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, $I=\frac{12}{4}=3 \,mA$