A cesium photocell with a steady potential difference of 60 V across it is illuminated by a small bright light placed 1 m away. When the same light is placed 2m away, the electrons crossing the photocell -

are one-quarter as numerous

$i∝\frac{1}{r^2}$

$\frac{i_2}{i_1}=(\frac{r_1}{r_2})^2=(\frac{1}{2})^2=\frac{1}{4}$

$i_2=\frac{i_1}{4}$

$n_2=\frac{n_1}{4}$