A cesium photocell, with a steady potential difference of 60 V across, is illuminated by a bright point source of light 50 cm away. When the same light is placed 1 m away, the photoelectrons emitted from the cell -
Answer & explanation
Correct answer: option 1
d1 = 50 cm
d2 = 1m = 100 cm
$d_1=2d_1(I∝\frac{1}{d^2})$
I2 = I1/4
$n_{ph_2}=\frac{n_{ph_1}}{4}$
$n_{e_2}=\frac{n_{e_1}}{4}$