A photosensitive surface is illuminated by a point source of light 1 m away. When the source is shifted to 2 m, then

number of emitted electrons is a quarter of the initial number

The correct answer is Option (4) → number of emitted electrons is a quarter of the initial number

Given: A point source of light is moved from 1 m to 2 m away from a photosensitive surface.

Since light intensity varies inversely as the square of distance,

$I \propto \frac{1}{r^2}$

Therefore,

$\frac{I_2}{I_1} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

Photoelectric current (number of emitted electrons per second) ∝ Intensity.

Thus, when the source distance doubles, the number of emitted electrons becomes one-fourth of the original.

Energy of each emitted electron depends only on frequency of light, not on intensity.

Final Answer: number of emitted electrons is a quarter of the initial number.