A photosensitive surface is illuminated by a point source of light 1 m away. When the source is shifted to 2 m, then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → number of emitted electrons is a quarter of the initial number
For a point source the intensity falls as \( 1/r^{2} \), so doubling the distance makes it one quarter. In the photoelectric effect the number of photoelectrons emitted per second is proportional to the intensity, while the maximum kinetic energy of each electron depends only on the frequency of the light (\( K_{\max} = h\nu - \phi \)), not on the intensity or the distance.
Option (1) each emitted electron carries one eighth of the initial energy. — Incorrect. The energy of each electron depends on the frequency, which has not changed.
Option (2) number of emitted electrons is half the initial number — Incorrect. The intensity falls to one quarter, not one half, so the number of electrons is not halved.
Option (3) each emitted electron carries half the initial energy — Incorrect. The energy of each electron stays the same, because the frequency is unchanged.
Option (4) number of emitted electrons is a quarter of the initial number — Correct. The intensity becomes \( (1/2)^{2} = 1/4 \) of its value, so the number of emitted electrons becomes a quarter.
NCERT (2026-27):
Experimental Study of Photoelectric Effect – “… The intensity of light is varied by changing the distance of the light source from the emitter.”
Effect of Intensity of Light on Photocurrent – “… This implies that the number of photoelectrons emitted per second is directly proportional to the intensity of incident radiation.”
Effect of Potential on Photoelectric Current – “… In other words, the maximum kinetic energy of photoelectrons depends on the light source and the emitter plate material, but is independent of intensity of incident radiation.”