In a photoelectric effect experiment, if the intensity of the incident light of frequency greater than the threshold frequency, is increased, then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → The number of photo electrons emitted by the surface in unit time will increase.
In the photon picture, each photoelectron is ejected by one photon of energy \( h\nu \). Raising the intensity at a fixed frequency only increases the number of photons per second, so more photoelectrons are emitted per second. Their maximum kinetic energy is fixed by Einstein’s photoelectric equation:
\( \displaystyle eV_0 = h\nu - \phi_0 \)
It has no intensity term, so the stopping potential depends only on the frequency and the metal.
Why the other options are incorrect
Option (1) Stopping potential for the emitted photo electrons will increase. — Incorrect. The stopping potential depends on the frequency, not the intensity.
Option (2) Photo electrons with higher energies will be emitted. — Incorrect. Brighter light gives more electrons, not faster ones.
Option (4) Stopping potential for the emitted photo-electrons will decrease. — Incorrect. The stopping potential does not change with intensity.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Note that the intensity of light of a given frequency is determined by the number of photons incident per second. Increasing the intensity will increase the number of emitted electrons per second. However, the maximum kinetic energy of the emitted photoelectrons is determined by the energy of each photon.”