When light of suitable frequency is incident on a photosensitive mate…

The correct answer is Option (4) → The saturation current will increase

By Einstein's photoelectric equation, the maximum kinetic energy depends only on the frequency of the light, while the intensity decides only the number of photons arriving per second.

    \( \displaystyle K_{\mathrm{max}} = h\nu - \phi_0 = eV_0 \)

More photons per second eject more photoelectrons per second, so the current at saturation rises; each photoelectron still gets the same maximum energy.

Option (1) The maximum kinetic energy of photoelectrons will increase — Incorrect. \( K_{\mathrm{max}} = h\nu - \phi_0 \) does not contain the intensity, so it stays the same.

Option (2) The stopping potential will increase — Incorrect. The stopping potential is fixed by \( eV_0 = K_{\mathrm{max}} \), which does not change.

Option (3) The stopping potential will decrease — Incorrect. For the same reason the stopping potential does not decrease either.

Option (4) The saturation current will increase — Correct. The saturation current is proportional to the number of photoelectrons emitted per second, which is proportional to the intensity.

NCERT (2026-27):
Effect of Frequency of Incident Radiation on Stopping Potential – “… (ii) For a given photosensitive material and frequency of incident radiation, saturation current is found to be proportional to the intensity of incident radiation whereas the stopping potential is independent of its intensity (Fig. 11.3). (iii) … Above the threshold frequency, the stopping potential or equivalently the maximum kinetic energy of the emitted photoelectrons increases linearly with the frequency of the incident radiation, but is independent of its intensity (Fig. 11.5).”