Stopping potential for photoelectrons:

The correct answer is Option (3) → depends on both the frequency of the incident light and the nature of the material of cathode

By Einstein’s photoelectric equation the stopping potential \( V_0 \) measures the maximum kinetic energy of the photoelectrons.

    \( eV_0 = h\nu - \phi_0 \Rightarrow V_0 = \displaystyle\frac{h\nu}{e} - \frac{\phi_0}{e} \)

Option (1) does not depend on the frequency — Incorrect. \( V_0 \) increases linearly with the frequency \( \nu \).

Option (2) does not depend on the material of the cathode — Incorrect. The work function \( \phi_0 \) depends on the material, and \( V_0 \) decreases as \( \phi_0 \) increases.

Option (3) depends on both — Correct. Both \( \nu \) and \( \phi_0 \) appear in the expression.

Option (4) depends on the intensity — Incorrect. The intensity only changes the number of photoelectrons emitted per second, that is the photocurrent, and not their maximum energy.

NCERT (2026-27):
Summary – “… The stopping potential (\( V_0 \)) depends on (i) the frequency of incident light, and (ii) the nature of the emitter material. For a given frequency of incident light, it is independent of its intensity. The stopping potential is directly related to the maximum kinetic energy of electrons emitted: \( eV_0 = \frac{1}{2}mv_{\max}^{2} = K_{\max} \).”