The stopping potential of photo-electrons depends upon:

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

By Einstein's photoelectric equation, the stopping potential measures the maximum kinetic energy of the photoelectrons.

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

\( V_0 \) depends on the frequency \( \nu \) and on the work function \( \phi_0 \), which is a property of the cathode material. Intensity does not appear.

Option (1) Frequency of incident light only — Incorrect. The work function term also changes \( V_0 \) when the cathode material is changed.

Option (2) material of the cathode only — Incorrect. For a given cathode, \( V_0 \) rises linearly with the frequency.

Option (3) both the frequency of the incident light and the material of the cathode. — Correct. Both \( \nu \) and \( \phi_0 \) appear in the equation.

Option (4) intensity of the incident light. — Incorrect. Intensity changes the number of photoelectrons (the saturation current), not their maximum kinetic energy, so the stopping potential is unchanged.

NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Using Eq. (11.1), the photoelectric equation, Eq. (11.2), can be written as

    \( \displaystyle eV_0 = h\nu - \phi_0;\ \text{for } \nu \ge \nu_0 \quad \text{or} \quad V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \qquad (11.4) \)

This is an important result. It predicts that the V₀ versus ν curve is a straight line with slope = (h/e), independent of the nature of the material.”

Electron Emission – “… The work function (φ₀) depends on the properties of the metal and the nature of its surface.”