JEE Physics 2025 Question with Solution

Given: The question asks for the correct statement about stopping potential in the photoelectric effect.

Find: The correct relation for the stopping potential.

In the photoelectric effect, the energy relation is

$$E = h\nu - W$$

where $h\nu$ is the energy of the incident photons and $W$ is the work function.

The stopping potential $V_s$ is related to the maximum kinetic energy of emitted photoelectrons by

$$eV_s = K_{\text{max}}$$

Rearranging,

$$V_s = \frac{K_{\text{max}}}{e}$$

Therefore, the stopping potential is $\frac{1}{e}$ times the maximum kinetic energy of the emitted photoelectrons.

The correct option is C.

Given: Light of frequency greater than the threshold frequency falls on a metal surface in the photoelectric effect.

Find: What the stopping potential represents.

The emitted electrons have a range of kinetic energies. The most energetic electrons have maximum kinetic energy $K_{\text{max}}$.

The stopping potential is the minimum negative potential applied to stop even these most energetic photoelectrons. At this potential, the photocurrent becomes zero.

Using energy conservation,

$$K_{\text{max}} = eV_0$$

Hence,

$$V_0 = \frac{K_{\text{max}}}{e}$$

So the stopping potential is numerically equal to the maximum kinetic energy per unit charge.

Therefore, the correct option is C.