JEE Main 2025 Jan 29 Shift 2, Physics Q4: Introduction and Photoelectric Effect

Einstein's photoelectric equation: $\frac{hc}{\lambda }=W+K_{\max }$, where $W$ is the work function and $K_{\max }$ is the maximum kinetic energy of the emitted photoelectrons.

Stopping potential $V_S$ is defined so that $e{V}_S=K_{\max }$, hence $V_S=\frac{K_{\max }}{e}$, i.e. $\left(\frac{1}{e}\right)$ times $K_{\max }$.

(1) is wrong: $V_S$ decreases with increasing $\lambda$ since $K_{\max }=hc/\lambda -W$. (2) and (4) are wrong: $V_S$ is independent of intensity.

$\therefore$ Option (3) is correct.