Q 12-11-057JEE MainJEE Main 2025 (29 Jan, Shift 2)Easy
In an experiment with the photoelectric effect, the stopping potential
Answer: (D) is $\left(\dfrac{1}{e}\right)$ times the maximum kinetic energy of the emitted photoelectrons
The stopping potential just stops the fastest electrons: $eV_0 = K_{max}$, so $V_0 = \dfrac{K_{max}}{e}$.
It does not depend on intensity, and since $eV_0 = \dfrac{hc}{\lambda} - \phi$ it **decreases** as wavelength increases.
Solution by Sreeraj P, M.Sc Physics