Q 12-11-064JEE MainJEE Main 2024 (6 Apr, Shift 1)Easy
In a photoelectric experiment, light of energy $2.48\ \text{eV}$ irradiates a photosensitive material. The stopping potential was measured to be $0.5\ \text{V}$. The work function of the photosensitive material is
Answer: (C) $1.98\ \text{eV}$
$\phi = h\nu - eV_0 = 2.48 - 0.5 = 1.98\ \text{eV}$.
Solution by Sreeraj P, M.Sc Physics