Q 12-11-090JEE MainJEE Main 2023 (10 Apr, Shift 2)Medium
The variation of stopping potential $V_0$ as a function of the frequency ($\nu$) of the incident light for a metal is shown in figure. The work function of the surface is
Answer: (B) $2.07\ \text{eV}$
The threshold frequency is where $V_0=0$: $\nu_0=5\times10^{14}\ \text{Hz}$.
$$\phi=h\nu_0=\frac{6.63\times10^{-34}\times5\times10^{14}}{1.6\times10^{-19}}\approx2.07\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics