Q 12-11-170JEE MainJEE Main 2019 (12 Apr, Shift 1)Easy
The stopping potential $V_0$ (in volt) as a function of frequency $(\nu)$ for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be: (Given: Planck's constant $h = 6.63\times10^{-34}$ J s, electron charge $e = 1.6\times10^{-19}$ C)
Answer: (D) $1.66$ eV
The line meets $V_0 = 0$ at the threshold frequency $\nu_0 = 4\times10^{14}$ Hz.
$$\phi = h\nu_0 = \frac{6.63\times10^{-34}\times4\times10^{14}}{1.6\times10^{-19}}\ \text{eV} \approx 1.66\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics