Q 12-11-153JEE MainJEE Main 2020 (4 Sep, Shift 1)Medium
The given figure shows a few data points in a photoelectric effect experiment for a certain metal, plotted as stopping potential against frequency. The minimum energy for ejection of electrons from its surface is: (Planck's constant $h = 6.62\times10^{-34}\ \text{J s}$)
Answer: (A) $2.27\ \text{eV}$
The stopping potential becomes zero at the threshold frequency, which from the graph (point $B$) is $f_0 = 5.5\times10^{14}$ Hz.
$$\phi = hf_0 = \frac{6.62\times10^{-34}\times5.5\times10^{14}}{1.6\times10^{-19}}\ \text{eV} \approx 2.27\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics