Q 12-11-164JEE MainJEE Main 2019 (9 Apr, Shift 1)Easy
The electric field of light wave is given as $\vec E = 10^{-3}\cos\left(\dfrac{2\pi x}{5\times10^{-7}} - 2\pi\times6\times10^{14}t\right)\hat x\ \dfrac{\text{N}}{\text{C}}$. This light falls on a metal plate of work function $2\ \text{eV}$. The stopping potential of the photo-electrons is (Given, $E\ (\text{in eV}) = \dfrac{12375}{\lambda\ (\text{in Å})}$)
Answer: (D) $0.48\ \text{V}$
From the phase, $\lambda = 5\times10^{-7}\ \text{m} = 5000\ \text{Å}$, so the photon energy is
$$E = \frac{12375}{5000} = 2.475\ \text{eV}$$
$$eV_0 = E - \phi = 2.475 - 2 = 0.475\ \text{eV} \Rightarrow V_0 \approx 0.48\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics