Q 12-11-048JEE MainJEE Main 2025 (23 Jan, Shift 2)Easy
In the photoelectric effect, an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is $2.14\ \text{eV}$ and the stopping potential is $2\ \text{V}$, what is the wavelength of the em-wave? (Given $hc = 1242\ \text{eV nm}$, where $h$ is Planck's constant and $c$ is the speed of light in vacuum.)
Answer: (A) $300\ \text{nm}$
$$\frac{hc}{\lambda} = \phi + eV_0 = 2.14 + 2 = 4.14\ \text{eV}$$
$$\lambda = \frac{1242}{4.14} = 300\ \text{nm}$$
Solution by Sreeraj P, M.Sc Physics