Q 12-11-046JEE MainJEE Main 2025 (22 Jan, Shift 2)Easy
A light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of $2\ \text{eV}$. If the same surface is illuminated by a light source of wavelength $\lambda/2$, then the maximum kinetic energy of ejected electrons will be (the work function of the metal is $1\ \text{eV}$)
Answer: (D) $5\ \text{eV}$
With wavelength $\lambda$: $\dfrac{hc}{\lambda} = K + \phi = 2 + 1 = 3\ \text{eV}$.
Halving the wavelength doubles the photon energy to $6\ \text{eV}$:
$$K_{max} = 6 - 1 = 5\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics