Q 12-11-159JEE MainJEE Main 2019 (10 Apr, Shift 1)Easy
In a photoelectric effect experiment, the threshold wavelength of light is $380\ \text{nm}$. If the wavelength of incident light is $260\ \text{nm}$, the maximum kinetic energy of emitted electrons will be (Given $E\,(\text{in eV}) = \dfrac{1237}{\lambda\,(\text{in nm})}$)
Answer: (C) $1.5\ \text{eV}$
$$K_{max} = \frac{1237}{260} - \frac{1237}{380} = 4.76 - 3.26 \approx 1.5\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics