Q 12-11-001NEETEasy
The energy of a photon of wavelength $400\ \text{nm}$ is approximately (take $hc = 1240\ \text{eV nm}$)
Answer: (B) $3.1\ \text{eV}$
$$E = \frac{hc}{\lambda} = \frac{1240}{400} = 3.1\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics