Energy of a photon corresponding to a wavelength of 600 nm is 2.08 eV. The energy of a photon of wavelength 400 nm will be:

3.12 eV

The correct answer is Option (2) → 3.12 eV

The energy of the photon is,

$E=hv=\frac{hc}{λ}$ [frequency = $v=\frac{c}{λ}$]

and,

$E_1.λ_1=E_2.λ_2$

$E_2=E_1\left(\frac{λ_1}{λ_2}\right)$

$=2.08×\frac{600}{400}=3.12eV$