The correct option is B Frequency of this light is 5×1014 s−1
Given,
Wavelength of the light = 600 nm
We know,
hν=E
∴Frequency ν=cλ=3×108600×10−9=5×1014 Hz
Again,
wave number −ν=1λ=1600×10−9=1.67×106 m−1
So energy of the photon,
E=hν=(6.6×10−34)(5×1014) J
=3.3×10−19 J
=3.3×10−19×6.2×1018 eV = 2.05 eV