Calculate the energy of excited state E1=Eg+7.7eV =10.5+7.7eV =18.2eV The energy of photon of wavelength 4960 A is calculated as, E=
hc
λ
=
12400
4960
=2.5eV Calculate the next possible spectral line emitting state energy. E2=E1−2.5eV =18.2−2.5eV =15.7eV Therefore, the energy of two states is 15.7eV and 18.2eV