Hint to find the same
We are required to find the energy required to excite doubly ionized lithium. We know that,
En = –13.6 Z2 / n2
Hence, the excitation energy = ΔE = E3 – E1 = –13.6 × (3)2 [1/32 – 1/12]
= +13.6 × (9) [1–1/9] = 13.6 × (9) (8/9) = 108.8 eV.
Wavelength λ = hc / ΔE
