For hydrogen like atom.

E_{n} \(\propto\) -\(\frac{Ƶ^2}{n^2}\)

E_{n} = -k\(\frac{Ƶ^2}{n^2}\)

for hydrogen atom-

E_{4} = -k\(\frac{(1)^2}{4^2}\)

E_{4} = \(\frac{-k}{16}\)

given, E_{4} = €

\(\because\) k = -€ x 16**----(1)**

For Li^{+2}

E_{1} = \(-\frac{k(3)^2}{1^2}\)

= -k x 9

using equation** (1) **we get

E_{1} = € x 16 x 9

E_{1} = 144 €