For hydrogen like atom.
En \(\propto\) -\(\frac{Ƶ^2}{n^2}\)
En = -k\(\frac{Ƶ^2}{n^2}\)
for hydrogen atom-
E4 = -k\(\frac{(1)^2}{4^2}\)
E4 = \(\frac{-k}{16}\)
given, E4 = €
\(\because\) k = -€ x 16----(1)
For Li+2
E1 = \(-\frac{k(3)^2}{1^2}\)
= -k x 9
using equation (1) we get
E1 = € x 16 x 9
E1 = 144 €