The electron in the n ^ th orbit of Li ^ 2 + is excited to ( n + 1...

The difference energy between two energy levels is

$∆ E = 2 . 18 \times {10}^{- 18}\left[1{n}_{1}^{2} - 1{n}_{2}^{2}\right]J$

$1 . 47 \times {10}^{- 17} = 2 . 18 \times {10}^{- 18} \times ( 9 ) \left(\dfrac{1}{( n{)}^{2}} - \dfrac{1}{( n + 1{)}^{2}}\right)$

$0 . 749 = \dfrac{1}{{n}^{2}} - \dfrac{1}{( n + 1{)}^{2}}$

$n = 1$

Hence, the transition is taking place between $1\overset{}{\to}2$, and it comes under Lymann series.

If we take $0 . 749 ≃ 0 . 75 = \dfrac{3}{4}$