Correct Answer - 12
Since electron goes the state where the path length is 5 times de-broglie wavelength
`implies 2pir=5lambda`
Also `(2pir)/n=lambda implies n=5`
Hence electron goes the `5^(th)` state.
2nd highest energy line will be `4to1`
If this photon is used for `Li^(+2)` then
`13.6(1-1/4^2)=13.6xxZ^2(1/n_1^2-1/n_2^2)`
`n_1" " n_2`
`implies 3to12` Hence final excited state =12