One of the lines in the emission spectrum of `Li^(2 +)` has the same wavelength as that of the second line of Balmer series in hydrogen spectrum. The electronic transition correspnding to this line is.

Explanation

`(1)/(lambda_(He^(+))) = (1)/(lambda_(Li^(+)))`
`RZ^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))] = RZ^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`2^(2) [(1)/(2^(2))-(1)/(4^(2))] = 3^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`4[(1)/(4)-(1)/(16)] = 9 [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`(4)/(9) xx (3)/(16) = [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`:. (1)/(n_(1)^(2)) -(1)/(n_(2)^(2)) = (1)/(12) n_(1) = 3, n_(2) = 6`
`(3 rarr 6)` will have same wavelength as that of second line of Balmer series in `He^(+)` spectrum.