What hydrogen-like ion has the wavelength difference between the first lines of the Balmer Lyman series equal to `59.3nm`?

Explanation

For a spectral line,
`1/lambda = RZ^2 [1/(n_1^2) - 1/(n_2^2)]`
For Lyman line,
`1/(lambda_("Lyman")) = RZ^2 [1/(1^2) - 1/(2^2)] = (3RZ^2)/4`
Or `lambda_("Lyman") = 4/(3 RZ^2)`
For Balmer line
`1/(lambda_("Blamer")) = RZ^2 [1/(2^2) - 1/(3^2)] = (5 RZ^2)/(36)`
or `lambda_("Balmer") = (36)/(5RZ^2)`
`lambda_("Balmer") - lambda_("Lyman") = (36)/(5 RZ^2) - (4)/(3 RZ^2)`
`implies 59.3 xx 10^(-7) cm = 1/(RZ^2) [36/5 - 4/3]`
`implies 59.3 xx 10^(-7) = 1/(109677.8 Z^2) [(108 - 20)/(15)]`
`= 59.3 xx 10^(-7)`
or `Z^2 = 1/(109677.8) xx 88/(15 xx 59.3 xx 10^(-7)) = 9`
or `Z = 3`. This corresponds to `Li^(2+)` ion.