What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition `n = 4` to `n = 2` of `He^(o+)` spectrum ?

Explanation

For `He^+` spectrum,
`1/lambda = RZ^2 [1/(n_1^2) - 1/(n_2^2)]`
Now, `n_1 = 2, n_2 = 4 and Z = 2`
`1/lambda = R xx (2)^2 [1/(2^2) - 1/(4^2)] = 3/4 R " " …(i)`
For H atom `1/lambda = R[1/(n_1^2) - 1/(n_2^2)]" " …..(ii)`
Since `lambda`, is the same, equating equations (i) and (ii)
`1/(n_1^2) - 1/(n_2^2) = 3/4`
Now, if `n_1 = 1 and n_2 = 2`.
Therefore, the transition from, `n = 2` to `n = 1` in H atom will have the same wavelength as the transition form `n = 4` to `n = 2` in `He^+`.