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

Explanation

For H - like particles in general `bar(v) = (2pi^92) m Z^(2) e^(4))/(ch^(3)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
`:.` For `He^(+)` spectrum, for Balmer transition, n = 4 to n = 2
`bar(v) = (1)/(lamda) = RZ^(2) ((1)/(2^(2)) - (1)/(4^(2))) = R xx 4 xx (3)/(16) = (3R)/(4)`
For hydrogen spectrum `bar(v) = (1)/(lamda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = (3)/(4) R or (1)/(n_(1)^(2)) - (1)/(n_(2)^(2)) = (3)/(4)`
which can be so for `n_(1) = 1 and n_(2) = 2`, i.e., the transition is form n = 2 to n = 1