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 Balmer transition, n=4 to n=2, for `He^(+)` spectrum, the Rydberg equation is,
`overline(v)=(1)/(lamda)=RZ^(2)((1)/(2^(2))-(1)/(4^(2)))=Rxx2^(2)xx(3)/(16)=(3R)/(4)" "[because" for He, Z=2"]`
For hydrogen spectrum,
`overline(v)=(1)/(lamda)=RZ^(2)((1)/(n^(2))-(1)/(n_(2)^(2)))=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))" "[because"For H, Z=1"]`
According to the question,
`R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=(3R)/(4)`
or `((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=(3)/(4)` . . . (1)
The equation (1) will be true if n=1 and n=2.
`therefore` For hydrogen spectrum transition is from n=2 to n=1.