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

Explanation

We know that for `He ^(+)` spectrum
`(1)/(lamda ) = RZ ^(2) [ (1)/( n_(1) ^(2)) - (1)/( n _(2) ^(2)) ]`
Now `n_(2)= 2, n _(2) = 4 and Z =2`
`(1)/(lamda ) = R xx (2) ^(2) [ (1)/( 2 ^(2)) - (1)/( 4 ^(2))]" "...(i)`
`(1)/(lamda ) = R [ (1)/( n _(1) ^(2)) - (1)/( n _(2) ^(2)) ]" "...(ii)`
Since `lamda ` is same in both equations
`therefore `form n = 2 n =1 is - H-atom will have the same wavelength as the transition from n = 4 to n =2 in `He ^(+).`