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 `He^(+)` ion the wave number `(barv)` associated with the Balmer transition , n = 4 to n = 2 is given by .
`barv = (1)/(lambda) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
Where ,
`n_(1) = 2`
`n_(2) = 4`
Z = atomic number of helium
`barv = (1)/(lambda) = R (2)^(2) ((1)/(4) - (1)/(16))`
`= 4 R ((4 -1)/(16))`
`barv = (1)/(lambda) = (3R)/(4)`
`implies lambda = (4)/(3R)`
According to the question, the desired transition for hydrogen will have the same wavelength as that of `He^(+)`.
`implies R (1)^(2) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] = (3R)/(4)`
`[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] = (3)/(4) .......... (1)`
By hit and trail method, the equality given by equation (1) is true only when `n_(1) = 1 ` and `n_(2) = 2`
`therefore` The transition for `n_(2) = 2` to n= 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of `He^(+)` spectrum.