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

Answer

`n = 2` to `n = 1`

Explanation

`barv = 1/lambda = R_H Z^2 (1/(n_1^2) - 1/(n_2^2))`
For `He^(+)` spectrum , `Z = 2, n_2 = 4, n_1 = 2`
`bar(v) = 1/lambda = R_H xx 4 (1/(2^2) - 1/(4^2)) = (3R_H)/(4)`
For hydrogen spectrum, `1/lambda = (3R_H)/4`
`:. bar(v) = 1/lambda = R_H xx 4 (1/(2^2) - 1/(4^2)) = (3R_H)/(4)`
For hydrogen spectrum, `1/lambda = (3R_H)/(4)`
`:. barv = 1/lambda = R_H xx 1 (1/(n_1^2) - 1/(n_2^2))`
`R_H (1/(n_1^2) - 1/(n_2^2)) = (3R_H)/(4)`
`1/(n_1^2) - 1/(n_2^2) = 3/4`
`:. n_1 = 1, n_2 = 2`
Thus, the transition is from `n = 2` to `n = 1`