Which transition in the hydrogen spectrum have the same wavelength as Balmer transition, `n = 4` to `n = 2`, of `H2^+` spectrum ?

Options

A. `n_1 = 1, n_2 = 2`
B. `n_1 = 1, n_2 = 3`
C. `n_1 = 3, n_2 = 4`
D. `n_1 = 2, n_2 = 4`

Answer

A

Explanation

(a) For `He^+` ion,
`(1)/(lamda) = Z^2 R[(1)/(n_1^2) - (1)/(n_2^2)] = (2)^2`
`R[(1)/(2^2) -(1)/(4^2)] = (3R)/(4)`
For hydrogen atom, `(1)/(lamda) = R[(1)/(n_1^2) -(1)/(n_2^2)]`
So, `(3R)/(4) = [(1)/(n_1^2) -(1)/(n_2^2)]`
or `(1)/(n_1^2) -(1)/(n_2^2) = (3)/(2)`
i.e., `n_1 = 1 and n_2 = 2`.