Which of the following transition in `He^(+)` ion will give rise to a spectral line which has the same wavelength as some spectral line in the hydrogen atom ?

Options

A. `n=3" to "n=1`
B. `n=43" to "n=2`
C. `n=4" to "n=2`
D. `n=4" to "n=3`

Answer

C

Explanation

In case of `He^(+)` ion the Rybderg constant happens to be four times that in case of hydrogen atom. Therefore, to have the same wavelength of some spectral line one must have
`[(1)/(n_(f)^(2))-(1)/(n_(f)^(2))]=4[(1)/(p_(f)^(2))-(1)/(p_(i)^(2))]`
where `n_(f)` are principla quantum numbers of final and intial orbits for hydrogen atom and `p_(f) and p_(i)` are those for `He^(+)` ion. This given `p_(f)=2n_(f) and p_(i)=2n_(i)`.