The photon radiated from hydrogen corresponding to `2^(nd)` line of Lyman series is absorbed by a hydrogen like atom 'X' in `2^(nd)` excited state. As a result the hydrogen like atom 'X' makes a transition to `n^(th)` orbit. Then,

Options

A. `X=He^(+), n=4`
B. `X=Li^(++), n=6`
C. `X=He^(+),n=6`
D. `X=Li^(++) , n=9`

Answer

D

Explanation

(D) Energy of `n^(th)` sate in hydrogen is same as energy of `3n^(th)` state in `Li^(++)` `:. 3 to 1` transition in H would give same energy as the `3xx to 1xx3` transition in `Li^(++)`.