The photon radiated from hydrogen corresponding to `2^(nd)` of Lyman series is abosorbed 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

Energy of `n^th` state in hydrogen is same as energy of `3n^(th)` state in `LI^(++) . Therefore 3 to1` transition in H would give same energy as the `9 to 3` transition in `Li^(++)`.