Photons of equal energy were incident on two different gas samples. One sample containing H-atoms in the ground state and the other sample containing H-atoms in some excited state with a principle quantum number 'n'. The photonic beams totally ionise the H-atoms. If the difference in the kinetic energy of the ejected electrons in the two different cases is `12.75 eV`. Then find the principal quantum number 'n' of the excited state.

Answer

4

Explanation

In H-atom, `4` lines are observed in Balmer series. So, electron is in `n=6(6 rarr 2, 5 rarr 2, 4 rarr 2, 3 rarr 2)`. In `He^(+)` ion, one line is observed in Paschen series. So electron is in `n=4 (4 rarr 3)`.
`(H)_(6 rarr 2)=(He^(+))_(12 rarr 4)`
`:.` electron in `He^(+)` will jump from `n=4` to `n=12`