Consider a hydrogen-like atom whose energy in `n^(th)` excited state is given by `E_(n) = (13.6Z^(2))/(n^(2))`
When this excited atom makes transition from excited state to ground state most energetic photons have energy`E_(max)` = 52.224eV and least energetic photons have energy `E_(min)` = 1.223eV. find the atomic number of atom and the state of excitation.

Explanation

Maximum energy is liberated for transition `E_(n) rarr ` 1 and minimum energy for `E_(n) rarr E_(n-1)`
Hence, `(E_(1))/(n^(2)) - E_(1)` = 52.224eV ... (1)
and `(E_(1))/(n^(2)) - (E_(1))/((n- 1)^(2)) = ` 1.224 e V .... (2)
Solving above equations simultaneously, we get `E_(1) = `-54.4eV and n = 5
Now `E_(1) = - (13.6Z^(2))/(1^(2))` = - 54.4eV.
Hence , Z = 2 i.e, gas is helium originally excited to n = 5 energy state.