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)to1` and minimum energy for `E_(n)toE_(n-1)`
Hence, `(E_(1))/(n^(2))-E_(1)=52.224eV" "…(1)`
and `(E_(1))/(n^(2))-(E_(1))/((n-1)^(2))=1.224eV" "…(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.