Consider a hydrogen-like atom whose energy in nth excited state is given by
`E_n (13.6Z^2)/n^2`
When this excited atom makes a transition from excited state to ground state most energetic photons have energy `E_(max)=52.224` eV and least energetic photons have energy `E_("min")` =1.224 eV. Find the atomic number of atom and the state of excitation.

Explanation

Maximum energy is liberated for transition `E_n rarr1` and minimum energy for `E_n rarr E_(n-1)`
Hence, `E^1/n^2-E_1=52.224eV " "....(i)`
and `E^1/n^2 -E/((n-1)^2)=1.244eV" "...(ii)`
Solving above equations simultaneously, we get
`E_1 = -54.4 eV 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.