Consider a hydrogen-like atom whose energy in nth excited state is given by
`E_(n) = - (13.6 Z^(2))/(n^(2))`
When this excited 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 initial state or excitation.

Explanation

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