Consider a hydrogen-like atom whose energy in n'h excited state is given by
`E _(a) =( 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 transiton `E _(n) to 1 ` and minimum energy for `B _(u) to E _(n -1)`
Hence, `(E ^(1))/( n ^(2)) - E _(1) = 52224 eV" "(i)`
and `(E ^(1))/(n ^(2)) - (E _(1))/( ( n -1 ) ^(2))= 1.224 eV" "(ii)`
Solving above equations simulateously, we get
`E _(1) = 54.4 eV nd n = 5`
Now, `E _(1) = ( 13.6 Z ^(2))/( 1 ^(2)) = - 54. 4 eV`
Hence, `Z =2`
i.e., get is helium originally excited to `n = 5` energy state.