A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectively Determine the values of n and Z. (lonization energy of H-atom = 13.6 eV)

Explanation

The electronic transition in a hydrogen -like atom from a state `n_(2)` to a lower state `n_(1)` are given by
`DeltaE=13.6Z^(2)(1/n_(1)^(2)-1/n_(2)^(2))`. For the transition from a higher state n to the excited state `n_(1) =2` , the total energy released is `(10.2+17.0) eV` or `27.2eV` and `n_(1) =2` and `n_(2)= n`.
we have `27.2=13.6Z^(2) [1/4-1/n_(2)]`.... (1)
for the eventual transition to the second excited state `n_(1) =3` , the total energy released is `(4.25+5.95)eVor 10.2eV`
thus `10.2=13.6Z^(2) [1/9-1/n^2]`.....(2)
Dividing the Eq.(1) by (2) we get `27.2/10.2=(9^(2)-36)/(4n^2-36)`
Solving We get `n^(2) =36` or `n=6`
Substitution `n=6` in any one of the above eqations,
we obtain `Z^(2)=9` or `Z=3` , thus `n=6` and `Z=3.`