A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies `10.20` and `17.00 eV` respectively. Alternatively, the atom from the same axcited state can make a transition to the second excited state by successively emitting two photons of energy `4.25 eV` and `5.95 eV` respectively. Determine the values of n and Z `("ionisation energy of hydrogen atom"=13.6 eV)`. Given answer `=n+Z`.

Explanation

The electronic transitions in a hydrogen-like atom from a state `n_(2)`1 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 first excited state `n_(1)=2`, the total energy released is `(10.2+17.0)` eV or 27.2eV.
Thus `DeltaE=27.2eV,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)eV` or `10.2eV`.
Thus
`10.2=13.6Z^(2)[(1)/(9)-(1)/(n^(2))]" "..,...(2)`
Dividing the Eq. (1) by Eq. (2) we get `(27.2)/(10.2)=(9n^(2)-36)/(4n^(2)-36)`.
Solving we get `n^(2)=36` or n = 6
Substituting n = 6 in any one of the above equations, we obtain `Z^(2)=9` (or) Z = 3, Thus n = 6 and Z = 3.