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 transitions in a hydrogen-like atom from a state `n_(2)` to a lower state `n_(1)` are given by
`Delta`E = 13.6`Z^(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 `Delta`E = 27.2 eV , `n_(1) ` = 2 and `n_(2)` = n.
We have `27.2 = 13.6 Z^(2) [ (1)/(4) - (1)/(n^(2)) ] ` .... (2)
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 = 13.6`Z^(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.