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

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 e V` or `27.2 e V`
Thus , for `Delta E = 27.2 e V, n_(1) = 2 and n_(2) = n`, we have
`27.2 = 13.6 Z^(2) [(1)/(4) - (1)/(n^(2))]` ...(i)
For the eventual transition to the second excited state `n_(1) = 3`, the total energy released is `(4.25 + 5.95) e V or 10.2 e V` Thus,
`10.2 = 13.6 Z^(2) [(1)/(9) - (1)/(n^(2))]` ...(ii)
Driving Eqs. (i) and (ii), we get `(27.2)/(10.2) = (9 n^(2) - 36)/(4 n^(2) - 36)`
Solving we get `n^(2) = 36` or `n = 6`
Substituting `n = 6` in any of the above equations, we obtain
`Z^(2) = 9` or `Z = 3`
Thus, `n = 6` and `Z = 3`