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

` Delta E = Z^(2) (13.6)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
The electron makes a transition from state `n_(2) ` is the first excited state `(n_(1) = 2)` releasing `10.2 = 17 = 27.2 eV` energy
`27.2 = Z^(2)(13.6)[(1)/(4) - (1)/(n_(2)^(2))]`........(i)
When the electron comes to the second excited state `(n_(1) = 3) ` the energy is `4.25 = 5.95 = 10.2 eV` Hence
`10.2 = Z^(2)(13.6) ((1)/(9) - (1)/(n_(2)^(2)))`......(ii)
ON solving equation (i) and (ii) we get `Z = 3 ` and `n_(2) = 6`