Q: 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 photons of energies 10.20 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 (ionization energy of hydrogen atom = 13.6 eV )

Sol: The electronic transitions in a hydrogen like atom from a state n_2 to a lower state n_1 are given by

$\large \Delta E = 13.6 Z^2(\frac{1}{n_1^2} – \frac{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 27.2 eV

Thus ΔE = 27.2 eV , n_1 = 2 and n_2 = n.

We have , $\large 27.2 = 13.6 Z^2(\frac{1}{4} – \frac{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)eV or 10.2 eV.

Thus , $\large 10.2 = 13.6 Z^2(\frac{1}{9} – \frac{1}{n^2} ) $ …(ii)

Dividing the eq. (i) by Eq. (ii) & 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.