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 eV and 17.00 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).
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Sol. The electronic transitions in a hydrogen like atom from a state n 2 to a lower state n 1 are given by
Δ E = 13.6Z 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 Δ E = 27.2 eV, n 1 = 2 and n 2 = n. We have 27.2 = 13.6Z 2 
For the eventual transition to the second excited state n 1 = 3, the total energy released is (4.25 + 5.95) eV or 10.2 eV. Thus 10.2 = 13.6Z 2 
Dividing the two equations, we get 
= 
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.
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