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 Zionisation energy of hydrogen atom=13.6eV)

Explanation

The electronic transitions in a hydrogen-like atom from a state `n_2` to a lower state `n_1` are given by:
`DeltaE=13.6Z^2[1/n_1^2-1/n_2^2]`
For the transition from a higher state `n_1=2` to the first excited state `n_1 = 2`, the total energy released is
(10.2+17.0) eV or 27.2 eV. Thus `DeltaE = 27.2eV, n_1 = 2 and n_2=n`. We have
`27.2–13.6Z^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) eV or
10.2ev. Thus 10.2 = `13.6Z^(2) [1/9-1/n^2] " " ...(ii)`
Dividing the Eq. (i) by Eq. (i), 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`