Ahydrogen-like atom (atomic number Z) is in a higher excited state of quantum numbern. 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=13.6 eV).

Explanation

The electronic transitions in a hydrogen-like atom from a state `n _(2)` to a lower state `n _(1)` are given by:
`Delta E = 13.6 Z ^(2) [ (1)/( n _(1) ^(2)) - (1)/( n _(2) ^(2))]`
For the transition from a higher state n to the first exicted state `n _(2) =2,` the total energy released is
`(10.2 + 17.0)eV or 27.2 eV. `Thus `Delta E = 27.2 eV, 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 exicted state `n _(1) = 3,` the total energy released is `(4.25 + 5.95)eV or 10.2 eV.` Thus `10.2 = 13.6 6 Z ^(2) [ (1)/(9) - (1)/( n ^(2))]`
Dividing the Eq . (i) by Eq . (ii) , we get
`(27.2)/(10.2) = ( 9n ^(2) - 36 )/( 4n ^(2) - 36)`
Solving we get `n ^(2) ~~ 36` Substituting `n = 6` in any one of the above equations, we obtain
` Z ^(1) = 9`
`or Z =3`
Thus `n = 6 and Z =3`