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 `10.2 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 2 photons of energy `4.25 eV` and`5.95 eV` respectively. Determine the value of `(n+z)`

Explanation

For the given excited state, `n_(2) = n`
1st excited state means, `n_(1) =2`
2nd excited state means, `n_(1) =3`
Energy difference between n = 2 and n = 3 will be `= (10 + 17) - (4.25 + 5.95) = 16.8 eV`
`:. 16.8 = 13.6 ((1)/(2^(2)) - (1)/(3^(2))) Z^(2) = 13.6 xx (5)/(36) xx Z^(2)`
or `Z^(2) = (16.8)/(13.6) xx (36)/(5) = 9 or Z = 3`
Energy difference between `n_(2) = n and n_(1) = 2` will be
`= 10 + 17 eV = 27 eV`
`:. 27 = 13.6 ((1)/(2^(2)) - (1)/(n^(2))) xx 3^(2)`
or `3 = 13.6 ((1)/(4) - (1)/(n^(2))) or (1)/(4) - (1)/(n^(2)) = (3)/(13.6) or (1)/(n^(2)) = (1)/(4) - (3)/(13.6) = (13.6 - 12)/(4 xx 13.6)`
or `n^(2) = (4 xx 13.6)/(1.6) or n = 6`