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)`

Answer

9

Explanation

`27.2xx1.602xx10^(-12)=R_(H)z^(2)hc[1/(2^(2))-1/(n^(2))]`……..(1)
`10.2xx1.602xx10^(-12)=R_(H)z^(2)hc[1/(3^(2))-1/(n^(2))]`…….(2)
After solving (1) and (2)
`n=6, z=3`
`n+z=9`