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

`10+2+17=13.6 Z^2(1/2^2-1/n^2) and 4.25+5.95=13.6 Z^2(1/3^2-1/n^2)`
solving the above two equations we get, Z=3,n=6