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

Total energy liberated during tranition of electron from nth shell to first excited state, (i.e., 2nd shell)
`=10.20+17.0=27.20 eV`
`=27.20xx1.602xx10^(-12)` erg
`:. (hc)/(lambda)=R_(H)xxZ^(2)xxhc[1/2^(2)-1/n^(2)]`
`:. 27.20xx1.602xx10^(-12)=R_(H)xxZ^(2)xxhxxc[1/2^(2)-1/n^(2)]` ...(i)
Similarly, total energy liberated during transition of electron from nth shell to second excited state, (i.e., 3rd shell)
`=4.25+5.95=10.20 eV`
`=10.20xx1.602xx10^(-12)` erg
`:. 10.20xx1.602xx10^(-12)=R_(H)xxZ^(2)xxhxxc[1/3^(2)-1/n^(2)]` ...(ii)
Dividing Equations (i) by (ii)
`n=6`
On substituting the value of n in Equations (i) or (i)
`Z=3`
So, `n+Z=6+3=9`