A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 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 two photons of energies 4.25 eV and 5.95 eV, respectively Determine the values of n and Z. (lonization energy of H-atom = 13.6 eV)

Answer

`6,3`

Explanation

From the given conditions:
`E_(n)-E_(2)=(10.2+17)eV=27.2 eV`
and `E_(n)-E_(3)=(4.25+5.95)eV=10.2 eV`
Equation (1)-(2) gives
`E_(3)-E_(2)=17.0eV` or `Z^(2)(13.6)(1/4-1/9)=17.0`
`implies Z^(2)(13.6)(5//36)=17.0 implies Z^(2)=9 implies Z=3`
From equation (1) `Z^(2)(13.6) (1/4-1/n^(2))=27.2`
`implies (3)^(2) (13.6) (1/4-1/n^(2))=27.2 implies 1/4-1/n^(2)=0.222`
`implies 1//n^(2)=0.0278implies n^(2)=36implies n=6`