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

`n=6, Z=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`
Equations (i)-(ii) gives.
`E_(3)-E_(2)=17.0 eV`
or `Z^(2)(13.6)((1)/(4)-(1)/(9))=17.0`
`rArrZ^(2)(13.6)(5//36)=17.0`
`rArr Z^(2)=9`
or `Z=3`
From equation (i),
`Z^(2)(13.6)((1)/(4)-(1)/(9))=27.2`
or `(3)^(2)(13.6)((1)/(4)-(1)/(n^(2)))=27.2`
or `((1)/(4)-(1)/(n^(2)))=0.222`
or `(1)/(n^(2))=0.0278`
or `n^(2)=36`
`:. n=6`