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)

Explanation

According to the question,

We, know energy of electron in nth is given by
`DeltaE=13.6Z^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
Case I
Energy emitted by photon, `DeltaE=10.20+17`
=27.2 eV
So, `" " 10.2+17=13.6Z^(2)((1)/(2^2)-(1)/(n^(2)))" " ( :.n_(1)=2, n_(2)=n)`
`" " 27.2=13.6Z^(2)((1)/(4)-(1)/(n^(2)))" " ....(i)`
Case II
Energy emitted by photons, `Delta 4.25+5.95=10.2 eV`
`4.25xx5.95=13.6Z^(2)((1)/(3^(2))-(1)/(n^(2)))" " ( :.n_(1)=3,n_(2)=n)`
`10.2=13.6 Z^(2)((1)/(9)-(1)/(n^(2)))" " ....(ii)`
On divinding Eq. (i) by Eq. (ii) we get
`(27.22)/(10.2)=(1//4-1//n^(2))/(1//9-1//n^(2))`
`rArr" "(27.2)/(9)-(27.2)/(n^(2))=(10.2)/(4)-(10.2)/(n^(2))`
`rArr" " (17.2)/(n^(2))=(27.9)/(9)-(10.2)/(4)`
`rArr" " n^(2)=36rArrn=6`
Substituding the value of n in Eq. (i) , we get
`27.2=13.6Z^(2)((1)/(4)-(1)/(n^(2)))=13.6Z^(2)((1)/(4)-(1)/(6^(2)))`
`=13.6 Z^(2)((1)/(4)-(1)/(36))=13.6 Z^(2)((9-1)/(36))=13.6Z^(2)xx(2)/(9)`
`Z^(2)=9rArrZ=3`