A gas of identical hydrogen-like atoms has some atoms in the lowest in lower (ground) energy level `A` and some atoms in a partical upper (excited) energy level `B` and there are no atoms in any other energy level.The atoms of the gas make transition to higher energy level by absorbing monochromatic light of photon energy `2.7 e V`.
Subsequenty , the atom emit radiation of only six different photon energies. Some of the emitted photons have energy `2.7 e V` some have energy more , and some have less than `2.7 e V`.
a Find the principal quantum number of the intially excited level `B`
b Find the ionization energy for the gas atoms.
c Find the maximum and the minimum energies of the emitted photons.

Explanation

(a) Six lines are emission spectrum
`(n(n-1))/(2)=6 implies n^(2)-n-12=0`
`(n-4)(n+3)=0`
n=4
That is atom goes to excited state corresponding to n=4. Since some emitted photons have energy lt 2.7 eV i.e. there should be an energy level between initial level and n=4. Hence quantum number corresponding to initial excited level =2.

`DeltaE=13.6 ^(2) ((1)/(2^(2))-(1)/(4^(2)))`
`2.7=13.6 Z^(2)((1)/(4)-(1)/(16))=13.6 Z^(2)xx(3)/(16) " " ....(i)`
`Delta E'=13.6 Z^(2) ((1)/(1^(2))-(1)/(oo))=13.6 Z^(2) " " ....(ii)`
`(ii)//(i)`
`(DeltaE')/(2.7)=(1)/(3//16) implies Delta E' =14.4 eV=13.6 Z^(2)`
Ionisation energy =14.3 eV
(c) `(DelaE)_(max)=13.6 Z^(2)((1)/(1^(2))-(1)/(4^(2)))=14.4xx(15)/(16)=13.5 eV`
`(DeltaE)_(min)=13.6 Z^(2) ((1)/(3^(2))-(1)/(4^(2)))=14.4xx(7)/(144)=0.7 eV`