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.

Answer

A, B, C, D

Explanation

The transition state of six different photon energies are shown

Since after absorbing monochronatic light , same of the emitted photons have less then `2.7 eV` , this indicates that the excited level `B` is the excited level then energy less then `2.7 eV` (this is because if `n = 3` the excited level then energy less then `2.7 eV` is not possible)
(ii) for hydrogen like atoms we have
`E_(n) = (- 13.6)/(n^(2))Z^(2) eV atom`
`E_(1) - `E_(2) (-13.6)/(16)Z^(2) eV = ((-13.6)/(9)Z^(2) = 2.7`
`rArr Z^(2)= (2.7)/(13.6) xx(4 xx 16)/(12) rArr I.E.= 13.6Z^(2)((1)/(l^(2)) - (1)/(prop^(2)))`
`= 13.6 xx (2.7)/(13.6) xx (4 xx 16)/(12) = 13.5 eV `
(iii) Max energy
E_(4) - `E_(3) = -13.6Z^(2)((1)/(4^(2)) - (1)/(3^(2)))`
`= 13.6 xx (2.7)/(13.6)xx (4 xx16)/(12) xx(7)/(9 xx 16) = 0.7 eV`