A gas of identical hydrogen-like atoms has some atoms in the lowest (ground) energy level A and some atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy level absorbing monochromatic light of photon energy 2.7 eV. Subsequently, the atoms emit radiation of only six different photon energies. Some of the emitted photons have energy 2.7 eV, some have energy more and some have less than 2.7 eV.

Kevin Nash
10 years ago
Hello Student,
(i) The transition of six different photon energies are shown.
Since after absorbing monochromatic light, some of the emitted photons have energy more and some have less than 2.7 eV, this indicates that the excited level B is n = 2. (This is because if n = 3) is the excited level then energy less than 2.7 eV is not possible)
(ii) For hydrogen like atoms we have
E­ = - 13.6 / n2 Z2 eV / atom
E4 – E2 = - 13. 6 /16 Z2 – (- 13.6 4) Z2 = 2.7
⇒ Z2 x 13.6 [1/4 – 1/16] = 2.7
⇒ Z2 = 2.7 13.6 x 4 x 16 / 12 ⇒ I.E. = 13. 6 Z2 (1/ 12 – 1/ ∞2)
= 13. 6 x 2.7 / 13. 6 x 4 x 16 / 12 14.46 eV
(iii) Max. Energy
E4­ – E3 13.6 Z2 (1 / 42 – 1 / 12)
= 13.6 x 2.7 / 13.6 x 4 x 16 / 12 x 15 / 16 = 13. 5 eV
Min. Energy
E4 – E3 = - 13.6Z2 (1/ 42 – 1 / 32)
= 13.6 x 2.7 / 13.6 x 4 x 16 / 12 x 7 / 9 x 16 = 0.7 eV
Thanks
Kevin Nash