Modern Physics> A gas of identical hydrogen-like atoms ha...

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.

Radhika Batra , 10 Years ago

Grade 11

1 Answers

Kevin Nash

Last Activity: 10 Years ago

Hello Student,

Please find the answer to your question

(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_n = - 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