InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A hydrogen like atom (atomic number Z) is uin a higher excleted atate of quantum n , The excited atom can make a two photon of energy `10.2 and 17.0 eV` respactively , Alernately the atom from the same excited state by successively eniting two photons of energies `4.25 eV and 5.95 eV` respectively Determine the value of n and Z (lonization energy of H- atom `= 13.6 eV`) |
|
Answer» Correct Answer - C For hydrogen like atoms E_(0) = (13.6)/(n^(2)) Z^(2) eV//atom` `Given E_(n) - E_(2) = 10.2 + 17 = 27.2 eV` ….(i) `E_(n) - E_(3) = 4.24 + 5.95 = 10.2 eV` `:. E_(3) - E_(2) = 17` `But E_(3) - E_(2) = (13.6)/(9)Z^(2) - ((-13.6)/(4)Z^(2))` `= - 13.6 Z^(2) [(1)/(9) - (1)/(4)]` `= - 13.6 Z^(2) [(4 - 9)/(36)] = (13.6 xx 5)/(36)Z^(2)` ` (13.6 xx 5)/(36)Z^(2) = 17 rArr Z= 3 `E_(n) - E_(2) = - (13.6)/(n^(2)) xx 3^(2) - [-(13.6)/(2^(2) xx 3^(2) ]` `- 13.6 [(9)/(n^(2)) - (9)/(4)] = -13.6 xx 9 [(4 - n^(2))/(4 n^(2))]` ....(ii) frequency (i) and (ii) ` -13.6 xx 9 [(4 - n^(2))/(4 n^(2))]= 27.2` `rArr 122.4 (4 - n^(2)) = 108.8 n^(2)` `rArr n^(2)= (489)/(13.6) = 36 rArr n = 6` |
|