I was reading about Bohr's atomic model in our textbook when came up with this question: Since the electron in hydrogen atom is most stable in the first orbit, why should it descend from higher orbits to any orbit except the first orbit(that means balmer series for example)?
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1$\begingroup$ en.wikipedia.org/wiki/Emission_spectrum and this one too avogadro.co.uk/light/bohr/spectra.htm $\endgroup$– ParaH2May 11, 2016 at 18:06
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1$\begingroup$ It says "This is as a result of an electron 'falling' from a higher energy level to one of lower energy." What I want to know is why"a lower energy level" and not the ground state as it is the most stable? $\endgroup$– Amir MoeiniMay 11, 2016 at 18:20
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1$\begingroup$ @Shadock The references cited here are more relevant ioffe.ru/astro/QC/CMBR/sp_tr.html Transition to the lowest energy is most probable, but others have a non-zero probability. I think QED is needed to explain properly. $\endgroup$– DavePhDMay 11, 2016 at 18:25
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$\begingroup$ I see...; thanks, the tables are although well clarifying. $\endgroup$– Amir MoeiniMay 11, 2016 at 18:32
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1$\begingroup$ @Shadock quantum electrodynamics $\endgroup$– orthocresolMay 11, 2016 at 20:26
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1 Answer
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The entire process has two parts: Excitation and relaxation. The first is clear enough, we excite an electron into a higher energy level.
Now this electron wants to fall back down. There is no roadsign or intrinsic selection rule that requires the electron to immediately go back to the ground state. Any state in a lower shell will be less energetic and by chance electron 1 happens to stop in state A while electron 2 of a different atoms stops in state B. Any electron that is not in a ground state will remain there — probably for a duration of about a few femtoseconds or less — and then relax further. The universe is not deterministic so nothing is telling the electron it has to take the fastest path.
Note that this gedankenexperiment does not make any prediction which path is most likely. You would need more theory (that DavePhD is likely to give) to explain that.
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1$\begingroup$ Well, for example if the H atom is in the 3s or any other s state, it will need to go to an intermediate state before going to the ground state, for electric dipole radiation. The 2s lifetime is therefore infinite for electric dipole radiation. Generally, the other lifetimes are in the nanoseconds. physik.uni-wuerzburg.de/EP4/teaching/WS2012_13/KM_ueb/… $\endgroup$– DavePhDMay 13, 2016 at 13:08