I have used data from a question regarding the reciprocal wavelengths observed when lithium transitions from excited states to a common ground state. The lowest energy transition is from n=3 to n=2. The question goes on to say that this lowest energy transition, in high resolution, has 7 peaks, and asks for an explanation as to why there are not 8 lines. Can I have some guidance on this? I think spin-orbit coupling is relevant. I am aware that the dipole moment of an orbiting electron interacts with a magnetic field (from the orbital angular momentum of the electron?) and so some orbitals are raised/lowered slightly, those with an "l" greater than 0.
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$\begingroup$ Well, do you think there could be 8 lines (in theory)? Can you enumerate the 8 transitions that might cover? $\endgroup$– Jon CusterJun 1, 2017 at 20:23
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$\begingroup$ Can you derive the term symbols for the relevant electronic states? Do you know the selection rules for electronic transitions? $\endgroup$– orthocresolJun 1, 2017 at 20:26
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$\begingroup$ all doublets. S(1/2), P(1/2), P(3/2), D(5/2) and D(3/2)? Selection rule is that angular orbital quantum number (l) must be + or - 1, and total angular momentum (j) 0, + or - 1? $\endgroup$– gamma1Jun 1, 2017 at 20:43
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$\begingroup$ Yes, those are the right term symbols for the excited state. If you find the relevant term symbols for the ground state (which aren't very different except for the lack of D terms, since there's no 2d orbital) you should be able to find the 7 allowed transitions. Personally I don't know what the "8th" one is supposed to mean, though. $\endgroup$– orthocresolJun 1, 2017 at 22:45
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1 Answer
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The energy levels are in increasing energy 3S, 3P, 3D and the final state is 2P. The P-P transition is not allowed by $\Delta L =\pm1$ selection rule. The two transitions are 3S-2P at approx 812 nm and 3D-2P at approx 610 nm. As you are asked about the lowest energy transition this must be the 3S-2P. (Li also has non-zero nuclear spin depending on its isotope which lead to hyperfine transitions and are ignored )
With some spin-orbit coupling the (Russell-Saunders) terms symbols are for the upper level $^2S_{1/2}, $ and for the lower level $^2P_{3/2}, \, ^2P_{1/2}$. The $M_J$ levels are assumed to be split so the degeneracy is 2 for the S and 2 or 4 for the P states as given by $2J+1$. The selection rules on $M_J$ levels are $M_J = 0,\, \pm 1$.
Thus there are four transitions possible from $^2S_{1/2}$ to $^2P_{1/2}$ starting in $ M_J = 1/2 $ in $^2S$ to $(1/2, -1/2)$ in $^2P$ and similarly $ M_J = -1/2 \rightarrow (1/2,\, -1/2)$, making 4 transitions in all as the splitting of the S and P levels will be different. (This is the same case as the splitting of sodium $D_1$ line but 'upside down' in energy to that case).
The other transitions are from $^2S_{1/2}$ to $^2P_{3/2}$. In this case there are 6 transitions as the P splits into 4 levels. The transition between the $\pm 1/2$ make four of these (as just outlined above) and in addition there are transitions between $M_J = 3/2 \rightarrow 1/2$ and $M_J = -3/2 \rightarrow -1/2$.
This makes ten transitions in all, thus it seems I have calculated the wrong transitions (or misunderstood completely), however, you should be able to follow the method. Note that the observed spectral lines may not correspond to transition as distinct transitions may have the same energy gaps.
I've added a sketch below, it is hand drawn but should be clear enough. The $S_{1/2} - P_{1/2}$ transitions are a subset of these as you can see.
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$\begingroup$ Very informative answer-so there are not in fact 7 lines? Also, M(J), does this denote the total angular moment part of the term symbol (bottom right) $\endgroup$– gamma1Jun 2, 2017 at 11:45
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$\begingroup$ OP wrote n=3 to n=2 so I interpreted it as 2S/2P/2D initial state and 2S/2P final states. I got seven lines so I guess that was what was intended. Can attach a picture of Grotrian diagram later (I'm out right now). But I agree that it might be a bit vague. $\endgroup$ Jun 2, 2017 at 11:56
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$\begingroup$ the $M_J$ is the range (projection in field) of the bottom right total angular momentum $J=L+S$ in the term symbol $^{2S+1}L_J$. J has substates $M_J$ which are split in a magnetic or electric field and range from $-J\cdots +J$ in unit steps, and there are $(2J+1)$ of these in total. $\endgroup$ Jun 2, 2017 at 11:59
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$\begingroup$ This is what I was going for: i.stack.imgur.com/64NDv.jpg $\endgroup$ Jun 2, 2017 at 12:49
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1$\begingroup$ ok, that picture makes sense, I just took the 'hight resolution' bit to mean that the levels were split into states. The question is ambiguous though. $\endgroup$ Jun 2, 2017 at 14:21
