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I'm having trouble with this question from a first year undergraduate chemistry paper. The question states

  1. How many lines will appear in the line spectrum of the Boron atom if the valence electron in the ground state is excited to the 3p orbital?

  2. If in another experiment, the electron of $\ce{B^4+}$ in $n=6$ is excited to $n=7$, how many lines will appear in this line spectrum?

Boron has an electron configuration of $\rm{1s^22s^22p^1}$, so this means it would have to jump from $2\rm{p}$ through $3\rm{s}$ to $3\rm{p}$. I don't understand how I would work out the number of lines that would be seen in the emission spectrum though.

Update:

The memo gives an answer of 6 possible lines - I think the reasoning behind this is that for the valence electron in the ground state (which is 2p) it could jump (depending on how much energy is given to the system) from 2p -> 3p or 2p->3s or 3s->3p but because we don't know the spin on the electron, whether positive or negative we say there are 6 possible lines (3 lines for each possible electron state). Not sure if that reasoning makes sense or not?

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  • $\begingroup$ What does 'the 5b atom' mean here? Can you clarify? $\endgroup$ – Todd Minehardt Mar 1 '16 at 13:25
  • $\begingroup$ Sorry, mistake on my part. It should just be B, Boron atom. Have corrected. $\endgroup$ – Blargian Mar 1 '16 at 13:34
  • $\begingroup$ Indeed, the first part doesn't make much sense. The 3p <- 2p transition is Laporte forbidden. I assume the intention of the question is to ask about spin-orbit coupling. I think you might be better off asking one of your tutors. $\endgroup$ – orthocresol Mar 1 '16 at 16:07
  • $\begingroup$ I will certainly ask my tutor. I have a feeling you are right with regards to the question relating to spin-orbit coupling. The memo gives an answer of 6 possible lines - I think the reasoning behind this is that for the valence electron in the ground state (which is 2p) it could jump (depending on how much energy is given to the system) from 2p -> 3p or 2p->3s or 3s->3p but because we don't know the spin on the electron, whether positive or negative we say there are 6 possible lines (3 lines for each possible electron state). Not sure if that reasoning makes sense or not? $\endgroup$ – Blargian Mar 1 '16 at 19:25
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    $\begingroup$ The 2p-3p transition is forbidden en.wikipedia.org/wiki/Laporte_rule Also spin-orbit coupling is not as simple as that. Search for "Russell-Saunders coupling" $\endgroup$ – orthocresol Mar 1 '16 at 19:50

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