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Helium atom in the $2^1$ $P$ and $2^3$ $P$ excited states

Now I'm guessing that 1 electron should be considered in the 1s state, but what about the other?

Should I consider the other as simply being in the $p$ orbital ($l=1$) and having the spin as either antiparallel (for the $2^1$ $P$) or parallel (for the $2^3$ $P$)

..because the 1 and 3 can't possibly refer to the azimuthal quantic number $l$... or could I be mistaken?

share|cite|improve this question
Hi vlad-ardelean - I've reopened the question, since after your edit the reason it was closed no longer applies. But the question is still unclear. In the title you use $s^1p$ and such, whereas in the body you use $2^1p$, but I'm not sure what the latter is supposed to mean. Could you check your notation and fix it where necessary? – David Z May 6 '13 at 1:44
thx for editing, didn't know about the special dollar syntax. – vlad-ardelean May 6 '13 at 10:08

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