He I, He II and He III questions

Uh, I know my question seems highly stupid, but I think I somehow confused the cause of spectrum of these things, I don't know why...I'm getting a lot of confusion these days...
He I (neutral He) has 2 electrons, thus the emission is caused by both of them or one of them go up and back down, right?
He II (Singly ionized He) has 1 electron, but one more electron can come from the outside and then come and get down with the He II atom, then emit light too.
I can't tell the difference between the 2 lights emitted...One get up and get down, and one just get down, but so what? they all get down to the same level and emit the same wavelength, or do they?
And if so, He II have only 1 electron, but there is another electron ready to come...should the states be singlet and triplet instead of doublet like Hydrogen?
Thank you very much.
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Please...someone help me... =.='. Is my question too hard to understand...?]
I'm not really sure what you are asking for.
Are you asking what transitions involving photons are allowed? Also, your title refers to He III, but your question does not.

Based on the last question, it seems like you are trying to consider an intial state with one electron and a final state with two electrons. This of course cannot happen with electrodynamics. You need to consider them both as two electron systems. The intial state may have only one bound electron, but there are still two electrons if you are considering interactions with a free electron to give you the final state with two bound electrons. Make sense?

So yes, a He+ ion with a single electron can be described with as a doublet state. But the whole system (He+ ion and a free electron) would be a singlet or triplet (or some superposition of course) as a whole. Take that initial state, throw on the operator you are interested in, and then put in the final state (He, having two bound electrons) ... this will give you the matrix element you appear to be looking for.

Does that help at all?
If not, can you please explain a bit more precisely what the question is.
Oh, Thank you so much. Your answer answer my question.
Now, My new question is: For drawing a energy level diagram of He+, do I draw it without the interaction of the "outside" electron, thus no effect of orbital angular quantum number l. Or should I draw it with the interaction with the "outside" electron, leading to the effect the orbital angular quantum number l make on the energy level? (I'm trying to draw an energy diagram explaining the spectrum of Helium)


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He+ contains just 1 electron, so the energy levels would be for that one electron only.

If you include a second outside electron, then you are actually talking about neutral He I.

He I energy level diagrams typically involve excitation of just one electron.
For singly ionized helium, it has only one bound electron. The other unbound electron is far away and does not affect the energy levels of the bound electron. In this case, the energy levels of the bound electron (helium has Z = 2 protons) are Z^2 = 2^2 = 4 times the energy levels of neutral hydrogen (the binding energy of the electron in the 1s state is about 4 x 13.6 eV = 54.4 eV).
For neutral helium, there is normally one electron in the 1s state, and the other electron (if the atom is excited) in a higher bound state. There are two sets of energy levels; one for parahelium, and the other for orthohelium (see http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/helium.html). In one case (ortho), the spins of both electrons are aligned, and because of the Pauli exclusion principle, the second electron cannot transition to the ground state (1s shell). For the other case (para), there are no restrictions based on the Pauli exclusion principal.

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