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Term Symbol for Carbon He2s^22p^2

  • #1
Term Symbol for Carbon Total Angular Momentum

1. Write the term symnbols for carbon atom He2s22p2 which has two equivalent p electrons taking into account the Pauli exclusion principle.



2....



3. Since the carbon has two equivalent p-electrons, we know the l1=1 and that l2=1 as well. Also the spins for both the electrons are 1/2. So the sum of the spin is either 1 or 0. I'm not quit sure where to go from here. l = 2, 1 or 0, meaning we have S, P and D terms. I tabulated a table of 15 elements permuting the various l, s terms for the two electrons making no two have the same numbers. Is there an easier way to do this?

So, Far I've gotten 1D2, 3P2,1,0, 1P1 and 3S1. Are there all the terms?
 
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Answers and Replies

  • #2
674
2
You need to use Hund's rules here. Unless you are supposed to list all possible choices.
 
  • #3
Hund's rules would just tell me which state is the lowest in energy of all the possible ones right? But the ones I listed are not Pauli forbidden states right and they are all possible?
 
  • #5
Ya i actually just read that, but I still don't understand.

You have a state where
electron 1: l = 0 s = +1/2
electron 2: l = 1 s = -1/2

Here L = 1, and S = 0
So doesn't this correspond to a state of 1P1.. why is this term not included?
 
  • #6
674
2
Lets assume you have the 2 electrons in that configuration. Then you know:

m_l = +1, and m_s = 0

But you also know, L=1 or 2 can give you m_l = +1. So you don't know which L it is. Also, you know S=0 or 1 can give you m_s = 0. You usually check the higher L's and S's first before you check the next lowest L or S. And since I can make states with m_l = +2, +1, 0, -1, -2 using m_s=0 only (I am unable to do so with m_s=+/-1), then I know L=2 and S=0, and I take those possible electron configurations away.

Then you might ask about,
e1: l=0, s=-1/2
e2: l=1, s=+1/2

Well, since I already checked L=2 with S=1,0 and only L=2 S=0 worked out. Now I will check L=1 and S=1,0. Turns out, L=1 S=1 can exist, which does have a state m_l = 1, m_s = 0, and I can also make m_l=1 and m_s = +/-1 states that weren't used yet. So I pull out that possible electron configuration.

Now I am unable to make anymore m_l=1, and m_s=0 states. So poor L=1, S=0 can't exist.

Basically it is a combination game, where you try all possible combinations of L,S that can use up all the possible electron configurations exactly.

Also, once you include 3 electrons this game can become very difficult very fast. But using Hund's rules makes it very easy to find the ground state with any number of electrons.
 
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