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\prob{31}
In a $^3S$ state of the helium atom, the possible values of the total electronic angular momentum quantum number are

  1. 0 only
  2. 1 only
  3. 0 and 1 only
  4. 0, 1/2, and 1
  5. 0, 1, and 2

Atomic}Spectroscopic Notations

Spectroscopic notation is given by ^{2s+1}L_j, and it's actually quite useful when one is dealing with multiple particles. L\in (S,P,D,F), respectively, for orbital angular momentum values of 0,1,2,3. s=1/2 for electrons. j is the total angular momentum.

Knowing the convention, one can plug in numbers to solve 3=2s+1 \Rightarrow s=1. Since the main-script is a S, l=0. The total angular momentum is j=s+l=1.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
eshaghoulian
2007-09-30 14:03:16
Whoa. You have to be careful. Given l and s,the possible values of j are, in general, j=l+s, l+s-1, l+s-2,...,|l-s|. Hund's rules tell you that j can only equal l+s or |l-s|, depending on whether the subshell is more than half-filled. In this case, l+s and |l-s| are the same number, so it does not matter.Alternate Solution - Unverified
Comments
deneb
2018-10-13 21:26:52
I\'m confused what the answer is describing. Is the s=1 referring to the entire helium atom? This particular helium atom has two electrons which each have s=1/2, so together s=1? And then since it\'s in the S orbital, l=0, then j=1? NEC
Poop Loops
2008-10-05 14:50:27
So is s = 1/2 or s = 1? Two different "s"? I understand S = 0 because of 0 orbital angular momentum.

But does s refer to spin or what?
gear3
2013-10-09 05:48:01
the spin of Helium is 1.That's it
NEC
bucky0
2007-11-01 14:09:11
to clarify, the correct answer is BNEC
eshaghoulian
2007-09-30 14:03:16
Whoa. You have to be careful. Given l and s,the possible values of j are, in general, j=l+s, l+s-1, l+s-2,...,|l-s|. Hund's rules tell you that j can only equal l+s or |l-s|, depending on whether the subshell is more than half-filled. In this case, l+s and |l-s| are the same number, so it does not matter.
Jeremy
2007-11-04 15:44:38
I don't think Hund's are involved in the solution. Hund's rules are only to determine the ground state, and problem isn't asking about that. I think the real point is that j can be \- l-s\- or any integer step above that, up to, and including, l+s. In this case, \- l-s \- = l+s =1, so j=1 is the only option.
DDO
2008-11-04 19:35:23
Jeremy that is Hund's 3rd rule.
iriomotejin
2010-10-06 08:42:25
Hund's rules deal with ground state, not with excited ones. The state of atom in this problem is obviously excited, so Hund's rules are irrelevant.
Alternate Solution - Unverified

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