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GR9277 #31
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Problem
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\prob{31}
In a  state of the helium atom, the possible values of the total electronic angular momentum quantum number are
- 0 only
- 1 only
- 0 and 1 only
- 0, 1/2, and 1
- 0, 1, and 2
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Atomic }Spectroscopic Notations
Spectroscopic notation is given by  , and it's actually quite useful when one is dealing with multiple particles. ) , respectively, for orbital angular momentum values of  .  for electrons. j is the total angular momentum.
Knowing the convention, one can plug in numbers to solve  . Since the main-script is a S,  . The total angular momentum is  .
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Alternate Solutions |
eshaghoulian
2007-09-30 14:03:16 | Whoa. You have to be careful. Given  and  ,the possible values of  are, in general, j=l+s, l+s-1, l+s-2,...,|l-s|. Hund's rules tell you that can only equal  or |l-s|, depending on whether the subshell is more than half-filled. In this case, and |l-s| are the same number, so it does not matter. |  |
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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? |  | 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
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| | bucky0
2007-11-01 14:09:11 | to clarify, the correct answer is B | | eshaghoulian
2007-09-30 14:03:16 | Whoa. You have to be careful. Given and ,the possible values of are, in general, j=l+s, l+s-1, l+s-2,...,|l-s|. Hund's rules tell you that can only equal or |l-s|, depending on whether the subshell is more than half-filled. In this case, 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 can be  or any integer step above that, up to, and including, . In this case,  , so  is the only option.
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DDO
2008-11-04 19:35:23 |
Jeremy that is Hund's 3rd rule.
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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.
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