GREPhysics.NET: GR9277 #31

GR9277 #31

Problem

GREPhysics.NET Official Solution

\prob{31}
In a $^3S$ 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

Atomic $\Rightarrow$ }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 $j=s+l=1$ .

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.

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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?

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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

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eshaghoulian
2007-09-30 14:03:16

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.

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to clarify, the correct answer is B

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