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GR9277 #31
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{31}
In a 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 , 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 .  Alternate Solutions
 eshaghoulian2007-09-30 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.Reply to this comment 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?
 gear32013-10-09 05:48:01 the spin of Helium is 1.That's it 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.
 Jeremy2007-11-04 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.
 DDO2008-11-04 19:35:23 Jeremy that is Hund's 3rd rule.
 iriomotejin2010-10-06 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.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$