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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. 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?     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}\$