GREPhysics.NET: GR9277 #59

GR9277 #59

Problem

GREPhysics.NET Official Solution

\prob{59}
The ground state of the helium atom is a spin

singlet
doublet
triplet
quartet
quintuplet

Atomic $\Rightarrow$ }Orbital

The ground state of Helium has $1s^2$ which is $l=0$ $n=1$ .

However, because both electrons are in the same l and n state, the Pauli Exclusion Principle (no two electrons can have exactly the same quantum number) requires that one have $s=1/2$ and the other has $s=-1/2$ for a combined total spin of $s=0$ , as in a spin singlet.

Thanks to user cakedamber for pointing this out.

(Compare with things in the p orbitals, which have $l=1$ , allowing for $m_l=-1,0,1$ .)

Alternate Solutions

cakedamber
2005-11-11 22:22:50

Sorry about the null post above, I was getting the hang of your interface. Anyhow, I'm sorry to say this, but your explanation above is flat-out wrong. The singlet state is when the total SPIN angular momentum is 0, i.e. when $S^2 = 0$ . You're absolutely right that in the $1s$ state, $l=0$ , but $l$ is ORBITAL angular momentum, not spin angular momentum, so that's irrelevant. The reason the ground state of helium must be a singlet is more complicated. Electrons are fermions, which means that the overall wavefunction for two electrons must be antisymmetric. In the ground state of helium, both electrons are in the same spatial state, meaning that their combined spatial wavefunction is symmetric. Therefore, in order to keep their overall combined wavefunction antisymmetric, their combined spin wavefunction must be antisymmetric -- meaning that they are in a spin singlet state, and thus (A) is the right answer.

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Comments

Maxwells_Demon
2008-09-28 04:17:20

How do you deal with the spin from the 2 protons from the He in your justifications? I'm a bit confused still...

Poop Loops
2008-10-25 15:52:20

I think it's because there are also only 2 protons and they are in the ground state, so they have to have 0 total spin. If they had the same spin, they would have to be in different states, so that's no longer the ground state.

I am just confused about Helium-3 because you end up with 1 neutron instead of 2, so I don't know what that does to the particle's angular momentum, but I guess that won't be on the test since it's more complicated.

PhyAnnie
2008-11-06 05:34:56

Actually I don't think there' s anything to do with the proton or neutron. We only need to focus on the two electrons and think of their states. (The wave function we've all been discussing is about electrons.)

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ewhite2
2007-10-27 12:41:27

The solution still isn't entirely correct. The spin triplet state also has a s = 0 possibility, so this narrows it down to the singlet and triplet state. You still need to take into account the symmetry of the spatial and spin wave functions as in cakedamber's solution.

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jax
2005-12-05 19:21:26

Just some additional info to help on this problem. The terms 'singlet', 'triplet' 'doublet' come from the multiplicity $2S + 1$ of the state.

So for states that we call singlets, $2S+1 = 1 \Longrightarrow S = 0$ (as in this case when we have spin $\frac{+1}{2}$ and $\frac{-1}{2}$ )

For doublets $2S+1 = 2 \Longrightarrow S = 1/2$ , etc.

agaliarept
2006-12-01 18:49:05

Thank you sir. Another needed post.

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astro_allison
2005-11-25 04:32:34

is it fair to say that the gs will always be a spin singlet? (no matter the Z; He, H, Li, ...).

yosun
2005-11-26 01:44:48

astro_allison: if $1s^2$ is a ground state, then it is a spin-singlet (see solution above for why). however, if $1s^1$ is a ground state, then it is not a spin singlet. So, in general, spin-singlet status depends on the number of electrons in a particular configuration... not sure what you mean by whether it's "fair"?

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cakedamber
2005-11-11 22:22:50

yosun
2005-11-11 22:50:42

cakedamber: thanks for pointer; the solution has been updated.

cakedamber
2005-11-12 16:08:57

Nice. That's much better.

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cakedamber
2005-11-11 22:14:41

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Post A Comment!

You are replying to:

Just some additional info to help on this problem. The terms 'singlet', 'triplet' 'doublet' come from the multiplicity $2S + 1$ of the state.
So for states that we call singlets, $2S+1 = 1 \Longrightarrow S = 0$ (as in this case when we have spin $\frac{+1}{2}$ and $\frac{-1}{2}$ )
For doublets $2S+1 = 2 \Longrightarrow S = 1/2$ , etc.

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