# Question about intrinsic angular momentum

1. May 19, 2007

### 6Stang7

1. The problem statement, all variables and given/known data
If electrons did not have intrinsic angular momentum, the states occupied by the electrons int he ground state of He would be
a) (n=1, l=0) and (n=1, l=0)
b) (n=1, l=0) and (n=1, l=1)
c) (n=1, l=0) and (n=2, l=0)
d) (n=2, l=0) and (n=2, l=1)
e) (n=2, l=0) and (n=2, l=1)

3. The attempt at a solution

From what I know, the intrinsic angular momentum for the ground state is limited to two electrons due to the exclusion princple. In this case, wouldnt the answer be a? He has only 2 electrons, and they are already in (n=1, l=0) and (n=1, l=0) (one spin up and one spin down).....right?

2. May 20, 2007

### malawi_glenn

exclusion principle forbidds two identical fermions to occopy the same quantum state. So if electron didnt posess spin the ground state would be (c), since (b) says that n=1 and l=1 is possible, witch is not.

But on the other hand, if electrons didnt posess spin, it would not be a fermion, and hence no exclusion principle to take care of..

ps why did you post this in calculus forum??

Last edited: May 20, 2007
3. May 20, 2007

### 6Stang7

So what you are saying is that with the exclusion principle, He would normally have its two electrons in (n=1, l=0) and (n=1, l=0) since you are allowed two electrons in the n=1, l=0 state. Without out the exclusion principle, one electron would instead move up to the n=2 state? Would this not be the lowest possible potential energy? Since the electrons would now be considered bosons, couldn't any number of them occupy the same quantum state?

PS: mistake; I just now realized it (been studying all day for finals)

4. May 20, 2007

### malawi_glenn

Yes, but withiout spin, they would be bosons and hence no exclusion principle to obey.

5. May 21, 2007

### 6Stang7

What I mean is since they would be bosons and would ignore the exlcusion princple, wouldn't they then want to obey "mother nature" and go to the lowest potential energy, i.e. occupie the same quantum state?

6. May 21, 2007

### malawi_glenn

yeah, but in your first post you didnt motivate for having 2(n=1; l = 0)

and i belive I never said that one electron would move to n=2 state, is that the answer given to you in the textbook or teacher?

7. May 21, 2007

### 6Stang7

In your first post, you said "So if electron didn’t possess spin the ground state would be (c),..." which is (n=1, l=0) and (n=2, l=0).

My thinking is that normally He has its two electrons in (n=1, l=0) (more specific, (n=1, l=0, s=-1/2) and (n=1, l=0, s=+1/2)). Without the spin, there is no exclusion principle to follow, as well as no s values for the electrons. In that case, both electrons would sit in (n=1, l=0), or (n=1, l=0) for e1 and (n=1, l=0) for e2. Therefore, the answer would be a.

You seem/seemed to be suggesting c was the answer (which might have been a misunderstanding on my part). If it was c, that would imply that one electron went from (n=1, l=0), to (n=2, l=0). In this case, one electron was promoted, thereby increasing the amount of energy it has. This would mean that it was not at the lowest potential, which is not that "mother nature" wants.

8. May 21, 2007

### malawi_glenn

"But on the other hand, if electrons didnt posess spin, it would not be a fermion, and hence no exclusion principle to take care of.."

was my final answer altough;) Beginning with saying "(n=1, l=0) and (n=2, l=0)" scince the first state would only be able to occopy one electron, only three quatum numbers (n,l and m_l). BUT without spin, no exclusion principle to follow.

And the answer would be (a), but i thought your motivation for that was not so clear.