What state has min angle between angular momentum & z-axis

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1. Homework Statement [/b]
What state has a minimum value of the angle between the angular momentum and the z-axis; the electron spin or the atomic n=3 state? [with l not equal to 0]

Homework Equations


L = \sqrt{l(l+1)}\hbar
Lz = m\hbar
S = \sqrt{s(s+1)}\hbar
Sz = ms\hbar

The Attempt at a Solution


For the atominc n=3 state..
Since n = 3 (and l isn't 0), l = 1, 2 ... and ml = -l ... l = -2,-1,0,1,2
So then when you draw it.. (I don't seem to be able to get the picture in here]
The Lz vector is pointing up along the z-axis and the L vector is some degrees to the right of it.

The angle between the z-axis and angular momentum can be written using cosine...
cos(\theta) = \frac{L_{z}}{L}
I understand up to there.

But then the next part the professor did is add in numbers and I'm not sure where the numbers came from. He wrote this:
cos(\theta) = \frac{2\hbar}{\sqrt{6} \hbar} = \frac{2}{\sqrt{6}}
The \sqrt{6} comes from L = \sqrt{l(l+1)}\hbar using l = 2. But why l = 2 and not l = 1?
And I don't know where the 2 on top came from.

Then he wrote another one.. cos(\theta) = \frac{1/2}{\sqrt{3/4}}
I'm assuming this second one is for the electron spin since that is the one we are comparing it to? But.. how do you draw the vectors for electron spin? I'm confused there.

So, I'm confused on where the numbers came from and how you take electron spin into account here.

If anyone could help, I'd appreciate it!
 
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mbradar2 said:
1. Homework Statement [/b]
What state has a minimum value of the angle between the angular momentum and the z-axis; the electron spin or the atomic n=3 state? [with l not equal to 0]

Homework Equations


L = \sqrt{l(l+1)}\hbar
Lz = m\hbar
S = \sqrt{s(s+1)}\hbar
Sz = ms\hbar

The Attempt at a Solution


For the atominc n=3 state..
Since n = 3 (and l isn't 0), l = 1, 2 ... and ml = -l ... l = -2,-1,0,1,2
So then when you draw it.. (I don't seem to be able to get the picture in here]
The Lz vector is pointing up along the z-axis and the L vector is some degrees to the right of it.

The angle between the z-axis and angular momentum can be written using cosine...
cos(\theta) = \frac{L_{z}}{L}
I understand up to there.

But then the next part the professor did is add in numbers and I'm not sure where the numbers came from. He wrote this:
cos(\theta) = \frac{2\hbar}{\sqrt{6} \hbar} = \frac{2}{\sqrt{6}}
The \sqrt{6} comes from L = \sqrt{l(l+1)}\hbar using l = 2. But why l = 2 and not l = 1?
No reason. Try plugging in l=1, and you'll see that you get a larger angle. Your professor knew this in advance, of course, and I'm guessing he wanted to save time by jumping straight to the right answer.
mbradar2 said:
And I don't know where the 2 on top came from.
What variable appears on top in the equation?
mbradar2 said:
Then he wrote another one.. cos(\theta) = \frac{1/2}{\sqrt{3/4}}
I'm assuming this second one is for the electron spin since that is the one we are comparing it to? But.. how do you draw the vectors for electron spin? I'm confused there.
It's exactly the same as for orbital angular momentum. Literally the only difference is that there are certain values allowed for the spin angular momentum that wouldn't be allowed for orbital angular momentum (namely, the half-integer spins).

For what it's worth, you don't really need to do a drawing to solve this problem. You can work directly from the equations.
 
diazona said:
No reason. Try plugging in l=1, and you'll see that you get a larger angle. Your professor knew this in advance, of course, and I'm guessing he wanted to save time by jumping straight to the right answer.

Ah, okay. Yes he does like to save time. Thanks for clearing that up.

What variable appears on top in the equation?
Lz = ml\hbar
ml = -l ... 0 ... l , which for my l=1 gives me ml = -2,-1,0,1,2
So I'm just supposed to do trial and error to see which one would give me the smallest angle?

It's exactly the same as for orbital angular momentum. Literally the only difference is that there are certain values allowed for the spin angular momentum that wouldn't be allowed for orbital angular momentum (namely, the half-integer spins).
Oh, okay. I guess I need to review that a bit more.

For what it's worth, you don't really need to do a drawing to solve this problem. You can work directly from the equations.
How do I do that? I need to know to use the cosine from the drawing, don't I?

Thanks for your help! :)
 
mbradar2 said:
Lz = ml\hbar
ml = -l ... 0 ... l , which for my l=1 gives me ml = -2,-1,0,1,2
So I'm just supposed to do trial and error to see which one would give me the smallest angle?
Sure, you can do that. Once you try all the possibilities once, you should understand how you can immediately pick the correct m in the future.
mbradar2 said:
How do I do that? I need to know to use the cosine from the drawing, don't I?
Well, you said you understood where
\cos\theta = \frac{L_z}{L}
came from, right? What would be the equivalent for spin?
 

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