# Magnetic moments of atoms

1. May 26, 2008

### abdul-pablo

We can easily write down configuration of electrons in an atom of given element. The question is how can I calculate the magnetic moment of given atom knowing configuration of electrons in that atom? Or maybe some other data is needed? Especially I am interested in calculation of magnetic moment of 5B, 11Na, 17Cl, 23V.

2. May 26, 2008

### Gokul43201

Staff Emeritus
To do this, you need to know the following:
1. Hund's Rules
3. Calculation of the Lande g-factor

3. May 27, 2008

### abdul-pablo

OK.
Let's take a 23Na atom. Its electronic configuration is 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>1</sup>.
We can omit talking about electrons from first and second shell as their total contribution to magnetic moment is 0. The only electron we consider now is that on the 3s subshell. Its spin projecton on specified axis is ms = 1/2*h_bar.
azimuthal quantum number (angular momentum) l = 0. So ml = 0 as well. Therefore the projection of the total angular momentum along a specified axis mj = 1/2*h_bar. I have a problem to find j and J. I understand I need this numbers. Is my procedure right up to this moment? What shall I do next?

4. May 27, 2008

### Gokul43201

Staff Emeritus
In the case, of Na (and all alkali metals), it becomes easier since $l=0$ (so the only contribution to the magnetic moment comes from spin) and since there is only one electron in partially filled shells (i.e., $S=1/2$).

The spin contribution to the magnetic moment is simply $M_{spin} = g \mu _B \sqrt{S(S+1)}$, where $g=2$ and $\mu _B$ is the Bohr Magneton.

If all you want is the electronic magnetic moment, you are done at this point. If you also care about nuclear magnetic moments, there's further to go.

Note, we had g=2, only because l=0. In general, the complete formula for the g-factor will need to be used.

5. May 27, 2008

### abdul-pablo

I understand that the same situation is in case of 5B, where there is also only one electron on the last subshell 2p. Therefore Na and B has the same electronic contribution to magnetic moment.

Let's leave now the calculation of magnetic moment of nucleus and take care of contribution of electrons in case of atoms of other elements.

Let's consider chlorium Cl. It's electronic configuration is 1s^2, 2s^2, 2p^6, 3s^2, 3p^5. I should consider only the 3p subshell as it is the only not fully occupied subshell. According to Hund's rules electrons on this subshell have the following quantum numbers:
1) n=3, l=1, ml=1, ms=1/2
2) n=3, l=1, ml=0, ms=1/2
3) n=3, l=1, ml=-1, ms=1/2
4) n=3, l=1, ml=1, ms=-1/2
5) n=3, l=1, ml=0, ms=-1/2
The unoccupied place can be decribed with ml=0 or 1 or -1 providing that other two are ocuupied.

Hence, S=1/2 + 1/2 + 1/2 - 1/2 - 1/2 = 3*1/2 - 2*1/2 = 3/2 - 1 = 1/2
L=1+1+1+1+1=5*1=5
and J= S + L (as well as j=s+l)
so J = 1/2 + 5 = 11/2

I have found
g_J =
g_L * {J(J+1)-S(S+1)+L(L+1)} / {2J(J+1)} + g_S * {J(J+1)+S(S+1)-L(L+1)} / {2J(J+1)}

where g_L = 1 and g_S = 2.

6. May 29, 2008

### Gokul43201

Staff Emeritus
Be careful. For B, the outermost electron lives in $l=1$, which is different from Na. So, you now have S=1/2 and L=1, making J=1/2.

Correct up to this point.

You are making a mistake here. You should have:
L=1+0+(-1)+1+0 = 1

The correction to L will now give you a different value for J

Recalculate g_J using the correct value of J.

Then it's down to the last step:

$$M = g_J \mu _B \sqrt{J(J+1)}$$

PS: Sorry for the delay. Hope this is not too late already.

Last edited: May 29, 2008
7. May 30, 2008

### abdul-pablo

So why there is J=L-S instead of J=L+S=1+1/2=3/2? An the other question why it is assumed that electrons firstly have spin 1/2 and then -1/2. I mean that if we take the absolute value of total spin of electrons it does not matter on which 'side' (- or +) they really are.

I guess that L is not sum of l of all electrons but the sum of ml numbers. Am I right? L is total orbital momentum or just its projection onto some axis?

So now let's consider 23V atom.
It's electronic configuration is [Ar] 3d^3, 4s^2.
The quantum of its electron on the 3d (which make contribution to magnetic moment)
1) n=3 l=2 ml=0 s=1/2
2) n=3 l=2 ml=1 s=1/2
3) n=3 l=2 ml=2 s=1/2
or
1) n=3 l=2 ml=0 s=1/2
2) n=3 l=2 ml=-1 s=1/2
3) n=3 l=2 ml=-2 s=1/2
or
1) n=3 l=2 ml=0 s=-1/2
2) n=3 l=2 ml=1 s=-1/2
3) n=3 l=2 ml=2 s=-1/2
or
1) n=3 l=2 ml=0 s=-1/2
2) n=3 l=2 ml=-1 s=-1/2
3) n=3 l=2 ml=-2 s=-1/2

Therefore
L=0+1+2=3
S=3 * 1/2 = 3/2

J=3 + 3/2 = 9/2

Then I calculate g_J basing on the formula from the previous post and put it into the formula for the magnetic moment, which you have provided.

Now, have a look at the magnetic moment of nucleus. Proton has the spin magnetic moment u = g_S * (e *s) / (2*m_p), where g_S=1, e is charge, s=1/2 spin, and m_p is mass of proton. Nuetron does not contribute any magnetic moment as it has no charge.
Now, I should find the spin configuration of protons in the nucleus of given atom (for the beginning let's take Na) and add spin magnetic moments together. Then the problem of magnetic moment caused by motion of protons arises. How can I calculate it?

P.S. It's still not too late, but I should hurry up to prepare myself to the exam

Last edited: May 30, 2008
8. May 30, 2008

### Gokul43201

Staff Emeritus
Don't have much time now to respond fully, but you need to know this:
Read Hund's Rules carefully - if the sub-shell is less than half-filled we use J=|L-S| and if it is more than half-filled, we use J=L+S.

9. May 30, 2008

### abdul-pablo

And for subshell excatly half-filled?

10. May 30, 2008

### Gokul43201

Staff Emeritus
For a half-filled sub-shell, L=0, so J = L+S = |L-S| = S

11. May 30, 2008

### abdul-pablo

I have noticed a mistake I made in the previous post when descibring the quantum numbers of V. They should be:
1) n=3 l=2 ml=0 s=1/2
2) n=3 l=2 ml=1 s=1/2
3) n=3 l=2 ml=-1 s=1/2
Therefore L=1+0-1=0 and S= 3 * 1/2 = 3/2
Following formulas you've provided J=S= 3/2
Now I should calculate g_J using the formula already mentioned in one of my posts

12. May 30, 2008

### Gokul43201

Staff Emeritus
Actually, l=1, ml = 1, 0, -1

The rest is correct.

13. May 30, 2008

### abdul-pablo

Contribution of electrons to the magnetic moment of an atom is found.

Now, have a look at the magnetic moment of nucleus. Proton has the spin magnetic moment u = g_S * (e *s) / (2*m_p), where g_S=1, e is charge, s=1/2 spin, and m_p is mass of proton. Nuetron does not contribute any magnetic moment as it has no charge.
Now, I should find the spin configuration of protons in the nucleus of given atom (for the beginning let's take Na) and add spin magnetic Moments together. Then the problem of
magnetic moment caused by motion of protons arises. How can I calculate it?

14. May 30, 2008

### Gokul43201

Staff Emeritus
Nuclear magnetic moments are somewhat messy (very messy in some cases), and are typically ignored when calculating the magnetic moment of an atom. Typically, the nuclear moment is about 3 orders of magnitude smaller than the electronic moment.

So, first I must ask you what course this is for, and whether you have covered nuclear physics (shell model of the nucleus, magic numbers, etc.) in this course. If not, you are done with what you have calculated so far.

In any case, there are some misconceptions in your understanding of nuclear moments. First of all, both protons and neutrons have spin, and both can contribute to the nuclear magnetic moment. But if the nucleus has an even number of protons and an even number of neutrons, then the magnetic moment is zero. If there is an odd number or protons or neutrons you only need to find the value of I for the outermost nucleon in the shell model. Only a handful of naturally occurring elements have an odd number of both neutrons and protons, and the calculation becomes very complicated for these nuclei.

Then you use the relation between angular momentum and magnetic moment for a nucleon: $$\mu_{nucl} = \frac{e}{2m_p}gI$$

This is used whether the unpaired nucleon is a proton or a neutron. The only difference in the two cases is the value of g, which is 5.59 for a proton and -3.83.

Last edited: May 30, 2008
15. May 31, 2008

### abdul-pablo

I thought that nuetron does not contribute to magnetic moment due to lack of charge, not spin. I did not know the formula for magnetic moment of nucleons. I guess that it may be somehow similar to the previous formulas, where the elementary charge of an electron apears.

I suppose that it is obligatory to aquire knowledge about calculation of nuclear magnetic moment in my course, because it goes too deep into the topic (considering that it is not the major course at my studies). Nevertheless I've got also sort of will to know how the world looks like and to develope myself in terms of subject which is interesting to me. (generally entire physics).

I understand that I in given formula is the angular momentum of the outermost nucleon.