# Finding a method to compute the magnetic moment of an even-odd nucleus

#### JD_PM

Homework Statement
Compute the magnetic moment of 57-Ni and compare it with its experimental value of $-0.8 \mu_N$. Note that I have computed two extra magnetic moments for odd-even nucleus since I wanted to understand why I was getting a wrong answer (I also got those wrong). This is extra work, since the original question is about $\mu$ of 57-Ni. I want to understand what I am doing wrong.
Homework Equations
$\mu_J=g_J\times j\times \mu_N$
I am having difficulties computing the magnetic moment for an even-odd (proton-neutron) nucleus.

The formula is:

$$\mu_J=g_J\times j\times \mu_N$$

I checked this helpful post: https://physics.stackexchange.com/questions/290110/when-calculating-nuclear-magnetic-moments-how-does-one-decide-between-l-frac

I worked out the magnetic moment for 57-Ni.

Based on the shell model we see that only the unpaired neutron contributes to a non-zero magnetic moment; all neutron shells can be filled up to $2p_{3/2}$. Thus the quantum numbers are:

$$j=\frac{3}{2}, s=\frac{1}{2}, l=1.$$

Thus we know that $j=l+\frac{1}{2}$, so to calculate $g_J$:

$$g_J=\Big(1+\frac{1}{2j}\Big)g_l+\frac{1}{2j}g_s$$

For neutrons: g_l = 0 and g_s =-3.8260837.

Knowing that we get for 57-Ni case:

$$g_J = -1.275$$

Thus:

$$\mu_J = -1.913 \mu_{N}$$

This result is wrong. It is far from its experimental value; $\mu_J = -0.8 \mu_{N}$. I have read (Krane page 126-127) that it is acceptable to get a theoretical value slightly different from the experimental one. That is not the case here of course.

I checked more similar examples:

a) 87-Sr (38 protons). $J^\pi=\frac{9}{2}^+$ and $j=l+\frac{1}{2}$ Applying the same method I get:

$$\mu_J = -1.913 \mu_{N}$$

Experimental value: $\mu_J = -1.093 \mu_{N}$.

Again a significant difference; something is wrong.

b) 91-Zr (40 protons). $J^\pi=\frac{5}{2}^+$ and $j=l+\frac{1}{2}$ Applying the same method I get:

$$\mu_J = -1.913 \mu_{N}$$

Experimental value: $\mu_J = -1.304 \mu_{N}$.

Again a significant difference; something is wrong.

Note that in all three cases I get the same mistaken theoretical value. There must be something I am missing.

To recap, this is the method that I've used:

1) Get $j$ based on the shell model (note that in some cases there are exceptions, but we are not concerned with that in this post).

2) Get $g_J$.

2.1)If $j=l+\frac{1}{2}$ meets your case then use:

$$g_J=\Big(1+\frac{1}{2j}\Big)g_l+\frac{1}{2j}g_s$$

Finally calculate the magnetic moment:

$$\mu_J=g_J\times j\times \mu_N$$

2.2)If $j=l-\frac{1}{2}$ meets your case then use:

$$g_J=\frac{1}{j+1}\Big[\Big(j+\frac{3}{2}\Big)g_l-\frac{1}{2}g_s\Big]$$

Finally calculate the magnetic moment:

$$\mu_J=g_J\times j\times \mu_N$$

What's wrong with it?

Thanks.

PS: Let me know if something needs to be added and I will do it.

I've asked this question at PSE as well, but it's not catching a lot of attention.

Last edited:
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#### JD_PM

Hi @mfb may you help me out with this one? I am basically stuck in how to compute the magnetic moment of an even-odd nucleus. I've found the above method but I am missing something (I am getting wrong answers for the three examples).

Thanks.

#### mfb

Mentor
With g_l = 0 your calculation simplifies to $\mu_J=g_J\times j\times \mu_N = \frac{1}{2j}g_s \times j\times \mu_N = \frac{1}{2} g_s \mu_N$ and that is the same for every nucleus.
I'm not sure which step went wrong, I didn't look at that for years, @Orodruin should be able to help.

• JD_PM

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