Method for calculating nuclear magnetic moments

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 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 how I understood the method:

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:

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

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

I believe your calculations are correct. Notice the piece-wise nature of the Schmidt lines in the plot attached - if you extend your examples to a few other nuclei (try for instance: $$^{75}$$Ge with J$$^\pi$$=1/2$$^-$$, $$^{47}$$Sc with J$$^\pi$$=7/2$$^-$$, $$^{147}$$Eu with J$$^\pi$$=11/2$$^-$$) you will get different values. In particular, you happen to have looked at only odd-neutron elements with j=l-1/2.