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: When calculating nuclear magnetic moments, how does one decide between $l+\frac{1}{2}$ and $l-\frac{1}{2}$ Schmidt lines?
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.
