Problem with shell model and magnetic moment of Lithium-6

[bznm]

I have a problem with the calculus of magnetic moment of Li-6.

The configuration of protons is $1p_{3/2}$, and the neutrons' one is the same.

I have to add the magnetic moment of uncoupled proton and uncoupled neutron.

I use the following formula for $J=l+\frac{1}{2}$ (J is the particle spin):
$\frac{\mu}{\mu_N}=g_lJ+\frac{g_s-g_l}{2}$

For the proton I have: $g_l=1; g_s=5.58 \rightarrow \frac{\mu}{\mu_N}=J+2.29=3.79$

For the neutron I have: $g_l=0; g_s=-3.82 \rightarrow \frac{\mu}{\mu_N}=-1.91$

So the total $\frac{\mu}{\mu_N}=3.79-1.91=1.88$, exactly 1 more than the correct value, 0.88!
What's wrong?

 

[PeterDonis]

Your value for the proton's magnetic moment is off by 1: it should be 2.79, not 3.79. Not sure where you are getting the values you are using to calculate it.

 

[bznm]

Your value for the proton's magnetic moment is off by 1: it should be 2.79, not 3.79. Not sure where you are getting the values you are using to calculate it.

It was told me that the proton's magnetic moment is 2,79 but, when I consider a proton in the nucleus, I have to consider the proton's J value.

In this case, the proton's J is 3/2 and, if you insert this value in the formula, you obtain 3.79.

 

[Bill_K]

Li-6 is an odd-odd nucleus, and therefore the magnetic moments predicted by the shell model are not in complete agreement with experiment.

Quoting from Preston, "Physics of the Nucleus", p323:

"Turning to odd-odd nuclides, the shell model would suggest simply adding the magnetic moments due to proton and neutron configurations, ignoring any interaction between the unfilled neutron and proton shells, except perhaps in the light nuclei in which neutrons and protons are filling the same shells and i-spin is a good quantum number. It may be argued that, in this latter case, the neutrons and protons have precisely the same spatial motion and orientation but g-factors of opposite sign, and therefore the corrections to their free-particle g-values are roughly equal and opposite. Hence, despite the occurrence of interconfiguration mixing and quenching the free-nucleon g-factors can be used, and μ is just the sum of the neutron and proton moments of the extreme single-particle model. For nuclides in which neutrons and protons are filling different shells, it would seem appropriate to take the values of g_p and g_n from neighboring odd nuclides, thus allowing for interconfiguration mixing. This works quite well, and whenever the value g_emp obtained from empirical g's differs from g_sp obtained from free-nucleon g's, the observed value is always much nearer g_emp. Some cases are shown in Table 12-1, where both μ_emp and μ_sp are calculated from the following formula, the only difference being the g-values used:

$$\mu = \frac{1}{2}\left[(g_p + g_n) + (g_p - g_n)\frac{j_p(j_p + 1) - j_n(j_n + 1)}{J + 1}\right]$$

The table entry for Li-6 has μ_sp = 0.6, μ_emp = 0.4 and μ_obs = 0.8.

The conclusion which can be derived from our discussion is that, for the nearly spherical nuclei, which we have mainly considered, magnetic-moment values are consistent with the shell model, but it is essential to include interconfiguration mixing in the ground state."

(BTW, I think it would have been appropriate for you to mention that you were simultaneously posting this same question to both PF and stackexchange!)

 

[PeterDonis]

In this case, the proton's J is 3/2

But the neutron's J is 3/2 as well, correct? If you're going to include the J term, it seems to me that you should include it for both the proton and the neutron; i.e., I don't see why you have $g_l = 0$ for the neutron. (And if you include both, as the quote Bill_K posted points out, you have to consider the sign as well; if the neutron's $g_l$ is opposite in sign to the proton's, the two J terms cancel each other.)