4
$\begingroup$

Lithium-6 isotope has an approximate magnetic momentum of $0.88\ \mu_N$ in its fundamental nuclear state. I'm trying to find its angular momentum and parity.

I found in a standard table: $I=1^+$ and while I get the parity part, I don't understand the rest. What I have done is filling the shells separately for protons and neutrons and finding that both have their outer particle in the $1 P_{3/2}$ state. Sum of these gives possible total angular momenta to be either $0$, $1$, $2$ or $3$.

So my question is: how do you choose between these to find the real value $I=1$?

$\endgroup$
2

2 Answers 2

2
$\begingroup$

Don't forget about the importance of nucleon-nucleon pairing. If you imagine the lithium-6 nucleus as an alpha particle and a deuteron bound in an $s$-wave state with positive parity and no angular momentum, you reproduce the spin and parity. Wikipedia tells me the deuteron has magnetic moment 0.857$\mu_N$, and of course the $\alpha$ has zero; quite close to your value for $^6$Li. You can explain the difference in a hand-waving way by remembering that both the $\alpha$ and deuteron both have some $d$-wave component which will mix in a more complicated way.

To give a more shell-model-friendly explanation, remember that as you fill electron orbitals you have to assign your atom (or ion) a total electronic spin; as you fill nuclear orbitals, you have to keep track of both total spin and total isospin, as permitted by the exclusion principle.

$\endgroup$
1
  • $\begingroup$ I've spelled out the shell-model analysis in my answer, and explained there why I don't like the analysis in terms of an alpha-deuteron cluster. $\endgroup$
    – user4552
    May 26, 2019 at 18:51
2
$\begingroup$

${}^6\text{Li}$ is known experimentally (Pyykkö 1991) to have a very small quadrupole moment, $-0.082\ \text{fm}^2$. This indicates a spherical shape. So as a start we could try just using the spherical single-particle levels, ignoring residual interactions, and filling levels up to the Fermi level for both protons and neutrons. Doing this, we find that the configuration of the valence particles would be $\pi p_{3/2}\otimes \nu p_{3/2}$. This would produce a degenerate multiplet of states with spin-parity values $0^+$, $1^+$, $2^+$, and $3^+$. We expect the degeneracy to be lifted by the residual interaction.

In light nuclei, the valence protons and neutrons occupy some of the same states, so neutron-proton interactions can be quite strong. This interaction is a particularly strong attraction when the intrinsic spins of the proton and neutron are coupled to spin 1; this is the reason that the deuteron has spin 1. The spatial overlap between the two $p_{3/2}$ states is maximized if their orbital angular momenta are coupled to spin 0, and this would be expected to maximize the attractive interaction. The 0 orbital angular momentum and the spin 1 from the intrinsic spins gives a total angular momentum of 1. So this would explain the reason why the $1^+$ state is so much lower in energy than the others, and is in fact the only particle-stable state in the nucleus -- the other members of the multiplet are resonances.

A couple of other answers (1, 2) talk about ${}^6\text{Li}$ as a cluster consisting of an alpha and a deuteron. The trouble with these clustering descriptions is that they always violate the Pauli exclusion principle, so I would prefer an explanation in terms of the vanilla shell model, which seems to work here.

Pyykkö, "The Nuclear Quadrupole Moments of the 20 First Elements: High-Precision Calculations on Atoms and Small Molecules," 1991 - https://www.degruyter.com/view/j/zna.1992.47.issue-1-2/zna-1992-1-233/zna-1992-1-233.xml (open access)

$\endgroup$
1
  • $\begingroup$ regarding $\pi p_{3/2} \otimes \nu p_{3/2}$ shouldn't we take in consideration the Pauli exclusion Principle that in this case would only allow $I= 0^{+}, 2^{+}$? or am I missing something. $\endgroup$ Jul 13, 2019 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.