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In my introductory nuclear physics course, the following question came up:

Consider the odd-odd nucleus $^{38}_{17}Cl$, which has 17 protons and 21 neutrons. Its 17th proton sits in the $1d_{3/2}$ orbit, while its 21st neutron is in the $1f_{7/2}$ orbital. Calculate the possible spins and parities of this nucleus.

so I calculated the spin and parity the way odd-odd nuclei spin and parity are calculated. The total nuclear spin $I$ should be $j_1+j_1 = 5$. Its parity is $(-1)^0 (-1)^3 = -$
So the nuclear spin and parity is $5^-$, in accordance with the value i looked up online.

The part I don't get is why there should be multiple possible spins and parities. Does it have to do with the total angular momentum ranging from $|j_1-j_2|$ to $|j_1+j_2|$? How does it relate to what i calculated?

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The part I don't get is why there should be multiple possible spins and parities. Does it have to do with the total angular momentum ranging from $|j_1-j_2|$ to $|j_1+j_2|$?

Yes. What you calculated was $|j_1+j_2|$, which is the maximum. In the absence of any interaction between the neutron and the proton, all of those spin couplings, from the minimum to the maximu, would give states with the same energy, and the ground state would be degenerate. They do interact, and apparently in this example the result is that the maximum spin gives the lowest energy. But there is normally no easy was to predict that kind of thing without doing a full shell-model calculation in which you diagonalize a large Hamiltonian.

For more on this sort of thing, see this question: How do you go about guessing the ground-state spin and parity of a nucleus?

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  • $\begingroup$ So it's impossible to say which one is the ground state, but all the states the nucleus can possibly be in are $2^-$, $3^-$, $4^-$ and $5^-$? $\endgroup$ – Jeroen May 28 at 21:57
  • $\begingroup$ It's not impossible, but normally it can't be done without a big numerical calculation. In the linked answer, I do give a couple of examples where it can be determined without a big calculation. $\endgroup$ – Ben Crowell May 28 at 22:46

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