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This is probably a very basic question about notation. If we have the following notation for Boron nuclei:

$$^{12}_5 B(1^+)\;\text{or}\; ^{10}_3B(3^+)$$

What does the number in the parentheses mean? I guess $1$ and $3$ correspond to a spin number and $\pm$ to the parity. But what spin is that exactly and how do you find it?

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    $\begingroup$ Minor comment (v2): presumably $\rm^{10}_3B$ is boron-10 with the proton number typo’d. $\endgroup$
    – rob
    Feb 5, 2022 at 15:02

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I think the nuclear spin or the total angular momentum. See Possible spins and parities of $^{38}_{17}Cl$

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    $\begingroup$ Correct. This database confirms to confirms those are the ground-state $J^\pi$ for boron-12 and boron-10. $\endgroup$
    – rob
    Feb 5, 2022 at 15:02
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    $\begingroup$ @GeorgePhys: please accept this answer, if it helps you $\endgroup$
    – MrQ
    Feb 5, 2022 at 15:15

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