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I found following reaction in my lecture: $$^{10}Be_{6} \rightarrow ^{10}B_5+e^{-}+\overline{\nu}_e$$

since Beryllium has an even number of protons and neutrons its nuclear spin is $0$. However, Boron has a nuclear spin of $3$, the electron $+\frac{1}{2}$ and the anti neutrino $-\frac{1}{2}$ if I am not mistaken.
(I deduced the spins using the shell model)

I always thought that the sum of the spins should be equal on both sides or is there something that I am missing?

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  • $\begingroup$ Doesn't charge conservation require a positron? $\endgroup$
    – DJohnM
    Jul 5, 2019 at 17:21
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    $\begingroup$ possibly related: physics.stackexchange.com/questions/346264/… $\endgroup$
    – fredwhileshavin
    Jul 5, 2019 at 17:21
  • $\begingroup$ Possible duplicate of Spin conservation in $\beta^+$ decay $\endgroup$
    – Jon Custer
    Jul 5, 2019 at 17:48
  • $\begingroup$ This is not a duplicate of the question linked to in the comments above. This example is different because the intrinsic spins can't be coupled to make the observed spins. $\endgroup$
    – user4552
    Jul 5, 2019 at 18:11
  • $\begingroup$ Alessio can you please post a link to your lecture/course? $\endgroup$
    – magma
    Jul 5, 2019 at 23:05

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The intrinsic spins of all the particles are not the only angular momenta involved. If they were, then you'd be right -- it would be impossible for this decay to occur, because you can't couple spins 3, 1/2, and 1/2 to make spin 0. (The + and - signs you stated in the question don't make sense, though.)

To get this reaction to occur, we need 2 additional units of orbital angular momentum to be contributed by the electron and the antineutrino. This is hard to do, which is why the decay has a half-life of more than a million years.

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  • $\begingroup$ I thought that parity would play a role but I am probably mistaken that is why I used $+\frac{1}{2}$ and $-\frac{1}{2}$ even though I am not sure if the anti neutrino has a negative sign $\endgroup$
    – Struggling_Student
    Jul 5, 2019 at 18:30
  • $\begingroup$ So you are saying that $$^{10}Be_6 \rightarrow ^{10}B_5 + 3e^{-} + 3\overline{\nu}_e$$ would be the correct way to write the reaction? $\endgroup$
    – Struggling_Student
    Jul 5, 2019 at 18:39
  • $\begingroup$ Parity is a separate quantum number, not a sign of the angular momentum. No, there are not 3 of each particle. The particles carry $3\hbar$ of orbital angular momentum. $\endgroup$
    – user4552
    Jul 5, 2019 at 18:44

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