# Beta decay of beryllium into boron: conservation of spin violated?

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?

• Doesn't charge conservation require a positron? Jul 5 '19 at 17:21
• possibly related: physics.stackexchange.com/questions/346264/… Jul 5 '19 at 17:21
• Possible duplicate of Spin conservation in $\beta^+$ decay Jul 5 '19 at 17:48
• 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.
– user4552
Jul 5 '19 at 18:11
• Alessio can you please post a link to your lecture/course? Jul 5 '19 at 23:05

• 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 Jul 5 '19 at 18:30
• 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? Jul 5 '19 at 18:39
• 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.