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?