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?

  • $\begingroup$ Doesn't charge conservation require a positron? $\endgroup$
    – DJohnM
    Jul 5, 2019 at 17:21
  • 1
    $\begingroup$ possibly related: physics.stackexchange.com/questions/346264/… $\endgroup$ 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

1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$ 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? $\endgroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.