# Spin, parity, etc. conservation in decays/reactions

I have a long list of short physics questions I'm reviewing, and there are several that ask about whether certain decays or reactions are allowed. Clearly these questions are testing my knowledge of the rules about spin and parity conservation, but apparently I don't have such knowledge. I've gathered a couple of the questions below as examples; what are the general principles I need to consider in order to answer questions like these?

• An excited nuclear state of ${}_4^8Be$ has spin 1 and sufficient energy to decay into two $\alpha$ particles. Is such a decay allowed?

• Can the reaction $\pi^-+\pi^+\to\pi^0+\pi^0$ take place if the $\pi^\pm$ come together with mutual orbital angular momentum $\ell=1$?

• Can a neutral particle with spin 1 and even intrinsic parity decay into two $\pi^0$, given that its mass exceeds twice the pion mass?

• If you are told that in a certain reaction, the electron comes out with its spin always parallel to its momentum, argue that parity conservation is violated.

• en.wikipedia.org/wiki/Isotopes_of_beryllium#Decay_Chains. In short $_4 Be^8$ decays into two alphas with a half-life or 6.7x10$^{-17}$ sec. – Sofia Dec 30 '14 at 12:21
• Sofia, thanks for the link. Unfortunately, that page doesn't have any information about spin. The question is whether that particular decay is permitted when the beryllium has a nuclear spin of 1. – thecommexokid Dec 30 '14 at 20:58
• where from did you take that the spin is 1? Please look in the table at the site en.wikipedia.org/wiki/Isotopes_of_beryllium#Table. It's written there that the spin is zero. The $^8Be$ is so unstable that is splits into two within cca. $7x10^{-17}$sec. I didn't see anywhere spin 1. You can look also at periodictable.com/Isotopes/004.8/index2.p.full.html – Sofia Dec 30 '14 at 22:31
• From the question. "An excited nuclear state of ${}^8_4$Be has spin 1 and sufficient energy to decay into two α particles. Is such a decay allowed?" Obviously the answer is going to be "No, it's not allowed" but I want to know why not. – thecommexokid Dec 31 '14 at 2:25