| MadSci Network: Physics |
OK, you're getting the spins of the neutrino and electron (S) mixed up with the angular momentum of the nucleus I. So if you flip a spin-1 nucleus because l=1 you have the same magnitude of I in a different direction, delta I=0. You can't have I(0->0) in this case because there's no angular momentum of the nucleus to flip. You have to think of the nucleus more like a spinning sphere than about the internal structure of it when you're considering I. In the case of I=0, I would have to go to 1. It's the change in the magnitude of I for the case where I is not 0 that allows delta(I)=0 for l=1. Keep that delta part in mind. Example: 15C beta decaying with a parity change to 15N. Spin I=1/2 is the same, so delta(I)=0. Fermi decay is allowed if you just flip the direction of the spin and the combined spin S of the neutrino and the beta are 0 but the total angular momentum is 1.
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