Determination of spin-parity of Hoyle state

[ridge1988]

I've completely forgotten how to do this/ not sure I ever really knew. I did some selection rules for in nuclear physics when you've got magnetic / electric transitions, but I don't think its the same and either way I don't remember it and am a bit lost.

The J-pi states of 8Be and 4He are 0+ and 0+.

The Hoyle state in 12C also has J-pi of 0+ . I've forgotten how you calculate what the allowed states are, i.e what the selection rules are for going from 8Be to 12C?

Thanks!

 

[lpetrich]

Selection rule has the details. You carry over the discussion there of atoms' electronic transitions to nuclei.

It's in error about multipoles being 2ⁿ poles; it's more like 2n poles. But the rest of it looks correct.

There are two kinds of multipole: electric and magnetic, which differ by their parity:
E(n): parity (-1)ⁿ
M(n): parity (-1)ⁿ⁺¹
for multipole n. From angular-momentum addition, the conditions on the initial and final angular momentum are:

|Jf - Ji| <= n <= |Jf + Ji|

The fastest transition, the electric dipole, E1, has change of J = -1, 0, or 1, with 0 -> 0 not allowed.

So there would be no way for Be-8 and He-4 to go to C-12's ground state or Hoyle resonance while emitting a single photon. It must therefore be a two-photon process:

(initial state) -(photon)-> (another excited state) -(photon)-> (final state)

I went to the Nudat 2 nuclear-physics database at Brookhaven, and looked up C-12.

The ground state is 0+
The Hoyle resonance is 0+, 7652.4 keV

However, there is an excited state in between: 2+, 4438.9 keV

A single-photon transition from the Hoyle resonance to this state would be E2, electric quadrupole, like a single-photon transition from there to the ground state.