We calculated the Q-value of the beta-delayed proton emission in this post. The decay channel is rare, because, when a neutron changes to a proton and emits an electron, the proton orbital must be empty, i.e. the nucleus is N > Z, and the proton emission threshold is larger than neutron threshold. After the beta decay, in most cases, N > Z for the daughter nucleus, so that to have a proton emission is rare.
The situation is different for the beta-delayed neutron emission. After the beta decay, the neutron shell could still have an excessive number of neutrons and undergoes neutron emission. The Q-value for beta-delayed neutron decay is
whenever , the beta-delayed neutron emission is possible. Nuclei with just 2 or 3 neutrons away from the stability village have beta-delayed neutron emission. For example, any isotopes beyond these nuclei can undergo beta-delayed neutron emission 9Li, 12Be, 13B, 16C, 17N, 22O, 22F, 26Ne, etc.. The 2-step process can be pictured like this:
The interesting thing about the beta-delayed neutron emission is the nature of the daughter nucleus right after the beta-decay, and the branching ratio for the gamma-decay and neutron emission. Normally, the daughter nucleus is bound and only goes to gamma decay to release excessive energy. When the daughter nucleus is unbound and neutron emission is possible, what is the nature of the nucleus? Since the neutron emission is not 100% but ~15%. This indicates the daughter nucleus is a compound nucleus, involving collective motion.
In particle configuration, beta decay is when a neutron converts to a proton in the same orbital, as if the neutron decay to proton + electron inside the orbital. For example, 16N beta decay, decays to the 6.14 MeV 3- state (68%) and to the ground state (26%) of 16O. The 3- state can be understood as a coupling of a proton p1/2 hole + a proton d5/2 particle, which can also couple to 8.87 MeV 2- state. The ground state is a sd-shell neutron to proton p1/2. The 7.11 MeV 1- is probably proton p1/2 + a proton s1/2 particle.
17N can populate 17O (Sn = 4.14 MeV ) unbounded state between 4.5 to 6.0 MeV [H. Ohm, W. Rudolph, K.-L. Kratz, Nuclear Physics A 274, 45-52 (1976)]. The 0.0 (5/2+) MeV, 0.87 (1/2+) MeV and 3.06 (1/2-) MeV are also populated by the beta-decay. The ground state can be understood as a sd-shell neutron converting to proton p1/2. The 0.87 MeV state can be imagined as follows, 17N should have a fraction of the sd-shell neutron pair in (s1/2)^2. A s1/2 neutron in 17N converts to proton p1/2 and filled the p-shell, leaving a single s1/2 neutron on top of the 16O core. For the 3.06 MeV state, one of the p1/2 neutrons converts to a proton in the p1/2 orbital, leaving a p1/2 neutron hole in 17O. The spectroscopic factor for the 16O(d,p) reaction of the 3.06 MeV state is only 0.032 and the SF for the same state from 18O(p,d) reaction is 0.88 with L = 1. How to understand the branching ratio? How doe sit related to particle configuration?
To be precise, the 4.55 (3/2-), 5.08 (3/2+), 5.39 (3/2-), and 5.93 (1/2-) MeV are populated. These 4 states have been populated by the 16O(d,p) and 18O(p,d) reactions, here are the spectroscopic factors
Energy [MeV] | J-pi | L | SF (adding) | SF (removal) | branching ratio |
4.55 | 3/2- | 1 | 0.23 | 0.14 | 36.6 +- 2.6% |
5.08 | 3/2+ | 2 | 1.25 | 0.13 | 0.6 +- 0.4 % |
5.39 | 3/2- | ?? | ?? | ?? | 55.5 +- 3.5% |
5.93 | 1/2- | ?? | ?? | ?? | 7.4 +- 0.5% |
We can see that, there is 55% to populate 17O 5.39 MeV state, but this state is not known to be populated using a single neutron-adding/removal experiment. It is not clear whether the lack of information is due to experimental limitations or whether those states are not neutron single-particle states. If it is later, the situation becomes very interesting. What is the 5.39 MeV state of 17O?