Can free protons show $β^+$ decay if provided with energy?

I'm studying introductory Nuclear Physics in school and we were taught that free protons never undergo $$β^+$$ decay since the mass of neutron is greater than mass of proton, so the $$Q$$ value of the reaction is negative.

However, if suppose we provide protons with the energy via a collision or some other means, can we expect them to undergo $$β^+$$ decay?

If so, does this extend to any non spontaneous nuclear reaction? Will sufficient energy lead to it happening?

1 Answer

The overall process for $$\beta^+$$ decay is:

$$p \to n + e^+ + \nu_e \tag{1}$$

To supply energy to the proton we need something to collide with it, but there is no other particle on the left hand side to supply that energy. What we could do is in effect add an electron to both sides of the equation. On the right hand side the electron and positron cancel out and we're left with:

$$p + e^- \to n + \nu_e$$

which is the closely related process of electron capture. Now we can supply the required energy by giving the incoming electron the required energy, and as it happens this process has been discussed in the question Is there a term for electron capture outside the nucleus? The problem is that this process has a very low probability and in practice it's hard to think of a situation where it could be observed. Nevertheless it is theoretically possible.

Another possibility would be to start with our initial equation (1) and add an antineutrino to both sides. This time the neutrino and antineutrino cancel on the right side and we get:

$$p + \bar\nu_e \to n + e^+$$

This is known as inverse beta decay and the reaction is a standard way to detect anti-neutrinos. The kinetic energy of the incoming anti-neutrino supplies the required energy.

So if you're prepared to accept electron capture or inverse beta decay as a form of $$\beta^+$$ decay then the answer to your question would that yes it is possible for a free proton.

There is another way energy could be supplied. Hadron have excited states, and the first excited state of the proton is the $$\Delta^+$$ particle. So we could imagine a process where a proton is excited to a $$\Delta^+$$ by a collision. The $$\Delta^+$$ certainly has enough energy to decay by $$\beta^+$$ emission, but the problem is that it has so much energy that strong force processes dominate and we typically get decay to a neutron and pion or a proton and pion. Weak force processes are so much slower that $$\beta^+$$ decay of a $$\Delta^+$$ would be fantastically unlikely.

• Similarly—and of interest to neutrino physicists—there is $p + \bar{\nu}_e \to n + \beta^+$ which is sometimes called "inverse beta-decay" though that is arguably a misnomer. It has a neutrino energy threshold of $0.8\,\mathrm{MeV}$ with the proton at rest. Commented Mar 20, 2019 at 0:35
• @dmckee cool, thanks :-) I've included IBD in my answer. Commented Mar 20, 2019 at 5:58