Free neutrons (outside the nucleus) are known to be highly unstable with a mean lifetime of around 14 minutes and 40 seconds. Inside the neutron, which is composed of 1 up quark and 2 down quarks, a down quark changes to an up quark by releasing a $W^-$ boson, and thus creating a proton.

$$d\rightarrow u+W^-$$

$$udd\rightarrow uud+W^-$$

$$n^0\rightarrow p^++W^-$$

The neutron has a decay energy of 0.782343 MeV. The energy differential is sufficient to create an electron (which just needs 0.511 MeV), an anti-neutrino and 0.269 MeV of energy.


$$n^0\rightarrow p^++e^-+\bar{\nu}_{e}+0.269 MeV$$

The remaining energy can be used as Kinetic Energy for expelling the electron. Or a gamma ray photon is produced which carries away the extra energy.

$$n^0\rightarrow p^++e^-+\bar{\nu}_{e}+{\gamma}$$

The electron require 13.6 eV to break free of the proton. In some cases the proton interacts with the electron, which is unable to achieve the 13.6 eV required, and as a result remain bound to the proton, forming a neutral hydrogen atom, with all the neutron decay energy being carried off by the neutrino. What determines whether the neutron decay results in production of a hydrogen atom? How can we ascertain whether the extra energy is carried away by the gamma ray photon, or by the neutrino or indeed used for the expulsion of the electron?

  • $\begingroup$ "How can we ascertain..." - are you asking how we can predict whether the energy is carried away by a gamma ray? Or are you asking how we can measure whether the energy is carried away by a gamma ray? $\endgroup$ Sep 30, 2020 at 19:28
  • $\begingroup$ Can we indeed predict (Yes !) whether hydrogen atom is produced, or the extra energy gets the electron expelled? And if the energy is not used by the electron (which binds with the proton resulting in the production of hydrogen), can we predict whether the energy is taken by the gamma ray photon or by the neutrino? Or is all these just a matter of chance? $\endgroup$ Sep 30, 2020 at 20:04
  • $\begingroup$ The current version of the question (v3) double-counts the electron mass when computing the $Q$-value of the decay. The endpoint of the beta energy spectrum is 780 keV, not 270 MeV. $\endgroup$
    – rob
    Sep 30, 2020 at 20:26

1 Answer 1


Ultimately, anything based on quantum mechanics (including quantum field theory) outputs probabilities that different events will occur, not deterministic outcomes. There is no factor, hidden or otherwise, that completely determines the outcome. Experiments demonstrating that the Bell inequality was violated eliminate any possibility that there's some hidden, deterministic local variable that's responsible for the outcome of specific events.

That said, the probabilities for these different processes can be predicted, using what we already know about quantum field theory and the weak interaction.

The interaction where a beta decay generates a photon is known as radiative beta decay. The mechanism is well known, and is depicted in the following three Feynman diagrams (from this paper):

enter image description here

In diagrams a and b, the photon is emitted as final-state radiation from either the electron or the proton, and in diagram c the photon is emitted as part of the interaction at the weak vertex. Radiative corrections for final-state radiation are a fairly standard procedure in any quantum field theory course, and they are done in much the same way here. The vertex calculations are a bit more tricky, but are still well within our understanding of the weak interaction in quantum field theory. For more details, see the paper referenced above. It turns out that roughly 4 of every thousand free neutron decays is a radiative free neutron decay.

The formation of a hydrogen atom from a free neutron decay, on the other hand, is a bit trickier, and fully explaining the derivation would require most of a research paper. Fortunately, someone has already published such a paper. This process is called bound-state beta decay, and the prediction, again, only involves applying the machinery from quantum field theory and our understanding of the weak interaction. The interaction $n\to H+\bar{\nu_e}$ is predicted to occur in roughly 4 out of every million free-neutron decays.

As a side note, a lifetime of 14 minutes is not "highly unstable" in terms of the Standard Model. The neutron is by far the most stable single hadron other than the proton; the next longest-lived hadrons are charged pions and kaons, which have a lifetime of a few tens of nanoseconds. Even the famously long-lived muon has a lifetime at rest of a few microseconds. Compared to most subatomic particles, the neutron's lifetime is an eternity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.