[liquidise]

> [isolated neutrons] decay into protons. During the process, each decaying neutron emits an electron and an antineutrino.

> ...detected sparks of light each time a neutron decayed.

Detecting a spark of light would also require photon(s) to be emitted, right? Is this not called out because it is a byproduct of the decay and not part of the decay reaction itself?

[sharikous]

No, they don't measure decays.

They let the neutrons decay for a while and then measure all the remaining neutrons in the trap by lowering there a detector.

This detector has a scintillator so when a neutron is captured it emits some photons and those are converted to an electric signal.

So no. There is no requirement for the neutron to emit an additional photon during the decay that is measured

[omgJustTest]

As the article says:” The team kept neutrons in the bottle for periods of between 20 seconds and nearly half an hour, and detected sparks of light each time a neutron decayed”

The decay products of neutrons is a proton, and electron and electron antineutrino. Light emissions would most likely be in the from acceleration or impingement of charge particles in the “bottle” magnetic field.

Clearly the article claims they are detecting “sparks”.

The reason the photons are not part of the primary emission is related to the momentum / energy balance in the decay. There are other conserved quantities such as lepton number.

As you say the light is a byproduct of the acceleration/ detections scheme of the charged particles emitted.

[LatteLazy]

Yes, the decay will also release some energy in the form of photons but they didn't mention it.

[a1369209993]

Neutron decay does not produce photons. Rather, the energy difference is carried as kinetic energy by the resulting proton, electron and (anti)neutrino. This can cause light to be emitted (very shortly) later when those particles slam into things, but as sharikous notes that doesn't seem to be they're talking about here.

[sharikous]

That's inaccurate, please see my reply to the same comment