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A delta resonance decays as given in http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/delta.html . I wonder, why is it not a radioactive decay? In principle, most/all decays should be radioactive as it is a quite broad description:

Radioactive decay (also known as nuclear decay, radioactivity or nuclear radiation) is the process by which an unstable atomic nucleus loses energy (in terms of mass in its rest frame) by emitting radiation, such as an alpha particle, beta particle with neutrino or only a neutrino in the case of electron capture, or a gamma ray or electron in the case of internal conversion.

https://en.wikipedia.org/wiki/Radioactive_decay

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The key point is that the technical definition is limited to atomic nuclei, the delta particle is definitely not a nucleus because it's lifetime is too short for playing any kind of role in an atom.

Said otherwise, it's impossible to form an atom out of a delta particle, and as such you don't consider the decay of a delta particle to be the decay of an atomic nucleus. It's really a matter of definition/convention.

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Usually we use the word "decay" for a state that's either metastable or that decays through the electroweak interaction. States that fall into these categories will tend to be relatively long-lived.

An example similar to yours that is not usually described using the word "decay" is neutron emission. Because there is no Coulomb barrier, a state that is unbound with respect to neutron emission is not metastable. (Cf. alpha decay, which is the decay of a metastable state, hindered by having to get out through the Coulomb barrier.)

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    $\begingroup$ It would not be unusual for a particle physicist to say that a Delta "decays" to to a nucleon and a pion, but that is about context. We wouldn't generally use the adjective "radioactive" in this case. $\endgroup$ – dmckee Mar 17 at 0:16
  • $\begingroup$ I think decay is used even for unstable states, but as dmckee says radioactive is specifically meant to be used for (meta-stable) nuclei of protons and neutrons only. $\endgroup$ – KF Gauss Mar 17 at 8:34

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