From: David Roberson
I believe that the term gamma ray is reserved for photons
that originate from the nucleus. The energy of these rays is not the
criteria.
One would suppose that the energy contained within the
radiation emitted by the nucleus is determined by the energy steps between
the stored quanta. Dave, Bob In modern usage, and as taught at Universities today (so we should use this convention on vortex) the gamma ray is the highest energy photon and its point of origin is not usually considered relevant, only its frequency, or energy. I will go into greater detail below, since this terminology is a source of continuing confusion on the internet; plus there is one notable exception to the rule above. Gamma rays typically have frequencies above 10^19 Hz and energies above 100 keV but the dividing line on the low-end is arbitrary - and 100 keV is considered the standard, below which we find x-rays, regardless of point of origin - but there is one exception. All radiation from radioactive decay is defined as gamma, no matter what the energy level. That is usually no problem since the lower limit to gamma energy derived from radioactive decay is often around 100 keV anyway, and only in a few situations is the energy of nuclear decay sometimes less than that - tritium decay and neutron decay. In short, the nucleus CANNOT normally emit wavelengths below gamma, due to its own small size, since it must act as an antenna in order to radiate (and its small circumference would determine that wavelength limit, if there was no QM). Nevertheless, it is fair to say that visible light or UV cannot be emitted by any nucleus. Another reason to end the association of gamma radiation with the nucleus is that most gamma radiation (cosmologically) originates outside of any nucleus; and on earth gamma radiation that is NOT associated with a nuclear origin can easily arise from electron-positron annihilation or other kinds of matter/antimatter interaction, pion decay, bremsstrahlung, inverse Compton scattering, and synchrotron radiation. Historically bremsstrahlung or "braking radiation" was reserved for x-rays, regardless of energy - since it is usually produced in inner electron orbitals - but in modern usage - if the radiation has energy larger than 100 keV it should be called gamma or gamma bremsstrahlung. BTW - If we wanted to help out "another" theory with a plausible scenario - i.e. to invent a kludge which would make the "pre-radiation of adequate UV photons before the actual fusion event [DD->He]" explanation work, especially in the context of antenna theory, this can be done. However, I doubt anyone will borrow this: This explanation would be that D+D occasionally forms incompletely, not as 4He but instead as a two proton core - the diproton species (2He) with neutrons only slightly bound to this core, and at a substantial distance away (in short as a "halo"). This species can be called the "diproton with halo" and could shed the full 24 MeV, which cannot be done via electrons. The 2He nucleus does have a short lifetime, which is possibly extended long enough by having a halo to do the following: the two neutrons become separated in a remote halo orbital, from whence the circumference is adequate for them to shed UV photons (possibly in the 100+eV range) which are easily thermalized. This species (which will be called the "diproton with halo") could then be positioned to shed the full 24 MeV in as a few as 250,000 sequential photons, at the same time as the halo orbital is shrinking down. This would all transpire sub-nanosecond. In the end, the two halo neutrons spiral down to collapse into the 2He core, forming an alpha, but with almost no excess mass. The falsifiability is a matter of documenting the EUV emission.
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