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If we consider a hypothetical situation in which $\beta^-$ deacay is happening without emitting any anti-neutrino, to be precise, a neutrinoless beta decay is happening. So, in this case, how the energy spectrum will look like?

I know the energy spectrum in case when the beta decay takes place along with emitting an anti-neutrino. But I do not have any knowledge what will the spectrum look like when the anti-neutrino is not being emitted. If anyone can provide the spectrum along with explanation, that will be a great help. I am attaching the spectrum when anti-neutrino is emitted.

Image Courtesy:https://www.radioactivity.eu.com/site/pages/Beta_Spectrum.htm

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    $\begingroup$ You may want to take a look at this article, where the details about neutrinoless double beta decay are discussed: “Neutrinoless Double-Beta Decay: Status and Prospects”, Michelle J. Dolinski, Alan W.P. Poon, Werner Rodejohann, Annual Review of Nuclear and Particle Science 2019 69:1, 219-251 $\endgroup$ Dec 1, 2021 at 3:18

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You would see a monoenergetic line. The fact that this isn't the case led Wolfgang Pauli to postulate the existence of the neutrino.

If the beta electron and the nucleus were all there was, then momentum conservation would dictate that after the decay, the two would fly apart in opposite directions. The entire decay energy (called the Q-value) of the decay would be split between the electron and the recoiling atom. Because of the huge mass difference (an electron weights 511keV$/c^2$, your example Bismuth-210 nucleus some 210GeV$/c^2$) that means that the electron flies away with approximately the Q-value of the decay.

By the way, experiments that search for neutrino-less double-beta decay do look for such a line in the combined energy of the two electrons coming out of the decay.

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  • $\begingroup$ So then if the neutrino is not emitted, it will become a two-body problem. So, as per my understanding, we will see a single line at the Q value i.e. the maximum kinetic energy in the curve and parallel to the Y axis. Am I right? $\endgroup$
    – Neutralino
    Nov 28, 2021 at 7:24
  • $\begingroup$ Right. Not exactly Q but Q minus the kinetic energy of the recoiling nucleus, but that's still almost Q. $\endgroup$
    – rfl
    Nov 28, 2021 at 11:51
  • $\begingroup$ Isn't a neutrinoless double-beta decay still a three-body final state? Why are the electrons monoenergetic in that case? $\endgroup$
    – rob
    Dec 1, 2021 at 1:46
  • $\begingroup$ Oh, I see. The question is about a hypothetical neutrinoless single-beta decay, and your answer about double-beta decay refers to a line in combined electron energies. Time for me to read more carefully. $\endgroup$
    – rob
    Dec 1, 2021 at 3:47
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It's hard to say what your hypothetical situation would look like because your hypothetical situation would break most of the laws of quantum mechanics. It would be a very different universe than what we have today (if there even would be a universe).

Instead of thinking of a hypothetical situation that could never occur, it might be more instructive to compare the beta decay to an alpha decay. In an alpha decay, there is no additional particle created, so the emitted alpha particle energy is a discrete value (not a distribution). The change in binding energy is distributed between the alpha particle and the final nucleus, and the distribution is determined by their relative masses.

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  • $\begingroup$ Actually this question was asked to me during my PhD interview. I also replied with the same logic that it would violate most of the laws of QM. But the professor was adamant that I answer this and draw the graph. He said that, think this is really happeing. Now draw it. Now what can I say? :P $\endgroup$
    – Neutralino
    Dec 4, 2021 at 12:28

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