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As discussed in the relevant Wikipedia article, beta decay via electron capture is possible in circumstances when positron emission is not.

From a simple (and perhaps naive) point of view, the two decay modes should be $$ ^A_ZX + e^- \rightarrow \ ^A_{Z-1}Y+ \nu_e $$ (for electron capture) and $$ ^A_ZX \rightarrow \ ^A_{Z-1}Y+e^+ + \nu_e $$ (in the case of positron emission).

Clearly, these two reactions are related through crossing. However, there are isotopes in nature which permit the first but not the second. This would appear to violate crossing symmetry since an allowed reaction is related to a forbidden reaction through crossing.

The usual heuristic explanation for why positron emission is sometimes energetically forbidden is that the reduction in nuclear charge necessarily requires releasing one of the bound-state electrons into the continuum (meaning that positron emission is really the emission of two particles not one). In that case, the reactions would look like $$ [^A_ZX]^{0+} \rightarrow \ [^A_{Z-1}Y]^{0+} + \nu_e $$ (in the case of electron capture) and $$ [^A_ZX]^{0+} \rightarrow \ [^A_{Z-1}Y]^{0+} + e^- + e^+ + \nu_e $$ (for positron emission), which are obviously not related through crossing.

But this way of viewing the reaction raises several questions: Would a bare nucleus be able to decay through positron emission more easily? Does the weak force still mediate these atomic decays which seem to actively involve the "spectator" electrons?

More seriously, what is preventing a "two step" reaction that begins with the emission of a lone positron but is followed by a separate "relaxation" reaction to release the electron? Alternatively, if the atom were surrounded by positive ions with which the negatively charged decay ion could easily form an ionic molecule, would electron-less positron emission become an allowed decay route?

There are many other related questions, but they all boil down to the same main issue:

Can the asymmetry between positron emission and electron capture be accounted for without violating crossing symmetry? And, if so, how?

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    $\begingroup$ Possibly related; also related. $\endgroup$
    – rob
    Commented Mar 19, 2023 at 0:47
  • $\begingroup$ Have a look hyperphysics.phy-astr.gsu.edu/hbase/Particles/parint2.html , and note that it is discussed, the symmetry applies with interactions and decays involving elementary particle states, not collective or nuclear ones. $\endgroup$
    – anna v
    Commented Mar 19, 2023 at 4:54
  • $\begingroup$ @annav The page that you linked doesn't say anything about applying only to elementary particles. In fact, it specifically says that "crossing symmetry applies to all known particles." Regardless, if what you say is right, then that opens up a serious can of worms since any interaction between composite systems should be able to be decomposed at some level into interactions between elementary constituents. If there is a specific principle at play here, I'd love to know what it is. $\endgroup$
    – Geoffrey
    Commented Mar 21, 2023 at 3:29
  • $\begingroup$ when there are many elementary particles present , to model them with the Feynman diagrams that have the crossing symmetry is an impossible task. New quantum models are needed to model the collection and the crossing symmetry cannot manifest. imo. $\endgroup$
    – anna v
    Commented Mar 21, 2023 at 4:47

1 Answer 1

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If you are interested in spectator electrons, it's helpful to include them in your decay relations:

\begin{align} \left[ Z e^- + {^A_Z X} \right]_\text{bound} &\to \left[(Z-1)e^- + {^A_{Z-1}Y} \right]_\text{with hole} + \nu_e \tag{$e^-$ capture} \\ \left[ Z e^- + {^A_Z X} \right]_\text{bound} &\to \left[Ze^- + {^A_{Z-1}Y} \right]_\text{bound}^{1-} + e^+ + \nu_e \tag{$e^+$ emission} \end{align}

This formulation makes it clear that electron capture has a final state with $Z-1$ charged leptons, while positron emission has $Z+1$ charged leptons in the final state. You can get into the weeds about the energies associated with the daughter ion from the positron emission, or the more-significant energies associated with the inner-shell electron vacancy in the daughter atom following capture. But the biggest effect on Q-value is that you have an additional rest mass of $2m_e$ in the positron-emission reaction. If two decays have all of the same quantum numbers, the one which releases more energy goes faster, so electron capture is nearly always preferred over positron emission.

Note that you're already tracking spectator electrons implicitly when you compute decay Q-values, because tables of atomic masses are for neutral atoms, and therefore include the masses and binding energies of the electrons in the neutral atom.

Would a bare nucleus be able to decay through positron emission more easily?

A bare nucleus would have zero branching ratio for electron capture, since there are no electrons to capture. The partial width for positron emission would remain (approximately) the same without the spectator electrons. You wouldn't get more positron emission from a bare nucleus, but positron emission wouldn't compete with electron capture.

A famous example is the electron-capture decay $\rm^7Be \to {}^7Li + \nu_e$, whose decay energy is less than $2m_e$, so that the positron-emission channel is forbidden. On Earth, beryllium-7 is unstable, but bare beryllium-7 nuclei may travel interstellar distances as cosmic rays.

Does the weak force still mediate these atomic decays which seem to actively involve the "spectator" electrons?

Yes. In order for the neutrino to appear in the final state, the decay must involve the charged weak current.

More seriously, what is preventing a "two step" reaction that begins with the emission of a lone positron but is followed by a separate "relaxation" reaction to release the electron?

The fate of a negative ion depends on its environment; you'd have to ask a condensed-matter person about that. Note that the positron is generally energetic ionizing radiation, which will lose energy by creating lots of electron-ion pairs as it travels through the medium. If you have a mega-eV decay in a material with a ten-eV ionization energy, you'll have about $10^5$ ions to worry about.

Alternatively, if the atom were surrounded by positive ions with which the negatively charged decay ion could easily form an ionic molecule, would electron-less positron emission become an allowed decay route?

Chemical bonds like this typically have eV-scale energies, which are irrelevant to mega-eV-scale decay processes.

Can the asymmetry between positron emission and electron capture be accounted for without violating crossing symmetry?

Crossing symmetry tells you that the matrix element for the interaction is the same. But computing a decay rate, or even whether a decay is allowed or not, also requires that you consider the initial and final energies. To get the same interaction probability from the same matrix element, you'd need additional energy in the initial state, by e.g. irradiating your radionuclide with fast electrons. But the fast electrons can participate in electromagnetic interactions which the slow electrons cannot, which breaks the symmetry on the initial-state side.

Crossing symmetry was relevant to the Homestake solar-neutrino experiment which used the reaction

$$ \nu_e + {\rm^{37}Cl} \to {\rm^{37}Ar^+} + e^- $$

to "reverse" the argon-37 electron-capture reaction. That works because the neutrinos which irradiated the chlorine can only participate in the weak interaction.

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  • $\begingroup$ This mostly makes sense to me, but part of your answer highlights my confusion. Using the semi-empirical mass formula, the mass difference between bare $^7Be$ and bare $^7Li$ is $\Delta m\approx 0.95 MeV/c^2$. This should be sufficient to allow for positron emission decay of the bare nucleus if only $m_e$ is needed but not if $2m_e$ is needed. If we observe deep space $^7Be$ cosmic rays, that suggests that the decay is still strictly forbidden. Why does this decay still not occur even though the capture of a free electron by the bare nucleus would obviously be allowed? $\endgroup$
    – Geoffrey
    Commented Mar 22, 2023 at 21:54
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    $\begingroup$ I think your data are wrong. The mass difference between neutral Be-7 and neutral Li-7 is about 0.9 MeV, according to the NNDC. Neglecting the sub-keV correction from electron binding energies, the Q-value for positron emission is about $-$0.1 MeV. $\endgroup$
    – rob
    Commented Mar 22, 2023 at 23:33
  • $\begingroup$ I think that I finally understand. If the neutral atoms differ by $<2m_e$ then the bare nuclei will almost certainly differ by $<m_e$ (since any realistically-sized atom will never have total electron binding energies that could account for enough of a difference). As a result, the decay of the bare nucleus will be forbidden by straightforward conservation of energy in the rest frame and, therefore, so will the atomic decay. For this reason, the decay doesn't actually involve the spectator electrons -- it's all just relativistic kinematics. Does that sound right? $\endgroup$
    – Geoffrey
    Commented Mar 23, 2023 at 14:43
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    $\begingroup$ That’s right. To find inner-electron binding energies, look up the energies of the K-shell x-rays emitted when vacancies in those shells are filled. For heavy nuclei, it’s hundreds of keV, but less than the electron mass. $\endgroup$
    – rob
    Commented Mar 23, 2023 at 16:37
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    $\begingroup$ I'm not comfortable speculating about weird edge causes in these comments. As a rule, the Q-value for positron emission is $2m_e$ less than the Q-value for electron capture. One of the "linked posts" on this page discuss an edge case in $e^-$ decay, and I think it links to some resources that you might mine for other edge cases if you're interested. $\endgroup$
    – rob
    Commented Mar 24, 2023 at 3:58

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