Talk:Electron capture

26 v 25
Surely 26 should become 25 not stay as 26.

no, the only mass change in &epsilon; is the loss of an electron. -- metta, The Sunborn  19:54, 6 April 2006 (UTC)

The Sun
If a positron were emitted from one of the protons in the proton-proton chain reaction (in the Sun), surely the proton would lose mass, not gain mass, because antimatter does not have negative mass, and furthermore, a neutron is more massive than a proton, so surely the proton must undergo electron capture.--Lukeelms 00:13, 21 October 2006 (UTC)
 * PP isn't really beta decay, because that would require first creating a diproton (Helium-2), which is impossible, which would then decay to deuterium (Hydrogen-2). However, in other respects, PP is the same as other forms of positron capture.
 * The key is that the total mass of the D nucleus is less than that of the two H nuclei, or even the two H nuclei plus the positron.
 * Or, to put it the way most physicists think of things, the binding energy of D is 2.2MeV, compared to 0 + 0 for two H (or compared to the hypothetical diproton, which would have negative binding energy). As the article says, whenever the difference in binding energy is > 1.022MeV, positron emission is allowed. This is explained in more detail in the article on beta decay. --69.107.75.113 08:24, 16 March 2007 (UTC)

Is it realy correct to say that electron capture is the inverse of beta emmission? Beta particals are emmited into the surroundings and can travel several metres, but electron capture involves the nuclius absorbing an electrom previously orbiting the atom.
 * Well, whether it's correct or not, physicists actually do call it inverse beta decay (just as beta decay itself is sometimes referred to as electron, or positron, emission).
 * Is it actually correct? Well, from a particle physics point of view they're inverse, but from an atomic/nuclear physics point of view they're not, so it depends on what you're focusing on. --69.107.75.113 08:24, 16 March 2007 (UTC)

Perhaps some refferance should be made to the schrodinger equation in this artical explaining that functions with an envolope inside the nuclius will sometimes give small, but non-zero probabilities?

Inner bremsstrahlung
It might be worth mentioning inner bremsstrahlung, a gamma photon being emitted at the expense of the neutrino, especially since this is mentioned in the double electron capture and double beta decay articles (although admittedly the prospect of neutrinoless double capture is more exciting than bremsstrahlung in normal beta decay, given the consequences for the standard model).

However, I have no idea how to fit this into the article. (P.S., why does using even a single German word seem to invariably lead to writing ridiculously-long sentences?) --69.107.75.113 08:55, 16 March 2007 (UTC)

Binding energy curve
It might be illuminating to mention that, because of the shape of the binding energy curve, electron capture is mostly seen at the middle of the periodic table (above the Fe hump), and mostly with the lighter isotopes. While the examples fit this pattern, it isn't exactly obvious.

Also, Co-57 might be a better example, just because cobalt is so much more familiar to people that krypton or rubidium--and because it's probably the most-studied example (thanks to the Mossbauer effect, and the fact that it's just over the Fe hump in the curve). --69.107.75.113 09:13, 16 March 2007 (UTC)

Link Broken
The first referential link is broken. Needs to be fixed. —Preceding unsigned comment added by Sean keevey (talk • contribs) 13:18, 24 March 2008 (UTC)

Examples query
Is there any reason why the three reactions which are written out towards the top of the article aren't listed as common examples lower down? It's not important, it just seems odd... Djr32 (talk) 20:58, 16 November 2008 (UTC)

Proton bombardement?
Is this the same thing? ie.: When a high energy-proton collides with an atom, it causes the ejections of an electron from the outer layer of the atom.? --CyclePat (talk) 16:09, 6 August 2009 (UTC)

Captured electron nucleon status
When a proton captures an "orbital" electron, we then have a neutral charged combination of a proton and an electron. Is that combination considered to be identical with a neutron based on the standard model concept?WFPM (talk) 17:05, 21 May 2011 (UTC)


 * Don't forget the neutrino. As it says in the intro, the electron capture changes the proton to a neutron AND simultaneously emits a neutrino. Yes, the neutron formed is identical to any other neutron. However it cannot be considered as a proton-electron combination because the neutrino carries away energy, momentum and angular momentum (spin).
 * Historically Rutherford did think of the neutron as a p-e combination in the 1920s, but this idea led to incorrect predictions. The neutrino emission was added to the theory (mostly by Fermi) in order to restore agreement with experiments. Dirac66 (talk) 19:17, 21 May 2011 (UTC)

Am I supposed to worry about the mechanics of how one of the up quarks of the proton changes to a down quark as a result of this capture process? 1: the proton reels in the electron, then we have a proton plus an electron. Then 2: the electron changes an up quark to a down quark?, plus a neutrino? And as individual entities, the neutron is more massive than the proton (plus the electron). Except inside of an atom, I guess. I'm specifically thinking about ec in the context of EO54Xe127 going spontaneously to OE53I127, which is the only reason there is any stable 53Iodine. Any comment?WFPM (talk) 13:04, 22 May 2011 (UTC) Note that to get to OE53I127 from OO53I126 involves an incremental mass increase of 0.998849 amu's,(for an incremental neutron addition), whereas to get there from EE54Xe126 only requires an increase of 0.99656 amu's for a proton addition. So it takes more energy to add on the neutron than it did for the proton (per the CRC handbook).


 * In the quark model, yes, the neutrino (ν) is still involved: u + e → d + ν. I don't think the process (or mechanics) can be divided into the two steps you suggest. Discussions I have seen involve quantum field theory and W bosons which are out of my depth, so I don't worry about them.
 * As for the mass (or energy), the replacement in the nucleus of a proton by a neutron would be endothermic for a free proton, but in a process such as Xe-127 to I-127 the changes in the strong interactions between nucleons must be exothermic enough to dominate the overall process. Dirac66 (talk) 18:54, 22 May 2011 (UTC)

Well, as V Smith said "It's complicated aint it?" And I'm thinking about posting a profile chart about the isotopes of Iodine like I did in Talk:Isotopes of lead except in the case of 53 Iodine, there might be some irregularities of data worth discussing. I think most people don't pay much attention to the implication of some of the charts' details and don't think much can be learned from a study of them; and I think a lot could be learned by paying attention and learning the implications of some of the "irregularities" in the details.WFPM (talk) 20:51, 22 May 2011 (UTC)And as for the "electron capture" situation I see their analogy to the 2 nucleon Deuteron to 2 neutron change situation, and I see no way that could be an exothermic reaction. And did you ever see the May, 1985 National Geographic article and wish to comment?WFPM (talk) 22:11, 22 May 2011 (UTC)
 * We know Xe-127 → I-127 is exothermic because this decay is observed. This clearly does not apply to d → 2n which is not observed, and which is endothermic from the observed masses. If you want a general method to predict which EC will be exo (possible) and which will be endo (impossible), then you need the theory of nuclear stability. And as I have explained to you previously, this requires quantum mechanics. Sorry but one can only understand so much without getting into QM. Dirac66 (talk) 23:39, 22 May 2011 (UTC)

Well I'm kind of a Science fiction fan/Engineer/free thinker and I go where my interests lead me. And I think that the best way to learn about a 3 dimensional real physical entity is to stick with the best description that you can achieve within the 3 dimensional space plus time continuum, and then try to deal with any unusual details. But I'm with Newton in that I'm partial to the simplest solution. And of course that involves a process of collecting and examining all the data, with particular attention to discrepancies. So when Feynman et al tells me that I should learn about QM including that is not understandable, I think that the time might be better spent rationalizing events in terms of what I think I know rather than trying to learn an alternative concept that he says in not understandable.WFPM (talk) 03:21, 23 May 2011 (UTC)

I'll give you another hypothetical question. Say I was able to build a tubular enclosure (pipeline) between me and a distant source of light. Is there any reason to doubt that I would be able to see the light source through the pipeline as the distance got longer?WFPM (talk) 02:00, 25 May 2011 (UTC)

units of half life for common examples?
In the Common Examples section, the electron capture half lives of some elements are given, but there is no explanation of the units. For instance, "55Fe - 2.6 a". What is "a"? This may be a standard unit for those with the specific domain knowledge, but lots of people reading this won't know it. 24.79.82.67 (talk) 15:10, 15 May 2012 (UTC)


 * d = day, a = an = year (in French). Évidemment! I don't know why French is used here, but for some reason this seems to be the standard abbreviation in nuclear physics. But I agree that this is confusing in an English encyclopedia for general readers, so I'll change the "a" in the table to "yr" which should be clearer. Dirac66 (talk) 16:25, 15 May 2012 (UTC)

Fluorine-18

 * If all the protons of OO9F18 were paired with neutrons, (as they should be to minimize the atomic mass value), then the occurrence of "electron capture" would require that one of the 9 deuterons would have to be broken up to create 2 extra neutrons. And I don't know of any reduced mass value for 2 paired neutrons. So for this to happen, where would the OO9F18 get the (2Mev} activation energy required to break up the proton-neutron pair? And is there any evidence that this decay mode of the OO9F18 is valid for the atom at a temperature of 0 degrees Kelvin? And why isn't that EC occurrence mentioned in the article?WFPM (talk) 23:17, 17 June 2012 (UTC)
 * According to the table at Isotopes of fluorine, F-18 decays not by electron capture but by β+ emission. So that is why this isotope is not mentioned in this article about EC. I will not try to analyze why it decays by β+ emission, but it clearly does since this isotope is the main source of positrons for medical PET. Dirac66 (talk) 01:03, 18 June 2012 (UTC)
 * This table only shows major decay routes. Of course, all isotopes that decay by positron emission also decay by EC (though sometimes at lesser probabilities). The reverse is not true if the isotope is proton rich but doesn't have enough excess energy for e+ decay, in which case only EC is possible. In the case of F-18, the branching ratio is 96.86 (19) % by beta plus (positron emission), and 3.14 (19) % by EC. As for WFPM's question, as usual, he's assuming that which is not in evidence -- in this case a model of F-18 in which is composed of 9 deuterons. It's not composed of 9 deuterons. That's all that need be said. If you think it is, WFPM, answer your OWN question! S  B Harris 01:45, 18 June 2012 (UTC)


 * I'm trying to!! But OEH3 Tritium goes the other way, and the incremental mass value for excess neutrons is skewed in favor of multiples of 2, as well as EE6C14 and OO7N16. So we've got this quick change of tendency between Nitrogen and Oxygen. And maybe you can visualize these atomic nuclei without seeing proton-neutron pairs but I can't. As you have explained it's a packing problem involving some magnetized particles and your images neglect the magnetic property and utilize spherical nuclei which don't indicate its physical orientation. A question? when you bombard a target with deuterons and get a merge, what do you use for the incremental deuteron mass value in the reaction calculation? I'll bet it's the deuteron mass value increment. And that implies that the merged composite atom didn't disassociate within the nucleus or at least you assumed it didn't.WFPM (talk) 02:48, 18 June 2012 (UTC)


 * Well, that's your problem. In light nuclei the N/P ratio tends to be about 1:1, so you can look at them as NP pairs, or you can look at them equally well has having been built out of helium/alphas, which of course they were. After Ca-40 there aren't any stable nuclei with N = P, so you need to have excess N. By the time you get to the heaviest nuclei the N/P ratio is about 3:2. I see no deuterons at 3:2. There's no mystery why N and P tend to be even: it's just the same in molecules where the Pauli principle tends to make electrons come in pairs since they can sit in the same orbital in a pair with no extra energy cost (one spin up, the other spin down). Neutrons and protons are the same way. Even-P elements are more abundant and have more stable isotopes. Even N isotopes are common for the same reason-- neutron pairing has nothing to do with deuterons-- it's just neutron pairing. No more mysterious than electron pairing, and happens for the same reason. S  B Harris 04:26, 18 June 2012 (UTC)


 * So you're in favor of neutron pairing. And I see extra neutrons as a balancing factor in the nucleus, which involves their being separately located. And I agree with you that pairing involves 1 upspin with 1 downspin. So the question comes up as to whether identical particles can pair up and we know that PN's can and PP's cannot. That might explain the OE1H3 instability, with the P's staying as far apart as possible. But you ought to like EE6C14 and OO7N16 if you're willing to pair N's. I have nothing against the stability of EE8O18, but I don't see any paired N's in there and the question is as to what's wrong with OO9F18? If the PN's are paired then why should 1 of them break up and where did the breakup energy come from? And in trying to balance the nuclear structure of unbalanced (odd Z) atoms I have to add a number of neutrons I notice that the even number of neutron additions is superior for achieving and/or maintaining stability but I still don't see any neutron pairs, because the neutrons have to be added in a balancing manner to the nucleus, which wouldn't be in associated pairs.WFPM (talk) 15:32, 18 June 2012 (UTC) And I suggest to you that it is more important to know that EE92U238 consists of 92 deuterons plus 54 extra neutrons than that the N/P ratio is 3.174/2.WFPM (talk) 16:38, 18 June 2012 (UTC)


 * C-14 and N-16 gain a lot of stability from neutron pairing-- C-14, with pair of N's AND pairs of P's, has a half-life of thousands of years even though it has too lopsided a P/N ratio for a nucleus of its size. Poor N-16 has paired neutrons but not paired protons. Neutron pairing, even with proton-pairing, won't save you from an infinite excess of N, you know-- and that excess is smaller with smaller nuclei. By the time you get up to oxygen at Z=8, you can take 2 extra N over your even P number, and stay stable. That's the first it can be done. That's not possible with lighter elements, is all. What's "wrong" with F-18 has been explained to you-- it isn't "composed" of 9 deuterons (P and Ns are NOT paired with each OTHER-- that's YOUR mad idea, not mine), and that's it. So it's not stable. There is some limited evidence that "alphas" exist in a sort of way in nuclei, particularly heavy ones, but deuterons-- no. Protons are not paired with neutrons in nuclei-- they don't even notice each other. They are in separate orbital-sets like electrons; one for all the protons, one other set for all the neutrons. Orbitals like to have NN and PP pairs, due to Pauli. F-18 is a perfect example of an odd-odd nucleus that could become even-even (with all P's and N's paired with other P's and N's, if it simply underwent positron emission or electron capture. This would allow pairing of both protons AND neutrons, even though they would not have equal numbers, but that wins out here, since, as noted, O-18 is the lightest stable element to have N-P = 2. It has one more neutron pair than proton pair, and gets away with it. I don't know what you think a "balancing" addition of neutrons is, but I assume you that pairs of neutrons are like pairs of electrons in an atoms. The "pair" is everywhere, with wavefunction spread out over a large nuclear volume. Nucleons don't simply sit there like marbles. And you might think of U-238 as 92 deuterons and 54 extra neutrons, but nobody else does. A nuclear physicist would use the nuclear shell model and guess that U-238 is blessed with 82, 8, and 2 protons in closed proton shells, and 126 and 20 neutrons in closed neutron shells. And that's why U-238 is very nearly stable, and other configurations (+/- a neutron or +/-a proton) are NOT. S B Harris 22:02, 18 June 2012 (UTC)

Well I certainly appreciate your attention and response to my comments about my concepts and I'll have to return to my Kaplan as well as Wikipedia to mull over some of what you tell me. But if I have to have the P's and the N's paired with each other and I'm going to have a hard time explaining why EE4Be8 should require an additional extra n to be stable. and what kind of pairs do you have in OO3Li6? And if equal particles pair in opposing spin conditions, and if closeness of association {packing) is related to reduced free energy content I would have a hard time melding your 3 domains of protons plus 2 domains of neutrons into a EE92U238 composite nucleus with close or closer spacing than Dr Urey's deuteron particle. And I better understand why you are not concerned with the order of presentation of the nucleons within an atom, and given the existence of these domains I don't see why the emission of an alpha particle (two from each domain) is so common. But thank you.WFPM (talk) 01:38, 19 June 2012 (UTC)
 * The emission of alphas is common because the neutrons and protons interpenetrate so they aren't in any separate "domains." Their wavefunctions are in orbitals and shells like electrons in atoms. The helium nucleus is a thing of beauty and joy forever, simply because it is the only bound structure of nucleons in which no nucleon has any orbital angular momentum. It's like the helium atom with two electrons in the 1s orbital, except in the nucleus there are 2 protons in their 1s and 2 neutrons in the same space in THEIR 1s. Add a fifth particle of any kind and it must go into a 1p and have orbital momentum and a lot more energy, so that is why no 5 nucleon nucleus is stable. Atoms spit out an alpha whenever they can, since the energy of binding of those 4 particles is so high, and that energy is available to drive the process. Be-8 cannot resist breaking into two alphas. The extra neutron inhibits this only because when the two alphas leave, they need to leave the free neutron with all the mass-energy a free neutron has and needs (this can't go with either alpha, for reasons explained) and leaving the free neutron with alimony and child support costs a ton of energy. So the alphas stay together for the sake of the little thing. Excuse the metaphor. S B Harris 01:51, 19 June 2012 (UTC)

And yet you still don't like a planar structure alpha concept. It has to have some kind of structure. And the problem comes when you try to bind 2 alphas together. So I assumed the neutron did that by binding one corner, because the core of my concept is around a cubic EE4Be8 atom, with added deuterons and balancing extra neutrons. I like Gamow, but I don't like his triple-alpha accumulation concept. And if I could sell the 4Be8 + 2 deuteron concept to get EE6C12 then I move all the alpha created deuteron accumulation processes out to the end of each series like in the Janet periodic table. And after that it's initially loosely bound by extra neutrons and subject to bombardment and unbalanced forces. So the deuterons are accumulated in layers, with each alpha particle being the last 2 deuterons forming an alpha particle on the top. Everywhere else you have side-bonded deuterons, which would be hard to convert to alpha particles. And they're all spinning in synchronous "contact?" with each other and the atom also is spinning. Like in gears and magnetized cylinders. You've got to admit that it's pretty as compared to the popcorn ball alternative.WFPM (talk) 13:48, 19 June 2012 (UTC) You might note Dr Pauling's comments about "hypothetical structures" in his 1969 "General Chemistry" book page 94 (ISBN 0-486-65622-5) (paperback) Also page 860.````


 * Once again, the charge density of an alpha has been measured. It is not planar. It is maximal at the dead center, and it falls off exponentially and symmetrically from there. It look exactly like the electron charge structure of the helium atom, and for the same reason (the protons are in 1s orbitals in the nucleus, just as the electrons are in 1s orbitals outside the nucleus). That is all. You can't argue with experiment. You keep asking "Why is nature like THIS, when my pet model demands THAT?" How many times do I have to say that it's because your pet model is wrong? No other explanation is needed. Oxygen-16 has no spin; the reason for that is obvious. Now, tell me why O-17, which is 0-16 plus one more neutron, has a spin of 5/2? And why F-17 has the same spin? Looks like either an extra P or N go into a new orbital with total spin of 5/2. But F-18 has a spin of 1. If you put in both a P and an N, the orbital spins cancel, and all you have left is the particle spins of 1/2, pointing in the same direction. Oxygen-18 has no spin at all, just like O-16. Why? The two extra neutrons go into the same neutron-orbital, cancelling orbital momentum, but now one can go in spin-up, the other spin-down, and the net is zero. S  B Harris 19:42, 20 June 2012 (UTC)

I don't understand nuclear orbitals, because I used 3/8"dia Neodymium cylindrical magnets to make some of my models and their properties and interrelationships are not compatible with the idea of orbital type motion within the dense nucleus. I also don't understand net spin values like EE4Be8 is Zero (Agree), EO4Be9 is (-3/2)? and OE5B10 is (+3)?. In my models, all the protons spin in one direction and neutrons in the other. However if you turn the model upside down, the noted direction of the spin reverses, and it's possible to have a confusion about spin direction due to that factor. So I can see how the addition of 2 neutrons might be added up to net zero spin.WFPM (talk) 04:00, 21 June 2012 (UTC) So when we talk about spin, we must admit that they're all spinning, and we're talking about the net spin based on some criteria of direction.WFPM (talk) 13:44, 4 July 2012 (UTC)

By Whom comment
This evening I added a "by whom" to the first paragraph under "Reaction Details"

After more browsing, I found a couple of references that might be relevant to the basic physics, but so far no references to the cosmological assertion.

H. Irnich et al., Phys. Rev. Lett. 75, 4182 (1995).

Yu.A. Litvinov et al., Phys. Lett. B 573, 80 (2003).

Update, same poster: I found a reference that might be suitable. Here's a raw URL: http://articles.adsabs.harvard.edu//full/1964ApJ...139..318B/0000335.000.html

I plan to inquire in the physics forums whether this will serve. — Preceding unsigned comment added by 76.115.88.202 (talk) 04:22, 9 September 2012 (UTC)

How is this possible
Hi, as an interested layman I would like to know how the capture of an electron can turn a proton into a neutron. Both particles consist only of three quarks. I suppose electron capture turns one up quark into a down quark, thereby quanging the hadron's nature. But where does the electron stay? In the quark? Probably not, quarks are not thought to be compound. In the neutron, between the quarks? That would mean neutrons consist of more than just their three quarks. Well, if neutrons contain an electron, that would explain both their neutral charge and their slightly bigger mass.

If the article is slightly confusing to me, it might very well be my fault. On the other hand, there are more people who want to find out more about these topics, so it might be worth while explaining them. Steinbach (talk) 23:17, 17 June 2014 (UTC)
 * in fact it is not correct to think with "classical" arguments. In fact electron and quark are not balls. The idea is just that, as shown in the diagram on the right, a W+ boson is created via the interaction of the electron with the up quark. The boson disintegrates almost immediately into one neutrino and a down quark. It is well described by the electroweak interaction. Pamputt (talk) 20:04, 19 February 2015 (UTC)

Some changes
I've rewritten the first paragraph to include the X-ray, Auger electron and gamma ray emissions. The distinguishing text at the top of the article was also made more specific, and the caption to the diagram was greatly expanded. The caption makes the diagram box large; this could be turned into a show more tab or removed. Why does the atom in the diagram have one electron each in three shells? Roches (talk) 03:35, 24 April 2015 (UTC)

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Quark, W-boson representation needed
This topic is incomplete without quark-boson representation, the full equations of how the reaction proceeds.

Asgrrr (talk) 22:02, 15 August 2017 (UTC)

Electron capture in "non-proton-rich" nucleus of an atom.
I have curiosity for this thing. If anyone knows, please answer in details. If there is a neutral or non-neutral atom, which is "not-proton-rich" in its nucleus, how the electron capture can occur or how it can be done?

Question from Milind Chatrabhuji from Vadodara, Gujarat, India.

Auger effect
"In the Auger effect, the energy absorbed when the outer electron replaces the inner electron is transferred to an outer electron. "

Should this not be "the energy released?" 165.255.60.165 (talk) 06:34, 15 October 2022 (UTC)


 * Maybe. But if you think of the energy as being absorbed by the atom and then getting transferred to an outer electron, it might be okay as it is. P.E of the first electron is converted to K.E. as it falls, I guess, which is not clearly absorption, nor release. Somehow that K.E, if that's the right thing to call it, ends up as K.E. of the outer electron, and eventually P.E. of the outer electron, to put it in classical terms. Not sure what this would be in QM terms. Polar Apposite (talk) 18:24, 13 July 2023 (UTC)

Electron capture in astrophysics
This page could use an "Electron capture in astrophysics" section.

Sadly, I'm unqualified to write it.

Web search for `"electron capture" astrophysics"' brings up a number of promising surveys which might help. 50.0.193.12 (talk) 06:25, 6 April 2023 (UTC)

EC representation
In the above as it stands, charge conservation is violated, either write Mg– in the right side, or do not write e– in the left side. 93.150.81.118 (talk) 16:36, 22 June 2023 (UTC)

Energy Difference
In the article it says:

If the energy difference between the parent atom and the daughter atom is less than 1.022 MeV, positron emission is forbidden as not enough decay energy is available to allow it.

But the rest mass of a positron is 0.511 MeV. 1.022 MeV is the minimum energy needed to create an electron-positron pair. — Preceding unsigned comment added by 213.80.51.126 (talk) 13:47, 27 June 2023 (UTC)


 * In comparison with an electron capture process, the positron emission one doesn't gain one electron mass and has to produce one electron mass, so the difference is exactly two electron masses. 93.150.81.118 (talk) 18:12, 30 June 2023 (UTC)

Decay energy
"If the energy difference between the parent atom and the daughter atom is less than 0.511 MeV, positron emission is forbidden as not enough decay energy is available to allow it, and thus electron capture is the sole decay mode. For example, rubidium-83 (37 protons, 46 neutrons) will decay to krypton-83 (36 protons, 47 neutrons) solely by electron capture (the energy difference, or decay energy, is about 0.9 MeV)." This seems to make no sense because 0.9 MeV is *more* than 0.511 MeV. Polar Apposite (talk) 18:13, 13 July 2023 (UTC)

Electron capture is broader than suggested in the opening paragraph - it also happens in neutron star formation
Previously someone else wrote: 'Electron capture in astrophysics. This page could use an "Electron capture in astrophysics" section.'

I concur. I'm not a physics graduate, so I'm also not volunteering to write it. MathewMunro (talk) 03:04, 7 January 2024 (UTC)

List of nuclides whose theoretical beta plus energy does not exceed 1.022 MeV
This table does not include those nuclides with theoretical beta plus energy greater than 1.022 MeV but with no positron emission observed according to NUBASE2020: 85Sr to 85Rb (decay energy 1.0647 MeV), 134Cs to 134Xe (decay energy 1.2333 MeV), 146Pm to 146Nd (decay energy 1.4712 MeV), 146Gd to 146Eu (decay energy 1.0291 MeV), 170Hf to 170Lu (decay energy 1.0563 MeV), 174mLu to 174Yb (decay energy 1.5449 MeV), 192Ir to 192Os (decay energy 1.0473 MeV), 227Pa to 227Th (decay energy 1.02558 MeV) and 252Es to 252Cf (decay energy 1.26 MeV).

No such nuclide exist for Z = 2, 5~17, 19, 21, 28, 29, 30, 35, 39, 41, 44, and for Bi and Po there are only isomers (210mBi and 212mAt). Pb and Cm are the only known elements with four isotopes (200,202,203,205Pb and 238,240,241,243Cm) with beta plus energy not exceeding 1.022 MeV; the other elements have at most three. Gd is a near miss, as the beta plus decay energy of 146Gd is barely higher than 1.022 MeV, so there would be no hope to observe its actual positron emissions.

In this table, proton excess means (atomic number of the given nuclide) - (atomic number of its isobar with the lowest energy). Characterizations of proton excesses:

Proton excess 1: Neutron-deficient odd-mass nuclides with the EC products being beta-stable. The only known exceptions are 164Ho, 180Ta, 236Np, and 242Am which are odd-odd.

Proton excess -1: Neutron-rich odd-odd nuclides sandwiched by two beta-stable even-even isobars. The only known exceptions are 90mY, 182mTa and 210mBi whose EC products are not beta-stable, because the energies have 90mY>90Sr>90Y, 182m2Ta>182Hf>182Ta, and similarly 210mBi>210Pb>210Bi.

Proton excess 2 or 4: Neutron-deficient even-even nuclides. The only known exceptions are 145Sm and 213Rn, as witnessed by the low energy difference between 145Sm and 145Nd (779.40 keV) and between 213Rn and 213Po (955.14 keV). Neither 213At → 213Po nor 213Rn → 213At has been observed due to the high alpha-instability of 213At and 213Rn; see here and here. Curiously, both of the neutron numbers (83 and 127) are one plus magic numbers. It is likely that 267Db would be the third such nuclide (see below).

Proton excess 0: Isomers of beta-stable nuclides. The only known example is 212mPo, and the energies have 212mPo>212Bi>212Po.

The only known listed nuclides with proton excess 4 are 152Dy and 164Yb. 266Sg, 270Hs and 272Hs could also be examples (see below). It is very likely that 266Sg, 270Hs and 272Hs are also in the list above, but more precise measurements of atomic masses are required to confirm. For none of these nuclides the process of electron capture has been observed. Their proton excesses:

Note that alpha decay is or is estimated to be not ignorable for some of these nuclides with N < 126, i.e., the following nuclides are not stable enough even fully ionized: 129.104.241.214 (talk) 23:54, 13 February 2024 (UTC)