Talk:Electron capture

Physics Mid‑importance

	Physics portal This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.PhysicsWikipedia:WikiProject PhysicsTemplate:WikiProject Physicsphysics articles
Mid	This article has been rated as Mid-importance on the project's importance scale.

Astronomy Mid‑importance

	Astronomy portal This article is within the scope of WikiProject Astronomy, which collaborates on articles related to Astronomy on Wikipedia.AstronomyWikipedia:WikiProject AstronomyTemplate:WikiProject AstronomyAstronomy articles
Mid	This article has been rated as Mid-importance on the project's importance scale.

26 v 25

Surely 26 should become 25 not stay as 26.

no, the only mass change in ε is the loss of an electron.
--metta, The Sunborn 19:54, 6 April 2006 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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).[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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 H arris 01:45, 18 June 2012 (UTC)[reply]

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)[reply]

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 H arris 04:26, 18 June 2012 (UTC)[reply]

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)[reply]

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 H arris 22:02, 18 June 2012 (UTC)[reply]

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)[reply]

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 H arris 01:51, 19 June 2012 (UTC)[reply]

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.````[reply]

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 H arris 19:42, 20 June 2012 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

@Steinbach: 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)[reply]

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)[reply]

External links modified

Hello fellow Wikipedians,

I have just modified one external link on Electron capture. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:

Added archive https://web.archive.org/web/20131226210737/http://www.ca.infn.it/~oldeman/resneu/physscr5_T59_030.pdf to http://www.ca.infn.it/~oldeman/resneu/physscr5_T59_030.pdf

When you have finished reviewing my changes, please set the checked parameter below to true or failed to let others know (documentation at {{Sourcecheck}}).

This message was posted before February 2018. After February 2018, "External links modified" talk page sections are no longer generated or monitored by InternetArchiveBot. No special action is required regarding these talk page notices, other than regular verification using the archive tool instructions below. Editors have permission to delete these "External links modified" talk page sections if they want to de-clutter talk pages, but see the RfC before doing mass systematic removals. This message is updated dynamically through the template {{source check}} (last update: 5 June 2024).

If you have discovered URLs which were erroneously considered dead by the bot, you can report them with this tool.
If you found an error with any archives or the URLs themselves, you can fix them with this tool.

Cheers.—InternetArchiveBot (Report bug) 07:22, 22 December 2016 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

EC representation

2613 Al

+ e - \to

2612 Mg

+ ν e

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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).

Nuclide	A	Z	N	Decay energy (keV)	Branching ratio (%)	Proton excess to the isobar with the lowest energy
⁵Li	5	3	2	292.65	0	1
⁷Be	7	4	3	861.893	100	1
³⁷Ar	37	18	19	813.873	100	1
⁴¹Ca	41	20	21	421.315	100	1
⁴⁴Ti	44	22	22	267.63	100	2
⁴⁹V	49	23	26	601.856	100	1
⁵¹Cr	51	24	27	752.576	100	1
⁵³Mn	53	25	28	596.837	100	1
⁵⁵Fe	55	26	29	231.212	100	1
⁵⁷Co	57	27	30	835.927	100	1
⁶⁷Ga	67	31	36	1000.76	100	1
⁷⁰Ga	70	31	39	654.56	0.41	-1
⁶⁸Ge	68	32	36	106.34	100	2
⁷¹Ge	71	32	39	232.506	100	1
⁷³As	73	33	40	340.83	100	1
⁷⁶As	76	33	43	923.543	0.02	-1
⁷²Se	72	34	38	335.4	100	2
⁷⁵Se	75	34	41	863.392	100	1
⁸¹Kr	81	36	45	280.801	100	1
⁸³Rb	83	37	46	906.91	100	1
⁸⁶Rb	86	37	49	518.554	0.0052	-1
⁸²Sr	82	38	44	179.82	100	2
^87mSr	87	38	49	106.33	0.3	0
^90mY	90	39	51	135.81	0	-1
⁸⁸Zr	88	40	48	676	100	2
⁹³Mo	93	42	51	404.78	100	1
⁹⁷Tc	97	43	54	320.33	100	1
¹⁰⁰Tc	100	43	57	168.08	0.0018	-1
¹⁰¹Rh	101	45	56	541.7	100	1
¹⁰⁰Pd	100	46	54	358	100	2
¹⁰³Pd	103	46	57	543.075	100	1
¹¹⁰Ag	110	47	63	888.6	0.3	-1
¹⁰⁹Cd	109	48	61	214.24	100	1
¹¹¹In	111	49	62	861.79	100	1
^113mIn	113	49	64	69.1	<0.0036^[1]	0
¹¹⁶In	116	49	67	469.37	0.23	-1
¹¹⁰Sn	110	50	60	631.1	100	2
¹¹⁹Sb	119	51	68	590.92	100	1
¹²⁴Sb	124	51	73	616.46	0	-1
^124mSb	124	51	73	616.46	0	-1
¹¹⁸Te	118	52	66	278.4	100	2
¹²³Te	123	52	71	52.22	N/A	1
¹²⁵I	125	53	72	185.77	100	1
¹³⁰I	130	53	77	419.03	0	-1
¹²²Xe	122	54	68	725	100	2
¹²⁷Xe	127	54	73	662.33	100	1
¹³¹Cs	131	55	76	355.42	100	1
¹³⁶Cs	136	55	81	86.43	0	-1
¹²⁸Ba	128	56	72	529.9	100	2
¹³³Ba	133	56	77	517.499	100	1
^133mBa	133	56	77	805.746	0.0096	1
¹³⁷La	137	57	80	620.6	100	1
¹³⁴Ce	134	58	76	382.7	100	2
¹³⁹Ce	139	58	81	278.88	100	1
¹⁴²Pr	142	59	83	745.75	0.0164	-1
¹⁴⁰Nd	140	60	80	443.5	100	2
¹⁴⁵Pm	145	61	84	163.37	99.99999972	1
¹⁴⁸Pm	148	61	87	541.51	0	-1
¹⁵⁰Pm	150	61	89	86.32	0	-1
¹⁴⁵Sm	145	62	83	616.03	100	2
¹⁴⁹Eu	149	63	86	695.36	100	1
¹⁵⁴Eu	154	63	91	717.22	0.02	-1
¹⁴⁸Gd	148	64	84	26.67	0	2
¹⁵¹Gd	151	64	87	464.18	99.9999989	1
¹⁵³Gd	153	64	89	483.64	100	1
¹⁵⁵Tb	155	65	90	822.7	100	1
¹⁵⁷Tb	157	65	92	60.052	100	1
¹⁶⁰Tb	160	65	95	105.69	0	-1
¹⁵²Dy	152	66	86	599.8	99.9	4
¹⁵⁹Dy	159	66	93	365.57	100	1
¹⁶¹Ho	161	67	94	858.29	100	1
¹⁶³Ho	163	67	96	2.555	100	1
¹⁶⁴Ho	164	67	97	986.22	60	1
¹⁵⁸Er	158	68	90	887.2	100	2
¹⁶⁰Er	160	68	92	329.6	100	2
¹⁶⁵Er	165	68	97	376.26	100	1
¹⁶⁷Tm	167	69	98	748.42	100	1
^167mTm	167	69	98	927.90	0	1
¹⁷⁰Tm	170	69	101	313.99	0.131	-1
¹⁶⁴Yb	164	70	94	865.7	100	4
¹⁶⁶Yb	166	70	96	305.4	100	2
¹⁶⁹Yb	169	70	99	909.65	100	1
¹⁷⁶Lu	176	71	105	106.77	0.45	-1
^176mLu	176	71	105	229.62	0.095	-1
¹⁷²Hf	172	72	100	337.8	100	2
¹⁷⁵Hf	175	72	103	686.85	100	1
¹⁷⁹Ta	179	73	106	105.622	100	1
¹⁸⁰Ta	180	73	107	852.2	86	1
^180mTa	180	73	107	929.3	0	1
^182m2Ta	182	73	109	144.88	0	-1
¹⁷⁶W	176	74	102	723.8	100	2
¹⁷⁸W	178	74	104	91.2	100	2
¹⁸¹W	181	74	107	187.68	100	1
¹⁸³Re	183	75	108	556	100	1
¹⁸⁶Re	186	75	111	579.35	7.47	-1
^186mRe	186	75	111	727.55	0	-1
¹⁸²Os	182	76	106	838	100	2
¹⁸⁵Os	185	76	109	1012.797	100	1
¹⁸⁹Ir	189	77	112	532.3	100	1
¹⁸⁸Pt	188	78	110	505.12	99.999974	2
¹⁹¹Pt	191	78	113	1008.45	100	1
¹⁹³Pt	193	78	115	56.794	100	1
¹⁹⁵Au	195	79	116	226.82	100	1
¹⁹⁸Au	198	79	119	325.57	0	-1
¹⁹²Hg	192	80	112	765.1	100	2
¹⁹⁴Hg	194	80	114	69.1	100	2
¹⁹⁷Hg	197	80	117	600.11	100	1
²⁰¹Tl	201	81	120	481.2	100	1
²⁰⁴Tl	204	81	123	344.27	2.9	-1
²⁰⁰Pb	200	82	118	804.8	100	2
²⁰²Pb	202	82	120	49.7	100	2
²⁰³Pb	203	82	121	974.62	100	1
²⁰⁵Pb	205	82	123	50.5	100	1
^210mBi	210	83	127	207.82	0	-1
^212mPo	212	84	128	659	0	0
²¹¹At	211	85	126	785.36	58.2	1
²¹³At	213	85	128	73.93	0	1
²¹⁶At	216	85	131	473.199	0	-1
^216mAt	216	85	131	634.199	0	-1
²¹³Rn	213	86	127	881.21	0	2
²¹⁵Rn	215	86	129	86.55	0	1
²¹⁷Fr	217	87	130	656.03	0	1
²²⁰Fr	220	87	133	869.48	0	-1
²¹⁶Ra	216	88	128	312.1	<0.00000001	2
²¹⁹Ra	219	88	131	775.87	0	1
²²³Ac	223	89	134	591.78	1	1
²²⁶Ac	226	89	137	641.11	17	-1
²²⁰Th	220	90	130	917.3	0	2
²²²Th	222	90	132	581.5	0	2
²²⁵Th	225	90	135	672.01	10	1
²²⁹Pa	229	91	138	311.46	99.52	1
²³²Pa	232	91	141	499.53	0.003	-1
²²⁸U	228	92	136	300.5	2.5	2
²³¹U	231	92	139	381.65	99.996	1
²³⁵Np	235	93	142	124.214	99.9974	1
²³⁶Np	236	93	143	933	86.3	1
²³⁸Np	238	93	145	147.32	0	-1
²³⁴Pu	234	94	140	393.1	94	2
²³⁷Pu	237	94	143	220.03	99.9958	1
²³⁹Am	239	95	144	802.11	99.99	1
²⁴²Am	242	95	147	751.295	17.3	1
²⁴⁴Am	244	95	149	75.4	0	-1
²³⁸Cm	238	96	142	972.8	90	2
²⁴⁰Cm	240	96	144	213.6	0.5	2
²⁴¹Cm	241	96	145	767.42	99	1
²⁴³Cm	243	96	147	7.48	0.29	1
²⁴⁵Bk	245	97	148	810.74	99.88	1
²⁴⁸Bk	248	97	151	687	0	-1
^248mBk	248	97	151	667±50	30	-1
²⁴⁴Cf	244	98	146	763.7	25	2
²⁴⁶Cf	246	98	148	123.3	0.004	2
²⁴⁷Cf	247	98	149	646	99.965	1
²⁵¹Es	251	99	152	377.58	99.5	1
²⁵⁴Es	254	99	155	651.2	0	-1
^254mEs	254	99	155	731.5	0.076	-1
²⁵⁶Es	256	99	157	150#	0	-1
²⁵⁰Fm	250	100	150	847	10	2
²⁵³Fm	253	100	153	335.84	88	1
²⁵⁷Md	257	101	156	406.73	85	1
²⁶⁰Md	260	101	159	<5	-1
²⁵⁶No	256	102	154	208.3	0	2
²⁵⁹No	259	102	157	486	25	1
²⁶²Rf	262	104	158	270#	0	2
²⁶⁷Db	267	105	162	630#	?	1 or 2?

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 (^{200,202,203,205}Pb and ^{238,240,241,243}Cm) 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:

Nuclide	Zagrebaev et al. prediction		KTUY prediction
Nuclide	Isobar with the lowest energy	Proton excess	Isobar with the lowest energy	Proton excess
²⁶⁷Db	²⁶⁷Rf	1	²⁶⁷Lr	2
²⁶⁶Sg?	²⁶⁶Rf	2	²⁶⁶No	4
²⁷⁰Hs?	²⁷⁰Rf or ²⁷⁰Sg^[2]	4 or 2	²⁷⁰Rf	4
²⁷²Hs?	²⁷²Sg	2	²⁷²Rf	4

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:

Nuclide	A	Z	N	Alpha decay energy (MeV)	Alpha decay half-life (yr) or estimation using Geiger–Nuttall law
Nuclide	A	Z	N	Alpha decay energy (MeV)	Estimation 1^[3]	Estimation 2^[4]
¹⁴⁵Pm	145	61	84	2.32	6.3×10⁹
¹⁴⁹Eu	149	63	86	2.40	5.0×10¹⁰	3.8×10¹⁰
¹⁴⁸Gd	148	64	84	3.27	86.9
¹⁵¹Gd	151	64	87	2.65	3.01×10⁷
¹⁵²Dy	152	66	86	3.73	0.2715
¹⁵⁸Er	158	68	90	2.67	8.4×10¹⁰	2.1×10¹¹
¹⁷⁶W	176	74	102	3.34	9.5×10⁷	3.2×10⁸
¹⁷⁸W	176	74	104	3.01	2.2×10¹¹	6.7×10¹¹
¹⁸²Os	182	76	106	3.37	1.2×10⁹	2.9×10⁹
¹⁸⁸Pt	188	78	110	4.01	1.07×10⁵
¹⁹²Hg	192	80	112	3.38	6.3×10¹¹	6.9×10¹¹

129.104.241.214 (talk) 23:54, 13 February 2024 (UTC)[reply]

References

^ "Adopted Levels for ¹¹³In" (PDF). NNDC Chart of Nuclides.
^ For an odd Z nuclide sandwiched by two beta-stable isotopes, it is equally likely that its isobar with one less proton or with one more proton has lower energy. The former happens for ³⁶Cl, ⁴⁰K, ⁶⁴Cu, ¹⁰⁸Ag, ¹⁵²Eu and ²⁴²Am, while the latter happens for ⁷⁰Ga, ⁸⁰Br, ¹²²Sb, ¹⁹²Ir and ²⁰⁴Hg. ²⁶⁹Db and ²⁷¹Db could both be beta stable in the prediction of Zagrebaev et al., so it does not tell us which of ²⁷⁰Rf and ²⁷⁰Sg has lower energy.
^ Using $\lg T_{\alpha }(\mathrm {yr} )={\dfrac {aZ+b}{\sqrt {Q_{\alpha }}}}+cZ+d$ with $a=0.831$ , $b=53.983$ , $c=0.248$ , $d=-73.565$ .
^ Using $\lg T_{\alpha }(\mathrm {yr} )={\dfrac {aZ+b}{\sqrt {Q_{\alpha }}}}+cZ+d$ with $a=2.073$ , $b=-39.813$ , $c=-0.511$ , $d=-15.842$ .

[1] "Adopted Levels for ¹¹³In" (PDF). NNDC Chart of Nuclides.

[2] For an odd Z nuclide sandwiched by two beta-stable isotopes, it is equally likely that its isobar with one less proton or with one more proton has lower energy. The former happens for ³⁶Cl, ⁴⁰K, ⁶⁴Cu, ¹⁰⁸Ag, ¹⁵²Eu and ²⁴²Am, while the latter happens for ⁷⁰Ga, ⁸⁰Br, ¹²²Sb, ¹⁹²Ir and ²⁰⁴Hg. ²⁶⁹Db and ²⁷¹Db could both be beta stable in the prediction of Zagrebaev et al., so it does not tell us which of ²⁷⁰Rf and ²⁷⁰Sg has lower energy.

[3] Using $\lg T_{\alpha }(\mathrm {yr} )={\dfrac {aZ+b}{\sqrt {Q_{\alpha }}}}+cZ+d$ with $a=0.831$ , $b=53.983$ , $c=0.248$ , $d=-73.565$ .

[4] Using $\lg T_{\alpha }(\mathrm {yr} )={\dfrac {aZ+b}{\sqrt {Q_{\alpha }}}}+cZ+d$ with $a=2.073$ , $b=-39.813$ , $c=-0.511$ , $d=-15.842$ .

[1]

[2]

[3]

[4]