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From evolutionwiki:

"Potassium 40 decays into argon 40 through a process known as electron capture. In electron capture, an electron from the innermost electron shell "falls" into the nucleus, causing a proton to convert into a neutron."

What's the implication of an electron "falling" into the nucleus, what happens to the electron? Is it the electron that subsequently will be emitted? Why does it not decay by emitting an alpha particle?

user1904218
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    It might be useful to read about the nature of s-shells and electron position distribution functions. Say http://physics.stackexchange.com/questions/135222/why-cant-electrons-fall-into-the-nucleus or http://physics.stackexchange.com/questions/20003/why-dont-electrons-crash-into-the-nuclei-they-orbit?. – dmckee --- ex-moderator kitten Jun 26 '16 at 19:40
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    Carbon-14 does not decay by emitting an alpha particle. It decays by emitting an electron and an electron antineutrino to become nitrogen-14. – user16622 Jun 26 '16 at 19:56
  • @user16622: crap, you're right. I've just started learning this stuff, and it can get confusing at times. – user1904218 Jun 26 '16 at 20:52

2 Answers2

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The decay of potassium-40 to argon-40 is either a $\beta^+$ decay in which what is emitted is not an electron but a positron $$ {}^{40}{\rm K} \to {}^{40}{\rm Ar} + e^+ + \nu_e $$ or, more frequently (if we have whole atoms), an electron capture that you mentioned in which no charged leptons are emitted at the end! About 11% of the potassium-10 decays proceed in this way. $$ {}^{40}{\rm K} + e^- \to {}^{40}{\rm Ar} + \nu_e $$ The remaining 89% of the decays of potassium-40 go to calcium-40 (the beta-plus decay is a small fraction of a percent). Note that the two reactions above differ by moving the positron from the right hand side to the left hand side so that the sign has to be changed.

Potassium-40 has 19 protons and 21 neutrons. Argon-40 has 18 protons and 22 neutrons. So if we focus on the "minimum part" of the nuclei, the reactions above may be reduced either to $$ p \to n + e^+ + \nu_e$$ or $$ p+e^- \to n + \nu_e $$ which are the standard reactions switching protons and neutrons. In particular, the second reaction displayed right above this line is the more "microscopic" description of the electron capture you're primarily interested it.

These reactions preserve the electric charge, baryon number, and lepton number. They also have to preserve energy. A free proton couldn't decay to the neutron and other two particles because it's lighter. Even a proton and a low-velocity electron wouldn't have enough mass/energy to produce the neutron (plus the neutrino) as in the second reaction.

But when the protons and neutrons are parts of whole nuclei, the energies of the initial and final nuclei are affected by the nuclear interactions. In particular, the argon-40 nucleus (and especially atom) is highly bound which means lighter and the reactions where argon-40 appears as a product are therefore "more possible".

To summarize, the electron capture (=falling of the electron) simply means that the proton has a nonzero probability to meet with one of the electrons – probably in the inner shells – and merge into a new particle, a neutron, plus a neutrino. This process can't occur in the vacuum due to the energy conservation but in the context of the nucleus, the interactions with other neutrons and protons make the final state with the new neutron favorable.

On the contrary, alpha decays are more rare. Among 24 isotopes of potassium, only potassium-36 may alpha-decay. Carbon-14 doesn't alpha-decay, either. Among the isotopes of carbon, only carbon-9 alpha-decays. Both of these alpha-decays must be preceded by a beta-decay. Usually just heavy enough nuclei (with too small an excess of neutrons) alpha-decay.

rob
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Luboš Motl
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  • "an electron capture that you mentioned in which no leptons are emitted at the end!" Maybe better to say "no charged leptons" because the neutrino is present in the final state. – dmckee --- ex-moderator kitten Jun 26 '16 at 19:38
  • Yup, your correction is right. – Luboš Motl Jun 26 '16 at 20:38
  • From the quark level Feynmann diagram ($u\to d+e^+ +\nu_e$), does $\beta^+$ decay process mean that initial u quark (in this case) is heavier than d quark? (I don't think so.) If not, how to draw the diagram for this process? I think the binding energy has to become some virtual particles (but what's this virtual particle?) and it hits u quark and then this virtual u quark decays? – luyuwuli Oct 04 '16 at 13:57
  • Dear @luyuwuli - the up-quark is indeed lighter in "isolation" (except that it can't be isolated) than the down-quark but only the total energy/mass is conserved. There are other contributions to it, from interactions of the quarks with each other and so on, and those change between the initial and final protons/neutrons/nuclei. When you look, you see that the rest mass of the initial nucleus exceeds the sum of the final rest masses so the energy conservation law doesn't ban this process. – Luboš Motl Oct 07 '16 at 08:47
  • All such beta-decays (regardless of signs) are a sign of the "weak nuclear interaction". In all the well-known ones, the virtual particle is a W-boson (with charge Q=+1 or Q=-1). The up-quark changes to a down-quark while emitting a positive W-boson. This positive W-boson is split into the positron-neutrino pair. (One could also have an antineutrino absorbing the W-boson and becoming a positron, or things like that.) It's just like the exchange of virtual photons except that the virtual W-boson carries a charge and changes one fermion species to a "similar one" but with a different charge. – Luboš Motl Oct 07 '16 at 08:49
  • See e.g. this picture https://en.wikipedia.org/wiki/Beta_decay#.CE.B2.E2.88.92_decay with a virtual W-boson in the beta-decay. It's analogous with the inverse one. – Luboš Motl Oct 07 '16 at 08:50
  • @LubošMotl Thanks for your detailed explanation. I totally agree what you've said. However, my question is to understand these extra contributions. To be specific, can these extra "contributions" be transformed to diagrams? $u\to d+W^+\to d+e^+ +\nu_e$ seems to claim that u is heavier than d and these extra contributions are not shown at all. There should be something beyond feynman diagram. Or since quarks are confined, the "confined u quark mass" is temporarily heavier than "confined d quark mass" due to these extra contributions? – luyuwuli Oct 08 '16 at 07:02
  • Hi, the W-boson is virtual so it does not have to obey $E^2 = m_0^2 c^4+ p^2 c^2$, and, consequently, it's simply not true that the mass of $u$ has to exceed the mass of $d$ plus the mass of $W$. It surely doesn't, $W$ is over 80 GeV, much heavier than either $u$ or $d$ quark. This point - what it means for the virtual particle to be virtual - would invalidate every conclusion of yours about any process with virtual particles so your mistake has nothing to do with the confinement or other special features of quarks. – Luboš Motl Oct 10 '16 at 08:10
  • @LubošMotl I was not informed by your comment, so it's been a long time after comment on this. Perhaps I still didn't express myself clearly. I'm not saying that $m(u)>m(W)$ because (as you've explained) $W$ is not on shell. My question is can the u d quarks in $u\to d+e^+ + \nu_e$ be on shell? If they are, how can this process happen if u is lighter than d? Thus I conclude that since quarks are always confined, the "on shell" condition is not well-defined. In some cases, u quark mass can be larger than d. – luyuwuli Nov 11 '16 at 08:38
  • @LubošMotl That's why I doubt that quark mass is changeable according to different surroundings or due to the special properties of quarks (confinement, because you never observe an on-shell quark), there is always spaces for us to define effective mass to change the mass relations. So I insist that it's wrong to say that $m(u)$ is always smaller than $m(d)$. – luyuwuli Nov 11 '16 at 09:28
  • Dear @luyuwuli - of course that in the process $u\to d+e^+ +\nu_e$, at least one of the particles has to be off-shell - or be a part of a broader system with some interaction energies - because the total sum of the products exceeds the mass of the up-quark. I talk about interactions because inside protons, neutrons, or nuclei, the "elementary reaction" is just $u$ becoming $d$, but the counting of the energy is different because the interactions of the quark with the environment changes, too, and such a process indeed becomes possible for many environments. That's what $\beta^+$ decays are. – Luboš Motl Nov 11 '16 at 16:24
  • $m(u)$ is always smaller than $m(d)$. These are really constants of Nature. At most, they may be said to depend on the renormalization energy scale $E$ but it's still true that $u$ is lighter than $d$ for any value of $E$. This in no way bans the beta decays because the energy of the nuclei etc. isn't given simply by $m(u)$ or $m(d)$. There are lots of interaction terms that you seem to overlook and this leads you to all your preposterous conclusions. – Luboš Motl Nov 11 '16 at 16:27
  • @LubošMotl Still, I agree with most of what you've said. But my point is one should compare $m(u,E_1)$ and $m(d,E_2)$ at different energy scale because of the change of the environment nuclei. Once one uses Feynman diagram, one implicitly assumes that the initial and final states are free, which doesn't agree with reality. However, one can still use Feynman diagram, because other interactions can be considered into the effective mass of the quarks (free quarks v.s. dressed quarks). – luyuwuli Nov 12 '16 at 09:46
  • Apologies, I don't understand what you're saying. If you evaluate the masses of up- and down-quarks at nearby renormalization scales that differ by a factor of two or something like that, it will still be true that the up-quark is lighter. The renormalization group running of the masses is slow. – Luboš Motl Nov 12 '16 at 16:53
  • Sorry for my bad English. I want to say that even though we believe the process is described by the Feynman diagram, the subtleties is to interpret the mass involved. Because one cannot shut down other interactions, one has to sum up these interactions into an effective mass. Just like the current quark mass v.s. constituent quark mass, the flavor is the same but the mass is really different. Constituent quark is a current quark+other complicated interactions. – luyuwuli Nov 13 '16 at 04:44
  • Same happens here in $\beta^+$ decay. In order to solve the dilemma between current quark mass relations of u d quarks and conservation of energy. Conceptually, it will be better to emphasize that the mass shows up in Feynman diagram is not current quark mass but some effective mass with other interactions considered. – luyuwuli Nov 13 '16 at 04:44
  • @LubošMotl Maybe the system didn't inform you about my comments and sorry for bothering you so long time. However, I still want to know what's your perspective about my last two comments? – luyuwuli Nov 21 '16 at 03:03
  • Dear @luyuwuli - when you consider interactions of composite objects such as nuclei, you should do something "harder" than just replace masses of quarks by some other (effective) masses. You should embed the elementary interactions of the quarks (a small Feynman diagram) as a bulk part of a larger Feynman diagram that also includes the other particles in the nucleus that participate in the process. Those do get a kick so it's not right to assume that the "decaying quarks" are the only ones that change in the interaction, and that's why just using effective parameters doesn't really do the job. – Luboš Motl Nov 21 '16 at 06:02
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    @LubošMotl Sure, treating the whole process as one quark's flavor change is oversimplified. I think since we agree on the fundamental process , it's time to end the discussion. Thanks for your patience and explanation. – luyuwuli Nov 21 '16 at 06:20
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Electron capture

Electron capture (K-electron capture, also K-capture, or L-electron capture, L-capture) is a process in which the proton-rich nucleus of an electrically neutral atom absorbs an inner atomic electron, usually from the K or L electron shell. This process thereby changes a nuclear proton to a neutron and simultaneously causes the emission of an electron neutrino.

p + e− → n + ν_e

K,L,M are the principle quantum number(n) of the orbitals of the electrons.

The orbitals of electrons, give the probability of the electron to be found at (x,y,z). For l=0 ( the angular momentum quantum number) the electrons have a probability of passing through the nucleus. If the nucleus is unstable with respect to another, i.e. there exists the extra energy needed for creating a neutron and an electron neutrino from a proton and and electron, one has electron capture and a lower by one Z nucleus appears.

anna v
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