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Beta decay of a neutron

  1. Mar 7, 2005 #1
    When an atom experience beta decay, will the atom become ion?
    This is my deduction:
    1.An atom will release an electron in beta decay.
    2.The electron is replaced by the electon produced from the decay of neutron.
    3.However, the proton number increase by one. It still need one more electron to be normal! :confused:
    Does the deduction contain mistakes?
  2. jcsd
  3. Mar 7, 2005 #2


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    1.A nucleus will release an electron in [itex] \beta^{-} [/itex] decay.
    2.Nope,the beta decay IS the decay of the neutron...
    3.Yes,it's not an ion in the common sense.It's a "weird" ion.Usually,we see ions as atoms losing/gaining electrons from their shells,but not altering the # of protons inside the nucleus.

  4. Mar 7, 2005 #3
    Beta decay is one process that unstable atoms can use to become more stable. There are two types of beta decay, beta-minus and beta-plus.

    During beta-minus decay, a neutron in an atom's nucleus turns into a proton, an electron and an antineutrino. The electron and antineutrino fly away from the nucleus, which now has one more proton than it started with. Since an atom gains a proton during beta-minus decay, it changes from one element to another. For example, after undergoing beta-minus decay, an atom of carbon (with 6 protons) becomes an atom of nitrogen (with 7 protons).

    During beta-plus decay, a proton in an atom's nucleus turns into a neutron, a positron and a neutrino. The positron and neutrino fly away from the nucleus, which now has one less proton than it started with. Since an atom loses a proton during beta-plus decay, it changes from one element to another. For example, after undergoing beta-plus decay, an atom of carbon (with 6 protons) becomes an atom of boron (with 5 protons).

    Although the numbers of protons and neutrons in an atom's nucleus change during beta decay, the total number of particles (protons + neutrons) remains the same.

    Hope this helps..
  5. Mar 7, 2005 #4
    This is a dangerous formulation that can easily be interpreted in the wrong way, for that reason i say your answer is wrong.

    The beta decay really announced the advent of QFT. In the beginning some scientists thought that the electron really came out of the nucleus, ofcourse this is wrong. What happens is this : the electron is created "out of nothing". This means that the energy involved in the beta decay is used to create this electron out of the vacuum. This kind of process is only possible in QFT and that is why beta decay was one of the first major breakthroughs of QFT.

    Also, keep in mind that negative beta decay (neutron ---> proton) is a particle decay mode while the positive beta decay (proton --> neutron) is a nuclear decay mode because the neutron is more heavy then the proton. This proton can only decay (due to energyconservation) when it is surrounded by many other protons in a nucleus...Part of the energy coming from proton-proton-interactions can also account for mass via E=mc².

    Beta plus decay commonly means the basic process p->n + e++v. It is a nuclear decay mode in that it can only happen if the proton is inside a heavier nucleus and the final state nucleus is more tightly bound; the process is forbidden in free space by energy conservation since a neutron alone is heavier than a proton

    Last edited: Mar 7, 2005
  6. Mar 7, 2005 #5
    What is QFT? :confused:
  7. Mar 7, 2005 #6
    QFT = quantum field theory

  8. Mar 7, 2005 #7

    If you are interested in an introductory article on QFT, read the string theory part 1 entry of my journal : https://www.physicsforums.com/journal.php?s=&action=view&journalid=13790&perpage=10&page=7 [Broken]

    This is just an introduction. In order to understand QFT you need a solid knowledge of QM and special relativity

    Last edited by a moderator: May 1, 2017
  9. Mar 7, 2005 #8
    Thank you very much sir for the link. I'm in the second year in my undergraduate course and I have just been introduced to QM. So I need to learn more in this field.

  10. Mar 8, 2005 #9


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    Last edited: Mar 8, 2005
  11. Mar 9, 2005 #10
    Reilly, the electron does not come from the atomic nucleus because there are no electrons in an atomic nucleus. It is that simple...

  12. Mar 9, 2005 #11
    Isn't this right?

    A neutron decays to a proton, an electron, and an antineutrino. This is called neutron beta decay. The term beta ray was used for electrons in nuclear decays because they didn't know they were electrons!
  13. Mar 9, 2005 #12


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    Oh,Marlon,i think you had something else on your mind at the Nucler Physics course,provided you had this course in your curriculum.


    P.S.Read first.Talk later.
  14. Mar 9, 2005 #13


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    My explanation of nuclear beta decay is the canonical one. Under normal circumstances there are no (real) electrons in a nucleus, but, in fact there are plenty, plenty virtual ones due to virtual pair production. Fermi, and the current approach tell us that for nucleii, beta decay is caused/described by a point interaction, necessarily inside the nucleus. Normally when computing beta decay probabilities, we use a standard scattering theory approach in which the electron and neutrino are in their asymptotic out state, as in plane waves.

    With time dependent perturbation theory, one can actually compute the probability for electron tunneling out of the nucleus -- remember that the electron inside a nucleus will see a a potential well created by the protons.

    Note, there are, to use your terms, no electrons inside neutrons. Well that's not really true because the neutron has as among component states, proton + electron+neutrino, proton, electron, pi0 meson, neutrino, electron-positron pair, quarks and pairs. and so on. It's not really quite so simple.

    Reilly Atkinson
  15. Mar 9, 2005 #14
    I think we are venturing into the realm of semantics. . .
  16. Mar 9, 2005 #15
    The electron from beta decay comes from virtual electron-positron pairs ?!! I have always thought that the beta particle and antineutrino was really the decay of the W boson itself.
  17. Mar 9, 2005 #16


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    I'm not so sure about that. One of the core issues surrounding the standard model is the extent to which fundamental particles are really fundamental. Where to products of beta decay come from go to those fundamental issues.

    We are definitely venturing into the realm of theory from the realm of phenomena, but the distinction is more than semantic, I think. We do not just have two people saying the same thing, while using different words for it.
  18. Mar 10, 2005 #17
    The main point of contention seems to be: Are there electrons in the nucleus?

    I think Marlon is saying there are no "real" electrons. That's where the semantics comes in.
  19. Mar 10, 2005 #18
    :rofl: :rofl: :rofl:
    Thank you dexter for providing us with the most "useful" post of this thread;

    From the very beginning you are again way off here since it is not nuclear physics that explains beta-decay (and the energy-distribution of the emitted electrons), yet it is QFT

    However thanks for the "input"

  20. Mar 10, 2005 #19


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    Relative to the typical energetics of nuclear beta decay, the W has sufficient mass to make the four field point interaction an extremely good approximation.

    My point in talking about virtual electrons was quite rhetorical, they are there and, along with photons and positrons, quite virtual ones, they do influence the beta decay process via radiative corrections, including mass and vertex contributions. Like, for example, the beta process includes diagrams that have a virtual photon emitted, which in turn produces an electron-positron pair, and the positron annihilates the "initial" electron. So, technically, the emergent beta decay electron could come from (virtual) pair production.

    Indeed, as I earlier noted, the only electrons inside a nucleus are virtual ones. But they can be made manifestly real through higher order electromagnetic interactions combined with the basic Fermi mechanism for beta decay. Is it live or Memorex?

    Semantics at issue? Of course.

    I'll set a problem, one that I think I might have used when teaching some nuclear theory in an Advanced QM course, a few years ago. Assume that the nucleons are confined by a square well potential. And, replace the attraction of electrons by protons as another square well.(OK to use non-rel wave funtions, but feel free to use Dirac's relativistic if you choose.) Use first order perturbation theory to calculate the beta decay transition probability. (Recall that such transitions are computed in the limit as t--> infinity. The problem here is quite similar to resonant scattering. If this does not make sense, I refer you to Goldberger and Watson's Scattering Theory)
    Now let's add a second part, designed for this thread. Show that if the electron's wave function is zero inside the nucleus, as in not there, the probability, of beta decay with good energetics is virtually zero -- non-zero only if there is any appreciable probability of finding the nucleons outside the nominal nuclear radius. I would hope this exercise might suggest that if the electron is not created in the nucleus, the resulting beta decay theory will be wrong.

    By the way, note that both nuclear physics and QFT are involved-- both are necessary to give a solid description of nuclear beta decay. For free neutron decay, QFT provides the appropriate description.

    Nothing virtual here. Only a straightforward application of basic QM.

    Reilly Atkinson
    Last edited: Mar 10, 2005
  21. Mar 11, 2005 #20
    I am getting more and more confused now. :confused: There are so many answer that I don't know which is true and which is not.
    Anyway, can somebody summarise the above replies in an easier way? I don't really understand QMT :shy: .
    Does the electron come from the nucleus, or from the virtual electron, like what Marlon have said?
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