Understanding Beta Decay: How It Affects Atoms and Ionization

[scilover89]

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!
Does the deduction contain mistakes?

 

[dextercioby]

1.A nucleus will release an electron in \beta^{-} 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.

Daniel.

 

[Reshma]

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!
Does the deduction contain mistakes?

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..

 

[marlon]

1.A nucleus will release an electron in \beta^{-} decay.

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

marlon

 

[Reshma]

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.

What is QFT?

 

[marlon]

QFT = quantum field theory

marlon

 

[marlon]

Reshma,

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

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

marlon

 

[Reshma]

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.

Thanks,
Reshma

 

[reilly]

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.
******************************

(RA) Of course the electron/positron comes out of the nucleus -- where else can it come from?

Yes, beta decay's specific mechanism is nucleon -> nucleon' + e/p + neutrino. Let's focus on standard beta deay in which a nucleus emits an electron, the neutron number goes down by one, the proton number goes up by one. So, indeed, beta decay can and does create an ion -- an ion really does not care where it came from -- the lifetime of which can vary considerably, dependent as it is on the environment of the nucleus. There is, in fact, a chance that the escaping electron could be captured -- probably the probability for this is small, and gets smaller the higher the electron's energy is.

A little history. Prior to his ground breaking work on beta decay, Fermi played a major role in developing QED, along with Dirac, Pauli, and Heisenberg. (And remember that Chadwick did not find the neutron until 1932, which meant no successful theory was possible for nucleii until 1932.) Fermi's 4-point interaction for beta decay was, in fact, inspired by QED -- for practical purposes, Fermi's approach was based on current-current interactions, and was first published in 1933. So Fermi's beta decay theory was at best a second triumph of field theory. But make no mistake about it, QED was the first field theory triumph -- all other interacting field theories are formally based on QED.

Again, the electron and neutrino come out of the decaying nucleus. First of all, the Born Approx works nicely for nuclear beta decay. This puts two things into play

1. The Pauli Principal says that all neutrons are at play. If the energetics are favorable, then a neutron will decay-- which one? Who knows.

2. And, when computing the Born Approx decay rate, the relevant integral is non-zero only within the nucleus, and appropriate antisymmetric wave functions must be used for initial and final states.

The structure of the 4-point interation, generically (N(x) v(x) P(x) e(x)+adjoint, where N is a neutron field, P a proton field, v a neutrino field, and e an electron field. I've not worried about adjoints, nor about the various gamma matrices involved. This pretty clearly says the leptons are not created from the vacuum -- conservation laws more-or-less preclude such a phenomena. Fermi's idea was the neutron transformed into proton, electron, neutrino. He's been more than vindicated, as in nuclear beta decay, the intermediate vector bosons basically contribute a 1/M*M, with M the vector boson mass, as the momentum/energy transfer is very small compared to M) This interaction allows the computation of the probability that a beta decay will occur at any place inside the nucleus, that is that the electron and neutrino do indeed emerge from the nucleus because they are formed inside the nucleus. The matrix elements are generally non-zero. but the decay can occur only if the energetics are favorable-- the neutron decays because the neutron mass is greater than the combined rest masses of the electron and proton. For nucleii, the initial nucleii must have a mass greater than the final nucleii and electron.

Regards,
Reilly Atkinson

 

[marlon]

Reilly, the electron does not come from the atomic nucleus because there are no electrons in an atomic nucleus. It is that simple...

marlon

 

[Reshma]

the electron does not come from the atomic nucleus because there are no electrons in an atomic nucleus. It is that simple...

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!

 

[dextercioby]

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

Daniel.

P.S.Read first.Talk later.

 

[reilly]

Reilly, the electron does not come from the atomic nucleus because there are no electrons in an atomic nucleus. It is that simple...

marlon

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.

Regards,
Reilly Atkinson

 

[juvenal]

I think we are venturing into the realm of semantics. . .

 

[misogynisticfeminist]

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.

Regards,
Reilly Atkinson

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.

 

[ohwilleke]

I think we are venturing into the realm of semantics. . .

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.

 

[juvenal]

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.

 

[marlon]

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

Daniel.

P.S.Read first.Talk later.

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"

marlon

 

[reilly]

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.

Regards,
Reilly Atkinson

 

[scilover89]

I am getting more and more confused now. 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?

 

[dextercioby]

Depends on what kind of decay you're talking.Free neutrons,or nuclei containing neutrons.In the latter case,it comes from the nucleus,no doubt about it.In the former,it comes from a W- boson which decays into an electron & an electronic antineutrino.The normal place where you find neutrons is the nucleus.However,this beta decay process of a neutron is a weak process and its Feynman diagram is Simple:

d---------->u+W^{-}---------->e^{-}+\bar{\nu}_{e}

For more details,read:pages 65->71 and 309->314 from D.J.Griffiths,"Introduction to Elementary Particles"

 

[Astronuc]

Here's a discussion on the neutron and neutron decay (aka beta decay).

http://hyperphysics.phy-astr.gsu.edu/hbase/particles/proton.html#c4

Interesting statement -

The decay of the neutron is associated with a quark transformation in which a down quark is converted to an up by the weak interaction.

As for where an electron in beta decay 'come from', as a nuclear engineer, I am only concerned that it comes from the nucleus. As physicist, especially particle physicist, I would be more specific as in 'quark transformation'.

http://hyperphysics.phy-astr.gsu.edu/hbase/particles/imgpar/ndecw.gif

Going back to the original post, the atom does briefly become an ion (on order of microseconds). When the nucleus increases by +1, it is an ion until it 'steals' an electron from a neighboring atom. The ejected electron travels mm or cm's away, meanwhile ionizing other atoms. There is a cascade effect, and eventually, all atoms are again neutral.

 

[marlon]

scilover,

The beta electron comes from the involved W-boson which is one of the force carriers of the weak interaction. This boson decays and yields the electron along with some other stuff, but let us keep it simple.

The fact that this W-boson exists really has to do with the energy involved in these decay-reactions. It is a bit like two particles sitting on a mattress. the mattress will vibrate and this vibration corresponds to a particle of certain mass and energy. This particle really is the exchange boson that represents the interaction between the two particles that sit on the mattress. This generated particle will carry momentum and therefore it really represents and force.


marlon

 

[marlon]

I would be more specific as in 'quark transformation'.

Completely correct. The weak interaction really transforms one quark into anothe (by changing the quark flavour). This interaction is expressed in terms of the three intermediate vector bosons (W+,W- and Z)

Z is neutral and the W's can change both electric charge and quarkflavour
marlon