Beta Decay Queries: Answers & Explanations

[Reshma]

I need to resolve a few queries on beta decay.

The standard explanation given in most texts is during beta decay, a neutron decays into a proton, an electron(or a positron) and a neutrino(or anti-neutrino).

This explanation however isn't completely correct since it can be easily shown using Heisenberg's uncertainity principle that it is not possible for an electron to exist in the nucleus. So where are the emitted electron/positrons and the neutrinos coming from coming from? Are they somehow formed during emission?

Another query on beta decay regarding electron capture. How is an electron captured by the nucleus since electrons can exist only in discrete energy levels, how does quantum mechanics permit an electron to be captured by the nucleus?

 

[Norman]

But the electron is not in the nucleus before the neutron decays.

the decay is:
$$n \rightarrow p^+ + e^- + \bar{\nu_e}$$
in the lowest order approximation (see Griffiths "Introduction to Elementary Particles", pp309-315):
$$n \rightarrow p^+ + W^- \rightarrow p^+ + e^- + \bar{\nu_e}$$
This slightly overestimates the average lifetime of a neutron by about 50% but that is excellent for only considering the weak interaction in the decay. You will have small effects due to the non-conservation of the axial charge, Cabbibo angle (from quark mixing, ie. strong interaction corrections) and a small Coulomb correction due to the attraction of the proton and electron.

I would really recommend you checking out Griffiths textbook for more information. It is a great introduction to particle physics- my personal favorite.

hope that helped.
cheers,
Ryan

 

[jtbell]

So where are the emitted electron/positrons and the neutrinos coming from coming from? Are they somehow formed during emission?

Yes, they are created in the emission process. They do not exist in the nucleus before the decay.

Another query on beta decay regarding electron capture. How is an electron captured by the nucleus since electrons can exist only in discrete energy levels, how does quantum mechanics permit an electron to be captured by the nucleus?

An atomic electron in a given "energy level" is not at a fixed distance from the nucleus. It has a quantum-mechanical probability distribution that is not zero at the location of the nucleus. So there is a small but non-zero probablity that the electron can be located inside the nucleus at any given moment. This gives the nucleus a chance to "capture" it.

 

[Reshma]

Thank you Norman and Jtbell for replying.

But using HUP, it can be shown that electrons cannot exist in the nucleus.
Let the uncertainity in position of electron be $\Delta x$ be of the order of nuclear size, 10^-14m. This will permit the electron to remain within the nucleus. We can calculate $\Delta p$, the uncertainity in the momentum using HUP,
$$\Delta p \Delta x \thicksim h$$
$$\Delta p = \frac{h}{\Delta x} = 6.62 \times 10^{-20} \mbox{ kg m/s}$$

Using the realtivistic relation between the total energy E_T and momentum p;
E_T² = p²c² +m₀²c⁴
Rest energy for electron is m₀c² = 0.511 MeV.
On substitution of the values taking $\Delta p = p$ we get the total energy of the electron as:
E_T = 124 MeV

This is a very high energy for an electron to have as it is known that emitted electrons have energies of the order of only few MeV. How is it possible to say that there is a probability for an electron to exist in the nucleus?

 

[Norman]

Thank you Norman and Jtbell for replying.
But using HUP, it can be shown that electrons cannot exist in the nucleus.
Let the uncertainity in position of electron be $\Delta x$ be of the order of nuclear size, 10^-14m. This will permit the electron to remain within the nucleus. We can calculate $\Delta p$, the uncertainity in the momentum using HUP,
$$\Delta p \Delta x \thicksim h$$
$$\Delta p = \frac{h}{\Delta x} = 6.62 \times 10^{-20} \mbox{ kg m/s}$$
Using the realtivistic relation between the total energy E_T and momentum p;
E_T² = p²c² +m₀²c⁴
Rest energy for electron is m₀c² = 0.511 MeV.
On substitution of the values taking $\Delta p = p$ we get the total energy of the electron as:
E_T = 124 MeV
This is a very high energy for an electron to have as it is known that emitted electrons have energies of the order of only few MeV. How is it possible to say that there is a probability for an electron to exist in the nucleus?

It is not existing in the nucleus. It is being ejected into the world around it. This is where you are going wrong. Beta decay would have not been found when it did if only the neutrino escapes. See: http://www.physics.isu.edu/radinf/beta.htm

The electron is not in the nucleus at the same time as the neutron. The energy of the neutron is used to create the decay products through interactions between the quantum fields.

Edit: for completeness sake when people talk about the beta particle, it is the electron they are referring to.

 

[jtbell]

But using HUP, it can be shown that electrons cannot exist in the nucleus. [...details omitted...] How is it possible to say that there is a probability for an electron to exist in the nucleus?

Visualize the electron's probability distribution as a fuzzy "cloud". This cloud is much larger than the nucleus. The nucleus is located inside the cloud.

The HUP says that the entire cloud cannot be contained inside the nucleus. It doesn't prevent the central part of a large cloud from overlapping the nucleus.

 

[Reshma]

OK, thanks Norman and Jtbell...I got it now!

 

[marlon]

Just to add to what Norman and Jtbell have answered. The fact that the electron is "created" using the energy of the "nucleus" (neutron) also explains why only QFT is able to explain this phenomenon. In QM (which explains both alpha and gamma decay) the number of particles must be fixed. Hence no particles can be created...

marlon

 

[Norman]

Just to add to what Norman and Jtbell have answered. The fact that the electron is "created" using the energy of the "nucleus" (neutron) also explains why only QFT is able to explain this phenomenon. In QM (which explains both alpha and gamma decay) the number of particles must be fixed. Hence no particles can be created...
marlon

This is a subtle (well I guess that depends on your viewpoint) and important point that Marlon made and should not be overlooked. Thanks for making it!

 

[Rade]

Just to add to what Norman and Jtbell have answered. The fact that the electron is "created" using the energy of the "nucleus" (neutron) also explains why only QFT is able to explain this phenomenon. In QM (which explains both alpha and gamma decay) the number of particles must be fixed. Hence no particles can be created...
marlon

A question If the neutron [N] has internal structure with charge, as observed at the T. Jefferson Accelerator (see Science News, April 27, 2002, Title: "Not so neutral neutron:clearer view of neutron reveals charged locales"), then QM also suggests that the "electron" observed in beta-decay of neutron to proton is pre-existing within the neutron. As explained in the Science News article, a slightly negative charge is present in the neutron near its surface, to balance a slightly positive charge near its core. This effect was observed using probes of cold deuterons [NP]. So, it would appear that either QFT or QM provides sufficient power of explanation for the electron released during beta-decay. I'm sure this has been published, but I do not have a citation.

 

[Astronuc]

A question If the neutron [N] has internal structure with charge, as observed at the T. Jefferson Accelerator (see Science News, April 27, 2002, Title: "Not so neutral neutron:clearer view of neutron reveals charged locales"), then QM also suggests that the "electron" observed in beta-decay of neutron to proton is pre-existing within the neutron. As explained in the Science News article, a slightly negative charge is present in the neutron near its surface, to balance a slightly positive charge near its core. This effect was observed using probes of cold deuterons [NP]. So, it would appear that either QFT or QM provides sufficient power of explanation for the electron released during beta-decay. I'm sure this has been published, but I do not have a citation.

Not quite

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html#c2

If one accepts that a d-quark transforms to an u-quark so that a neutron becomes a proton, which is along the lines of what Norman was showing.

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

and

http://hyperphysics.phy-astr.gsu.edu/hbase/particles/qrkdec.html#c1 -

The quarks exist in the nucleons, not electrons and neutrinos.

 

[Physics Monkey]

The neutron is composed of quarks (1 up and 2 down) which are charged. The measurements you refer to are measurements of the electric form factor of the neutron. This factor tells you something about the charge distribution in the neutron. There are however no electrons inside to begin with. Before an electron can be produced, the down quark has to decay to an up quark and emit a negatively charged W boson. The W then quickly decays into the electron and anti-neutrino.

 

[Rade]

The neutron is composed of quarks (1 up and 2 down) which are charged. The measurements you refer to are measurements of the electric form factor of the neutron. This factor tells you something about the charge distribution in the neutron. There are however no electrons inside to begin with. Before an electron can be produced, the down quark has to decay to an up quark and emit a negatively charged W boson. The W then quickly decays into the electron and anti-neutrino.

OK, thank you. Clearly, the "negatron" is not pre-formed within the neutron, but the "potential" to form the negatron, via decay of the W-boson, is "inside the neutron"--seems like sematics to me and depends on if you observe (e.g., probe with different levels of energy) beta-decay at the macroscopic or microscopic level of organization.

 

[Norman]

OK, thank you. Clearly, the "negatron" is not pre-formed within the neutron, but the "potential" to form the negatron, via decay of the W-boson, is "inside the neutron"--seems like sematics to me and depends on if you observe (e.g., probe with different levels of energy) beta-decay at the macroscopic or microscopic level of organization.

This is plain wrong. It is not semantics. If an electron existed inside the neutron, very different decays of the neutron would be seen. If the electron exists inside the nucleus, so must the proton AND the neutrino- according to your reasoning. Scattering experiments tell us that we see 3 pointlike "partons" in the nucleus that interact with projectile. These "partons" which we now call quarks, are massive, electromagnetically charged constituents. This makes absolutely no sense in your picture since the neutrino is only weakly interacting (not electromagnetically) and has such a small mass. In addition, the proton has substructure (confirmed experimentally) and that would show up under close examination.
And what is a "negatron"? Quit making up particles.

 

[Astronuc]

And what is a "negatron"?

Actually the term 'negatron' is synonymous with electron. It is used in conjunction with positron. Negatron means 'negative electron' just as 'positron' means 'positive electron'.

I heard and saw the term used several decades ago, but perhaps it is not used too much now.

 

[Norman]

Actually the term 'negatron' is synonymous with electron. It is used in conjunction with positron. Negatron means 'negative electron' just as 'positron' means 'positive electron'.
I heard and saw the term used several decades ago, but perhaps it is not used too much now.

I have never heard the term before. Even in the older literature from the 60s that I am currently digging through. Learn something new everyday! Thanks for the correction astronuc!
Cheers,
Ryan

ps:
On a complete sidenote, I know the older term for GeV (giga-electron volt) is BeV (from looking at these papers from the 60s), but what does the "B" in BeV mean?

 

[Astronuc]

In BeV, B stands for Billion.

Lawrence Berkeley Lab had an accelerator called the Bevatron.

The Bevatron was a particle accelerator at Lawrence Berkeley National Laboratory. It began operating in 1954, and the antiproton was discovered there in 1955, resulting in the 1959 Nobel Prize in physics for Emilio Serge and Owen Chamberlain. It accelerated protons into a fixed target, and was named for its ability to impart energies of Billions of eV. It was finally decomissioned in 1993.

from Wikipedia - http://en.wikipedia.org/wiki/Bevatron

I don't think negatron was commonly used. Although I saw the term, it was not very common. Electron has apparently been preferred.

 

[Norman]

In BeV, B stands for Billion.

Doh! So obvious.
Thanks again!

 

[jtbell]

And "billion" can of course be a source of confusion because it means different quantities on opposite sides of the Atlantic, which is why we now talk about "GeV" instead of "BeV". If we were designing that machine today (and hadn't already gone well beyond it) we'd probably call it the Gevatron.

 

[Rade]

Actually the term 'negatron' is synonymous with electron. It is used in conjunction with positron. Negatron means 'negative electron' just as 'positron' means 'positive electron'.
I heard and saw the term used several decades ago, but perhaps it is not used too much now.

The term "negatron" comes from the perspective of the macroscopic view of the nucleus--well before quarks were realized. It was recognized lexicon of nuclear physicist after WW II, as evidenced by this 1953 review of beta-decay by Charles Coryell at MIT:
Title: beta-Decay Energetics
Authors: Coryell, C. D.
Journal: Annual Review of Nuclear and Particle Sciences, vol. 2, pp.305-334 Publication Date: 00/1953
Coryell defined the negatron as "negative beta-particle".