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Beta decay and electron capture.

by theCandyman
Tags: beta, capture, decay, electron
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theCandyman
#1
Sep27-06, 02:26 PM
P: 395
Recently the radation physics class I am taking reviewed decay and I wondered how beta minus decay and electron capture can both emit neutrinos. I tried asking my professor, but I do not think I explained my question well enough to him.

To explain my mindset, imagine you have only one radioactive atom in the middle of a few electrons, nothing else exists that can influence them. The atom undergoes beta decay, emits an antineutrino, and captures a nearby electron to remain neutral. Now the atom undergoes electron capture and emits another neutrino. You are back where the cycle started, but down the energy of the two neutrinos, a small amount - but it is missing. Eventually that energy will add up if the cycle continues, so where does it all come from?

What is wrong with my reasoning above?
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mathman
#2
Sep27-06, 02:50 PM
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In general, you would not have both reactions possible with the same nuclei involved unless outside energy was supplied for one of the reactions.
jtbell
#3
Sep27-06, 06:29 PM
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Quote Quote by theCandyman
What is wrong with my reasoning above?
As mathman pointed out, your problem is with the assumption that both electron capture and beta decay can occur with the same nucleus. Your reasoning is a good proof by contradiction that this is not possible.

If you can find a nucleus that does behave this way, using the
Interactive Chart of Nuclides, I'll send you a cigar (or other poison of your choice).

Astronuc
#4
Sep27-06, 08:37 PM
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Beta decay and electron capture.

Recently the radation physics class I am taking reviewed decay and I wondered how beta minus decay and electron capture can both emit neutrinos. I tried asking my professor, but I do not think I explained my question well enough to him.

To explain my mindset, imagine you have only one radioactive atom in the middle of a few electrons, nothing else exists that can influence them. The atom undergoes beta decay, emits an antineutrino, and captures a nearby electron to remain neutral. Now the atom undergoes electron capture and emits another neutrino. You are back where the cycle started, but down the energy of the two neutrinos, a small amount - but it is missing. Eventually that energy will add up if the cycle continues, so where does it all come from?
There seems to be some confusion here regarding 'electron capture', which is usually the term reserved for capture of a K-electron by the nucleus, and very rarely by an L-electron.

When a nucleus undergoes beta decay, it's charge will increase from Z -> Z+1, and the atom (which is a new element) will attract a nearby electron to remain neutral, and nearby atoms will find other electrons, until somewhere that loose beta-particle slows down and becomes an atomic electron. Neutrality is preserved. The nucleus does not undergo electron capture.

Atoms with a deficiency of neutrons (or excess of protons) with respect to the line of stability, which one will find in the interactive chart of nuclides (link provided by jtbell), have a choice of positron emission or electron capture, but it's usually one or the other, and EC is prefered by most radionuclides above the stability line.

I strongly recommend perusing the Interactive Chart of Nuclides (courtesy of Brookhaven National Lab - bnl) and developing a 'gut feeling' of the trends of the properties of the radionuclides. Put the chart into decay mode to get an idea of the preference in decay modes.
theCandyman
#5
Sep28-06, 08:02 AM
P: 395
Quote Quote by jtbell
As mathman pointed out, your problem is with the assumption that both electron capture and beta decay can occur with the same nucleus. Your reasoning is a good proof by contradiction that this is not possible.

If you can find a nucleus that does behave this way, using the
Interactive Chart of Nuclides, I'll send you a cigar (or other poison of your choice).
I looked at the chart, and maybe I am misunderstanding something, but can Cl-36 undergo beta decay to Ar-37, then Ar-37 decay by electron capture to Cl-36 again?
Morbius
#6
Sep28-06, 08:25 AM
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Quote Quote by theCandyman
I looked at the chart, and maybe I am misunderstanding something, but can Cl-36 undergo beta decay to Ar-37, then Ar-37 decay by electron capture to Cl-36 again?
Candyman,

Cl-36 doesn't beta-minus decay to Ar-37; it decays to Ar-36.

In a beta-minus decay, the two ejected particles are an electron and an anti-neutriino.
[ One, the electron is a lepton, and the anti-neutrino is an anti-lepton. Therefore, the
total number of leptons remains the same; it has to. Conservation of Lepton Number
appears to be a fundamental law just as Conservation of Energy.]

The mass represented by the electron and anti-neutrino are very small. The rest
energy of an electron is 0.51 MeV. For comparison, the energy equivalent of an
atomic mass unit is 931.1 MeV. So the electron is about 1/2000-th the mass of
a single atomic mass unit. The anti-neutrino energy is even smaller.

Therefore the atomic mass number of the parent nuclide remains unchanged during
a beta-minus decay.

Dr. Gregory Greenman
Physicist
theCandyman
#7
Sep28-06, 11:10 AM
P: 395
Quote Quote by Morbius
Cl-36 doesn't beta-minus decay to Ar-37; it decays to Ar-36.
I must have been thinking A -> A+1 in my head. But choose another element, K-40 to Ca-40 looks like it might fit my example.
Astronuc
#8
Sep28-06, 12:14 PM
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Beta decay increases Z by 1, but A remains the same.

Positron emission and electron capture (EC) decreases Z by 1, but A remains the same.

On the Chart of Nuclides (with Z going vertically up the chart, and A increasing horizontally to the left), beta decay moves the nuclide on a diagonal lower left to upper right (think vertical angle).

Positron emission and EC move the nuclide upper right to lower left.

Alpha decay move Z down 2 and A down 4 (He nucleus is Z=2, A=4), and nuclide identity shifts upper right to lower left by two squares.

See - http://hyperphysics.phy-astr.gsu.edu...radser.html#c1
(note on hyperphysics, that Z increases horizontally and A increases vertically).

Unfortunately different systems can cause confusion.
Morbius
#9
Sep29-06, 09:04 AM
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Quote Quote by theCandyman
I must have been thinking A -> A+1 in my head. But choose another element, K-40 to Ca-40 looks like it might fit my example.
Candyman,

In any of these reactions; you just have to look at the energy involved.

If the reaction is exothermic; that is - it "releases" energy, then the reaction will
happend, sooner or later, since they are all stochastic.

If the reaction is endothermic; that is - it "requires" energy; then the nuclide is stable
against that - modulo the ever present quantum fluctuations in energy.

If a reaction is exothermic; then its inverse reaction is endothermic.

Dr. Gregory Greenman
Physicist
jtbell
#10
Sep29-06, 12:12 PM
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Quote Quote by theCandyman
K-40 to Ca-40 looks like it might fit my example.
K-40 undergoes beta-decay to Ca-40, correct.

Ca-40 does not undergo electron capture to K-40, but rather, double electron capture to Ar-40, which is stable. This is an exceedingly rare process. The half-life is on the order of [itex]10^{21}[/itex] years!


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