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Potassium-40 decay modes

 
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Jun18-12, 10:50 AM   #1
 

Potassium-40 decay modes


1)Is Potassium-40 dangerous material?How dangerous is it?Does it belong to restricted materials?
2)Why Potassium-40 decays in such different modes such as beta decay,electron capture,
and positron decay?What could be done to prevent it decay in other ways with exept beta decay?Or what could be done to prevent at least gamma radiation from it?
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Jun18-12, 03:35 PM   #2
 
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See above. With a half life that long it wouldn't be dangerous. Modes of decay are what they are, detailed nuclear physics would explain it. There is nothing that can be done to change it or to prevent gamma rays from being emitted.
Jun19-12, 08:26 AM   #3
 
It depends upon how much of it you have. a single molecule of the toxin in botulism wouldn't, the scheme of things, be all that dangerous. The same applies to K-40 (you have some inside your body). However, gather several curies and you're in dangerous territory.
Jun19-12, 10:37 AM   #4
 

Potassium-40 decay modes


http://www.nndc.bnl.gov/chart/decays...uc=40K&unc=nds
A 238 keV beta+ is considered strongly ionizing radiation.
Because it is a positron, it will also cause two 511 keV Gamma rays; again, strongly ionizing radiation, with even greater penetration.
The other decay modes are much less worrisome; the 560 keV beta- primary decay mode has a flux to dose rate of 0.6e-13 or so; the lambda is 1.8e-17, so you'd need about 1e30 atoms to get a 1Sv dose in a second, or about 1.75 kg to get a 1Sv dose in an hour.
The problem is that the body puts extra potassium in the bone marrow; then you have very targeted radiation that will stay with you for years, not hours. 2 grams will give a 1Sv dose in a year. The risk coefficient for bone marrow is 2e-3; what this means is that 1 gram of K40 absorbed into the bone marrow will essentially guarantee cancer.
Jun19-12, 11:33 AM   #5
 
How is it possible that the same atom exibit so different decay modes?
How is it determined?
Sep1-12, 05:33 AM   #6
 
Quote by Stanley514 View Post
How is it possible that the same atom exibit so different decay modes?
How is it determined?
1) Positron and electron capture

ALL nuclei which emit positrons also have decay mode by electron capture. Electron capture is basically skipping the intermediate, so it is always an option. Many nuclei only have energy for electron capture and cannot emit positrons, but if there is enough energy for positron emission, there are always two decay modes.

However, I do not know what determines the branching ratio.

2) Electron capture and emission

Potassium 40 has odd number of neutrons and protons. This makes it a high energy state because of the unpaired nucleons. BOTH the even isobars - calcium 40 and argon 40 - are lower energy - so potassium can decay by either electron emission or electron capture. Since the electron capture has enough energy, potassium can also decay by positron emission. Making a total of 3.

Again, I do not know what determines the branching ratio.
Sep3-12, 07:04 AM   #7
 
Quote by Stanley514 View Post
How is it possible that the same atom exibit so different decay modes?
How is it determined?
The prediction of decay modes is imperfect. The best idea is that the wave functions of the quarks and leptons have oscillations which can work in harmony or counter force, and when a group of quarks has more internal cohesion than cohesion to the main body, it will be ejected. This is why Alpha decays are the most likely for large nuclei.

See Half-life predictions for decay modes of superheavy nuclei

The decays by electron capture verses positron usually only vary in the amount of energy, although ejection of proton is often another decay choice. It is expected that a stray neutrino is needed for the positron decay, but the flux of neutrinos is high enough that it is the dominant decay mode.

The choice of beta verses neutron ejection is similar: will an anti-neutrino interact with a down quark in a way which will allow the beta to escape, before a udd cluster in the composite waveform is more stable alone than in the nucleus?

In stable nuclei, it is expected that neutrinos interact, but the resultant positron or electron is recaptured before the particle can travel out of the nucleus.
Sep4-12, 03:26 AM   #8
 
Quote by wizwom View Post
It is expected that a stray neutrino is needed for the positron decay, but the flux of neutrinos is high enough that it is the dominant decay mode.
Nonsense:

A stray neutrino cannot cause positron decay (violation of electron charge) - only a stray antineutrino could. But if it does, the whole energy of the neutrino goes to the positron.

Actually, positron decay is spontanous just like beta decay, and involves emission of a new neutrino completely irrespective of the presence of any antineutrinoes.
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