# Beta+ Decay

by Antigone
Tags: beta, decay
 P: 36 I am trying to learn what a Beta+ decay really is, and I dont really understand. For instance, how on Earth can a proton be turned into a neutron, when it has less mass? Someone once told me that in a Beta+ decay 511 kev is being emitted in forms a electromagnetic light. Is it those "511 kev" that are named "neutrinos" and "positrons"? I am really confused.
 P: 45 You are right: a free proton cannot beta decay. This can only happen in a nucleus where the binding energy of the child nucleus may compensate the loss of energy through the decay.
 P: 1,028 What Hawkwind said about the proton's stability... Now as for the Beta+: The Beta+ process is the proton going to neutron+positron+neutrino of electron. $p -> n + e^{+} + v_{e}$ The reason why you get the 511keV peak (that is quiet sharp eg in Na22) is because the positron annihilates with the electrons around it, and emit photons - it's the energy of that photons you measure. So, it has to do with positrons, but not with neutrinos. Neutrinos are totally different (you don't measure them) and they appear as "missing" energy or momenta. In a two point interaction like: $p -> n + e^{+}$ (no neutrino), the energy spectruum would be discrete in the Center of Mass frame system. However that's not the case, and we have a continuous spectrum. For that reason the idea of neutrinos was introduced, to save the conservation of energy and momenta, because a three point interaction as the 1st I wrote, allows continuous spectra,.
P: 36
Beta+ Decay

 Quote by ChrisVer What Hawkwind said about the proton's stability... Now as for the Beta+: The Beta+ process is the proton going to neutron+positron+neutrino of electron. $p -> n + e^{+} + v_{e}$ The reason why you get the 511keV peak (that is quiet sharp eg in Na22) is because the positron annihilates with the electrons around it, and emit photons - it's the energy of that photons you measure. So, it has to do with positrons, but not with neutrinos. Neutrinos are totally different (you don't measure them) and they appear as "missing" energy or momenta. In a two point interaction like: $p -> n + e^{+}$ (no neutrino), the energy spectruum would be discrete in the Center of Mass frame system. However that's not the case, and we have a continuous spectrum. For that reason the idea of neutrinos was introduced, to save the conservation of energy and momenta, because a three point interaction as the 1st I wrote, allows continuous spectra,.

So we can measure that a Na-22 loses 511kev. Now, when the Na-22 turns to Ne-22 during this process, a proton has turned into a neutron. But it also must have lost a electron a long the way. I say this because the Na-22 would have turned in to a negative ion if it had to many electrons. What is has lost is a electron and a proton.

I just have one question. We say that a electron have "kev 511". But electrons have different energylevels depending on where in the atom they are located (electron configuration). Does that mean that a valence electron, in theory, would have more "kev" than an electron that is "closer" to the nucleus?

That is a question I really would like to know!

Thank you so much
 P: 1,028 you should be more careful in what you type though. It doesn't only lose 511keV energy. To see the energy loss you have to take the binding energy of the Na22, the energies of the daughter products and take the difference. 511keV just correspond to the positron's (or electron's) rest mass (it's a way to measure the rest mass of electron). The electron-positron interaction can just happen anywhere (it has nothing to do with the weak interaction). The electron has 511keV in its rest frame (it's mass is 511keV). The energies corresponding to electrons' energy levels around the nuclei, is of the order of eV, so it doesn't really play a role in this case - I think. I am not really sure about it, somebody might answer it more precisely or correctly.
P: 36
 Quote by ChrisVer you should be more careful in what you type though. It doesn't only lose 511keV energy. To see the energy loss you have to take the binding energy of the Na22, the energies of the daughter products and take the difference. 511keV just correspond to the positron's (or electron's) rest mass (it's a way to measure the rest mass of electron). The electron-positron interaction can just happen anywhere (it has nothing to do with the weak interaction). The electron has 511keV in its rest frame (it's mass is 511keV). The energies corresponding to electrons' energy levels around the nuclei, is of the order of eV, so it doesn't really play a role in this case - I think. I am not really sure about it, somebody might answer it more precisely or correctly.

1. Chris, I dont have the knowledge to calculate that. How much energy would it lose? If anyone could help me with that, I would be greatful.

Is "511keV" a estimation because Na-22 loses a electron, or what do you mean? Do we know excactly what happens in beta+ decay?

Another thing. An Ne-22 has lost a proton and a electron, but gained a neutron. When it all comes down to it, that would be a more simple explanation of what has accually happend. In electron capture a electron and a proton becomes a neutron. When a Na-22 decays to a Ne-22 it will have a electron and a proton less, but gained a neutron.
 P: 1,028 for the last thing you said- it's just another process... as for the first, you just need to take the formulas and do calculation to see how much energy is released. ~500keV peak is what we measure, and it's due to the electron-positron annihilation- two photons leave. In the rest frame of the electron+positron, the energy is $2mc^{2}$. So each photon carries the $mc^{2}$ energy that corresponds to the mass energy of an electron/positron. That's why you get the 500keV peak. You get something like that: http://friendlyveggie.files.wordpres...D300%26h%3D167 Now as I would explain it, the energies of the atomic electrons are of the order of eV, so in that scale you won't be able to distinguish it (the scale is of 0.5MeV), even the distribution covers "more" energies... In the picture I sent it's around 100keV...
P: 36
 Quote by ChrisVer for the last thing you said- it's just another process... as for the first, you just need to take the formulas and do calculation to see how much energy is released. ~500keV peak is what we measure, and it's due to the electron-positron annihilation- two photons leave. In the rest frame of the electron+positron, the energy is $2mc^{2}$. So each photon carries the $mc^{2}$ energy that corresponds to the mass energy of an electron/positron. That's why you get the 500keV peak. You get something like that: http://friendlyveggie.files.wordpres...D300%26h%3D167 Now as I would explain it, the energies of the atomic electrons are of the order of eV, so in that scale you won't be able to distinguish it (the scale is of 0.5MeV), even the distribution covers "more" energies... In the picture I sent it's around 100keV...

So, basically what you are saying is this (and I really am trying to understand):

A positron collides with a electron, causing an electron-positron annihilation. This makes the atom produce two photons. These photons gives us the "500keV peak". So, the electromagnetic light is gamma rays (because thats what we see in positron-electron collissions?)

All I am saying is this. In electron capture we know that a K-Shell electron is being "captured" by a proton. A electron neutrino is emitted, and a neutron is created. What if a valence electron, for instance, would have been absorbed by the proton. In theory, if it had more energy, and the proton only can sustain the energy-amount of a k-Shell electron (and not even that much), wouldnt we have seen photons leave the atom? Could its amount of energy, which is greater than that of the k-Shell electron, be that which we see as two photons?
 P: 1,028 As I already said about the electron capture, it's just another process that might happen, which is somewhat equivalent to beta plus (instead of an outgoing positron you have an incoming electron). Of course I am not sure it is the main process that eg Na22 would follow, because it is a rather light nucleus, while electron capture tends to happen in heavier nuclei than that. In the electron capture, if it was the only one occuring, you wouldn't see the peak at energy ~500keV.
 P: 390 There is a lot of things that can happen with a positron. A fast positron can collide with an electron elastically - in which case the electron is accelerated and positron slowed. The fast positron can collide with an electron inelastically and emit photons. In that case the positron and electron still exist, and positron is slowed. The fast positron can undergo elastic or inelastic collisions with nuclei. And the positron can annihilate while fast. If a positron is not annihilated while fast, then a slow positron can annihilate. It may form a free positronium atom - or positronium compounds. A free positronium atom can, yes, only annihilate into two or more photons. This is because of conservation of energy and momentum - you can only conserve them if you either have at least two particles in final state, or your initial state happens to be on exact absorption line. Since positron and electron are 511 keV each, a stationary positronium can only conserve momentum on decay into two photons if these two photons have exactly equal energy of 511 keV and go in exactly opposite directions. Sometimes this is possible because it also conserves angular momentum. There are lots of alternatives. If a positron annihilates while fast, and can annihilate to two photons, then these photons cannot be both 511 keV. Because the kinetic energy of positron has to go somewhere. For example the photons could be emitted to the sides of positron movement, both of them over 511 keV, and not in opposite directions of each other but at an angle. Or one photon might be emitted ahead and the other behind positron movement - in which case the photon emitted behind would I think have less than 511 keV, though not sure - the photon emitted forward surely would have more. A positron and electron can only decay to two photons and conserve angular momentum if they happen to have opposite spins. If they have parallel spins then they cannot decay to 2 photons - they can only decay to 3, 5 or a bigger odd number of photons. But if even a stationary positronium decays to three photons then these will not have 511 keV energy - they are all in the same plane, but can be at any angle to each other so long as no 180 degrees is empty, and can have any energy under 511 keV. But 2 or more particles can be provided elsewhere! If positron annihilates with an electron that is in an atom then the nucleus is a third party that remains. Then the positron could annihilate with electron into just one photon, and the nucleus might take the recoil. And if the positron annihilates with an electron in an atom that has two or more electrons, there is no need to emit a photon. The remaining electron and the nucleus are the necessary two particles to conserve final state energy, so the annihilation energy might be spent to accelerate the electron and give the recoil momentum to the nucleus.
 P: 1,028 snorlack, I am skeptic in that you say about the photons' energy...Maybe because of my little knowledge on experiments - how could your idea explain the high peak of example Na22? except for if somehow, the positron produced by the beta decay has almost the same momentum with the corresponding electron and that should be statistically enhanced...
P: 390
 Quote by ChrisVer snorlack, I am skeptic in that you say about the photons' energy...Maybe because of my little knowledge on experiments - how could your idea explain the high peak of example Na22? except for if somehow, the positron produced by the beta decay has almost the same momentum with the corresponding electron and that should be statistically enhanced...
Perhaps because slowing down and then annihilating at nearly rest is a process which does happen fairly often?
Mentor
P: 12,055
 Quote by snorkack Perhaps because slowing down and then annihilating at nearly rest is a process which does happen fairly often?
To extend that answer: the annihilation cross-section is larger for small energies. It can happen that the positrons annihilate with a high energy, but it is a very unlikely process. Usually they slow down before they annihilate.
P: 390
 Quote by ChrisVer As I already said about the electron capture, it's just another process that might happen, which is somewhat equivalent to beta plus (instead of an outgoing positron you have an incoming electron). Of course I am not sure it is the main process that eg Na22 would follow, because it is a rather light nucleus, while electron capture tends to happen in heavier nuclei than that. In the electron capture, if it was the only one occuring, you wouldn't see the peak at energy ~500keV.
There are some light nuclei that cannot emit positrons at all, but can capture electrons. Beginning with beryllium 7, the next is chlorine 36.

And then electron capture competes with positron emission even when it is possible. For example K-40 can emit positrons, yet it captures 10 000 electrons per each positron emitted. On the other hand, fluorine 18 emits positron 97 % of time, and captures electron 3 % of time.

What are the branching ratios for sodium 22?
 P: 1,028 that's why I used the phrase "tends to", it's not a general principle, just a "rule" let us say which tells that middle nuclei decay via electron capture. I don't know the branching ratios, but I'm sure Na22 prefers to decay via normal beta+.