Particle Physics: Unstable Nuclides, Beta & Positron Decays Explained

vinter
Messages
77
Reaction score
0
Did I say Particle Physics? I shouldn't have said that. I don't know much about it.
But still, what I came upto recently was this :-


I read somewhere that unstable nuclides can disintegrate in several ways, beta decay and positron decay being two of them. In beta decay, a neutron gets converted into a proton and an electron and in positron decay, a proton gets converted to a neutron and a positron. Now, also, an electron and a positron annihilate each other giving lots of energy. So... If I carry a beta decay with a neutron, I will have a proton and an electron. Then if I carry a positron decay with this proton, I will get my original neutron back, plus a positron. So basically, the situation amounts to getting a pair of an electron and a positron from nothing. Now if I make them react, I will get energy for free! Isn't it a perpetual motion machine then? If not, what's the flaw?
 
Physics news on Phys.org
From what I understand about these processes, you can't undergo the reaction you're describing. If you're talking about free protons and neutrons (that is, outside of a nucleus), then you can get a neutron decay, just as you described, but free protons don't decay into a neutron and positron (though GUTs think they might decay into other things).

Now, if we look at a whole nucleus (with its protons and neutrons), then there are circumstances in which each of the processes you described will occur. However, they will not occur in succession. That is, a nucleus will not undergo beta decay, followed only by positron emission. This is because you need to think of the nucleus as a composite state with a certain energy. A beta or positron decay can occur when there exists a lower energy state for the nucleus in which one of the nucleons is switched (proton to neutron or the other way around), but you can't decay back to the original state because it has a higher energy.

There may, however, be circumstances in which you can undergo a beta decay, some series of other decays, and then a positron decay, but I'm not familiar enough with the various reactions to know for sure.
 
Last edited:
The answer, I believe, is in your question vinter. The key word here is 'unstable'. A neutron requires more energy and so is less stable than a proton, hence beta decay in free neutrons. It would make no sense for the proton to decay into a neutron again as it is already in its most stable state. However, a proton in a large nucleus packed with other protons is not in its most stable state thanks to the additional EM energy doing its best to overcome the strong interactions between the nucleons. To get to a more stable state, the proton decays into a neutron, positron, neutrino and released energy. With neutral charge and this hapless EM energy gone, the nucleon is in its most stable state. To decay into a proton again would require MORE energy to overcome the EM interactions, so no help with perpetual motion there, though if you do solve the problem I want royalties for helping you avoid the wrong route, you hear?
 
hih... and I thought I was going to become rich with this money making formula.
 
Vinter - a nucleus which undergoes beta decay (electron emission) has an excess of neutrons, while a nucleus which undergoes positron emission has a relative excess of protons (although it will have still more neutrons than protons in general). They are two different nuclides.

Most beta and positron emitters are artifically made, so it take energy to make them, more than one will recover from the annihilation from a positron-electron pair.

The 'spontaneous' decay of a proton is unlikely. As El Hombre Invisible pointed out, the proton is extremely stable, otherwise we would be without a lot of hydrogen.

A free neutron is unstable with a half-life of about 10.3 minutes, but in a nucleus it is much more stable. Nuclei with an excess of neutrons may be unstable - see Chart of the Nuclides - http://wwwndc.tokai.jaeri.go.jp/CN04/index.html

Notice the position of the positron and electron emitters with respect to the stable nuclides.
 
Last edited by a moderator:
Astronuc nucleus...

Astronuc said:
...a nucleus which undergoes beta decay (electron emission) has an excess of neutrons, while a nucleus which undergoes positron emission has a relative excess of protons (although it will have still more neutrons than protons in general). They are two different nuclides.

A free neutron is unstable with a half-life of about 10.3 minutes, but in a nucleus it is much more stable.

Astronuc, what is the Half-Life of a 'Deutronium' nucleus?
[/color]
 
Last edited:
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top