$\beta^+$ decay

Question

We've been discussing radioactive decay at school, and I grasped everything except for $\beta +$ decay. When I googled radioactive decay, I immediately found out they dumbed down radioactive decay for us, which is probably why they didn't care to explain what they did, they just showed some calculations. We have never discussed neutrino' s and antineutrino's, they leave that out of the equation, which is no problem since they have negligible mass and no charge.

So we've been taught that a proton and electron form a neutron, which I have also discovered is not true (I'm discovering a lot of new things :P). I learned that this is caused by the spontaneous (?) change from up to down quarks and vice versa.

However, at school I must keep to the 'rules' and by those rules I don't really understand $\beta +$ decay. I see $\beta -$ decay as follows:

An electron flees a neutron and leaves a proton. That's why you get an atom with a higher atomic number.

However, how would this work with beta + decay? Can it even be dumbed down to this kind of high school thinking?

score 6 · Answer 1 · answered Apr 07 '13 at 17:43

As a kid, I believed that a neutron "was" composed of a proton, electron, and a (negligible) antineutrino as well. But that's really wrong. The decay genuinely transforms the identity of a neutron to that of proton. The neutron and proton are equally elementary and equally composite, in fact. Microscopically, a down-quark gets changed to an up-quark (in proton) and electron and antineutrino. Again, the down-quark isn't "composed" of the three products: it's equally elementary as the up-quark.

The $\beta^+$ decay is a bit of a problem for the wrong "composite neutron" picture. But the only "coherent" picture is that you must simultaneously imagine a proton to be a composite of a neutron, positron, and a neutrino. This picture sort of contradicts the original $\beta^-$ composite picture for the neutron (unless you are ready to tolerate an infinite hierarchy of compositeness) but because there is really a sort of symmetry between the proton and the neutron, and between the $\beta^+$ and $\beta^-$ decay, the composite proton (containing a neutron and a positron) is the only legitimate picture you may add for the $\beta^-$ decay.

The problem for these composite models is that due to the Heisenberg uncertainty principle, the electron would have a kinetic energy greater than the rest mass of the neutron. — , Apr 07 '13 at 19:44

score 2 · Answer 2 · answered Apr 07 '13 at 22:23

This is an excellent question and here is a school level fudgy answer.

Observe that $\beta^{+}$ decay never occurs for free particles, but for matter inside a nucleus. In this way, the lighter proton can effectively steal any energy that it might need to decay to the heavier set of neutron/lepton/neutrino. Protons are very, very stable particles. Their direct decay has never been observed. So $\beta^+$ and $\beta^-$ decay are essentially different.

score 0 · Answer 3 · answered Nov 15 '16 at 19:51

Neutron mass ~ mass of proton + 2 electrons. Beta- decay, by Einstein's formula, the remaining mass is converted into kinetic energy, the speed of the proton and electron.

Beta- is an exothermic reaction; it gives off energy. Beta+ is endothermic; it needs energy to happen, the kind of energy that is available inside a nucleus or the heart of a star or a reactor.

score 0 · Answer 4 · answered Apr 07 '13 at 22:05

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It works with the same ideas, however there is the condition that since the proton is a lighter particle it needs more energy to be able to emit a positron(this might be in kinetic energy, or more likely, in some nucleus it may have lots of excess energy (its excited)). You should remember that everything wants to lose energy to be in a more stable state, so a ball at the top of the hill has an excess of gravitational potential energy and loses it when it rolls down. Similarly a neutron has a tiny more more energy than it would like and so would like to lose this by beta minus decay when it decays to the very stable proton. For the proton, if for some reason it is given lots of energy, it will have more energy than is contained in a neutron, so it wants to lose energy to be in a more stable state (in case of beta plus a neutron).

answered Apr 07 '13 at 22:05

Dmist

217

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It is not true that nuclei that decay by $\beta^+$ re necessarily excited. It is enough that they can get to a lower state by doing so. – dmckee --- ex-moderator kitten Apr 08 '13 at 03:27
But when the proton does decay, it does have an excess of energy? – Dmist Apr 08 '13 at 08:40
You can't really say that the proton does or does not have an excess of energy at all. The parent nucleus has more energy (mass) that the daughter nucleus plus the positron plus the neutrino (and including their kinetic energy). – dmckee --- ex-moderator kitten Apr 08 '13 at 12:23
Cheers that makes sense – Dmist Apr 08 '13 at 13:52

voix · Answer 5 · 2013-04-13T23:25:34.260

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$\beta^+$-decay can be explained as a result of nuclear protons collision:

$p^+ p^+\to p^+ p^+ W^- W^+ \to (p^+ W^-)\: p^+ W^+\to n\: p^+ e^+ \nu_e$

Compare to $\beta^-$-decay:

$n \to p^+ W^- \to p^+ e^- \bar{\nu}_e$

edited Apr 13 '13 at 23:25

answered Apr 13 '13 at 23:10

voix

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$\beta^+$ decay

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