Talk:Positron emission

Untitled[edit]

This article claims (as does Beta decay), that a positron is also called a 'beta plus particle'. Why is this, seeing as a positron is positively charged? Can someone confirm this?

Answer: Historical reasons. The electron, when ejected from a radioactive nucleus, was called a beta particle. When it was realized that beta particles were just electrons, the term "beta-minus decay" was used. Thus, by analogy, positron emission is therefore "beta-plus decay".

--24.80.110.173 04:30, 5 August 2005 (UTC)[reply]

Merger proposal[edit]

Positron emission is another term for beta plus decay. This page is too small to be a separate article. I'd like to suggest merging this with beta decay, which already contains information about beta plus decay.
- SkyLined (talk) 16:48, 22 March 2008 (UTC)[reply]

Good idea. -Lone Deranger (talk) 22:44, 24 March 2008 (UTC)[reply]

good idea as a use of beta decay —Preceding unsigned comment added by 86.139.239.206 (talk) 10:24, 11 May 2008 (UTC)[reply]

I believe this is a bad idea. The radioactive decay is confusing as is, and since positrons and electrons are different particles and have very different practical consequences (anihilation processes for positrons), I think it would be much clearer to keep the sections separate. —Preceding unsigned comment added by 213.136.49.117 (talk) 06:25, 4 September 2008 (UTC)[reply]

Do not merge them since they are two different decays in their path —Preceding unsigned comment added by 216.161.40.1 (talk) 19:26, 5 September 2008 (UTC) Do not merge the articles. Positron emmission is a term referred to by non physicists. I was using this page in studing nuclear cardiology and found this he[reply]

I believe that this is a good idea, since the page is too small. We should put it under a separate subsection titled "Positron Emission" and redirect positron emission to that section under the beta decay.

I suggest merging the two articles, seeing as beta decay already contains almost as much information on positron emission as positron emission does itself. Also, positron emission is a form of beta decay and should thus be included on that page.Luqavi (talk) 14:29, 28 January 2010 (UTC)[reply]

I'd say merging them is a bad idea. They are two totally separate decay types with separate products. --The High Fin Sperm Whale 17:46, 3 May 2010 (UTC)[reply]

I think that it's a bad idea, because the beta + decay occurrence isn't one of the three natural (alpha, beta-, and gamma) decay processes noted to be the natural atomic decay processes, and that the beta+ decay process occurs to the proton, whereas the beta- decay occurs to the neutron. And thus any discussion of this would complicate the beta- decay discussion.WFPM (talk) 09:03, 21 September 2010 (UTC) Also why wasn't the beta+ decay of 4Be9 (by Fermilab) mentioned in the article?[reply]

Error?[edit]

Shouldn't the energy be on the other side of the equation? Positron emission requires an input of energy as the neutron is heavier than the proton. 138.38.148.5 (talk) 21:34, 14 April 2008 (UTC)[reply]

If the energy were on the other side, the decay would not occur! Only exothermic decays happen. Other reactions are possible but require energy inputs as in an "atom smasher" or other collision initiator such as a cosmic ray interaction. I believe the key here is that Carbon 11 is unstable because of the unpaired neutron, but the decay energy depends on the whole ensemble of protons and neutrons present and should not be viewed as "decay" of a proton into a neutron, which, as you noted, is energetically impossible. Even in ordinary (non-nuclear) chemistry, reactions and are often driven by the combinations of atoms that occur on either side of the reaction equation, because bonds can be stronger or weaker depending on the combinations that occur. Carrionluggage (talk) 15:10, 15 April 2008 (UTC)[reply]

You are correct that a free proton is lighter than a free neutron, but the carbon-11 nucleus is heavier than the boron-11 nucleus. I hope that clarifies what carrion is saying about the context of the decay. —Preceding unsigned comment added by 150.209.42.212 (talk) 19:32, 11 June 2009 (UTC)[reply]

Yes but then the former EO6C11 nucleus (minus a positron) becomes OE5B11 and has to get rid of the 6th electron plus its kinetic energy, so wouldn't it wind up being less massive that it would have if it had just caught an orbital electron?WFPM (talk) 21:08, 17 April 2011 (UTC)[reply]

change of energy released[edit]

Someone has listed a change from 0.45 to 0.96 MeV but it does not show (in my browser - I still see 0.45) The 0.96 MeV value is the total energy released, positron and neutrino. The latter is very hard to detect so this number is not of much practical use. It is the endpoint of the positron spectrum - unattainable in practice. The value 0.45 MeV is the mean positron energy - useful in medical applications etc.

Why can't I see the change except in the history? Carrionluggage (talk) 21:59, 22 April 2008 (UTC)[reply]

Via what weak force boson?[edit]

"In beta plus decay, a proton is converted, via the weak force, to(...)"

Via what? How does the weak force act upon the proton to make it a neutron? Via a Z⁰, a W^- or a W⁺? Does it vary? And, the proton "gains" a weak boson or what? oO? Thγmφ (talk) 14:33, 6 April 2009 (UTC)[reply]

The proton contains two up quarks and a down quark. The neutron contains two down and one up. So to get from proton to neutron, you have to convert an up quark into a down quark. The up can be converted to a down and a W+, with the W+ decaying into a positron and a neutrino. It can't be a W- or Z0 instead of the W+, because those wouldn't carry the right electric charge. Always in quantum field theory you can draw more complicated diagrams with closed loops and such, but that's the simple (tree-level) one. -- Tim314 (talk) 01:04, 16 July 2011 (UTC)[reply]

I agree this point needs to be addressed. The proton decay page specifies there is no evidence of a proton breaking apart to any other particles, so how is it "converted" to a neutron and positron? —Preceding unsigned comment added by 71.53.203.123 (talk) 02:41, 14 January 2010 (UTC)[reply]

A free proton doesn't decay (or at least decays very, very rarely -- strong experimental limits have been placed on this.) A free proton can't decay into a neutron, because neutrons are heavier than protons. But inside an atomic nucleus, you can convert a proton to a neutron + positron + neutrino -- the extra energy can come from the other particles in the nucleus and the interactions between them.

Similarly, a free neutron isn't stable (it decays into proton + electron + anti-neutrino), but clearly they're stable inside the nucleus, or inside neutron stars.-- Tim314 (talk) 00:58, 16 July 2011 (UTC)[reply]

I want to add into this section a comment/question... If I understand what's above, it seems that in stellar nucleosynthesis, free protons do collide to form diprotons, and unlike billiard ball collisions, they don't just bounce off one another, but instead they interact according to various possible Feynman diagrams. In the vast majority of interactions, a diproton will quickly decay back into two free protons, but rarely, a positron will be emitted instead, leading to a deuterium nucleus. It seems that the positron emission requires a bit of energy input that comes from the other proton? I also wonder what the time scale is for this interaction, and what is the percentage of interactions that yield the deuterium nucleus... that's probably in the proton-proton chain article.... Thanks for any discussion! ~~cellodont~~

History?[edit]

I think a section discussing when such positron-emitting isotopes were discovered and or predicted would be of interest. — Preceding unsigned comment added by 69.164.56.1 (talk) 03:08, 11 March 2012 (UTC)[reply]

Occurrence of positron emission[edit]

"Positron emission occurs only very rarely naturally on earth, when induced by a cosmic rayor from one in a hundred thousand decays of potassium-40, a rare isotope, 0.012% of that element on earth." But there is also lanthanum-138, which has a probability of 66.4% for beta-plus decay (which means that it has a shorter partial half-life for beta-plus decay than potassium-40). 129.104.241.25 (talk) 12:43, 1 October 2023 (UTC)[reply]

The symbol β⁺ in the tables of isotopes is ambiguous[edit]

It means positron emission + electron capture for most isotopes, but only position emission for ⁴⁰K, ⁵⁹Ni, ⁹¹Nb, ^91m1Nb, ¹⁴⁰Pr, ¹⁴³Pm, ¹⁴⁴Pm, ¹⁴⁰Eu, ¹⁴²Eu, ¹⁵²Eu, ¹⁴⁰Gd, ¹⁴²Gd, ¹⁴²Tb, ¹⁴²Dy and ²⁰¹Pb. Of course there is actually no confusion caused because when β⁺ and EC appear at the same time, we know that the former refers only to positron emission. 129.104.241.214 (talk) 14:42, 28 December 2023 (UTC)[reply]