Talk:Beta decay

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I think we should say something about alpha decay. it doenst fit right in with beta and suctch, but I think it would be helpful for students.

I think the comment about beta decay being the same as neutron decay should be removed. Muon decay to electrons is beta decay, and in some sense the conversion of a proton to a neutron in some nuclei (with the emission of a positron rather than an electron) is also beta decay. I think it would be a good idea to say instead that neutron decay is a type of beta decay.

Following the established standard theory, is not clear at all the reason why an anti-neutrino should be bounded together to an electron and a proton into the neutron particle, other than to preserve conservation law in beta decay. At present there is another concurrent theory, that is able to predict exact values for the neutron particle and beta decay, that not contemplate the existence of neutrino at all. It is called Hadronic Physics, and it may be considered as a Quantum Mechanics' generalization that is applicable, instead of Classical Quantum Mechanics, on non-local, non-linear and non potential-derived interactions.

These are conditions that happen when there is significative spatial superposition between particles' wave packet functions.

Other informations can be found at www.neutronstructure.org

Hmmm

Shouldn't we also say something about double beta decay?--Deglr6328 04:18, 22 May 2005 (UTC) ===Definition---Beta decay can be defined as that physical phenomena that occurs when the gravitational energy driven process of matter accumulation into an atom requires that one of the mass constituents (Neutron or Proton) be converted the other for increased stability (lower free energy) purposes. In this instance of physical phenomena there are 3 Theorized possibilities. 1: The conversion of a Neutron into a Proton plus an electron plus a gamma ray (energy) emission. 2: The conversion of a Proton into a Neutron plus a positron plus a gamma ray. 3: The conversion of a Proton into a Neutron by the aquisition by the Proton of an "orbital electron". WFPMWFPM (talk) 14:13, 1 October 2008 (UTC)[reply]

Negatrons

To clarify on negatrons, consider the OED:

negatron, n. 2. An ordinary electron (as distinct from a positron). Now disused.

1933 Lit. Digest Dr. Anderson suggested also that the familiar negative electron be re-christened ‘negatron’, but it seems unlikely that this will be accepted.

And it wasn't. -- Xerxes 16:13, 2005 May 30 (UTC)

Interestingly, Ehmann and Vance in their book Radiochemistry and Nuclear Methods of Analysis actually do use the term "negatron".

--24.80.110.173 04:34, 5 August 2005 (UTC)[reply]

Confusing

Right. Neutron turns into proton and electron in Nucleus. That's fine. Beta minus particle ejected at high speed from aforementioned nucleus. Hang on. My A level education said that the nucleus is positive, and an electron is negative, and we've been told that opposite charges attract. So why does a negative electron get repelled by the positive nucleus, when they should be attracting? Cheers.

Well, the answer is in the question. The electron is ejected at high speed. So high that the attraction to the nucleus cannot stop it from escaping. -- Xerxes 20:24, 14 December 2005 (UTC)[reply]

Almost ever. But sometimes an electron can be emitted with such low energy that it is captured to one of the atomic electron shells (the probability is very small). It is the beta decay to bounding state. V1adis1av 14:42, 23 December 2005 (UTC)[reply]

Ok. But what makes the electron move in the first place?

It gains a lot of kinetic energy from the disintegration of the nucleus. It's the same energy as in nuclear fission; the mass of the original nucleus is converted into energy plus a lighter nucleus. -- Xerxes 15:48, 15 December 2005 (UTC)[reply]

Sorry. Me again. But doesn't the neutron in the nucleus turn into a proton, remaining the same weight, and so no energy is created or destroyed? And the whole matter into energy, is that e=mc² territory?

That's the beauty of the whole thing. Neutrons are ever so slightly more heavy than protons. Just heavy enough that they can decay into a proton, electron and electron neutrino. Also, when the neutron is bound inside a nucleus, you have to take into account the binding energy between it and the rest of the nucleus, which will also change after its decay into a proton. The slight mass difference can turn into quite a lot of energy, since (as you point out) it gets multiplied by c², a very large conversion factor. -- Xerxes 22:50, 15 December 2005 (UTC)[reply]

Ahhh. Interesting. So, one final question. Why is the mass turned into kinetic energy? Why not heat or anything?

Heat is a property of bulk matter. It doesn't make sense to talk about the heat of a single particle. Actually, heat is what you get when a bunch of particles have kinetic energy and they're not moving in the same direction but just bouncing around all over. For a single particle, kinetic energy is just about the only degree of freedom available. -- Xerxes 15:41, 16 December 2005 (UTC)[reply]

Ahh. Makes perfect sense now. Thank you Xerxes, may the wiki God shine upon you for ever.

There are only really two types of enerqy, kinetic, and potential, heat, movement, sound, evey light (i thinK?) is just an expression of this kinetic energy] - oxinabox1

Need a bit of help

What is the maximum possible energy an electron can acquire from natural beta decay? The typical energy? Such as the beta decay expected from fission daughter products. Surely max beta KE must be a property constrained by parameters of the weak interaction though this is waaay beyond my capablility to determine. I have tried looking for a table of beta decay energies for various isotopes but they don't seem to exist. I need to know this so that I can make edits to another article where cerenkov light production via beta decay is discussed. IANAP so if I could get the energy in KeV or MeV so that I neen't do conversions that would be very helpful. Thanks!--Deglr6328 00:21, 29 December 2005 (UTC)[reply]

Use this: http://www.nndc.bnl.gov/nudat2/indx_dec.jsp. The highest I found was 20.6 MeV from Boron-14. I don't see any reason why there would be an upper bound tho. -- Xerxes 03:58, 29 December 2005 (UTC)[reply]

Great! Thanks very much! Now if I could trouble you for just one more thing.... Could you determine, in KeV, what an electron at 99.97% c would be? Again, many thanks! --Deglr6328 07:07, 29 December 2005 (UTC)[reply]

About 20.3 MeV = 20300 keV. -- Xerxes 20:20, 29 December 2005 (UTC)[reply]

Thank you so much!! (though I think I could've handled the MeV to KeV conv. :oD) You're help is very much appreciated!--Deglr6328 00:44, 30 December 2005 (UTC)[reply]

Little insight

When the parent nucleus decays, does it form a daughter atom that is an ion?

Momentarily, yes. -- Xerxes 21:05, 14 May 2006 (UTC)[reply]

Momentarily? Exactly where would it get a new electron from to match its new proton? It had one closeby, but as explained under the Confusing section above, that one gets tossed away at high speed. -- Milo

The universe is awash in electrons. The ion will pick one up sooner or later. -- Xerxes 02:21, 26 June 2006 (UTC)[reply]

until it gained that electron wouldn't it be extremely chemically reactive (this is just my logic) - oxinbox

Feynman Diagram

I believe the feynman diagram shown is wrong. The convention I have been taught is to represent the W in the same way as the photon, with a wavy line, and there shouldn't be an arrow on the gauge boson. The arrows on the other two particles should have one towards the vertex, and one away, not both away. Does anyone disagree? LeBofSportif 18:46, 18 May 2006 (UTC)[reply]

What you describe is identical to the diagram found at Feynman_diagram#Beta_decay, which is closer to what I recall learning as well (though I don't have a reference at the moment). It's fine with me if you change it. You may also want to talk to WarX who appears to have created the present diagram. The diagrams in the two different articles should agree. -MrFizyx 19:30, 18 May 2006 (UTC)[reply]

Neither of these diagrams is especially pretty, but either is correct. There are no really hard rules when it comes to drawing these things up, and the conventions on the W are especially loose. -- Xerxes 22:57, 18 May 2006 (UTC)[reply]

Hey Xerxes. If you're a lattice gauge theory guy, then you know better than I. How about the backwards arrow on the anti-neutrino? -MrFizyx 00:40, 19 May 2006 (UTC)[reply]

Ah, right. That's totally wrong. It should look like Neutron Decay. I'll fix up a nicer version in the morning. -- Xerxes 03:19, 19 May 2006 (UTC)[reply]

OK, I added a new diagram. Let me know if it needs any further modification. -- Xerxes 17:44, 19 May 2006 (UTC)[reply]

Hi, I think the vertex on the right-hand-side should either have an electron and an anti-neutrino both going away from the vertex, or a neutrino going into the vertex and an electron going away, so that lepton number is conserved? Not an anti-neutrino in and an electron out? I may of course be confused. Spinosaurus87 13:59, 12 June 2007 (UTC)[reply]

Beta-plus decay and proton-proton fusion

The article says:

So, unlike beta minus decay, beta plus decay cannot occur in isolation, because it requires energy, the mass of the neutron being greater than the mass of the proton. Beta plus decay can only happen inside nuclei when the absolute value of the binding energy of the daughter nucleus is higher than that of the mother nucleus. The difference between these energies goes into the reaction of converting a proton into a neutron, a positron and a neutrino and into the kinetic energy of these particles.

I don't think this is strictly true--or, at least, it's a bit misleading. The positron emission in proton-proton chain fusion requires the exact same energy, for the exact same reason (because it's the exact same reaction), and gets it in the exact same way (the binding energy of D), and yet it does not "only happen inside nuclei." (Well, it happens inside an H nucleus, but not in isolation. The point is that the two H nuclei don't fuse into a diproton before one of them emits a positron. And if "only inside nuclei" includes H nuclei it doesn't mean very much anyway.)

This may all sound nitpicky, but given the confusion on the talk pages of multiple articles about how PP can work given that a neutron is more massive than a proton, I think it makes a difference.

One possible change would be to say something like "... can only happen in nuclear reactions when...", but that sounds even worse. A better solution would be to just note the relationship between PP fusion and beta decay in some clear way (which isn't occuring to me at the moment). --69.107.75.113 08:43, 16 March 2007 (UTC)[reply]

1.022MeV

The article dances around the fact that a minimum amount of energy is required for beta-plus (e.g., "In all the cases where β⁺ decay is allowed energetically"), but never gives the value of 1.022MeV. Since that value is in the electron capture article, and it's if anything more relevant here, it should be incorporated somewhere. --69.107.75.113 08:49, 16 March 2007 (UTC)[reply]

Radiation/decay

Excuse me? beta radiation redirects to this page? could anyone write a page on beta radiation, or possibly add a header here (altough i'm pretty sure it deserves its own page)

What exactly would you like to see discussed?

You could be arguing from historical difference. True, there were a few decades when we knew a bit about beta radiation and didn't know it was beta decay from nuclear reactions (a history that the article explains, although not in the stub History section). But there were centuries when we knew a bit about the Evening Star and didn't know it was the planet Venus; that doesn't mean Evening Star gets a separate page.

You could also be arguing from a technicality. Given the traditional definition, an electron beam fired from a particle accelerator, or for that matter the cathode ray in your TV (well, your parents' TV), counts. And for medical purposes (see particle radiation), a beam of electrons hitting you has the same effect no matter what the cause. But most people wouldn't call it radiation unless it came from... well, a radioactive source. Meaning a collection of beta-decaying nuclei.

So, in either case, I don't think the argument is very good. --69.107.71.101 17:29, 6 April 2007 (UTC)[reply]

I just checked, and Evening Star does have a separate page. But it's just a disambiguation page, to distinguish between Venus and the train, paper, pub, Judas Priest song, etc. named after it. There is no link to a separate article on the Evening Star; just a link to Venus. So, if Judas Priest writes a song called "Beta Radiation" (and it's notable), we'll need a separate Beta Radiation disambig page, and "beta radiation" will redirect there with a link back to here. But this probably isn't what you wanted. --69.107.71.101 17:31, 6 April 2007 (UTC)[reply]

Sorry, I meant to point out that beta decay is a process, which releases electrons. A beam of these electrons would be described as beta radiation, and an article describing uses, dangers, penetrative strength and history (for example) would I think be seperate from the beta decay article. I was recently looking for the thickness of aluminium that would shield you from beta radiation, but that information is not available. Hope that makes sense! --Doctorp9999 15:38, 15 April 2007 (UTC)[reply]

OK, good point. I'm not entirely sure it needs a separate article, but that information should be somewhere easy to find. (As a matter of fact, most of it already is on Wikipedia, but scattered in poorly-crosslinked articles on beta decay, radiation, particle radiation, radioactivity, radiation damage, radiation poisoning, and probably half a dozen more. Someone (like you) could perform a good service by organizing it.

As for your specific question, it depends on the energy of the beta particles. The 20.6MeV decay from boron-14 is a little different from the 1.311MeV from potassium 40. To get the numbers, you have to calculate or look up the stopping power of aluminum at the appropriate energy and do the math. Or, better, let someone do the math for you. Try http://physics.nist.gov/PhysRefData/Star/Text/ESTAR.html for a start. (If you're concerned about radiation safety, do you really want to trust Wikipedia, or follow it to a primary or secondary source?) Also, are you concerned with radiative stopping power? Effective safety shielding? Something else?

If I remember correctly, about 5mm of aluminum should stop beta radiation from almost any nuclear decay. (But don't take my word for that; I don't want to be responsible for your cancer or infertility, or for your spending 100x as much as necessary on shielding....) Also, remember that you have to consider bremsstrahlung X-rays and (for beta+ radiation) annihilation gamma rays, or you may end up with much worse damage from the photons than you would have gotten from the original electrons. --76.203.75.30 14:22, 30 April 2007 (UTC)[reply]

Don't panic! It was just for a plan for an experiment measuring te shielding of aluminium,but i wasn't sure about the range of distances. I'd be happy as one of my first edits to consolidate the information. Within this article?Doctorp9999 22:58, 3 July 2007 (UTC)[reply]

Speed of the beta particles

Beta particles move at a speed of 180,000 km/s, around 0.6c.

Surely this depends entirely on the energy of the emitted particle and should therefore vary depending on the structure of the parent nucleus? Zapateria (talk) 22:32, 18 February 2008 (UTC)[reply]

Yeah, that looks wrong. Tagged for fact-checking, will remove soon if no source/clarification given. Hqb (talk) 16:26, 19 February 2008 (UTC)[reply]

Merging with Positron emission

Seems to me that the positron emission page is too small to be a seperate page, I suggest moving all information from there to this page and merge it with the beta plus decay section. Please discuss at Talk:Positron emission#Merger proposal
- SkyLined (talk) 16:54, 22 March 2008 (UTC)[reply]