Talk:Neutron/Archive 2

Geometry
This article: http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=PRLTAO000099000011112001000001&idtype=cvips&gifs=yes states that the neutron has a negatively charged exterior, a positively charged middle, and a negative core. The negatively charged exterior of the neutron gives an intuitive explanation for why more neutrons are required in atoms with large numbers of protons, as the neutrons negatively charged surfaces attract the positively charged protons to stay clumped together in the atom.

Unit of Spin
When stating the spin, shouldn't there be a unit (ex. A neutron's spin is 1/2__)


 * That unit would be h_bar. Though anyone who is familiar with the unit h_bar, knows that a half spin implies 0.5 [h_bar] -guest

Anon comment
Neutrons and protons both are attached to the nucleus!!

--24.108.178.156 03:50, 6 February 2006 (UTC)


 * And the nucleus is made from ... ? ... said: Rursus ( m bork³ ) 12:18, 26 November 2009 (UTC)

resonances
I'm not sure the description of particles in a nucleus as resonances between protons and neutrons by exchange of pions is up to date. I remember it was one of the first strong interaction models that appear, but that was before quarks and QCD. Is that description still valid in a quark context?

Yep. QCD provides a little more detail to the picture, of course, but you still get triples of up-downs which correspond to proton-neutrons, and since these are colorless and gluons very sticky you tend to get quark-antiquark pairs connecting them, and these are pions.
 * I have doubts about whether the statement is correct as written, and even stronger doubts about whether it belongs in this particular place in the article. First of all, I have no idea what the word "typically" is supposed to indicate here. Also, I'm not sure why you'd want to call them "resonances." You could say that they're eigenstates of the three-quark system, which would be incomprehensible to the typical reader. You could use the term "resonances" to indicate that without using words like "eigenstates," but it sounds to me more like a slangy usage, and the general reader is no more likely to understand it. It's true that it's only an approximation to describe protons and neutrons in a nucleus as having permanent, unique identities. In a completely correct, relativistic description of the nucleus based on  QCD, the protons and neutrons would only arise as certain correlations among quarks. Well, unfortunately we don't have any such model that's capable of producing useful predictions about nuclei. It strikes me as extremely pedantic to have this statement in here, and it has nothing to do with the stability of the neutron, which is supposed to be the topic of this section of the article.--207.233.84.39 01:25, 28 March 2007 (UTC)

Here is the thing, My name is Jessi and I am in 8th grade we just started learning about Atoms,Protons,Neutrons,Elctrons,etc, and i am just haveing some trouble understanding it all. It is so comfusing

Aren't we all!! Particularly about subdividing the electrostatic charge on the electron!WFPMWFPM (talk) 03:44, 8 June 2008 (UTC)

Importance in chemistry
I did a Google search and I got a page saying that neutrons are not important in chemistry. Please complete this sentence:

Although neutrons are not important in chemistry, they are important in... 66.245.84.123 00:53, 28 Sep 2004 (UTC) (No, my IP address is not a part of this sentence.)


 * Ummm... It's not actually true, although many people who should know better do say that it is. The number of neutrons in its nucleus doesn't make much difference to the chemical behavior of an atom, but it does make some. If you drank enough heavy water to make it replace somewhere between a third and two-thirds of your body water (nobody has tried it), your hair would fall out and any fast-growing cancers would go into remission, owing to the slowing down of some very critical chemical processes in your body. And much chemical research uses radioactive tracers. And depending what you mean by important in chemistry, there's the little detail that the only stable nucleus without any neutrons is hydrogen (or more specifically, protium)! So without neutrons, chemistry would be a little on the boring side, if indeed there were any chemists to do it. Andrewa 01:40, 26 Oct 2004 (UTC)Landorjal danced with her cousin while explainig the meaning of a Neutron.


 * It is mainly important for hydrogen where the speed of the reaction changes by about a factor of 6 under most circumstances where hydrogen is important. Otherwise, it isn't so important... none the less most living things do show significant preference for certain isotopes Pdbailey 04:09, 23 Nov 2004 (UTC)

Also, neutron are far more important in atomic chemistry (I think thats how you say it). In fusion and fission reactions they play a major part in the process, such as the weak interaction where they release electrons or absord them... well look it up.

Re neutrons in chemistry, consider the stable isotopes of 53I Iodine. It has 2 stable? (long halflived) isotopes, and this is because of a triffling difference in the difference in energy (Q value) between that of 53I127 (126.904473 amu) and 54Xe127 (126.905184 amu) in both cases. so if the energy level (Q value) had been only slightly lower for 54Xe127 and 54Xe129, there wouldn't have been any stable isotopes of 53Iodine.WFPM (talk) 03:53, 1 April 2010 (UTC)

how many neutrons are in helium?
Depends on the isotope. Helium-3 and helium-4 have one and two neutrons respectively and are the only stable isotopes of helium. Helium-6 and helium-8 (4 and 6 neutrons) are beta decayers with half-lives under a second. -- Xerxes 20:35, 2005 Jan 22 (UTC)

Half-life of free neutrons
I've noticed that there seems to be a very wide range of values given for the half-life of free neutrons. A quick Google search showed up values ranging from around ten minutes to as much as 17 minutes. Even Wikipedea's two entries on the neutron and the free neutron currently give values of 886s and "about ten minutes" respectively.

I have found a reference which gives a measurement with a range of error: http://prola.aps.org/abstract/PRD/v5/i7/p1628_1 The value they give is 10.61 ± 0.16 min (about 636.6 +/- 9.6 seconds).


 * That source is of 1971, and can be safely ignored. -- Frau Holle 18:51, 11 October 2005 (UTC)

I've not edited any entries to reflect this as I have not found another reference confirming this value and the error range. Perhaps someone will be able to confirm the figures.

I assume that the wide range in published values is a reflection of how tricky it is to make the measurement. It would be interesting to hear if this is the case.


 * Indeed these conflicting half-lives can certainly lead one into dubious waters. This simply comes down to the instrumentation used by laboratories independent to each other. One lab might use a gas chamber detector while another might use a scintillation detector with less dead time.  I understand the consensus among many nuclear and particle physicists is that the half-life lies somewhere between 10.2 and 10.6 minutes. &mdash; oo64eva (AJ) (U @  14:53, Apr 15, 2005 (UTC)

State of the art as of 2003: 886 +- 2 s. -- Frau Holle 18:51, 11 October 2005 (UTC)

Please note that using the term half-life is not applicable in case of free neutrons. Scientist woulde be very happy if it would have been half-life, since they could have taken ultra cold neutrons home in a so called neutron bottles (to cook some neutron cakes:). The term half-life involves something fusing by half during half-life period. In that case neutrons would have fused with different times what is actually not happening since every single neutron will fuse in 886 sec (or about that time). And I'm actually not sure if the term mean lifetime is aplicable in that case as well since I'm not a specialist in particle physics. I'm doing some neutron spectroscopy. -- 195.83.126.10


 * Huh? Sorry Mr. Unsigned IP Address, that's crap.  Neutrons do have a well defined half-life.  They do not all decay at the same time, but probabilistically like all other particles.  Moreover, decaying is not "fusing".  "Fuse" means "combine". --Strait 17:55, 28 August 2006 (UTC)

I recommend that we use the Particle Data Group's figure for the lifetime. http://pdg.lbl.gov, in particular http://pdg.lbl.gov/2006/tables/bxxx.pdf , page 4) It is updated regularly and is a trusted name in physics. --Strait 17:55, 28 August 2006 (UTC)

It's important not to confuse "Half Life" with "Lifetime". Half life is defined as the time after which the particle has a 50% chance of decaying. Mean lifetime is longer than half life by a factor of 1.44 (see the article Mean_lifetime) since the average includes particles that live for much longer than the half life. I'm sure this is causing the apparent discrepancy. So for the neutron, mean lifetime is 886 seconds, but the half life is only 886 * ln(2) = 614 seconds. Lokster 13:26, 4 June 2007 (UTC)

Added the free neutron mean lifetime from the PDG (mine is 2006) (on the net at: http://pdg.lbl.gov) As a particle physics grad student, I strongly recommend using PDG values whenever possible, as it is a common standard. naturalnumber (talk) 10:47, 23 April 2008 (UTC)

Note that all this discussion does not relate to a reason as to why a physical particle with no external electrostatic field should suddenly decide to create one. The theory evidently is that it the particle is internally divided into parts which have quantities of opposite charges which sum up to a net positive or negative charge and which can instantaneously change to the opposite for some unexplained cause. And so nature has decided to reaccummulate these particles from smaller into larger accumulations, which are limited by the physical laws of the accumulation process, which says that as the accumulation size gets larger, it's easier and thus more likely for causative factors to not only disrupt and delay the process, but also to destructively breakup delicately already accumulated nucleons. How's that for a wordy discussion of the subject matter.WFPM (talk) 19:49, 15 April 2010 (UTC)


 * Yes it indeed does not relate to that. But why should it? Headbomb {talk / contribs / physics / books} 19:57, 15 April 2010 (UTC)

A leading question if I ever heard one. Because beyond just the accumulation of matter as a goal of nature, we have the existence of a diversity of atoms and their chemical compounds, and with a complexity of existence of this mish-mash that amounts to the milieu of our existence and should be what be what we are mentally trying to organize and understand rather than just saying "That's cool! So what's next!" And of course we're not going to accomplish much in the natural universe. But as long as we stay on the planet earth we should try to understand it and manage it. And now I'm beginning to sound like a Liberal!!!WFPM (talk) 20:52, 15 April 2010 (UTC)


 * Please keep discussions on topic. This has nothing to do with the neutron article. Headbomb {talk / contribs / physics / books} 23:48, 15 April 2010 (UTC)

Excuse me. But the topic in question is the possible range of occurrence and existence of accumulated free neutrons, which are the major constituent of most atoms. So with reason we need a good and authoritative (well referenced?) report on same. And we wind up quibbling about estimates of around 10 minutes. Which doesn't allow them to travel very far or accumulate very much, and implies that they cant cause certain atomic events to happen that would be possible with a slight difference in their properties. And that leaves us with a controversial existence of the neutron as a variable factor in nuclear and/or atomic processes without much agreement as to their true nature. And the only solution that I see to this problem is to not to be so narrow minded as to exclude all alternative ideas other than the prevailing opinion and hope for the best.WFPM (talk) 01:37, 16 April 2010 (UTC)

Category Glitch
What's up with the categories appearing on the top of the page? ‡  Jarlaxle   21:42, July 25, 2005 (UTC)

News! teleronci:Newly calculated elementary particle of a rest mass = 1.15819171.10^-30 kg which is included as an integer value in the rest mass of proton and neutron (Source: Meissner, R.: das Teleronki-Modell..., Aachen: Shaker-Verlag 2001.)

NPOV
I've removed the NPOV from the article and placed it here - since the anonymous inserter didn't see fit to discuss it first. Are there any grounds for believing the article is NPOV?? Thanks, Ian Cairns 13:13, 30 September 2005 (UTC)


 * Looks like an unusual form of vandalism to me. Tho I haven't been very pleased with some recent edits to this article, I wouldn't call them POV. -- Xerxes 15:57, 30 September 2005 (UTC)


 * I removed the tag altogether so it won't show up in the NPOV disputes category. -- Kjkolb 08:16, 9 October 2005 (UTC)

Making a Neutron
Does anyone know if there are any theories to the possibility that the Down Quark is composed of an Up Quark and an electron? I once heard that when a Proton and Electron mix it produces a neutron. So if this is true then if you mix an Up Quark and an electron it must form a down quark...I am not sure though. - BlackWidower

To answer your question, quarks are supposed to be fundamental particles - nothing inside. A down quark can decay into an up quark; this transition creates a W- boson, which quickly decays into an electron and anti-neutrino (and these are detected). This does NOT mean the electron is inside the quark to begin with! Check out the wikipedia article W and Z bosons. Hope this helps. Lokster 13:33, 4 June 2007 (UTC)

Last Guy Took Out Sievert
Neutron radiation is especially damaging to organisms due to it's penetrating qualities, and ability to damage DNA. The Sievert weights radiation exposure by it's damage to human physiology. A person may feel that exposure to 5 rads of Neutron radiation isn't a big deal (5 rads of photons is not that horrible), but it is, and the Sievert radiation scale shows this clearly. Please don't take it out again. --Wiki Tiki God 09:59, 2 November 2005 (UTC)


 * No offense intended by its removal. I understand what a Sievert is, and it's not directly pertinent to neutrons. If people want to read about neutron radiation, they'll click on the neutron radiation link. Seeing as how the See Also section for this article is extremely long already, I think we need to adopt a strong standard of relatedness for additions. But, of course, I'll defer if others disagree. -- Xerxes 16:21, 2 November 2005 (UTC)


 * Actually, tho I still think everything I say above is true, I think Sieverts looks good where it is, so I'm not recommending we should take it out. -- Xerxes 16:24, 2 November 2005 (UTC)

Neutrons, stability, origin, etc.
What part do neutrons play in maintaining the stability of the nucleus? How do neutrons ever find their protonic mates in the first place, ie. especially when making new atoms? (That reminds me, when making artificial matter like I don't know, element 114, is it mandatory to find some neutrons to collide with the protons in order to produce an atom?) -- Natalinasmpf 12:13, 5 November 2005 (UTC)


 * The creatoms of atoms is evidently a process involved with the accumulation of previoucly created nucleons in accordance with gravity force and other physical laws. That some of the nucleons have a property of electrostatic charge and associated electrostatic and electromagnetic force is deduced from theoretical and experimental data related to those forces. That the occurrence of electrostatic properties might be the result of the atom's accumulation process has not been evaluated. The atomic nucleus evidently can achieve a certain size by pairing protons and neutrons and then adding 'extra' neutrons to some maximum physical stability level, with the mode of instability being the change of the last added neutron into a proton. WFPMWFPM (talk) 00:30, 6 September 2008 (UTC)


 * Neutrons enhance the stability of a nucleus by increasing the amount of strong nuclear force felt by the nucleons. Both protons and neutrons generate the roughly the same amount of attractive strong force. However, neutrons are not charged, so they do not generate repulsive electric force like the protons do. Thus, a nucleus with neutrons is more stable than one made entirely of protons. In fact, there are no stable nuclei without neutrons except a single proton.


 * Sorry to butt-in, but in the side panel thingy, it says that neutrons are apparently 'impervious' to the strong interaction haha! (I'm fixing it, thats a pretty fundamental flaw, right?) 110.175.8.243 (talk) 02:29, 31 October 2010 (UTC)

If A=2Z are combinations of deuterons, where does the 9F18 atom find the energy to spontaneously break up a deuteron into 2 nucleons, and then change the individual proton into a neutron to make 2 extra neutrons.WFPMWFPM (talk) 01:07, 27 August 2008 (UTC)
 * Neutrons find their protonic mates during fusion reactions. These typically take place deep in the hearts of stars like the sun. Fusion may also take place in the shockwaves of supernova explosions at the end of a large star's life. Some elements, like helium and some lithium were fused during the first few minutes after the Big Bang, when the entire Universe was hot enough to sustain fusion.


 * When scientists build heavy nucelei out of smaller nuclei, they try to start with two smaller nuclei that have as many neutrons as possible. That way, the resulting larger nucleus will have enough neutrons to be more stable. It's important to note that they don't put together huge nuclei from individual protons and neutrons, but just from two smaller large nuclei. -- Xerxes 17:57, 5 November 2005 (UTC)


 * Sounds like crashing two small cars together to make a big car?. And when they are looking for a stable big atom they need to be selective in the potential stability probability of the desired big nucleus, (plus the estimated energy carrying spin off particles. That's why EE possibilities are better than others, and filled series, like EE102No? would be better than that, because it would probably be more stable with less excess extra neutrons. WFPMWFPM (talk) 20:33, 28 September 2008 (UTC)


 * Wouldn't neutrons start to be counter-productive though, if regarding stability for large atomic masses, since adding more mass as an already heavy atom (ie. bigger than uranium) gets larger would contribute to instability? You completely answered my previous question though, it just evokes my curiosity further. Why do large atoms become unstable in the first place: do the forces pulling them apart overcome the strong nuclear force at larger masses? -- Natalinasmpf 23:17, 5 November 2005 (UTC)


 * There's a tradeoff involved. Although neutrons increase the amount of strong binding energy, they are heavier than protons and subject to beta decay. Once your nucleus gets too many neutrons in it, some of the neutrons will beta decay into protons. For very large nuclei, the whole nucleus just becomes so big that you get spontaneous fission. It splits roughly in half.


 * You can look at a chart of how various nuclei decay at NuDat2 -- Xerxes 00:51, 6 November 2005 (UTC)

Working your way up from a lower number of extra neutrons toward a higher number in unstable areas you usually get first B+ emission, then electron capture, then alpha particle emission, and then fission, and finally B- emission; so fission is a kind of stability situation which can be destroyed by thermal neutrons in EO and OE isotopes and by fast neutrons in EE isotopes.see Talk:Nuclear model WFPMWFPM (talk) 04:13, 29 June 2008 (UTC)WFPMWFPM (talk) 12:58, 29 June 2008 (UTC)WFPMWFPM (talk) 13:01, 29 June 2008 (UTC)

Carcinogen?
Anyone know why "neutrons" are listed under List of IARC Group 1 carcinogens. Umm, what? —The preceding unsigned comment was added by 130.71.96.23 (talk • contribs).
 * Clarified that to "neutron radiation". Femto 13:55, 29 May 2006 (UTC)

Position of diagram
The diagram is just schematic; it is not a "picture" of a neutron. It deserves no more prominence than the quark-model content, which is where I think it belongs: in the table with the quark content. -- Xerxes 20:41, 8 June 2006 (UTC)

Radius
Is neutron really 10x smaller than proton?

Here neutron: Radius variable, about 1x10^-16. At proton page: Diameter: about 1.5x10^-15.

Either of these values is bad, or it will be usefull to note such discrepancy...


 * No, they are about the same size. This is true because the strong force, which is most of what holds them together, doesn't distingish significantly between up and down quarks.  Let's try to untangle what's going on here...


 * I have the Review of Particle Physics 2006 in front of me. It says that the proton "charge radius" is 0.875±0.0068 fm.  While the RPP does not define charge radius, Perkins' "Introduction to High Energy Physics"  says that it is the root mean square of the charge distribution.  Now, this is only a precise way of characterizing the size of the proton if what you are interested in is its electromagnetic interactions, since one could easily imagine that the RMS of its color charge distribution could be somewhat different.  In any case, twice this radius gives 1.75 fm, which, I guess, is "about 1.5 fm".


 * The RPP gives the neutrino "mean-square charge radius" as -0.1161±0.0022 fm2. What this means  is that the core of the neutron is somewhat positive and the outside is somewhat negative.  Clearly, this isn't a good measure of its size at all, because if you calculate the RMS, it's not real.  If one charges boldly ahead and ignores the minus sign, one gets 0.341fm, so I don't think that's where the 0.1fm the article currently has came from.


 * I'm having trouble finding a source that will parameterize the size of the neutron in any other way while also giving a precise value. Povh et al's "Particles and Nuclei" says that "nucleons have radii of about 0.8 fm."  Let's use this value until we can find something better. --Strait 23:52, 6 September 2006 (UTC)


 * In the liquid drop model, there is a parameter that can be interpreted as the radius (or volume, or whatever) of a neutron or proton. There are various versions of the liquid drop model, and in some of the fancier ones, with tons of free parameters fitted to all the known binding energies and so on, it's possible that they parametrize it separately for the neutron and proton. In any case, nuclear matter is somewhat compressible, so it's not really valid to think of neutrons and protons as having fixed, specific sizes regardless of the nuclear environment they find themselves in. The electromagnetic moments have the advantage that, they have precise definitions, and, at least in theory, could be measured to arbitrary precision for the free particles. However, that really tells you next to nothing about the physical size of the particle. The thing that comes closest to being interpretable as a size of the nucleon is the liquid drop parameter. The electromagnetic moments are important more as tests of QCD. The radius appearing in the liquid drop model is more like 1.1 fm, so that's really what should be in the article -- or maybe just 1 fm, since to define it with two sig figs of precision, you really have to be careful in explaining what you mean. (In de Shalit and Feshbach, a best-fit value of 1.12 fm is given, but you don't want to take the extra sig figs too seriously -- it depends entirely on what kind of data you're fitting.)--207.233.84.39 01:41, 28 March 2007 (UTC)

A neutron walks into a bar and orders a drink. The bartender says, "For you neutron: no charge!"

Bada bing. I can feel the love in this room. --Christofurio (talk) 02:53, 5 September 2008 (UTC)

It no longer says anything in the article about the diameter of a neutron!!! Obviously it's not a point particle, but the article doesn't mention anything about its size as mentioned here as measurable from the liquid drop model!!! :( Please can comebody with more knowledge than me add this in??? 49.193.186.124 (talk) 07:04, 25 November 2011 (UTC)

Proton to Neutron??
"Inside of a bound nucleus, protons can also transform via beta decay into neutrons. In this case, the transformation may occur by emission of a positron (antielectron) and neutrino (instead of an antineutrino):"

Is this really true? how can a proton increase its mass by decay? has someone been messing with the article? Toby Douglass (talk) 12:05, 21 September 2008 (UTC)


 * An Atom in EC or Beta+ decay does not increase in mass, but rather decreases it. All Beta decay processes involve a mass decrease on the part of the decayed particle. WFPMWFPM (talk) 01:24, 23 September 2008 (UTC)


 * That's my point. So how does a proton (lighter than a neutron) convert to a neutron by beta decay, e.g. by loosing mass?  Toby Douglass (talk) 17:52, 26 September 2008 (UTC)
 * That's what I thought too, until I looked up the incremental mass value increase for the last added proton in the CRC handbook, and it was always more for the added proton than for an alternative neutron. The one that particularly irritates me is the case of EOO17, with reported mass (87th edit.) of 16.9991317. This is then reported to increase to 17.999161 when you add a neutron, (at EEO18), but adding a proton takes you to OOF18 at 18.000938, which is .001777more. Then the isotope OOF18 is reported as being unstable and to decay spontaneously back to EEO18. WFPMWFPM (talk) 11:00, 27 September 2008 (UTC). Sounds like a cheap way to change a proton into a neutron. WFPMWFPM (talk) 15:29, 27 September 2008 (UTC)


 * Perhaps you may want to look at it this way: protons and neutrons in nuclei are all (on average) far lighter than protons alone, due to the mass lost with the binding energy. Thus, free neutrons are 1839 electron-massess and protons 1836 electron-masess. But in the tightest-bound nuclei, protons and neutrons average as little as 1820 electrons mass. It's impossible to tell which is which, but one can change into the other with no problem, so long as the average mass goes down in the process (which provides the energy). For example tritium has 3 nucleons which average 1832.3 electrons in mass--far less than a proton. It decays by beta decay to He-3 which averages 1831.9 electron-mass/nucleon. You'll find the same for positron decay: they all start out lighter than if made entirely of proteons, and wind up lighter still. S  B Harris 01:50, 4 December 2008 (UTC)


 * It is interesting to note that as the result of all this theorizing we are in the position of not having a concept of a basically neutral mass particle, since our concept of the neutron is that of a particle with mass that can emit an electron and become a positively charged proton. Now can this proton then emit a positron and again become a "neutron"? That sounds like a real complex gadget.WFPM (talk) 16:24, 10 March 2010 (UTC)


 * Sbharris is right: you have to be careful about what state the particle is in before you talk about its mass. For example, if you just add up the masses of the 3 quarks that comprise a neutron (or proton), you will find a lot less mass than you expect.  Most of the mass is theorized to be derive from relativistic speeds of the quarks zooming around each other.  CosineKitty (talk) 19:30, 10 March 2010 (UTC)


 * Yes but how do you rationalize the electrostatic charge change when a neutron becomes a proton using the quark theory? That's a charge 1 change. And the quarks individually only have a fractional charge. And both the proton and the neutron are supposed to have 3 quarks. So do all of them change, or what?WFPM (talk) 03:18, 12 March 2010 (UTC)


 * No. A proton is made of two Up quarks with +2/3 charge, and one Down with -1/3, total +1. In beta (positron-type) decay, one Up changes to a Down, giving a neutron with ONE Up of +2/3 and TWO Downs of -1/3 each, total zero. In the electron-type of beta decay where neutron goes to proton, a Down changes to an Up, for the reverse process.


 * The idea of a particle with charge +2/3 may be even stranger than one with charge +1/3, but there is a baryon particle called Δ++ which really does have a charge of +2. It's made of 3 Up quarks, each with +2/3 charge. S  B Harris 16:18, 1 April 2010 (UTC)


 * Well let's try the delta mass problem for 1H3 (Tritium) with mass 3.016049278 (87thCRC) changing to 2He3 with mass 3.016029319 by B- emission. does that sound reasonable?WFPM (talk) 03:36, 7 April 2010 (UTC)


 * Try it yourself. He-3 has 2 electrons and you can pretend one of them is the beta decay electron. So the difference between the nuclide weights is entirely the decay energy. Subtract one from the other and get a number very close to 2e-5 amu. Multiply by 938.494 MeV per amu and obtain 0.0185917 Mev. Which is just the decay energy of 18.590 kev to 4 decimal places. Actually, since we only have the decay energy to the nearest 10 eV, there's plenty of room to put the missing helium 26 eV electron binding energy. There's nothing missing here, to the limit of measurement. S  B Harris 06:25, 7 April 2010 (UTC)

Yes but starting with the 1H3 atom with only 1 electron and the 3.016049278 amu's don't I have to lose the mass of an electron when I emit it? So how am I going to wind up with a two electron 2He3 mass of 3.016029319?WFPM (talk) 14:21, 7 April 2010 (UTC) And in the same place it says that when a neutron decays by B- to a proton plus electron, the mass loss is from 1.008664916 t0 1.007825032 or 0.000839884 amu, which times 938.434 kev per amu gives 0.7882 Mev or 778 kev.WFPM (talk) 21:31, 7 April 2010 (UTC)PS okay I get it, The so called B- electrons are now the new orbital electrons, and we're talking about the reduction in the free energy level of the new electrons. And there's actually no emitted electron particle. Is that maybe it?WFPM (talk) 22:26, 7 April 2010 (UTC)


 * Of course there's an emitted electron, but its rest mass doesn't count in the equation, because you just have to add an electron back to the decay product, to get THAT neutral mass (which it picks up from the environment). So it's the same as if the emitted tritium electron went into the helium atoms' orbital, even though of course that doesn't happen, since the 18 KeV beta energy far, far over the ionization energy. If you want to do it the hard way, just note that tritium actually decays to an electron PLUS an ionized He-3+ atom, which doesn't have the neutral atom mass you quote, but is less than that, by the mass of an electron. You see? It's the same with the decay of the neutron, which has more mass than the proton by about 2.5 electron masses. But the decay energy is about 1.5 electron masses, since obviously you have to account for the rest mass of the beta-decay electron out of the total mass difference. S  B Harris 23:58, 7 April 2010 (UTC)

Well I certainly appreciate your trying to explain it to me! My concept of B- emission is for the decay neutron to physically emit a negative electron which would be detected plus one of Pauli's neutrinos. I have never understood this as being logical because it's one of the few areas where my favorite science author, Dr. Asimov, was unable to convince me about it. Have you read his 1966 Avon book on the neutrino? He really managed to boggle mind on that one.WFPM (talk) 01:17, 8 April 2010 (UTC)


 * I actually have read it, and few books have been published which managed to say so little, at book-length. Anyway, it's thoroughly outdated. Yes, neutrinos accompany beta decay, but since they have very little rest mass, they can carry off any amount of decay energy from essentially-zero, to quite a lot. That results in a spectrum of energy for the electrons, with the max at the decay energy, where the electron gets a kinetic energy of essentially all the available decay energy (here 1.5 electron masses worth at E=mc^2) and the neutrino gets essentially nothing.  S  B Harris 01:30, 8 April 2010 (UTC)

Well, in comparison to my Kaplan and Linus Pauling and the EB I still think he's the better explainer and maybe he gets paid by the word. And in accordance with your theory, the decay process would involve the temporary occurrence of a (neutron minus an electron) particle, which I hadn't thought about or heard described.WFPM (talk) 16:38, 8 April 2010 (UTC)


 * A "neutron minus electron particle??" No. when an up quark flips to a down (or vice-versa) the proton immediately becomes a neutron (or vice versa), and a W± particle is emitted. For beta-neg decay it's W- and for positron decay, W+. Then THAT particle further decays to the positron or electron, and a neutrino or antineutrino. S  B Harris 18:18, 8 April 2010 (UTC)

We sure seem to be theorizing in circles. Now when a neutron emits an electron, which it doesn't have, but makes up by changing the electrostatic charge on a quark, and we emit it, together with the excess free energy that it wants to get rid of; we are left with a particle with a positive charge, like a proton, and a respectably diminished mass. So now we wait until a loose electron is attracted and captured by it, and brings back enough spare free energy to get us back to square 1. And the net result is the increase in the Z number of the atom by +1, and if you're counting excess neutrons, reducing the number of unpaired (excess) neutrons by -2. But by increasing the number of protons by 1 and decreasing the numbers of neutrons by 2 we have also theoretically increased the accumulative electrostatic field strength of the atom as well as reducing the localized neutron strong attractive force component and thus reduced the stability of the atom. And this was caused by the accumulated of an additional excess neutron in the first place. And this event preferentially takes place in atoms with an odd number of neutrons, but eventually in the case of all neutrons above a certain level of excess neutrons. And in the case of heavy atoms the accumulation process leads to the emission of an alpha particle from the less unstable atoms and even to the fission of the less electrostatically and/or dynamically balanced OE or OO atoms. And this occurrence is occasioned but not caused by the (random?) change in the charge on a quark? And you would say that the two protons of 2He3 are identical. And in the case of EE4Be8 that it is an increase in stability for the last proton of EEBe8 to have a quark change and then be more stable with 2 excess neutrons than it was as an EE A=2Z atom, which by the way is the opposite way of change of quark as occurs in the 1H3 atom. And I must of course defer to your arguments, because you have much better references than I do. So C'est la vie!WFPM (talk) 22:41, 8 April 2010 (UTC)


 * It's really not as bad as you think going from tritium to He-3, although it's BARELY energetically possible, which is why the energy is so low. 18.6 kev is almost nothing-- one 50,000th of an amu. In tritium there are no "excess" neutrons. There are two very happy paired neutrons in a nice 1s orbital, no angular momentum, the best you can get. And one proton in the same. When it decays to He-3 you get the same situation: two happy paired protons in the same orbital, but now only 1 neutron in that same orbital (remember, the different particles don't "see" each other from the Pauli principle, because they're different, so there's room for 2 neutrons and 2 protons in that orbital, which is why He-4 is so stable). So why should a neutron in H-3 change to a proton, when (all other things being equal, which they are), you end up having to deal with extra electrostatic repulsion from 2 protons? Well, the answer is that neutrons are heavier. You have the 788 keV to play with, just from the difference in mass between a neutron and a proton+electron. That makes up for all the extra electrostatic repulsion, but 18.6 keV of it. And so it can happen. Here's a fun exercise: How close do two protons have to come to each other to have 788-18 = 770 kev repulsion? Well, r = kQ^2/E = 9e9*(1.6e-19)^2/(7.7e5*1.6e-19) = 1.87e-15 m = 1.87 fm. The size of a He-3 nucleus is about 1.25 fm * sqrt(3) = 2.16 fm. So it looks like there's plenty of energy available in the loss of the neutron mass to allow two protons to get as close as they get in He-3. S  B Harris 23:01, 8 April 2010 (UTC)

So now within the 1H3/2He3 space volume we have independent forces of 1:universal gravitational attraction, 2: An Electrostatic force of repulsion between protons, and 3: A strong force of attraction between the neutron and the adjacent protons which diminishes to zero in midspace at a short distance and then becomes a repulsive force, and in the existence of a very small and stable nucleus when an additional neutron is added. And it is theoretically made up of two spin 1 deuterons who are combined to result in a zero net spin atom, though, of course, they are all spinning either up or down in one axis of direction. But they can't be in contact with each other. And the means of communicating these force tendencies including magnitude and direction is by the existence and action of force and energy carrying relativistic particles whose magnitude of mass/energy existence cannot exceed the value of planck's constant. I bet Mother Nature had a hard time planning this all out.WFPM (talk) 00:47, 9 April 2010 (UTC)

Gee1 by now I thought you would have told me that it is much worse than I thought, because I don't know even half the details.WFPM (talk) 16:46, 10 April 2010 (UTC)


 * It is worse than you've put above, because your details are wrong. The He-4 nucleus is NOT made of two deuterons. That's YOUR model, not the standard one. The standard one is that the 2 protons and 2 neutrons occupy "1 s" orbitals very much like the 2 electrons in helium do. These have exponential differential volumetric probabilities that are maximal in the center and decay exponentially going outward. Since the Pauli exclusion does not opperate, all 4 particles occupy the same volume, and interpenetrate. However, this volume (as in superfluid helium where the same thing happens) is NOT that of one nucleon, but that of about 4 of them, as though 4 blobs of water had coalesced into one spherical blob of 4 times the volume. For larger numbers of nucleons the Pauli principle does tend the keep individual particles away from each other, however. No collection of 5 nucleons is even stable to decay-- they all come apart almost immediately. As I've said before, you can scoff at this all you like, but the only alternative I've seen from you involves having the He-4 nucleus composed of two little positive hard spheres and two little neutral ones. Which is cute, but does NOT explain the EXPERIMENTAL fact that electron scattering experiments show the charge in He-4 nuclei is a smooth exponential function, maximal in the center and tailing off outward, just exactly like the negative electron-charge distribution in a helium atom (and for the same reason). There's no way THAT happens with any combination of 2 touching positive spheres. S  B Harris 17:17, 10 April 2010 (UTC)

I'm certainly not scoffing, because I've spent over 30 years of my spare time trying to develop a rational concept of this subject matter and I certainly appreciate discussing the subject with someone who considers it seriously. But I have serious difficulty with some of the concepts like the ability of two objects to occupy the same space at the same time, Which Clerk Maxwell in the 9th EB regarded as being a "vulgar opinion". And I believe in some physical entities a lot stronger than others, because I have direct sensory perception of those, although I could be fooled as to the perception. And like he stated that nobody has ever seen a force, but rather deduced it from ensuing activity. So I muddle along, As Senator Eagleton once said and try to contribute and find out as much as I can. But You've got to admit that it sure sounds like a complicated plot organization; as compared to Isaac Newton's comentary that the lord did nothing in vain, and more is in vain when less will serve.WFPM (talk) 17:52, 10 April 2010 (UTC)


 * But less will NOT serve, if you want to have (say) chemistry. You can see these effects almost directly with electrons. A helium atom is the same size as a hydrogen atom (smaller, even, by a third). Why? Hydrogen atoms are paramagnetic, but helium atoms aren't. Then the next element, lithium, is a LOT larger (factor of 5), just from adding one more electron. And paramagnetic again. WHY? Beryllium is a little smaller than lithium. Nature's trying to tell you something, here. Electrons don't like each other, but in PAIRS, they do. A pair of electrons in an orbital takes up a lot less volume than twice what one electron does. S  B Harris 19:47, 10 April 2010 (UTC)

Sounds like part of the BCS theory for superconductivity. But I didn't theorize that electrons repel each other. People have to keep telling me that. I'm willing for them to be linked together in order to float through the matrix. I've merely got the each electron associated with a particular proton at the end of its tether, so that they can be emitted or accumulated as conditions (energy level) merit. And I don't need an electron for every one of the protons including those buried within the nucleus so I'm not even sure about that. I guess it's true that the more you learn the less you're sure of. As you're a physician I doubted that you would go along with with Newton's first rule and so do I. So I've paraphrased it to be "Nature does everything for a reason", which allows for a certain amount of cause and effect, and there really ought to be a reason.WFPM (talk) 22:42, 10 April 2010 (UTC)

I've even got an associative theory, where in the odd Z conductive materials, the electrons can link up in associative pairs and form a two electron unit that "walks" down the voltage gradient in the direction of the positive voltage potential, with the resistance factor being related to that required to release the lagging electron of the linked electrons. In the even Z elements, like 16Sulfur, we still have electrons all over the place, which as Dr Gustave Le Bon says "rub off very easily" but we don't have the required interaction between the atoms that facilitates this walking process. So that's what you get for thinking and maybe reading too much of Dr Asimov's science fiction.WFPM (talk) 18:14, 11 April 2010 (UTC)

Moving
Not that I wish to move this page, but is this page move protected? Because if it is, perhaps we should use the little green lock at the top-right hand corner? Gopal81 ChatMe!ReadMe!! 01:25, 4 December 2008 (UTC)

Liverpool?
In 1932 James Chadwick worked in Cavendish laboratory at Cambridge and moved to University of Liverpool in 1935. So how come that "Finally, in 1932 the physicist James Chadwick in the George Holt building at the University of Liverpool performed a series of experiments...He suggested that in fact the new radiation consisted of uncharged particles...These uncharged particles were called neutrons"? Avihu (talk) 18:26, 1 January 2009 (UTC)

Role of neutrons in reactors
In the production section, under nuclear reactors, it is therein stated that the role of the neutron is to sustain the chain reaction. While this is certainly one of the top criteria considered in fission reactor design, would it not be more to the point of fact to say that the presence of these fission produced neutrons is merely exploited to that (maintenance of fission chain reaction) end? Paraballo (talk) 03:23, 2 April 2009 (UTC)


 * The Thermal neutrons provide the kinetic energy to fission the EO nuclear fuel material, and they also aid in the conversion of some of the EE material into fussile nuclear material by being captured and being converted to protons to make the heavier EO nuclear fuel materials.WFPM (talk) 03:47, 7 April 2010 (UTC)


 * If you consider a unit cube of 1000 92U atoms, you note that the the occurrence of the U235 atom is approximately 7.2 atoms, or let's say 8 atoms. So if we divide the 1000 unit cube into 8 octants we would usually have a 5 by 5 by 5 cube with 125 92U atoms and only 1 U235 atom. And in that cube the U235 atom can be visualized as being in the center of a 3 x 3 x 3 or 27 atom cube, with the rest of the 98 U238 atoms being at least 1 atom distant therefrom. This is a non reactive concentration of U235 atoms. And the enrichment effort is carried out in an effort to raise the concentration level such that more of the U238 atoms are located within a 27 unit cube containing a (hopefully centered) U235 atom, and which equals a U235 concentration of 1/27 or 3.7%. So evidently, the fission energy of a U235 atom is not strong enough to to trigger the fission of the more remote U238 atoms.WFPM (talk) 04:12, 7 April 2010 (UTC)


 * Uh, no. The kinetic energy of thermal neutrons (which is miniscule in atomic terms) has NO role in fission. The fact that the neutrons are slowed down means they are more likely to be captured by U-235, but their kinetic energy is unimportant. They'd be captured even if they were cryogenically slowed. The U-235 pulls them in, and the nuclear force provides all the energy necessary to fission a U-235 nucleus. For U-238, however, that's not so, and the neutron needs to bring in some kinetic energy to get the fission job done, since the activation energy for this is far higher. So U-238 fission needs very hot neutrons, and don't even happen in a reactor from the "fresh" 2 or 3 Mev fission neutrons that come out of U-235 before they're moderated. It does (however) happen in fission bombs with 14 MeV neutrons. However, the 2 to 3 Mev "new" fission neutrons in a reactor are enough to convert U-238 to U-239, which quickly decays to Np-239 then Pu-239. That isotope is fissionable again by slow neutrons (fissile). S  B Harris 00:16, 8 April

I wasn't talking about the thermal kinetic energy of the thermal neutron, but on the ability of the neutrons from the fissioned U235 particle to result in the fissioning of the reduced concentration and consequently more remote U238 atoms, with the geometric analogy kind of indicating that you had to be almost adjacent to a fissioning U235 particle in order to be caused to fissioned by it's neutrons.WFPM (talk) 20:23, 8 April 2010 (UTC)


 * But hot neutrons from U-235 fission are very unlikely to be captured by neighboring U-238 atoms, which have so little cross section for that process that fission neutrons have plenty of chance to bounce off U-238 atoms and escape dense fuel elements and interact with the moderator. Thus, most of neutron loses in a large reactor core are in interacting with the moderator, not from U-238 capture. Enrichment of U-235 is soley to make up for moderator loses, not fuel element U-238 loses. Use the right moderator, and natural uranium works fine in a reactor (CANDU, Hanford, etc). S  B Harris 20:58, 8 April 2010 (UTC)

But no amount of moderator will make the EE's work if you dont have at least some of the fissile OE's or even EO's to fission spontaneously and start the process. And since a considerable proportion of the U235's emitted neutrons originate in the 1Mev energy category, maybe we're both partly right with respect to this process.WFPM (talk) 14:33, 12 April 2010 (UTC)

Secondary neutrons
One of the goals of the Mars Science Laboratory is to characterize the broad spectrum of surface radiation, including galactic radiation, cosmic radiation, solar proton events and secondary neutrons. I am not able to find a definition or description of secondary neutrons; I asume they are neutrons produced by cosmic radiation on an exposed surface, but I am not sure of this and I found no references. Could someone please add a line about it so I can link it to the MLS article? Thanks, BatteryIncluded (talk) 01:36, 7 August 2009 (UTC)


 * Since neutrons have a half-life of 10 minutes or so, they are not a primary component of cosmic rays, since they cannot live long enough to travel through all those light-years of space, not even at relativistic speeds. Nevertheless we observe neutrons coming from space, but these were created by secondary processes, by gamma rays striking molecules close to the earth. So they are called secondary neutrons, IIRC. Rwflammang (talk) 20:44, 14 April 2010 (UTC)


 * Yes. Actually by cosmic rays, which are not gamma rays, but rather bare atomic nuclei, moving very fast. S  B Harris 02:29, 15 April 2010 (UTC)

So our atmosphere is being disturbed by high energy bombarding particles. But consider the detail with which we are still observing individual energy sources which are 25,000,000 light years distant like from the Whirlpool Galaxy. And it can be obscured like through the plane of the Milky Way. So space must be pretty transparent huh? Except that we still have Olbers'parodox to think about.WFPM (talk) 12:15, 15 April 2010 (UTC)
 * Even if space were filled with dust we'd have Olbers' paradox, since the dust would eventually heat to the temperatures of all the surfaces of all the stars, if that was all you saw in every direction. Olbers' paradox is solved by the expansion of space (the universe) and consequent red-shifting and energy-loss of light from far-away stars, until the point that the edge of the observable universe, moving away at the speed of light, gives no light at all, so it's dark. Beyond that is a region of space which is moving away at MORE than the speed of light (Einstein permits this for space) so we see nothing from it, either. S  B Harris 20:40, 2 October 2010 (UTC)

Okay! So we've solved the darkness of space (which turns red at longer distances) and are left with our billionth of the universe which includes the Milky Way and Whirlpool galaxies, which evidently have plenty of <10 second old neutrons and protons to accumulate as well as presumably similar processes of accumulation, I guess. But it's hard to see how this process of accumulation is dominated by the disappearing ingredient, unless, As Dr Le Bon says, that the atom is only a temporary phase of occurrence of whatever Matter is really made up of. But I'm concerned with a best estimate of the actual nature of the atom and appreciate all the correlative information that I can acquire, and appreciate having discussions with you re this matter.WFPM (talk) 23:56, 2 October 2010 (UTC)
 * Which disappearing ingredient do you mean? All the neutrons had been absorbed into helium-4 (plus small amounts of D, He-3, Be, and B) in the first 3 minutes of the Big Bang, which didn't give the free neutrons time to decay. See Big Bang nucleosynthesis. All these isotopes/nuclides are indefinitely stable. They come directly to us, after 13.7 billion years (especially the D) with no problem. (some Be and B are made by cosmic ray spallation). Any neutrons that didn't get taken up into nuclei in the first half hour of the universe, of course decayed almost immediately after this phase. S  B Harris 00:51, 3 October 2010 (UTC)

Boy!!! Talk about a quick process recipe! All this in only 3 minutes!! And that gets us to the A = 4-6 level and then all we need is a second miracle like the Bang-Bang triple alpha accumulation process to get up to A = 12+ where we can start a more leisurely process. But I read somewhere that Gamow was sometime in favor of what I call a "hang around" deuteron concentration situation such that my deuteron addition accumulation process to the double alpha particle would be able to able to proceed in a more leisurely manner. But I guess the jury has decided about the facts and who cares about the miscellaneous details.WFPM (talk) 15:57, 3 October 2010 (UTC)
 * The triple alpha process didn't happen in the Big Bang, because there's not enough time for it during the short time between when helium can be formed and survive, but is still hot enough and dense enough, to fuse to carbon. Hoyle, who never believed in the big bang (yet actually named it-- derisively!) developed the triple alpha process as a way to make carbon from helium, IN STARS. He even predicted a resonance in C-12 which had to exist for this to happen, and that resonanace was FOUND LATER, as a result of Hoyle's prediction. It's a damned good theory that predicts things, rather than retrodicts things. The Big Bang theory predicted the microwave background long before it was discovered. And (although not as impressively) it retrodicts the concentrations of elements found in population II (first generation) stars. Just as advertised, they have very little carbon or any heavier elements. Those things are mostly made in supernovas, long, long after the big bang. Our own solar system condensed only 4.6 billion years ago-- with plenty of time for it to contain carbon from previous exploding stars long after the Bang. S  B Harris 19:03, 3 October 2010 (UTC)

My problem has always been as to how the nucleon was caused to occur in the first place. Because I could see the processes of nucleon to atom in the Whirlpool Galaxy and behind that the Black Hole that was going to swallow it all up, And now you're telling me we have to have stars to have the triple alpha C12 creation process! And how long did it take from a homogeneous gaseous mess to get to the starry heterogeneous state of the Whirlpool Galaxy? Only 500,000 years! And where did the creation of the nucleon take place, with all it's present day conceived complexities? I've always suspected that the creation of the nucleon was an exothermic process that had to be capable of expanding in multiple locations within the expanding spacial volume of contained and contracted matter, and that makes the stars the most likely place to look. But you say that that we have to have the nucleons before that and that they have to clump into stars before they can accumulate into atomic configuration quantities. And the more we complicate the concept of the different properties of the neutral and positively charged (proton) nucleons, the harder it is to rationalize any reasonable initial concept. And so the only thing we can do is to try to link together the knowledge that we have about the determinable properties of the known material entities in some reasonably consistent chronological manner. So I guess I'll go along with Fred Hoyle about the Big Bang process and hope that some new scientific information will crop up that will explain the Red Shift et al in a manner that allows some of the alternative accumulation processes to be considered maybe even including mine.WFPM (talk) 21:00, 3 October 2010 (UTC)


 * The Whirlpool Galaxy (M51) contains no nucleons forming. The stars you see there are mostly very young-- 10's to 100's of millions of years old, not billions. . Yes, all the nucleons (99.9999..%) or whatever, were formed in the first few minutes after the Big Bang. You could read that article, you know. For 400,000 years or so that was plasma consisting of protons, electrons, and helium nuclei (alphas). About 400,000 years ago those combined into neutral atoms, and that flash is what we see as the microwave background (redshifted a lot). Neutral atoms have no trouble collapsing into stars, as gravity simply pulls the gas in. A little shock is all it takes to get this going. The first galaxies formed less than a billion years after the big bang, but the ones you see near you are anywhere from 1 to 10 billion years old. Read Galaxy formation and evolution. Stars in individual galaxies come in all ages. Carbon-poor halo stars of galaxies can be 10 billion years old or more-- almost the age of the universe! And there are waves of younger stars in the arms, that were born a million years ago or less, some in stellar nurseries where we can see it happening, from clouds of hydrogen, and the dust scattered into space, from previous novas. That dust is heavy atoms (carbon or heavier). Naturally, younger stars tend to contain a lot more elements Z > 5, from such dust. Our own star is only 30% of the age of the universe. 98% of it (and our own solar system) is H, D, He, Be, and B atoms from the Big Bang, BUT the other 2% of our atoms (the "metals" with Z of 6 or larger) are from stars that formed long after that. They are >5 billion but < 13 billion years old. Our own star incororates dust from previous stars. But all tne nucleons of all the atoms, whether the atom is 6 billion or 13.7 billion years old (or 10 minutes old, for that matter), are 13.7 billion years old. The only exceptions are the very few nucleons we've made ourselves in accelerators, since the 1950's. And if you want to count "new" protons made from 13.7 billion year old neutrons that we have ejected from atoms by various means, then those are "new", too. Also "new" protons from beta decay. But all these processes conserve baryon number, so the total number of nucleons (with a minus for antimatter) doesn't change (so far as we know). S  B Harris 23:26, 3 October 2010 (UTC)

Well, I certainly appreciate the extended commentary, and I understand most of it, but I thought that the Whirlpool was pretty much like the Milky Way and that most of the stars were like our G2 and thus about 40% along their lifespan. And I don't know how you can say that no new nucleons are forming unless you know how they did form as well as what's going on in the centers of gravitational attraction. Then we have a few other problems like what the property of the electrostatic charge (and resulting force) really is such that you can isolate it into a small volume and even out of a point of matter. That was what Maxwell was trying to figure out, of course. And the 400,000 year point of neutral atom creation sounds like a good point for the creation of some simplex deuterons which could why not cause a process of increase of the A = 4 atoms to A = 12 during the next few million years? I wish I could remember where I saw about Gamow's Deuteron rich environment concept so I could use that for support. And thank you again for the commentary.WFPM (talk) 03:26, 4 October 2010 (UTC)


 * See here. It's all quantitative. The CALCULATED ratios of neutrons to protons at one second, and ultimately helium to hydrogen at 3 minutes, in an expanding and cooling universe, is just what we see.   S  B Harris 03:33, 5 October 2010 (UTC)

Most interesting and informative! And you can see where the uncertainty and argument began about how much deuterium can be created as a result of the process. They want to start out at a 1:6 neutron/proton ratio before anything else can happen, and then for (practically) all of them to go to 2He4, starting at 100 seconds. So they (practically all) wind up up as 2He4 in 10,000 seconds. And we wind up with 25% 2He4. They're willing to consider some 4Be7 but don't want to talk about 4Be8 or A=5 nuclides, and the 4Be7 was probably a 2He4 + 2He3 combination. So the argument is about how much neutrons to 1D2 and whether some 5B10 and 6C12 would be possible from the 4Be8 + D process. And I'm afraid I don't understand what the initial conditions (neutron/proton ratio) were such that the condition should "freeze out" to the 1:6 ratio. I kind of thought that it all started out as p + e, and then they must have some rate of e capture process to get to that ratio, but they don't show that. So first we have to create the neutrons and then store them before they decay back to what they were? Sounds pretty ambiguous. And that's all that I can think of for now, but will continue, and I wish they had had my magnetic models to work with while they were making all these decisions.WFPM (talk) 05:21, 5 October 2010 (UTC)

Notice that they don't mention the existence of 5B10 in the discussion, but they show a respectable amount of 5B in the Nucleosynthesis article, so maybe they didn't consider the possibility of 4Be8 + 1D2 to make 5B10 as a possibility for further accumulation while they had a lot of 1D2.WFPM (talk) 16:34, 5 October 2010 (UTC)

James Chadwick
I see that there was some back-and-forth reversion regarding the institutional affiliation of James Chadwick, who is credited with the discovery of the neutron. From what I have been able to find out, he graduated from the University of Manchester in 1911 and continued there until 1913. After World War 1, he joined the Cavendish Laboratory at Cambridge University, where he discovered the neutron in 1932. At the time he received his Nobel Prize, he was a professor at the University of Liverpool. I updated the article to say that he was at Cambridge when he did his experiments.

However, an earlier version of this article states that the neutron was discovered in a specific building on the University of Liverpool campus. If you can provide a supporting citation, or show where (specifically) one of the existing citations says this, we can put University of Liverpool back into the article. 69.251.180.224 (talk) 01:14, 16 January 2010 (UTC)

Isospin vs I_3
Should there be an extra property of I3 in the particle properties as then the usefulness of Isospin would actually be clear? 137.205.21.29 (talk) 13:10, 6 May 2010 (UTC)

Magnetic moment
How is explained the nonzero magnetic moment of the neutron, unusual for a electrically neutral particule?--MagnInd (talk) 10:24, 6 August 2010 (UTC)


 * Neutrons are not fundamental particles. They have electrically charged quarks inside of them. Dauto (talk) 15:25, 30 August 2011 (UTC)

Feynman diagram typo?
Shouldn't the arrow on the antineutrino be pointing in the other direction? Or perhaps instead it should just be a neutrino? 128.30.16.143 (talk) 16:35, 13 October 2010 (UTC)ms
 * No, from what I understand, the convention is that antiparticles have their arrows point the opposite direction (which is what the diagram shows) because they travel backwards in time (I think). Chris857 (talk) 23:48, 4 May 2011 (UTC)