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This is an old revision of this page, as edited by 72.178.12.19 (talk) at 05:13, 30 October 2009 ("Destroying" deuterium). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

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Big Bang abundance

If memory serves, isn't the deuterium abundance in the universe one of the supporting arguments for the big bang? I seem to recall reading that deuterium is not produced in stars, only consumed, so it had to come from somewhere else. But I could be wrong which is why I haven't added it to the article - MMGB Does that mean that A=2Z atoms are made up as accumulations of deuterons?WFPMWFPM (talk) 01:28, 27 August 2008 (UTC)[reply]

Correct. I added the relationship between big bang and deuterium in the article -- User:Roadrunner How about 2 deuterons fusing to become an alpha particle?WFPMWFPM (talk) 12:18, 27 August 2008 (UTC)208.191.143.56 (talk) 12:15, 27 August 2008 (UTC)[reply]

I was under the impression that deuterium was formed during the p-p chain of fusion occuring in stars. I thought most of it ended up following the entire chain to become helium but some of it remained as deuterium.

- C. Irvine

The image on the Nuclear Fusion page clearly shows two protons forming a deuterium nucleus. I don't know what percent is used in the rest of the reaction chain and how much eventually escapes the star and forms planets, but some of it must get away before becoming helium.

And from what I can remember of college physics, the hydrogen in light water fission reactors occasionally absorbs a neutron and becomes deuterium. If that's true, then it should happen any time hydrogen is subjected to neutron radiation, not just in man-made reactors. So the claim that "there are no known natural processes..." would be false then.

208.67.10.157 (talk) 09:01, 8 August 2008 (UTC)[reply]

WW II and H-bomb

I suspect that the text of this article is misleading in juxtaposing the account of concerns during the World War Two about the availability of heavy water to the German military nuclear research program with the link to the hydrogen bomb. A hydrogen bomb is a fusion weapon which was beyond the immediate aspirations of the research programs of all the belligerent nations. These weapons programs were aimed at creating a fission weapon. Heavy water is useful in this project through it being the ideal moderator. It depletes a neutron flux of its energy while absorbing very few newtrons. Its deuterium outperforms the cheaper substitutes, namely normal hydrogen (in ordinary water or parafin wax) and carbon (in graphite), in this role.

- Alan Peakall

Hello Alan. Thanks for your comments -- but you can improve the article directly. See Wikipedia:Welcome, newcomers. Feel free to edit the article yourself. -- Tarquin 18:20 Oct 15, 2002 (UTC)

Tracer usage

I am not clear if deuterium molecule being used as a tracer is in the liquid or gaseous state. You tell people in the sciences diatomic molecule, and immediately they think 'gas'. The deuterium is almost surely not solid. In case you are a laymen reading this and you don't already know, the deuterium would require a very low temperature to reach the solid state.

- desolderthis

I worked with modelling deuterium ice, and the freezing point is between 17 and 18 kelvins. I also added some data for deuterium at that temperature because I had to use it so often. -Mike

Mass value

My physics handbook says the mass is 2.01410178.

2002 CODATA recommended values : 2.013 553 212 70 (cf CODATA Value)

The number given in the article is incorrect. The CODATA page gives the mass of the "deuteron", which is just the bare nucleus. Add the mass of an electron, to get the atomic mass of deuterium. Trust your physics handbook before you trust wikipedia.  :-) DonPMitchell (talk) 21:54, 25 May 2009 (UTC)[reply]
You are quite right. The article has the mass of deuteron only, and you need to add the electron mass in u to get the isotope mass. The electron is .000554857991 u, for a total mass of 2.014 101 7926 u, which is very close to the physics handbook value. I'll change it. SBHarris 22:50, 25 May 2009 (UTC)[reply]
To many digits of accuracy, I would also check to make sure the mass defect from ionization energy is accounted for. DonPMitchell (talk) 23:14, 25 May 2009 (UTC)[reply]
Aha. 13.6 eV is 1.4 e-8 amu. Subtracting THAT now gives exactly the physics handbook value above, at 9 sig digits. SBHarris 23:41, 25 May 2009 (UTC)[reply]

Natural occurance

From the article:

It occurs naturally as deuterium gas, D2 or 2H2.

Wouldn't it be most common in the form DH (or 2H1H), with D2 being rather more rare? Bryan 05:30, 14 Jul 2004 (UTC)

-

That does make sense. I did some modelling involving D2 and DT. When it was in the DT form, I had to account for the weight so that there would be more TT at the bottom, most of the DT in the middle and more DD at the top. So according to that, it would make sense that Deuterium would not bind to only Deuterium and because Hydrogen is more common, it should bind with Hydrogen more often. (This same idea should apply to the heavy water comment, which would mean that most heavy water is actually HDO instead of D2O) -Mike

Hello Mike40033, if I correctly understand your table it was saying that tritium (hydrogen-3) is sable. That's not right, with a half-life of 12.4 years it beta decays as:

31H → 32He + 0-1β + 00ν

followed by some very soft X-rays from shell rearrangement. Since you already had this on the helium page, I assume it was a typo. Cheers, Securiger 08:10, 23 Sep 2004 (UTC)

How is deuterium extracted from water?

How is heavy water extracted from water, and how is deuterium extracted from heavy water?

Please e-mail me...

-superawesomepenguin@yahoo.com

Deuterium and Heavy Water should be merged!

Too confusing. They are not the same! Jclerman 21:14, 16 February 2006 (UTC)[reply]

Watch the units for density as published

I'm no expert, but the density of deuterium at STP as cited appears very, very wrong. Remember that this is a close relative to water. The original definition of the kilogram was the weight of 10 cm^3 of water, which would lead to a water density of 1000 kg/m^3. Assuming that the density of deuterium would rise as the ratio of the atomic mass to that of water, then the density of deuterium would be 1112.6 kg/m^3. Someone should check that before publishing it authoritatively, but I guarantee you that I'm at least in the correct order of magnitude.

-benedias@yahoo.com

Deuterium is D2 not D2O. D2 is a gas like hydrogen. D2O is called heavy water. Therefore the density looks OK. pstudier 20:25, 20 October 2005 (UTC)[reply]

info on these tables, please

Where can I read an explanation of the boxes in this type of table posted for each isotope?


Lighter:
Hydrogen-1
Deuterium is an
[[Isotopes of Hydrogen|isotope]] of [[Hydrogen]]
Heavier:
Hydrogen-3
Decay product of:
None
Decay chain
of deuterium
Decays to:
Stable

Jclerman 19:50, 2 February 2006 (UTC)[reply]

history

  • Urey was known as Harold Urey.
  • The thing manufactured in Norway was heavy water, not deuterium.

Jclerman 14:01, 18 February 2006 (UTC)[reply]

which substance?

can i purchase this amazing substance

neutrinos and heavy water

I know they use heavy water in the detection of neutrino's, but I really don't know how, can one of you smart guys tell me --Psychobrat 14:50, 14 March 2006 (UTC)[reply]

deuterium is NOT HEAVY WATER 69.9.31.103 14:51, 8 May 2006 (UTC)[reply]
It's not a dumb question. The largest tank of heavy water on Earth is used to detect all flavors of neutrinos in Sudbury. Both the deuterium AND the fact that the deuterium is present as heavy water, are necessary for the function of the detector, so it's difficult to know where to put the section on SNO. It finally wound up in heavy water, since it's the largest collection of it, but might be more appropriate for deuterium since other light-water detectors (like DUMAND and IceCube) have been proposed or built, but pure deuterium, per se, to my knowledge (as a bubble chamber material) has not been used to look at neutrinos. It could be in theory, but it's too difficult to keep that much liquid hydrogen around in a deep mine. Sbharris 17:13, 8 May 2006 (UTC)[reply]

oceanic heavy water

does oceanic heavy water occur more frequently for example at depths with high pressure. Posted by 66.173.105.130

deuterium is NOT HEAVY WATER 69.9.31.103 14:51, 8 May 2006 (UTC)[reply]

deuterium arc lamp

It's not mentioned in the article. What's it's relevance? Explain or delete, please. Also explain the meaning of the "counts" axis. Jclerman 04:08, 15 May 2006 (UTC)[reply]

I didn't add it, but the making of such UV lamps is an important application of deuterium. I just wish I knew why it worked better than protium, so I could put that in the article. I've posted a question about it in the Talk page of the guy who made the spectra. He seems to know a lot of such physics. Be patient. Sbharris 05:04, 15 May 2006 (UTC)[reply]

Radioactive?

If deuterium sometimes reacts to neutrinos, as indicated by the article, wouldn't it be radioactive? It could decay to Helium-2 by neutrino capture (if neutrinos can turn neutrons into protons), and decay back to deuterium, or decay by proton emmision and decay to Hydrogen-1. If it was radioactive, it would have a very long half-life. It sounds possible, because there are a lot of neutrinos. AstroHurricane001 00:49, 11 December 2006 (UTC)[reply]

The term "radioactive" (see radioactive decay) is reserved for things which decay on their own, not after being hit by something else (that process is induced radioactivity, like fission). D can be "ricochet-fissioned" by neutrinos into proton and neutron (as in SNO), or it can be turned into either He-2 or a dineutron (both of which immmediately disintigrate) by the appropriate neutrino or antineutrino. But the latter reactions are rare and not particularly characteristic of D vs other isotopes (although the direct ν-induced photofission-like breakup by neutral current Z's, does appear to be a special feature of D in the SNO, and is due to D's weak binding energy). Neutrino or antineutrino absorption can make many elements radioactive if it happens, just as neutron absorption does. SBHarris 09:01, 11 December 2006 (UTC)[reply]
Thanks for responding. Although He-2 probably exists all the time (particularily in deuterium oceans and the sun), it is not mentioned in the isotopes of helium article. If a dineutron disintigrates immediately, does it break up, completely dissapear out of existance, turn into deturium, or something else? I would think neutrinos are more common than anti-neutrinos, but if the have practically no charge (a neutrino, as in neutral), how could they change neutrons into protons, and vice versa? AstroHurricane001 18:27, 11 December 2006 (UTC)[reply]
I don't know where you're getting the weird ideas about He-2. Deuterium oceans?? He-2 is just two protons rammed together and there's no reason to think they stay together at all. Even the dineutron is unbound (not merely "unstable"), and the diproton has all the dineutron's problems plus two positive charges ripping it apart. In the sun the process of beta decay after a proton-proton collison must be essentially instantaneous. This process is so fast and with such a bad cross section, I don't even think it's ever been observed in the lab (in accelerators, fusion experiments, etc).

Forget the charged-current mechanism of neutrino induction of quark flavor change (you can read about it in the neutrino wiki, though). Just think of the decay of a neutron: it goes to proton, electron and antineutrino. If you fire in an antineutrino with enough energy, the proton gets changed back to a neutron and a positron. Similarly, neutrinos with enough energy are capable of changing neutrons to protons+electrons. In all cases the total charge is conserved; merely separated. SBHarris 19:47, 11 December 2006 (UTC)[reply]

Thanks for the clarifications, and when I said "deuterium oceans", I probably meant bodies of matter containing deuterium, like the ocean's heavy water, and in some cometary or planetary bodies. If the disintigration of dineutrons and diprotons occurs so quickly that they are not usually observed in the lab, it probably has a half-life of somewhere on the order of less than 10-25 - 10-50 seconds. That is fast. AstroHurricane001 00:41, 12 December 2006 (UTC)[reply]

I am using the kinetic energy of water to spin turbines. I plan on mixing Deuterium with my water to take advantage of the properties of heavy water. But I am unsure of how much Deuterium I will need to add to One gallon of water. Any help with this equation would be appreciated.

Doug douglassb@cox.net

"depleted" deuterium

The "appearances in pop culture" section makes a reference to "depleted deuterium" rounds in the "Warhammer" game. While the game may in fact use this terminology, the term itself doesn't make much sense. We talk about "depleted" uranium to refer to uranium metal that consists of the more stable U238 isotope, rather than U235. But in the case of deuterium, a) we're talking about an isotope that is already stable, and b) deuterium itself is a light gas (its "heaviness" notwithstanding), not a metal...and thus one can't make a bullet or shell from it. —The preceding unsigned comment was added by Spincycle (talkcontribs) 04:56, 5 April 2007 (UTC).[reply]

In the game, the depleted deuterium rounds have a core of depleted deuterium so you could make a bullet from it, but I dont see much point in making the core a light gas, perhaps its because deuterium is flammable? And you can't deplete deuterium, depleted deuterium would be deuterium consisting of a stable isotope of deuterium except that deuterium a) already is stable and b)is an isotope. --Curtis95112 (talk) 02:09, 22 January 2008 (UTC)[reply]

From the description of the Warhammer round, it apparently is a shell used in relatively small weapons for general use. The deuterium core apparently has fictional effects, as gas or water wouldn't contribute much to a small weapon's effects. A somewhat more realistic use is in Bolo (tank)#Bolo offensive systems as a fusion weapon, if someone was able to make deuterium fuse well in the described environment. But from the Warhammer description, its shell is not an explosive fusion weapon. -- SEWilco (talk) 02:55, 22 January 2008 (UTC)[reply]

Technically, you could use deuterium as a filler for incendiary shells, because it's highly flammable. I don't see the point of it, though, because hydrogen would work just as well, (e.g. Led Zeppelin). But then again, if you're making a video game, you don't have to be realistic, do you? 76.21.37.87 (talk) 05:52, 26 June 2009 (UTC)[reply]

Spin question

If you only have 2 spinning nucleons and the spin of either is +/- 1/2 then woulcn"t they both have to have the same orientation of spin for the combined spin of the two to be +/- 1? So if they were hypothetically touching you would say they were end to end connected like cylinders? Conversely if they had a combined zero spin could you say they could be side to side connected like cylinders?WFPMWFPM (talk) 16:28, 27 August 2008 (UTC)[reply]

Electronegativity

Is there any significant difference in electronegativity between deuterium and ordinary hydrogen? In other words, would a HD molecule be polar? Stonemason89 (talk) 14:59, 13 September 2008 (UTC)[reply]


Deuterium Spectral Series

Where can I get or how can I calculate the spectral series for this isotope of hydrogen. Here is an example of what I am looking for,(http://en.wikipedia.org/wiki/Hydrogen_spectral_series) this link is for protium. Gravitroid (talk) 07:41, 25 December 2008 (UTC)[reply]

Actually, the formula in the link does not give the correct values for protium because the value of the Rydberg constant is wrong. For protium the Rydberg constant should be taken to be around 109680 cm-1. For deuterium the formula is the same except that the Rydberg constant should be taken to be slightly larger, around 109710 cm-1. In general, for a hydrogenic atom, it should be taken to be Rinfinity mN/(mN + me), where Rinfinity is the Rydberg constant for an infinitely massive nucleus (109737.31568527 ± 0.0073 cm-1 [1]), mN is the mass of the nucleus, and me is the mass of the electron. Spacepotato (talk) 08:38, 25 December 2008 (UTC)[reply]

Anti-Deuterium needs a check

As of 2005? It's 2009, this section of the article needs addition of new information about recent advancements in research from chemists in the past few years since the section was written on Deuterium and Anti-Deuterium (maybe more compare and contrast as well). Dangerousd777 (talk) 09:36, 14 March 2009 (UTC)[reply]

Should there be a Category:Deuterium compounds as a subcategory of Category:Hydrogen compounds? AFAIK, chemically these compounds can be quite different. Albmont (talk) 15:13, 4 August 2009 (UTC)[reply]

Yes, but most of them exist as NMR solvents or labeled biologicals where that isn't the case. Most of the usefulness of having the cat tag is to note the few cases where the chemistry is quite different, and for the rest it's more a matter of noting commercial availability. Which is a problem, inasmuch as WP is NOT supposed to be a catelog. Some of the same problem exists for commercial perfluorinated analogs of C-H organics. SBHarris 22:53, 19 August 2009 (UTC)[reply]

"Destroying" deuterium

The sentence in the first paragraph caught my attention and I think it needs to be changed:

"There is little deuterium in the interior of the Sun, since thermonuclear reactions destroy it."

I'm under the impression that matter cannot be created or destroyed, but only changed in form with energy either being consumed or produced. —Preceding unsigned comment added by TechnoDanny (talkcontribs) 22:23, 14 September 2009 (UTC)[reply]

Matter can be destroyed-- it's mass that cannot be. In any case, deuterium is destroyed in this process, but the protons and neutrons that make it up, are not. SBHarris 03:24, 9 October 2009 (UTC)[reply]
By "destroyed", it means the assembly of one proton and one neutron stops existing, fusing with itself into a two proton and two neutron complex. 2D --> 4He. I'll clear up that sentence. 72.178.12.19 (talk) 05:13, 30 October 2009 (UTC)[reply]