Talk:Deuterium/Archive 1

Jupiter Deuterium Abundance
I think that the Galileo probe data on Deuterium abundance in Jupiter is correct and the other measurements on spectra are wrong. I flagged it as dubious. The Hubble and the Galileo probe both give 600 ppm ratio for atoms of D/H.Trojancowboy (talk) 21:51, 6 December 2010 (UTC)

Article is Not Up to Date, Needs Section on Ultra-Heavy Deuterium
Ultra-heavy deuterium is a solid with massive weight and extreme density. Is has been artificially created in the laboratory as a potential fusion reaction fuel. I don't have time to add it to the article but here is a non-wikipedia science article on the subject: http://www.sciencedaily.com/releases/2009/05/090511181356.htm

This merits its own section as well.

Sean7phil (talk) 20:15, 29 November 2009 (UTC)


 * But it's not appropriate as something in this wiki. Compressed deuterium only exists for a tiny, tiny fraction of a second, while it's under gigantic pressures from the laser. But any element can be compressed to thousands of times its normal density this way. Including ordinary hydrogen (protium). So this is not a property of deuterium per se, but of everything. SO why put it here? They're just doing it for deuterium because they need to do it for fusion [Protium doesn't fuse very well, even at these densities-- protium fusion in the sun only works because there's no time limit, and even then the energy production is less than your body makes (per volume)]. S  B Harris 00:09, 30 November 2009 (UTC)

This 'ultra-dense' section is invalid. Citing three papers with a common authors doesn't meet any reasonable requirement for sourcing and is highly suspicious. I'm going to remove this section in short order if it isn't already removed by the time I check back again. Fooburger (talk) 07:32, 6 March 2010 (UTC)

No other elements have been compressed this way neither has it been sought. This is why it is in the deuterium article. please make a new page for it before deleting it. 119.224.82.11 (talk) 08:14, 31 May 2013 (UTC)

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)

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)

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)

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 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)


 * 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. S  B Harris 22:50, 25 May 2009 (UTC)


 * 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)


 * Aha. 13.6 eV is 1.4 e-8 amu. Subtracting THAT now gives exactly the physics handbook value above, at 9 sig digits. S  B Harris 23:41, 25 May 2009 (UTC)


 * The mass at the top of the article is different thatn the one at the bottom; I'll change it to the "textbook value." —Preceding unsigned comment added by 76.115.86.124 (talk) 22:28, 16 October 2010 (UTC)
 * Thanks, you've got us. To 7 sig digits, the mass is indeed 2.014102 u. It was fixed in the infobox but not the article body. S  B Harris 23:26, 16 October 2010 (UTC)

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 &rarr; 32He + 0-1&beta; + 00&nu;

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)

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)

info on these tables, please
Where can I read an explanation of the boxes in this type of table posted for each isotope?

Jclerman 19:50, 2 February 2006 (UTC)

history
Jclerman 14:01, 18 February 2006 (UTC)
 * Urey was known as Harold Urey.
 * The thing manufactured in Norway was heavy water, not deuterium.

which substance?
can i purchase this amazing substance
 * Sure, just buy it from Sigma-Aldrich. ^_^ Double sharp (talk) 08:30, 18 May 2017 (UTC)

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

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)

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)
 * 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)

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)


 * 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. S  B Harris 09:01, 11 December 2006 (UTC)


 * 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)


 * 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. S  B Harris 19:47, 11 December 2006 (UTC)


 * 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)

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 (talk • contribs) 04:56, 5 April 2007 (UTC).

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)


 * 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) 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)

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)

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)

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)


 * The dipole moment of HD was measured back in 60th and is < 10^-3


 * Yes, there will be a dipole moment because of asymmetric nuclear vibration. Link. Double sharp (talk) 06:26, 13 September 2016 (UTC)

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)
 * 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 ), mN is the mass of the nucleus, and me is the mass of the electron. Spacepotato (talk) 08:38, 25 December 2008 (UTC)

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)

Also, this section contradicts itself. The first sentence says it has been made by two labs in 1965, and the second sentence states that it has not been produced. —Preceding unsigned comment added by 72.33.134.154 (talk) 21:45, 18 June 2010 (UTC)
 * There is no contradiction. The antideuteron, i.e., the nucleus of an antideuterium atom, has been produced.  The whole atom has not been produced. Spacepotato (talk) 22:27, 18 June 2010 (UTC)

Category:Deuterium compounds?
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)


 * 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. S  B Harris 22:53, 19 August 2009 (UTC)

"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 (talk • contribs) 22:23, 14 September 2009 (UTC)


 * 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.  S  B Harris 03:24, 9 October 2009 (UTC)


 * 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. 72.178.12.19 (talk) 05:13, 30 October 2009 (UTC)


 * The fusion of 2 deuterons into an alpha particle is not necessarily a process of the "destruction" of a deuteron, but rather a process of incorporating the 2 deuterons into a more compact and thus less massive alpha particle structure.WFPM (talk) 16:20, 12 October 2012 (UTC)

Lead Section
The lead section really doesn't seem appropriate to me. Why does the prevalence of deuterium, particularly its distribution across the solar system, warrant the vast majority of the lead? Should most of this information be moved into another section below, with a more generic introduction? Compare, for instance, the leads of any of the element articles (c.f. Nickel or Boron).

I am quite hesitant about trying to edit this myself, as I really don't have enough current science knowledge to feel comfortable doing so...anyone up to the task?Qwyrxian (talk) 04:19, 2 April 2010 (UTC)


 * I'll see what I can do, if nobody else objects in a few days. Perhaps it shouldn't have quite this much emphasis, but the reason it's different from other isotopes is that the mere concentration of deuterium constitutes a large fraction of human interest in it (larger than for any other isotope I can think of). It is used in isotopic tracing, climate temperature inference, and the primordial concentration sets severe constraintss on conditions in the big bang, and on how the elements that formed the earth came into being. I suppose the real problem is that the LEAD needs to say how common the stuff is, and this answer varies from place to place. This necessitates either a very, very vague and large-bounded answer, or else a lot of qualifications. S  B Harris 23:01, 18 June 2010 (UTC)

Links
There are numerous internal links to heavy water in this article. According to Wiki policy, only the first instance should be linked. WP:LINK --Janke | Talk 08:05, 3 November 2010 (UTC)
 * I'm removed the excessive links and also some of the "(see heavy water)" references, which I felt were similarly unnecessary.    — SkyLined (talk)

Table at end not clear
This is in the article after the "See Also" section without a section heading, and it is not clear (to me at least) what it is saying. Initially I thought that it was saying that Deuterium had a decay chain, and that Hydrogen-3 (tritium) was stable, because I read the table columns down instead of the rows across. If it had a section heading of "Deuterium isotopic production and decay" or something like that then I think it would be more clear what it is saying. —Preceding unsigned comment added by 69.231.150.254 (talk) 02:51, 26 January 2011 (UTC)

Drinking heavy water.
Combining "Small doses of heavy water... are routinely used as harmless metabolic tracers in humans and animals." with "these differences are enough to make significant changes in biological reactions." leads to the question.... does heavy water taste any different to "normal" water? Old_Wombat (talk) 10:47, 22 November 2011 (UTC)
 * Not much, if any. I can't taste the difference. Some people claim to be able to detect a slight sweetness in heavy water. If it's there, it's very faint. Nobody has done a double blind test that I know of. There is a separate article on heavy water, you know, if you haven't read it. S  B Harris 20:18, 22 November 2011 (UTC)

Wow, in < 24 hours a reply from someone who'd done it. I'm happy to take your word for it and move on. THanx for your quick reply. Old_Wombat (talk) 06:23, 23 November 2011 (UTC)

U.S. Department of Health & Human Services on "Consumption of Deuterium-Depleted Water" https://search.nih.gov/search?utf8=%3f&affiliate=nih&query=deuterium+depleted+water&commit=Search Tampasailor (talk) 13:52, 4 July 2017 (UTC)
 * Isn't that the opposite of heavy water (water enriched in deuterium)? Double sharp (talk) 13:54, 4 July 2017 (UTC)

Pycnodeuterium Section
This section states:

"Deuterium atoms can be absorbed into a palladium (Pd) lattice. They are effectively solidified as an ultrahigh density deuterium lump (Pycnodeuterium) inside each octahedral space within the unit cell of the palladium host lattice. It was once reported that deuterium absorbed into palladium enabled nuclear cold fusion.[23] However, cold fusion by this mechanism has not been generally accepted by the scientific community.[24]"

Reference no 24 has a 2005 publication date. Today, there is a great deal of interest in this process, which is now called LANR (Lattice Activated Nuclear Reaction) or LENR (Lattice Enabled Nuclear Reaction). Even DARPA and MIT are taking it seriously. This section needs to be expanded and updated, but I wouldn't be the best person to do this because (a)I only have access to popular media, and (b) I wouldn't understand the physics if I could get hold of it to read.

Would perhaps someone with a bit more knowledge than I have be willing to take this on?

--MolecularDolphin (talk) 05:42, 16 January 2012 (UTC)

Some notes on tetrahedral deuterium fusion "theory" (warning: original research, but more or less WP:CALC)
Some guy named Takahashi has suggested the four deuterium atoms in a tetrahedral latice just collapse into a configuration in which the nuclei are only 5 fm away from each other and then they fuse into Be-8. See the review paper in this book by Akito Takahashi "Progress in Condensed Cluster Fusion Theory".

I haven't been able to find an on-line copy of the paper in which actually performs this calculation (it's his 2 man paper), but he gives the results in both of the cites I gave above, and he gives 57.6 keV for this electrons at the minimum distance after 1.4 fs and a distance of 2.8 fm from the D's to the center of the tetrahedron, for an effective base length of 11.2 fm. But this is wrong by a factor of about 5,000 (that is, 4 electrons fitting into such a small volume must have Fermi energies of 250 MeV or so).

This is just the Heisenburg uncertainty theory applied to electrons, which you can't abritrarily compress to distances of 10 fm or so (which they need to be to screen D nuclear charges from each other) without the consequence that the electron kinetic energies have to be huge for their wavelengths to be that small. Compressing electrons to near neutronium cannot be done by adding equal charges to them. Electrons are probability clouds—they are gas. Adding positive charges to gas does not make it any less gassy.

The wavelength lambda of an electron is h/p. Which is not h/mv, but hc/E for an ultrarelativistic electron. So E = hc/wavelength.

The simple first pass math is that for a 5 fermi electron you have an energy of hc/lambda = 6.626e-34 (3e8)/[5e-15*1.6e-19] = 2.5e8 eV = 250 MeV (500 times rest energy, so certainly ultrarelativistic). One would expect that such an electron confined in such a box would have approximately this energy, as a minimum. Note that it’s 5 times what you’re going to get out of 4 D fusion, and this is just for ONE of the 4 electrons in the configuration. In real honest to god hot fusion in a thermonuclear weapon, of course, the electrons are never squeezed into spaces this small—there’s literally not enough energy to do it!

You can be fancy and calculate Ef = [Fermi energy] for an electron gas in Takahashi’s box of 4 deuterium atoms. The only assumption is that the gas is composed of 4 electrons in a cube of side 5 fm. Energy levels climb a bit, since electrons 3 and 4 have to go in at quantum number 2 (the first two having taken the ground state). Here’s the relativistic Fermi energy equation for the last of N electrons in a cubical box of length L. Note that Wikpedia doesn’t have this formula for the ultrarelativistic limit, but it’s here:

Ef = hc/2L  * [3N/pi]^1/3

In this case, N = 4:

Ef = (6.626e-34)(3e8)/ [2*(5e-15)] * [12/pi]^1/3 = 3.1 e-11 J = 194 MeV

Ef (ave) = ¾ Ef = 145 MeV.

The first figure is electrons of the highest energy, and the last is the mean energy, since of course with fermions the energy of the first two is lower than the next two, etc. But you see it’s all basically the same math and gives about the same numbers. 4 electrons in a 5 fm edge length box are going to need 4*145 = 380 MeV energy to simply confine them, which is 8 times what you get from the hypothesized fusion. I’m not saying it’s impossible (if you had some outside compressive force like pressure from a star). I’m saying it's mighty unrealistic as an "adiabatic" process, where the thing just happens sort of spontaneously because the charges cancel. The point is that cancelation of charges gives you no help, since this would be the energy needed to confine electrons to these volumes even if electrons had no charge at all.

This is the derivation for the energy of a (relativistic) Fermi electron gas, independent of whatever else is in it (protons or no protons, deuterons or no deuterons). It works for the density of white dwarf where E(fermi) is about 300 keV. Add positive charges and the energy goes down a bit, but not enough (that’s where you need the pressure of a star). The potential between a proton (or deuteron) and an electron 5 fm away is 2*9e9*1.6e-19/5e-15 = 576 KeV = 0.58 MeV. It’s not enough by a factor of 145 MEV/0.58 MeV = 250. So Takahashi can handwave all he likes, and his charge-charge “screening” is never going to give him more than 1/250th of the energy he needs to confine electrons into the space he needs to put them, in order that they screen the positive charges of D nuclei that are only 10 fm from each other. The electrons simply need so much energy to get into a space that small, that their kinetic energy alone is far higher than the energy you get from deuterium fusion.

And I will further note that the fermion part is not that important either. Electrons could be a boson gas and they are so low-mass that they still would exert a HUGE pressure at those tiny distances. For D4 you get 4 electrons which must occupy 2 energy levels in pairs, whereas if they were bosons they could occupy the same state. So the energy would go down by 50% or something, but in the last analysis, it's the wave nature of electrons and their small masses that are resisting you. Electrons could even be uncharged bosons, and you'd STILL need 100 MeV or so per electron, to squeeze them into that box along with their nuclei. Combine electrons with protons and turn them into neutrons first (inverse beta decay), as happens in the Sun, and that problem goes away since you are not dealing with the small electron mass. But now we're far from Takahashi, and it no longer works for deuterium (since in the D fusion mechanism to He-4 you don't need to get rid of any electrons-- you're stoichiometrically stuck with the same number you start with).

So, there are no deuterium electrons squeezed to 5 fm apart from each other in Pd at any point, or else they’d come whipping out at 150 MeV, when they were let go. So you can fuse D nuclei all you like, but what you cannot do, is take their electrons to those small internuclear radii with them. But this particular cold fusion theory requires that you do that.

So, to put it succinctly, these cold fusion theories with 4 deuteriums have forgotten a simple law of physics. There's no place to put those squeezed-together electrons that are supposed to be screening the deuteron charge. S B Harris 06:29, 16 January 2012 (UTC)

Deuterium in river water or rain water?
Hi, the article gives a ppm for deuterium in ocean water... the article on Heavy Water states "(in ordinary water, the deuterium-to-hydrogen ratio is about 156 deuterium atoms per million hydrogen atoms)" where "ordinary water" refers to Vienna Standard Mean Ocean Water. Do these 2 pages refer to ocean water because the VSMOW is one of the most common references of a "standard water", or do rain and rivers contain significantly less? if so why? distillation? Also if different, what ppm of deuterium should be present in rain or rivers?

I also read that since the last tests done right before the nuclear test ban treaty somewhere in the fifties I think, oceanographers found much use in the radioactive contamination from these tests to study oceans. However thats about tritium, what fraction of deuterium in ocean water originates from these tests, and hence wouldnt be a natural abundance?

What "ppm" is in use here? I read in talk page about why DHO should be more frequent than say DDO.. Is the correct interpretation 156 D's per million (H or D)'s? or per million molecules? — Preceding unsigned comment added by 83.134.177.206 (talk) 21:32, 17 February 2012 (UTC)
 * D's a bit less in rain water than ocean, due to fractional distillation. Hydrogen bombs don't produce any net D, since all D in them was extracted from oceans anyway. H-bombs use up D. But the amount lost to too small to measure. A few kg per nuclear weapon won't be missed. As to your question about what ppm means, it's answered in the first line you quote above. Read it again. S  B Harris 23:16, 17 February 2012 (UTC)

Deuterium bombardment
When deuterons are used to bombard a target atom, the mass of the proposed composite is calculated as being the sum of the target nucleus plus that of a deuteron. This implies that the accumulated deuteron must have been accumulated into the atom with the same reduced mass value as it had as an individual deuteron, and accordingly must remain paired. And if a variability analysis is carried out on the increased mass value caused by the increased "deuteron content" of the A = 2Z atom series, it is noted that the incremental mass increase for the EE created atoms is significantly less than than the mass increase for the OO created atoms, (by greater than .005 amu), which indicates that the dynamic balance of the structure is affected by the structural location of these deuteron additions to the compound nucleus.WFPM (talk) 00:33, 10 October 2012 (UTC)
 * I'm not aware that the mass of A= 2Z atoms (same as Z=N atoms) is calculated by adding the mass of a deuteron. The masses of these atoms are determined by locating the nearest atom with a measurable mass, and subtracting decay energies. Of course even-even (EE) nuclides of this type (which are all stable up to Ca-40) have lower energy per nucleon-- all their protons and neutrons are paired, and their spins are zero. Each odd-odd (OO) nuclide in this series shows signs of instability or unpairing. H-2, Li-6, B-10 and N-14 are all stable but only because their decay products would be too unbalanced in Z/N ratio. But they have spins of 1,1,3, and 1 respectively, as befits 1 unpaired proton and 1 unpaired neutron in each, spining in the same direction. The rest (F-18, Na-22, Al-26, P-30, Cl-34 and K-38) are all unstable to decay, with spins of 1,3,5,1,0, and 3. All these qre quite understandable except the spin of zero for Cl-34. I do not understand why the odd neutron and odd proton here spin in opposite directions to add to zero, when it is almost always more stable for them to spin in the same direction, which would give a net spin. But this state is the lowest energy for this nuclide, by 146 kev. Interestingly, it has a far shorter half life (about a 1.5 seconds) than the metastable Cl-34m (which has that extra 146 keV) where the 1 odd neutron and 1 odd proton are indeed unpaired with spins of 3/2 and both spining in the same direction, which causes this nuclide to have a spin of 3 and last 32 minutes. In any case you need to give up the idea that there are "deuterons" anywhere inside Z=N nuclei. The high spins and instability should warn you that there are no such things. S  B Harris 02:57, 10 October 2012 (UTC)


 * I admit that I'm hooked on the indications of my magnetized cylindrical models and that they tell me that side-by-side bonded nucleons have to be counter-rotating. And when 1 is side bonded to 2, the position of the 1 is across the junction of the 2. And of course everybody else says they can't be in contact. And you say that the protons and neutrons are paired in the EE's, but not in the A = 2Z OO's, but if they're not paired, then the incremental mass value of the 2 unpaired nucleons would be higher by approximately .0024 amu in each instance. But when I'm increasing from an EE nucleus to an OO nucleus, I'm not doing anything to the previous nucleus but just adding on 2 nucleons at the best balancing location (nearest to the axis of rotation of the nucleus). And it appears that if instead of a nucleon pair (a deuteron) just a neutron is added, then a situation of unbalance is created such as to foul up the ability of the nucleus to accumulate further paired nucleons, but only allow it to accumulate a few additional individual neutrons up to some limiting value. And the beta- occurrences of the high excess neutron nuclei are evidently due to the ability of some of the excess neutrons to change to a proton and pair up with adjacent other excess neutron and thus increase the atomic number of the atom. In the article about PET, it says that the atom OO9F18 is created from EE8O18 by a (p,n) "knock out" reaction. I assume that this means that the O18 atom is hit by a proton which drives out a neutron and leaves a residual 9F18 nucleus. But it doesn't say whether the new proton and the remaining extra neutron are paired. And the question becomes as to what would be the stability of the OO9F18 nucleus if they could become paired and then reduced to a ground state temperature of 0 degrees Kelvin. I've never been able to get anyone to answwer my question concerning that point. But thank you very much for your attention to my comment.WFPM (talk) 15:17, 10 October 2012 (UTC)
 * You really should read a text on nuclear physics to see how modern models explain nearly all of this. Yes, O-18 is hit by a proton and loses a neutron to become F-18. That radionuclide looks like O-16 (no spin, all protons in 4 pairs and neutrons in 4 pairs) but with an extra neutron and proton, neither of which are paired (neutrons don't like to pair with protons, see deuterium). These spin in the same direction (each with spin 1/2) to give a net spin of 1. O-18 looks like 0-16 but with an extra pair of neutrons which are paired with each other, so its spin is zero. O-17 and F-17 are isobar nuclides with the same mass number. Each looks like O-16 (no spin), but with with an extra added nucleon (proton for one, neutron for the other). Both have spins of 5/2, which is entirely due to the extra "lone" proton or neutron. S  B Harris 23:36, 10 October 2012 (UTC)


 * Well I'm stuck with the basic concept of a 2 planar stacked alpha particle EE4Be8 nucleus with some spin incompatibility problem that cab be overcome by a 1 to 2 side bond in one corner like in my magnets so as to make a stable EO4Be9 nucleus, or by an additional side bonded deuteron on one side to make OO5B10, and then of course by an additional side bonded deuteron (on the other side) to make EE6C12. Then it's extra deuteron wraps around the additionally stacked central alpha particles (in increments of 4 additional deuterons per series up to a stack of 8 alpha particles at No 120. And I can't see side bonded protons or neutrons, because side bonded nucleons have to have opposite magnetic polarity and spin. So it boils down to how does the EE4Be8 nucleus have 0 spin, and my models have all the protons spinning in one direction and the neutrons in the other. And of course I disagree with my Kaplan about the spin of the deuteron because my deuterons all have 0 spin, because they're side bonded to each other. When you're allowing a concept of spin compatibility with potential contact you have to come to that conclusion. And I can build a EE8O18 nucleus and see how it has a slightly more balanced structural condition than the OO9F18 nucleus. But I can't see why a deuteron should spontaneously become disassociated in order for this to happen. So my question still stands as to the stability of a OO9F18 nucleus in its ground state at 0 degrees kelvin.WFPM (talk) 09:38, 11 October 2012 (UTC)


 * There is another questionable situation of this nature in the case of EO54Xe127, with 5 alpha particles plus 44 deutrons and 19 extra neutrons, but which is preferentially able to capture a stray electron and change to the only stable isotope of 53 Iodine, namely OE53I127, and which incidentally presumably involves the unpairing of one of the 44 deuterons in favor of the creation of an additional extra neutron, and which is the opposite of what happened in the case of the indium-tin situation. And these are interesting structural condition considerations calling for a better understanding of the structure of the nucleus as a real physical entity and not as a particle in the middle of an "electron cloud" of "orbital?" electrons. And I think that the cognitive aid of a albeit simple model is helpful in better understanding some of these situations, and in particular those associated with the indicated chemical and physical periodic series periods of the Janet Periodic table.WFPM (talk) 19:57, 11 October 2012 (UTC)

Ultradense deuterium
The article has a section on ultradense deuterium. This raises the same questions that have been discussed extensively on the talk page of Rydberg matter. This is fringe science promoted by a guy named Leif Holmlid, who co-authors papers along with cold fusion kooks such as Hora and Miley. Holmlid's papers all reference his own papers extensively, but there is little evidence that anyone else in condensed matter physics believes his work. Three of the four references in this section are to Holmlid's papers.--75.83.76.23 (talk) 21:57, 29 May 2013 (UTC)
 * Yeah, this is pretty bad. To make an ultradense collection of normal matter you must pressurize it (which adds energy so that each electron can be confined in a smaller space). The cold fusion people seem to feel that normal matter should just collapse in on itself without energy input, with the electrons remaining between the nuclei to shield their charges, but without any notice taken of the Heisenberg uncertainty principle and the wave-particle duality, which dictate that electrons confined to a smaller volume must be higher in momentum and thus kinetic energy. Normal ground-state atoms are the size (r) they are, for a reason! Their scale is set by:


 * $$r = h/[2\pi\sqrt{2Em}]$$


 * where E and m are electron kinetic energy and mass. To simply ignore this physics is to ignore one of the most basic rules of quantum mechanics. Extraordinary claims (such as this one) require extraordinary evidence. Which we don't have. I'm going to be WP:BOLD and simply delete the two offending fringe science sections. S  B Harris 00:53, 30 May 2013 (UTC)
 * I'd also suggest deleting Rydberg matter, but I'm not going to do it myself because I'm not that active on WP, and condensed matter isn't my field.--207.233.87.87 (talk) 20:53, 30 May 2013 (UTC)
 * I don't see the obvious problem with Rydberg matter. It simply suggests that Rydberg atoms (on the edge of ionization) might be able to contact and coexist in some circumstances, and many groups have reported this (not fringe groups). Whether they they can or can't, and how big these collections can grow, but I can't see that this contravenes basic physics. Greatly delocalized electrons are seen in many circumstances. Rydberg matter is less dense than orginary matter and in some ways represents an excited state which is inhibited from collapsing into deexitation by various physical barriers. But that's how a great many processes in nature work; there is nothing strange here. Nobody suggests that Rydberg matter can be ultradence or be used for cold fusion, or is anything like the infamous collapsed hydrogen that cold fusion people have postulated. If anything, Rydberg matter is treatly puffed-up, not condensed or collapsed. S  B Harris 21:08, 30 May 2013 (UTC)
 * It's possible that Rydberg matter is partly good science and partly pathological science. You deleted the section on ultradense deuterium, which was probably the right thing to do; it certainly contradicted what you're saying above. There is certainly a pathological-science aspect to it, since, as I pointed out above, Holmlid has coauthored papers with cold-fusion kooks.--75.83.76.23 (talk) 01:13, 6 June 2013 (UTC)

Comprehensibility alert
Section Chemical symbol reads:
 * confers non-negligible chemical dissimilarities with protium-containing compounds

A [clarification is needed] is malplaced, since the language is unambiguous but too convoluted for easy comprehension. The normal treatment of such sentences in Swedish (NB: another language!) would be to consider removing double negations non- and dis- and to "verbize substantivizations", such as perhaps confers ... dissimilarities => differs from. Or some such.§ Rursus dixit. ( m bork3 !) 09:08, 24 December 2014 (UTC)

Start of second paragraph
"Because deuterium is destroyed in the interiors of stars faster than it is produced".... should that be "...as fast as it is produced"? Seems that destroying it faster than it is produced would not be long sustainable unless 1) there was a lot to start with and/or 2) it's only destroyed a very tiny bit faster than it is produced. Herostratus (talk) 17:17, 14 January 2015 (UTC)
 * This is completely true: once you have D inside a star, it is a sitting duck for fusion. Notice how unimpressive its binding energy is. Essentially all the D we see dates from the big bang. Unlike 3He (a decay product of cosmogenic T), there is no radiogenic D either because the other isobars of mass 2 are not particle-bound. Double sharp (talk) 06:30, 13 September 2016 (UTC)

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science fiction use
it is mentioned in star trek at the very least. the article should have a section that reveals what science fiction serieses it is that refers to it. — Preceding unsigned comment added by 84.212.111.156 (talk • contribs) 11:23, 28 October 2017 (UTC)
 * Rather irrelevant here - no need for such trivia. Vsmith (talk) 12:15, 28 October 2017 (UTC)
 * Agreed. We don't need to mention every work of fiction in which a substance (or any other word) is mentioned.  Even dilithium has only a link to the fictional substance.    D b f i r s   12:22, 28 October 2017 (UTC)
 * dilithium have a link to a star trek article at the very least while this does not. 84.212.111.156 (talk) 12:42, 28 October 2017 (UTC)
 * That's because "dilithium" is very important to Star Trek. It's a fictional substance.    D b f i r s   13:09, 28 October 2017 (UTC)

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Atomic Radius and Van Der Waals
Can someone please post the Atomic Radius and Van Der Waals contact distance for deuterium? TheValveboy (talk) 03:02, 22 March 2018 (UTC)

Symbol D
There's no D on the periodic table. 2600:1700:1DC0:8CC0:C2CB:38FF:FE11:20F8 (talk) 03:53, 8 August 2018 (UTC)
 * That is true. There are lots of things that have shorthand that are not standard element symbols (deuterium is certainly not a standard element!). See IUPAC Red Book (in the 2005 edition it is in section IR-3.2.2 "Isotopes of hydrogen"). DMacks (talk) 06:14, 8 August 2018 (UTC)

Nuclear spin and isospin section
The section on nuclear spin and isospin reads more like a lecture script or a textbook than an encyclopedia. In fact it is very hard to find (and understand) the conclusion, among that sea of derivations. To make it "encyclopedic", the conclusion should be given first, and a short QM/QCM justification then should be given for those readers who may need (and can understand) that information. --Jorge Stolfi (talk) 16:13, 24 April 2019 (UTC)


 * This comment is not very clear. Are you referring to the two sections Spin and energy and Isospin singlet state of the deuteron? Which sentence(s) do you consider to be the conclusion which is hard to find and understand? And what does QCM mean? Quantum classical mechanics??? Dirac66 (talk) 02:08, 26 April 2019 (UTC)


 * I am having trouble even figuring out what a conclusion would be like. But, for example, does deuterium D2 have ortho and para forms, like common hydrogen? Should't the "spin" section answer that? The conclusion of the "isospin" session seems to be that the whole discussion is kinda silly because there is only one answer.  Like asking why is 2+2 4, and then answering that, if 2+2 was 5, then 4 minus 2 would be 1, which is absurd.  Or am I exaggerating? QCM should have been QCD, sorry. All the best, --Jorge Stolfi (talk) 18:01, 27 April 2019 (UTC)


 * I have now added a mention of nuclear spin isomers (ortho and para forms) of D2. However because this article is more about D than D2, I have added (today) more details only as a section of the article on Nuclear spin isomers which explains H2 in detail, with a link from this article. This article does now at least say that the ortho and para forms exist as you requested. Dirac66 (talk) 01:57, 1 May 2019 (UTC)

Deuterated alcohol
What would happen if you drank an alcoholic drink that contained deuterated alcohol would it be toxic like heavy water — Preceding unsigned comment added by Wizzlemuss McToot (talk • contribs) 18:32, 13 July 2019 (UTC)
 * Neither heavy water nor deuterated alcohol are particularly toxic. -- Ed (Edgar181) 18:45, 13 July 2019 (UTC)
 * There are some biological effects and even toxicity if the amount of heavy water is high enough. See Heavy_water. Presumably heavy ethanol would be similar. Dirac66 (talk) 19:15, 16 July 2019 (UTC)

Mass in the infobox
There is a discussion near the top of this Talk page regarding the mass of the deuterium v. deuteron. The former is 2.01410 u, the latter is 2.013553 u, the difference being whether the electron is included or not (0.511 MeV, etc.). I write this here because the top info box is confusing - it says "Nuclide data", but then lists "Isotope Mass". I just was misled in a calculation for a time because of the difference - is there a way the infobox can be revised to better indicate the different masses of the isotope and the nuclide? Bdushaw (talk) 13:16, 17 November 2017 (UTC)

(Should the top part of the table list "Electrons 1"?) Bdushaw (talk) 13:18, 17 November 2017 (UTC)


 * I believe the article and the infobox are both correct now on this point. Deuterium is an atom including the electron, so its mass is 2.01410 u. The deuteron is the deuterium nucleus without the electron, so the deuteron mass is 2.013553 u. The words isotope and nuclide both include the electron(s); to specify that electrons are excluded, one should use the word nucleus (or isotopic nucleus). Dirac66 (talk) 18:21, 23 December 2017 (UTC)


 * This does not look correct to me. What is given is the deuteron mass not the deuterium mass even though the article title is "Deuterium". I would suggest giving both masses in the infobox. This would avoid much confusion. If no one objects I will make the change... Davidofithaca (talk) 16:59, 25 November 2019 (UTC)
 * The correct mass, per AME2016 II, is 2.014101 u. It is simply given as isotopic mass, and is from the same source as values in other isotope infoboxes. Mentioning both in the article (but not the infobox) couldn't hurt, though. ComplexRational (talk) 19:21, 25 November 2019 (UTC)


 * This article is on deuterium so it should have the mass for the neutral deuterium atom. The problem now is that the infobox has been changed in the last year. When I wrote above on 23 December 2017 that the article and the infobox were correct "now" (then), the infobox at that time said "Isotope mass 2.01410178 u" (see edit of 22 December 2017). However this was changed on 31 January 2019 to 2.013 553 212 745, which cites the NIST value ... but for the deuteron, which belongs in the Nuclear Properties section! I will now change the Infobox value to the correct deuterium mass quoted here by ComplexRational. Perhaps one of you could insert AME2016 II as the source. Dirac66 (talk) 00:30, 27 November 2019 (UTC)

How to understand observed electric quadrupole moment of deuteron?
Deuteron is p-n so naively should have zero electric quadrupole moment. However, experimentally it turns out quite large: 0.2859 e⋅fm^2 from this article.

It explains it by adding l=2 angular momentum states - should we imagine it as a hidden dynamics?

Maybe as oscillations between 'pn' and 'np' by some pi+ exchange? (but shouldn't it make it a linear antenna producing EM waves?)

To describe e.g. deuteron-proton scatterings they neglect quark structure, but require three-body force ( https://en.wikipedia.org/wiki/Three-body_force ) - would including quarks into considerations allow to focus only on two-body forces?

But what happens with quarks when biding proton and neutron into deuteron? I am working on soliton particle model suggesting that there is a shift of charge from proton to neutron for binding of deuteron, like uud-udd slightly shifting quark u toward right, d toward left - is such explanation of quadrupole moment allowed (e.g. by QCD)? Jarek Duda (talk) 04:51, 19 June 2020 (UTC)
 * This is not a forum. Cuzkatzimhut (talk) 09:55, 19 June 2020 (UTC)
 * I as asking for clarification of information in this article: saying that it is due to angular momentum l=2 means some dynamics - so what dynamics is it? Jarek Duda (talk) 10:03, 19 June 2020 (UTC)
 * Rhetorical questions related to personal research are veering towards forum chats. Cuzkatzimhut (talk) 15:42, 19 June 2020 (UTC)

I wouldn't class this as entirely a rhetorical question. The practical question is whether the section on electric and magnetic multipoles will mean anything to a Wikipedia reader, and whether it's written in encyclopaedic style. At the moment it's a piece of maths with no context or justification. It's reasonable for an encyclopaedia to list the features of something, like its magnetic dipole moment, but if it's necessary to show how this is derived, then the Wikipedia article should refer to an appropriate textbook or review. Actually putting the maths in the article without reference is confusing, making it unclear to the reader where it came from, or even whether it's original research. If the paragraph cannot be put in context, I would be inclined to remove it altogether - it's not that it has no value, it's just that this isn't its natural home! 79.64.118.122 (talk) 11:27, 23 June 2020 (UTC)
 * The rhetorical question referred to is the baiting call: "so what dynamics is it?" it responds too. (You of course appreciate the identical bountied question in the PSE forum.) The encyclopedic style permits mentioning a basic mainstream consensus without going into textbook details, or even steering readers to  a specific source which might, or might not, satisfy them.   The reason no chapter-and-verse citations are provided is precisely to forestall tendentious complains of the "but this does not convince me!" type. It is a routine complaint in WP by readers disinclined to do due diligence.  Deletion and then reinsertion with someone's personally favorite non-standard attribution is a no-no stunt.  Cuzkatzimhut (talk) 14:29, 23 June 2020 (UTC)
 * My apology, I just was not satisfied with the explanation I have found here, thought that somebody might share my concerns, maybe be able to improve the description. However, I am still searching in various sources, and it seems the problem is that we just don't really understand it, only use mathematical trick to swipe the problem under the rug. Jarek Duda (talk) 21:39, 24 June 2020 (UTC)