Talk:Deuterium

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)]. SBHarris 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)

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 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)

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)

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. SBHarris 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. SBHarris 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 → 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)

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?

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)

history

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

Jclerman 14:01, 18 February 2006 (UTC)

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)

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. SBHarris 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. SBHarris 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)#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)

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)

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 R_infinity m_N/(m_N + m_e), where R_infinity is the Rydberg constant for an infinitely massive nucleus (109737.31568527 ± 0.0073 cm^-1 [1]), m_N is the mass of the nucleus, and m_e 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. SBHarris 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. SBHarris 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)

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. SBHarris 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

Lighter: hydrogen-1	Deuterium is an isotope of hydrogen	Heavier: hydrogen-3
Decay product of: See proton emission	Decay chain of Deuterium	Decays to: Stable

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. SBHarris 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)

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" [2].

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: [3]

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 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. You can fuse D all you like, but what you cannot do, is take their electrons to those small radii with them. The this 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. SBHarris 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. SBHarris 23:16, 17 February 2012 (UTC)