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Is there a series of transmutations through nuclear decay that will result in the stable gold isotope ${}^{197}\mathrm{Au}$ ? How long will the process take?

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Get a nuclear reactor, wait some million years, and you will get about $1 worth of gold! Yay! –  Anonymous Pi Mar 16 at 19:46

5 Answers 5

up vote 8 down vote accepted

I guess you are really looking for this wikipedia page : http://en.wikipedia.org/wiki/Synthesis_of_noble_metals#Gold .

In short, there are gold synthesis technique, but they apparently all need some external energy (either $\gamma$-ray or neutron capture) and are not restricted to nuclear decay. One of them has for intermediate step the nuclear decay${}^{197}Hg\rightarrow{}^{197}Au+e^+$ with a 2 days half life. The unstable ${}^{197}Hg$ is obtained from a stable $Hg$-isotope by $\gamma$-ray irradiation (${}^{198}Hg+\gamma\rightarrow {}^{197}Hg +n$.)

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Natural gold exists, so the answer to the first part of your question is unambiguously "Yes". 'Cause all those heavy elements get made by transmutation in supernovae.

I can't answer the time scale thing because I haven't a table of the isotopes in front of me right now.


Checking with http://ie.lbl.gov/education/isotopes.htm I find that $^{197}$Pt has a beta decay branching fraction of about 3% and a halflife of about 95 minutes, and $^{197}$Ir decays by beta with 5 minute halflife and $^{197}$Os decays by beta with a 64 hour halflife...

Anyway, you can chase this as far as you care to.

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Thats not a full answer to the OP, he was strictly about nucear decay, i.e. does some isotope which decays into Gold exist? I don't know the answer, although I suppose if you don't put a limit of the halflife that several should exist. Supernovas build up heavy elements via neutron capture, and probably endothermic nuclear fusion, so Gold's existance does not prove it is a decay product. –  Omega Centauri Dec 1 '10 at 21:32
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This logic in this answer is incorrect. Gold is not produced by nuclear decay in supernovae, but rather by nuclear fusion at extremely high temperatures! –  Noldorin Dec 2 '10 at 16:46
    
Also the logic of the correction must be corrected... More precisely, Gold (an excited state of it, decaying gamma only) could form straightly from nuclear fusion -and then the logic of the answer is incorrect, as existence of Gold does not imply existence of the decay path- but actually it also forms from multiple decay chains from various excited levels of various elements. Really I can not point an element unable to be produced from beta or EC of an excited state of a close element. –  arivero Jan 17 '11 at 10:35
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@Noldorin and everyone else so convinced about fusion: The elements heavier than about nickel formed from core-collapse SNe were formed from radioactive decay, usually beta decay. Combining two very positive nuclei endothermically is very hard to do thermally, and is rather rare. Instead, there is rapid neutron addition to build up mass neutron-rich nuclei, which then decay to more stable admixtures of protons and neutrons. Gold is in fact right in the middle of the second r-process abundance peak. –  Chris White Jan 2 '13 at 7:45
    
@ChrisWhite: Yes, fair point, you are right of course. But the point remains that fusion produces elements necessarily heavier than gold (greater in number of nucleons) that are then able to decay into Au atoms. –  Noldorin Jan 2 '13 at 19:40

The following wikipedia link has a table of all the known isotopes of all the known elements. Gold has Z=79. According to the table, there's only one stable isotope of gold, $\;^{197}Au$ :

http://en.wikipedia.org/wiki/Table_of_nuclides_%28complete%29

From this you can figure out which isotopes decay into gold by considering the three possible decays (ignoring fission) beta decay, inverse beta decay, and alpha decay.

Beta decay means a neutron decays into a proton, an electron and an anti neutrino. The electron and anti neutrino go away. So the result is that the number of neutrons ("n" in the above table) decreases by one while the number of protons ("Z" in the above table) increases by one. The result is that you move in the table one square diagonally towards the top right.

Inverse beta decay is the opposite of beta decay (more or less). The result is that you move one square diagonally towards the bottom left.

Alpha decay means the loss of a helium nucleus of 2 protons and 2 neutrons. So you move two squares diagonally to the top left.

So if you want to look for chains that end with gold, do the following. Look at the isotopes 2 squares down and to the right (for alpha decay) or diagonally one square up right or down left (for beta decays). Now take a look at that isotope to see if it decays in the manner you've assumed. For example, $\;^{195}Hg$ is in a position to beta decay to gold.

To find out whether the isotope decays the way you've assumed, go to this website:

http://ie.lbl.gov/toi/perchart.htm

and click on the element, then the isotope. Clicking on Hg and then the isotope 195 shows that it does indeed decay to gold.

Note that this isotope can decay to gold in something like 46 ways. The different ways involve various excited nuclear states for either the mercury or the gold or both. Such excited states may also decay by emission of a photon (gamma ray).

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atom.kaeri.re.kr is perhaps more complete that lbl toi. In any case, the problem, to me, is that the barrier of alpha decay makes, for this specific zone of the periodic table, lifetime to be a more important parameter than energy balance. So some energetically possible transitions fail because competing lifetimes favour other reaction paths. –  arivero Jan 17 '11 at 10:46

Last time I did the sum, 201Hg to 197Au plus 4He did not need external energy:

200.970277 - (196.9665516 + 4.0026032) = 4.0037254 - 4.0026032 = 0.0011222.

Caveats:

  1. I did not check the error term of atomic weigths, and
  2. The first decay, alpha to 197Pt, should be awesomely slow. In fact
  3. 201Hg is considered stable in all the listings.

But the point is that the external energy would be recovered in this case.

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In fact, time ago I calculated (here, conjeturas.blogia.com/2006/061601-mercurio.php), as an amusement, that the nuclear burning of natural Hg into stable isotopes of Pt and Au would produce about 80 KwH per gram. (Again, I am not 100% sure if I read correctly the masses from kaeri, you are welcome to check it) –  arivero Jan 16 '11 at 4:58

the alchemists have dreamed about the production of gold (Z=82) from some cheap material and lead (Z=79) was their favorite choice. They were just using a wrong science - namely primitive chemistry instead of nuclear physics. But otherwise their choice of lead was OK. And indeed, lead became the element that was transmuted into gold for the first time sometime in 1980 (and maybe even in 1972). See

http://chemistry.about.com/cs/generalchemistry/a/aa050601a.htm

One has to remove three protons which costs a lot of energy. Needless to say, the transmutation remains economically unacceptable. That's true for other strategies, too.

Best wishes Lubos

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Actually, lead is atomic number 82 and gold atomic number 79. –  Steven Devijver Jan 17 '11 at 12:07
    
It seems that Pb, "Saturn" was used, due to nomenclature problems, as a reference to other materials, particularly Stibnite in a lot of recipes. Thus Sb and Hg were more interesting, to serious researchers, that Pb. For instance Starkey pbs.org/wgbh/nova/newton/alch-guide.html works with antimony and quicksilver to vegetate the gold. –  arivero Jan 17 '11 at 16:59
    
I wonder if it could be better to do the distinction between economically unacceptable, energetically unacceptable (exothermic vs endothermic), and physically forbidden. 201Hg->197Pt+4He is energetically acceptable, but I'd be not surprised if it is forbidden by parity or some other principle. –  arivero Jan 17 '11 at 17:06

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