Turn lead into gold via radioactive decay

Question

I was looking at the isotope's table and noticed that one of lead's isotopes can actually turn into stable gold through the following mechanism:

$$\ce{_{82}^{197}Pb -> _{81}^{197}Tl -> _{80}^{197}Hg -> _{79}^{197}Au}$$

I know (or at least guess) that such a process must be awfully ineffective. Still, I was wondering:

• How could we get a great quantity of $\ce{_{82}^{197}Pb}$?
• How long would it take for it to turn into gold? Can that be accelerated?
• Once most of it has been turned into gold, how can we extract the gold (there are still remaining $\ce{Pb,Tl,Hg}$ molecules)?
• What would be the yield of the process? How much money would it approximately cost compared to how much we win in gold?

Answer 1

Interesting idea, but it has already been done, and not cheaply - read on.

How could we get a great quantity of $\ce{_{82}^{197}Pb}$ ?

There would be two problems with getting a large amount of $\ce{_{82}^{197}Pb}$. First, the parent nuclide of $\ce{_{82}^{197}Pb}$ is $\ce{_{83}^{197}Bi}$ which is unstable and has a half-life of only 9.33 minutes - so you can't get a large quantity of $\ce{_{82}^{197}Pb}$'s precursor to begin with. Second, once $\ce{_{82}^{197}Pb}$ is formed, it has a half-life of 8.1 minutes, so it transmutes quickly to $\ce{_{81}^{197}Tl}$.

How long would it take for it to turn into gold?

$\ce{_{82}^{197}Pb}$ half-life = 8.1 minutes

$\ce{_{81}^{197}Tl}$ half-life = 2.84 hours

$\ce{_{80}^{197}Hg}$ half-life = 64.14 hours

After 10 half-lives, ca. 0.1% of the starting material will be left $$\ce{(1/2)^10 = $0.0009766$}$$ The last step is the slowest by far, so after about 641.4 hours (26.73 days), you should have something around 99.9% pure gold.

Can that be accelerated?

Unlike chemical reactions that can heated, catalyzed, etc., this type of nuclear transformation keeps a set schedule.

Once most of it has been turned into gold, how can we extract the gold ? (there are still remaining Pb,Tl,Hg molecules)

As noted above, you can get whatever purity you desire, just wait.

What would be the yield of the process?

It would be high for the 3 nuclear transformations you listed. Each of the elements you listed decays directly and only to the daughter isotope you've shown. However, as noted above, you can't start with $\ce{_{82}^{197}Pb}$, you generate it from $\ce{_{83}^{197}Bi}$, the decay of which adds some impurity along with lead. And then since the bismuth isotope is not long-lived you'd probably start with its precursor, and so on until you find something that has a long enough life that you could assemble a reasonable quantity.

Back around 1980 Glenn Seaborg actually transmuted bismuth to gold, but only a few thousand atoms (see this reference also).

How much money would it approximately cost compared to how much we win in gold?

The Wikipedia article I referenced directly above notes, "the expense far exceeds any gain." There are other ways (fission and fusion) to produce gold, but at least with the methods available today, the cost would be astronomical.

Answer 2

One thing Ron didn't cover is the side effects of all that decay. All that short-half-life stuff will be rather radioactive, to the point where you might not survive to see the results. It's also possible that if you get enough together to be profitable that it will simply vaporize itself. The stable gold should eventually condense so as long as you contain the vapor it might work. Or you'll have a very pretty gold-lined flask.

The stable isotopes of lead are 204, 206, 207 and 208, you'd have to knock a lot of protons and neutrons out to get down to 197.

Shame! Knock a proton off of lead and it turns into thallium. But as that's the point of the exercise just bump off two more and we're done.

Answer 3

I can’t guess at the cost, but consider this: if it were profitable, many people would do it, and then the value of gold would quickly reduce to a level consistent with the cost of producing it.

Answer 4

You can't really die from the radiation while the process is happening, if you do it right. So you will need a fat, thick lead box to protect yourself from the alpha, beta and gamma rays/particles. Also, the finally condensed gold will collect on the bottom of the box or on the sides. So technically I calculated that to get about $\pu{450 g}$ (one pound) of 99.9% pure gold, you will need to start with approximately 3 tons of the radioactive bismuth isotope. And to buy that amount, you will need somewhere about $43000.5 \times 10^6$ dollars to do that. So what is the trade-off? It is kind of a bad deal, so it will just be better to mine for gold.

Answer 5

The best way I can think of doing it would be to use lead as a cyclotron target, I would need to bombard it with very high energy protons. My intention would be to form Bi-197. By making a proton enter the nucleous of the lead, if the resulting nucleous has a high energy it would emit some neutrons or alpha particles or even some protons.

This process is spallation, this would form an unholy mixture of things with differnet atomic masses. I would expect to form as well as Bi-197 some Bi-205 (radioactive half life 15 days) and Bi-202 which will decay to long lived Pb-202- So the cyclotron target will stay radioactive a long time.

In the mean time the Bi-197 will have undergone a series of positron emissions to form Au-197. I would then need to dissolve the cyclotron target in acid and then use solvent extraction to separate the gold from the mercury, thallium, lead and other things in the target.

I am sure that the electricity needed for the cyclotron, the chemicals needed for the separation and the time of the workers needed will be more valuable than the gold that they foem.