Turn Mercury into Gold Cheaply?

[ben m]

If mercury is placed outside the exclusion area so that it can't amalgamate with the metal but is still unshielded from some gamma rays, since it is economical to extract high-priced elements like shown in quotes below, then would it be economical to recover gold from mercury?

There is no area "outside the exclusion area" that is exposed to gamma rays.

"Silver is produced as result of nuclear fission in small amounts (approximately 0.1 %). Because of this, extraction of silver from highly radioactive fission products would be uneconomical, but when recovered with palladium, rhodium and ruthenium (price of silver in 2005: about 200 €/kg, rhodium and ruthenium: about 300,000 €/kg) the economics change substantially. Silver becomes a byproduct of platinoid metal recovery from fission waste."

There's a big difference. The palladium group metals are actual fission products---they're something that results from the uranium falling apart, which is what you wanted it to do anyway. The whole process of Pd/Ag/Ru/Rh production is occurring in the middle of your reactor, and all of the fissions, beta-decays, gamma emission, etc., that occurs along the way is contributing to the energy that you're selling for $0.1/kWh. You make your money selling that electricity; if you get a few kg a year out of the Pd sales, they're really just icing on the cake.

The Pd/Ag/etc. metals are produced in fairly large quantities; they're produced via a decay, not via waiting for one particle to hit another particle. Putting a lump of Hg into the middle of the reactor would *not* turn it into Au anywhere near as effectively as U turns itself into Ag. It so happens that Au is *not* produced by U decay; it's too heavy, the heaviest thing in my book with a nonzero U235 fission yield is 162Dy.

Using the reactor to irradiate Hg would interfere with the electricity production. You'd be taking some of the neutrons that were supposed to help spin your moneymaking turbines, and shunting them off to a bucket of cold mercury. The actual Au production rate would be extremely low, for all of the reasons outlined repeatedly.

 

[7up]

There's a big difference. The palladium group metals are actual fission products---they're something that results from the uranium falling apart, which is what you wanted it to do anyway. The whole process of Pd/Ag/Ru/Rh production is occurring in the middle of your reactor, and all of the fissions, beta-decays, gamma emission, etc., that occurs along the way is contributing to the energy that you're selling for $0.1/kWh. You make your money selling that electricity; if you get a few kg a year out of the Pd sales, they're really just icing on the cake.

The Pd/Ag/etc. metals are produced in fairly large quantities; they're produced via a decay, not via waiting for one particle to hit another particle. Putting a lump of Hg into the middle of the reactor would *not* turn it into Au anywhere near as effectively as U turns itself into Ag. It so happens that Au is *not* produced by U decay; it's too heavy, the heaviest thing in my book with a nonzero U235 fission yield is 162Dy.

Using the reactor to irradiate Hg would interfere with the electricity production. You'd be taking some of the neutrons that were supposed to help spin your moneymaking turbines, and shunting them off to a bucket of cold mercury. The actual Au production rate would be extremely low, for all of the reasons outlined repeatedly.

I included the wiki quote only as a comparison to recovering gold from mercury to the economics of extracting other precious metals from uranium, not as an attempt to get gold from uranium.

What I was trying to ask is since gammas don't reflect well then some of them will be lost to shielding anyway, so couldn't there be an area shielded from the uranium, metal, etc but still be exposed to stray gamma rays before they have been fully absorbed by shielding?

eta
It would seem that even though "photonuclear interactions in heavy nuclei will be < 1/1000 as likely as electron interactions" there would be so many gamma rays striking Hg that 1/1000 doesn't seem to be a very limiting factor.

 

[ben m]

What I was trying to ask is since gammas don't reflect well then some of them will be lost to shielding anyway, so couldn't there be an area shielded from the uranium, metal, etc but still be exposed to stray gamma rays before they have been fully absorbed by shielding?

Nope. In a reactor, the gamma emitting U is immersed in water and contained in a thick-walled steel containment vessel. Most of the gamma radiation is meant to be absorbed in the fuel itself (probably not the source rod, but one of the neighboring rods); failing that, in the water. The steel vessel itself sees comparatively little gamma radiation; areas outside the vessel see almost none.

It would seem that even though "photonuclear interactions in heavy nuclei will be < 1/1000 as likely as electron interactions" there would be so many gamma rays striking Hg that 1/1000 doesn't seem to be a very limiting factor.

Every gamma ray that successfully makes a gold atom is worth (at current gold prices) 1.5*10^-20 dollars. Every gamma ray that heats up your reactor working fluid is worth (at current energy prices) 0.8x10^-20 dollars.

 

[Hindmost]

If mercury is placed outside the exclusion area so that it can't amalgamate with the metal but is still unshielded from some gamma rays, since it is economical to extract high-priced elements like shown in quotes below, then would it be economical to recover gold from mercury?

"Silver is produced as result of nuclear fission in small amounts (approximately 0.1 %). Because of this, extraction of silver from highly radioactive fission products would be uneconomical, but when recovered with palladium, rhodium and ruthenium (price of silver in 2005: about 200 €/kg, rhodium and ruthenium: about 300,000 €/kg) the economics change substantially. Silver becomes a byproduct of platinoid metal recovery from fission waste."

http://en.wikipedia.org/wiki/Synthesis_of_precious_metals#Silver

The entire nuclear island would be a mercury exclusion zone and that isn't going to change. Even with a high N16 gamma flux outside the reactor, there just isn't enough to make significant amounts of gold.

glenn

 

[7up]

How long would it take to turn Hg to Au above earth's atmosphere where there is a lot of gamma ray radiation?

 

[ben m]

How long would it take to turn Hg to Au above earth's atmosphere where there is a lot of gamma ray radiation?

There *isn't* a lot of gamma ray radiation in space. There may be a middling-large by human survival standards, but practically zero by inside-a-nuclear-reactor standards.

7up, I humbly suggest that you have access to Google, Wikipedia, and enough numbers that you can answer some of these questions yourself.

 

[7up]

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[7up]

Ben,

What I read said the earth's atmosphere blocks a lot of gamma radiation, but I get your answer.

 

[Hindmost]

How long would it take to turn Hg to Au above earth's atmosphere where there is a lot of gamma ray radiation?

This is starting to get a bit trite...launching stuff into space is very expensive. It would probably be easier to steal the gold in the rocket.

 

[7up]

This is starting to get a bit trite...launching stuff into space is very expensive. It would probably be easier to steal the gold in the rocket.

I Cubic Foot of Gold would weigh 548.18 Kilograms or 1206 pounds

http://wiki.answers.com/Q/How_much_does_one_cubic_inch_of_gold_weigh

548.18 kg of gold = 19336.480 oz

http://www.metric-conversions.org/weight/kilograms-to-ounces.htm

$1500/oz gold * 19336.48 oz = $28,995,720 (not bad)

I read fuel doesn't contribute much to the cost of the trip.

But if there aren't enough gamma rays up there then doesn't matter. I only asked because what I'd read made it sound like there were lots of gamma rays above earth's atmosphere.

It's just a question. No need to get upset. I guess I picked the wrong forum.