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Radioactive Heat?


KFC
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yep. faster the rate of decay, faster the rate of heat produced.

 

Technically only if you hold the energy per decay constant. P = QA, so you can get more power if the Q of the reaction compensates for the lower Activity.

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Nobody is going to let you have enough of any radioactive material to produce any useful heat.

he didn`t ask that!

 

that`s Twice now I`ve had to call you on this, either answer the question that was ASKED or say nothing.

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Alpha decay is the easiest to block, so that is what I suggest you use. I think it produces the least amount of energy per decay, though. How much heat (watts) do you need to generate?

 

Easiest to block and also probably the most efficient, since the alpha will deposit most/all of its energy in the material, rather than escaping. As for energy per decy, that will depend. There are high-Q alpha decays and low-Q beta and gamma decays. You just have to find the right ones.

 

As to "not getting enough material to generate heat" I'd suspect that there are people on the boards that can empirically contradict that. Further, I recall a calculation in a thread started by YT did showing how much/little radium you'd need. http://www.scienceforums.net/forum/showthread.php?t=2608

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What chemical would turn Alpha decay into heat? Would Americium be a good source of alpha particles?

 

Alpha decay is the easiest to block, so that is what I suggest you use. I think it produces the least amount of energy per decay, though. How much heat (watts) do you need to generate?

 

Enough to keep 30 grams of water at 60C.

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What chemical would turn Alpha decay into heat? Would Americium be a good source of alpha particles?

 

Isotopes, not chemicals, are what you should be looking at. Americium is hard to get as it is synthetic, and can decay via fission, resulting in much pollution and uncertainty. So no, it would be a bad source.

 

Enough to keep 30 grams of water at 60C.

 

Then it would depend on how much insulation you have. Also, why do you want to heat water with radioactive materials rather than conventional sources? I thought you wanted to make a nuclear battery or somesuch.

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Hmm, split water to combine it later in a fuel cell? I suppose it would function as both generator and battery if you can do that. The way NASA does (or did) power generation from radioisotopes was by using a thermocouple, but those are inefficient. The page I linked to has a list of possible radioisotopes, but they are geared towards space, so the list is more restrictive than if you only wanted to use it on earth.

 

Oh, and it seems I was wrong about Americium. Apparently, it does have very little penetrating radiation and would make a good fuel source.

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Americium is radioactive; it doesn't need a catalyst to generate heat.

 

I think the heat is to make a catalyst decompose water.

 

Coating it shouldn't be necessary, engineering-wise, though from a safety standpoint you'd want to seal it in something.

 

(FYI, I get .114 W/g output for Am-241)

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Am has similar chemical properties to Magnesium, so unless you planned on making a soln of it, a coating would be best.

 

and I can`t be 100% sure, but I think you Do need a specific license for all transuranic elements above a certain amount too.

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I think the heat is to make a catalyst decompose water.

 

Coating it shouldn't be necessary, engineering-wise, though from a safety standpoint you'd want to seal it in something.

 

(FYI, I get .114 W/g output for Am-241)

 

 

 

There is a lots of other particles(main alpha as stated earlier) that come out of Am is there anyway to turn those in to heat for maximum efficiency?

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There is a lots of other particles(main alpha as stated earlier) that come out of Am is there anyway to turn those in to heat for maximum efficiency?

 

Alphas tend not to penetrate. The deposition of their energy in a material is the heat. You can maximize the chance that the alphas will remain in the material by choosing the right shape (sphere or cube; nothing with a large aspect ratio) and making it larger. If you encased the sample (in metal, or a ceramic or glass, depending on where you are going to put it) the alphas would not escape, even if emitted from near the surface. Unless there are gammas emitted, all of the decay energy will show up locally as an increase in temperature.

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OK the original question was "Which ones produce more heat then other decay(gamma, alpha, beta), and more importantly that will not hurt me and less risky?"

There's a clear indication that he's concerned about the risk. That risk is the reason why he won't get hold of any radioactive material that will meaningfully generate heat.

Feel free to say it's not relevant. Also, unless you can prove otherwise I think it's fair to say it's true. You certainly won't get hold of it legally and I don't think this site condones breaking the law.

"he didn`t ask that!

 

that`s Twice now I`ve had to call you on this, either answer the question that was ASKED or say nothing."

 

This "Alpha decay is the easiest to block, so that is what I suggest you use. I think it produces the least amount of energy per decay, though. How much heat (watts) do you need to generate?" also doesn't answer the question.

 

Nor does this "okay, i'll quantify that with 'for any specific material the rate of heat produced will be proportional to the decay rate.'"

 

Nor (rather ironicaly) does this "he didn`t ask that!

 

that`s Twice now I`ve had to call you on this, either answer the question that was ASKED or say nothing."

 

All I was doing was pointing out that this thread was (and still is) a wild goose chase. (Obviously, I have heard of the internet. Perhaps someone would like to give me the address of a site selling large amounts of high activity radioisotopes? or perhaps they would like to keep quiet.)

 

Incidentally, "(FYI, I get .114 W/g output for Am-241) " means that a smoke detector has enough Am in it to produce something like 23 nanowatts of power; I wouldn't like to have to measure a heat flux that small.

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(FYI, I get .114 W/g output for Am-241)

It's been a while since my last physics class, but I think I know how to do it. P=QA, right? I don't have a chart of nuclides handy, so I can't do the calculations. Can you tell me what a random chunk of Uranium gets?

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It's been a while since my last physics class, but I think I know how to do it. P=QA, right? I don't have a chart of nuclides handy, so I can't do the calculations. Can you tell me what a random chunk of Uranium gets?

 

Yes, P = QA = [math]Q\lambda N[/math]

From http://atom.kaeri.re.kr/ton/ I see that Q= 4.27 MeV for U-238, and the half-life is 4.468e9 years. Let's ignore the other isotopes, which should have a very small effect.

 

So that's 6.83e-13 J * 4.92e-18/s * 2.53e21/g = 8.50 nW/g

 

Since Am-241's half life is ~seven orders of magnitude shorter, that's consistent. This, of course, demonstrates why nobody uses natural uranium decay as a thermal source. You use Pu-238, with a half-life of under 100 years.

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