Hot Hot Heat

A few days ago we calculated the radiative power output of a very radioactive source like cobalt-60. CCPhysicist suggests we also calculate its surface temperature. Sounds like a plan!

A radioactive plutonium pellet, glowing red hot. Still much less radioactive than our hypothetical cobalt.

It would be difficult to calculate the temperature of the sample by itself, as we have no clean way of determining how much energy escapes in the form of the various nuclear decay processes without heating the sample. Most but not all of the gamma rays will escape, very few of the beta rays will, and effectively all of the neutrinos will. Those that don't escape will end up absorbed, heating the sample. While we could probably manage to extract a decent estimate, in this case we can simplify the problem by making it more interesting.

Let's wrap it in radiation shielding. We have to if we want to get anywhere near it or use it as a power source. Cobalt-60 is about as holy Toledo dangerous as radioactivity gets. It's so dangerous (How dangerous is it!)... well, why not run the numbers? The unit of radiation dose is the Sievert. It's equal to 1 joule of energy absorbed per kilogram of body tissue. Not all radiation is created equal, some is more dangerous than others and thus there's a multiplier effect depending on the type of radiation. Conveniently for gamma rays that multiple is 1.

Our sample emits some 40 kW of power. Not all of them are in the form of gamma rays, but more than 90% is. We're not unjustified justified in pretending the absorption figure is 100%, especially considering much of the remainder is also going to be absorbed and cause damage. If you weigh 160 pounds and were to absorb the entire dose, you're receiving about 550 sieverts per second. Now this is a worst-case estimate since you can't possibly absorb all the radiation. Even if you're holding the source, a good fraction of the radiation won't be emitted in your direction and some of the radiation that is won't be pointed at you. But it doesn't matter. When there's this much radiation that factor of maybe 10-100 reduction is nothing.

One sievert will make you seriously ill. Four seiverts means survival is a toss-up. Ten sieverts is invariably fatal. The maximum allowable dose for workers in the nuclear industry is 50 millisieverts per year. Clearly we had better come up with some good shielding.

550 sieverts per second divided by 50 millisieverts per year gives us the reduction factor required to bring our radiation source down to a statutorily safe level. It turns out to be about 3.5 x 10¹¹, meaning we can only let around one part per trillion of the radiation through. Fortunately this isn't as hard as you might think. Yesterday's post discussed the exponential effectiveness of shielding, and we can use that knowledge to calculate the thickness of the necessary shield.

Grabbing a convenient chart from Wikipedia, we see that the absorption thickness of lead is about 1cm for gamma rays near the ~1-2 MeV range produced by Co-60 decay. The logarithm of the absorption factor we require is about 27. Therefore at least in theory 27 centimeters of thickness ought to stop enough radiation to be safe.

Adding that to the 1.5 inch radius of the cobalt lump, our spherical source plus shield has a radius of 0.308 meters, for a total surface area of 1.2 square meters. If we assume that all the heat is dissipated by thermal radiation (a decent assumption, if not a perfect one), we can use the Stefan-Boltzmann law to find the temperature. The equation is thus:

And the temperature comes out to a toasty 602 degrees Celsius, or 1,116 degrees Fahrenheit. Hot, perfect for all kinds of energy-hungry activities. Or at least it would be if this weren't just above the 600 C melting point of lead. Oh well. Slap a heat sink on it and call it a day.