As a student I know that toward the end of the semester it's sometimes a struggle to pay attention and retain motivation. As a teacher I have to try to fight that tendency in whatever way I can. It's not necessarily possible to make a physics quiz entertaining, but sometimes they can at least be made interesting.

You have a baseball-sized lump of metallic cobalt-60. It's radioactive, and is thus spitting energy in all directions mostly in the form of beta and gamma radiation. Wrap it in a sufficiently thick lead shell and that radiative energy is going to be absorbed and converted to thermal energy. What's the thermal power output?

(As I posed it in class, the question was much easier. But this open version is more fun.)

Forget all the radioactivity stuff for the moment. How much cobalt is in a baseball-sized lump? Well, a regulation baseball is supposed to be 3 inches in diameter. Using the formula for the volume of a sphere and putting things into the metric system, a baseball has a volume of 232 cubic centimeters. Cobalt has a density of about 8.9 grams per cubic centimeter, and so multiplying the two together we have 2.06 kilograms of cobalt. We might as well round it to an even 2 kg.

We're going to find the energy per atomic decay, and so I expect we'll need the number of atoms in the sample at some point in our calculation. Why not find it now? As it happens, there are 58.9 grams of cobalt per mol. If it's been a while since you took chemistry, a mol is a unit for counting, just like "dozen". Instead of 12 though, a mol is 6.022 x 1023, a fairly gargantuan number. We need it though, because atoms are so numerous. Doing the division, we find there are about 35 mols worth of cobalt in our sample. This is about 2.1 x 1025 atoms.

All right. Now that that's done, we can figure out how much energy a decaying cobalt atom releases. A cobalt-60 atom has a mass of 59.9338 u. It beta decays to a nickel-60 atom of mass 59.9307 u. That leaves a difference of 0.0031 u unaccounted for. Some of that goes into the mass of the electron that's being released, but the mass of an electron is only 0.00054 u - almost negligible in comparison. The missing mass must have been transformed into energy, so now we get to use the most famous equation in physics:

Plugging in our missing mass, we get 4.63 x 10-13 J. Not a whole lot, but then we have a lot of atoms emitting that energy. The last thing to do is find the rate at which the atoms are decaying. It turns out to be just the number of atoms in the sample divided by the average time for an atom to decay. For cobalt-60 this lifetime is 7.6 years.

Doing the math and converting years to seconds, we have decay at an initial rate of 8.76 x 1016 per second. Multiplying that by the energy per decay and we get a power of 40.5 kW.

That's no slouch; it's enough to power dozens of houses. Unfortunately cobalt reactors buried in the back yard are not forthcoming. Cobalt-60 is ludicrously dangerous, and at any rate all the energy has to come from somewhere. Cobalt-60 is not naturally occurring in any substantial quantity due to its short half-life, and has to be generated at considerable difficulty and great expense in nuclear reactors.

On the other hand if you need a long-lived autonomous source of energy far away from inhabited areas (say, in an unmanned space probe), this might be just the thing. Many probes from Cassini in orbit around Saturn to the Voyager probes beyond Neptune have been and are being powered by exactly this process, though using radiation sources other than cobalt. The USSR even used these radioisotope thermoelectric generators in earthbound applications including powering remote lighthouses.

In the foreground of this picture you can see the black heat sinks of the RTG on the New Horizons probe currently en route to Pluto:

Its current billion mile separation from earth renders it fairly safe.

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Pu-238 RTG (~20 W/g thermal, ~0.57 W/g electrical, 87.7 year half-life), projected 200 watts electrical at Pluto. US RTGs depend on Russia for Pu-238. Enviro-whiners go nutso every time an RTG is boosted into space.

Are these reactors designed with elements that will decay to produce daughter elements that will also decay and produce useful energy, or do the first-generation daughter elements just become dead weight?

Anon: Ni60 is stable so it would be just "dead weight". The heavy stuff like Pu or U have decay chains that keep producing for very long times. For example, the decay products of depleted uranium are more radioactive than the depleted uranium so if you buy a lump of it, the lump's energy production (and radioactivity) increases steadily for hundreds of thousands of years. Those interested in these sorts of things from a nuclear engineering standpoint can google "secular equilibrium" for more information.

By Carl Brannen (not verified) on 30 Jul 2009 #permalink

Other earthbound uses include powering T-800's, flux capacitors, and Stephen Hawking's new exoskeleton, Hawkerra. It looks like a giant turtle.

You forgot to calculate the equilibrium temperature of the lead wrapper. That is quite a lot of power.

Pu-239 is much longer lived so it is not as "hot", but it produces enough heat to (reportedly) feel as warm as a baby's bottom. Among stories in the bomb histories is the detail that simple thermal expansion is an issue in fitting the pieces of an atom bomb together.

BTW, many people don't realize that all of the waste heat has to be radiated away (as IR photons) when you are in space, whether in the case of that RTG in the picture or for the Shuttle or ISS.

Strontium-90 has also been used for this purpose, I think. This decays to Yttrium-90, which has a higher beta-energy than its parent. (An example for anon).

Sr-90 is also easier to shield than Co-60. Co-60 is used in radiotherapy, precisely because it can produce a gamma-ray beam that is reasonably penetrating.

By ColonelFazackerley (not verified) on 30 Jul 2009 #permalink

Matt,

Would a significant portion of the energy be carried away by anti-neutrinos? Such elusive particles account for, on average, 70% of the energy involved in tritium decay.

By Tim Gaede (not verified) on 31 Jul 2009 #permalink

Where did you get 7.6 years?
The T1/2 for co-60 is 5.27 years.

Also, how did you account for the logrithmic decay of heat generation?

The lifetime is the time required for Co-60 to decay to 1/e of its original number. It's longer than the half-life by a factor of about 1.44. The heat generation will of course fall off with time, but here I'm just interested in the initial heat generated.

Excuse me for coming quite late to this thread, but I'm afraid your calculation is a little off. 60-Co decays to an excited state of 60-Ni, which emits two gamma rays of 1.117 and 1.33 MeV with almost 100% efficiency. That should just about double the power, I believe.

By Mister Troll (not verified) on 15 Aug 2009 #permalink

Oops... and I correct myself within minutes. How embarrassing!

The gamma ray energy of course is included in the energy released by the initial decay. My mistake was thinking in terms of the energy of the beta particle, rather than the Q of the reaction... sigh.

By Mister Troll (not verified) on 15 Aug 2009 #permalink

Just out of interest, what would you have done if the student was not aware how big a baseball was? I for one, would not have a clue without reading that answer..

I fell into this trap when I was covering a business studies lesson. The discussion was on market cycles or something and I mentioned an LP as an example.. pop-culture FAIL!

By Donalbain (not verified) on 17 Aug 2009 #permalink

Every reactor could have a blanket of sulfur 34. With the much larger number of reactors necessary to avoid financial collapse, sulfur-35 could replace lithium ion batteries in autos. You would buy the car and never have to fuel it till it was ready to scrap. This would produce more income for reactors than the electrical generation, and would finally make them cost effective. A small personall aircraft would have no distance limit, other than pilot fatigue.

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