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 10^{23}, 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 10^{25} 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 10^{16} 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.