Swedens first Radioactive Boyscout? Or, how to boil water with smoke detectors.

On August 1, 2011, in Swedish on-line news paper Helsingborgs Dagblad, a story was told about a 31 year old man who tried to build his own nuclear reactor using Americium-241 from smoke detectors and other chemicals. The story reminds us of the story from the 1990's about the "Radioactive Boyscout", David Hahn. There is another story from the same source with an interview with Professor Janne Wallenius, Royal Institute of Technology in Sweden, who tries to explain the impossibility of the mans endeavour.

I would like to present to you a small exercise in physics and show the impossibility from another perspective. A nuclear reactors fundamental purpose is to heat water. Most of the time that heated water is used to generate momentum in a turbine which generate electricity so that you can read this blog post. Sometimes that heated water is used as a direct source of heat. But, how many smoke detectors would we need to boil a bucket of water?

Energy needed to boil water

Let us start with a bucket of 1 dm³ of water (i.e. one liter) at 25° Centigrade at normal atmospheric pressure. In order to make it boil we need to raise its temperature with 75 degrees to 100°.

Using measurements performed by others, we know that the specific heat capacity of water at 25° is about 4.2 Joule per Gram per Kelvin. That is, it takes 4.2 Joules of energy to heat one gram of water one degree.

To heat one liter (i.e. 1000 grams) of water 75 degrees from room temperature would take about

4.2 (J/g/K) * 75 (K) * 1000 (g) ≈ 315 kJ

So, using smoke detectors, we need to somehow transfer 315 kJ of energy into our bucket of water.

Energy from smoke detectors

An ordinary smoke detector contain about 1 µCi or 37 kBq of ²⁴¹Am. This unit means that we get about 37000 decays of ²⁴¹Am per second. From one decay of ²⁴¹Am we get about 5.5 MeV of energy. This is the same as about

5.5 (MeV) * 1.602⋅10−13 (J/MeV) ≈ 8.8⋅10−13 Joule

So an ordinary smoke detector generates about

8.8⋅10−13 (J) * 37⋅103 (1/s) ≈ 325⋅10−10 Watt

Conclusion: Boiling water with smoke detectors

Lets say we have patience. We can wait one year for our bucket of water to boil. One year is about 315360000 seconds. So, we have this many seconds to generate 315 kJ. To do this we need

315 (kJ) / 315360000 (s) ≈ 0.001 Watt

Each smoke detector gives 325⋅10⁻¹⁰ Watt so we need the following number of smoke detectors:

0.001 (W) / 325⋅10−10 (Watt/detector) ≈ 30⋅103

I.e. we would need some 30000 smoke detectors to boil our bucket of water if we can wait one year for it to boil. We could not drive a turbine with this.

Say we were in a situation were we need to boil water quickly. For instance if we were lost in the forest and need to get a hot cup of tea made from the sprouts of spruce. Then we only have about one hour and we would need 24*365 times the number of smoke detectors, or about 250 millions.

Clearly, it is impossible to boil water with smoke detectors. You can't find 250 million smoke detectors laying around in the woods.