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I was curious about a radioactive decay battery. My thought was to place an americium source from a smoke detector in a vacuum sealed borosilicate glass vessel. The Americium source would be "aimed" at a piece of Titanium 2 mm's away. The vacuum sealed glass vessel would have 2 wires coming out of it. 1 wire would be attached to the Source. 1 would be attached to the Titanium.

If I attached a voltmeter to the leads, What would the approximite voltage be?

As I understand it, the current of such a device would be in the nanoamp range (2 or 3).

I was very curious what the voltage would be? I understand that the answer would be approximite, but I was still curious. Would it be closer to 5 Volts? or 50,000 Volts?

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1 Answer 1

I couldn't find details of the idea on the net. There is a paper at http://arc.aiaa.org/doi/abs/10.2514/1.25837?journalCode=jpp but it's behind a paywall. So don't take this answer as gospel.

Anyhow, as I understand it the americium 242m battery is basically a capacitor. Alpha particles emitted by the americium leave it with a net negative charge and produce a positive charge on the shell around it. If you connect a wire between the two then electrons will flow from the americium to the shell just like a conventional battery.

Starting from zero, the voltage difference between the americium and the shell will rise linearly until either you start drawing off electricity or the voltage difference gets so big that it repels the alpha particles from the americium and the charging stops. According to Wikipedia the emitted alpha particles have an energy of 5.64 MeV, so the maximum voltage would be about 2.8 MV (half the energy because alpha particles carry a +2 charge)

However you specifically ask about americium from a smoke detector. This is americium 241 not 242m, but americium 241 does emit alpha particles and the energy is similar to 242m at 5.4MeV so in principle it should still work. The maximum voltage would be about 2.7 MV.

However according to this article a smoke detector emits about 37,000 alpha particles per second. An alpha particle has twice the charge of an electron, or $3.2 \times 10^{-19}$ coulombs, so 37,000 of them is about $10^{-14}$ coulombs and your current is therefore $10^{-14}$ amps. Not only is this too tiny to measure easily, but you'd need extremely careful conditions to stop leakage of electrons discharging your battery faster than the alpha particles could charge it.

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BTW was it you asked a similar question at en.allexperts.com/q/Physics-1358/2012/11/americium-battery.htm ? You should have aksed here first :-) –  John Rennie Nov 8 '12 at 18:21
Any gamma activity in your source (from the progenitor, the decay products or any other isotopes present) will tend to cause leakage by the photoelectric effect resulting in a lower peak voltage and less sustained current. I haven't checked on these particular isotopes but essentially all heavy radioactive isotopes have some gamma activity in there decay chain and most have rather a lot. –  dmckee Nov 8 '12 at 19:18
@dmckee Well yeah, but you have to consider the geometry. Your alpha particle collector (the positive terminal) stops the alphas in a very short distance, while the gammas take a much longer distance to interact, so most of them miss the collector altogether or hit in a location where the secondary charged radiation doesn't make it to the surface. Unrelated: the 2.7 MV is several times common HVDC line voltage. Those are used for practical energy distribution, but the challenges related to electrical discharge are significant. –  AlanSE Nov 8 '12 at 23:29

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