Voltage on Radioactive Power Source

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. 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? Thank You, Paul
- Paul (age 54)
Orlando, Fl

Nice question. In principle, one could devise a contraption in which the voltage between the electrodes was just big enough to almost, but not quite, stop the decay particles from crossing the gap. That would be a big voltage. In practice, a device like the one you describe would have closer to 5 V than 50 kV across it, at least on a linear scale. If the voltage got too big, there would be some sort of surface breakdown on the glass, giving a current leak.  You can make a rough guess by looking at the voltage ratings of various insulating spacers for wires. I don't think that ones a few mm thick are usually rated more than a few tens of volts. (Correction, see below- this can be hundreds of volts.) The radiation might also make problems for the glass insulation. So though 5 V sounds conservative, 50 kV sounds way too big.

Mike W.

The voltage of your "battery" will also depend on what kind of load you have attached.  With a maximum current of 2 or 3 nano-Amperes, the voltage will drop down to V = IR  --> 2 or 3 milli-Volts  for a 1 megOhm load.    LeeH

(published on 11/08/2012)

Thank you so much for your answer. I had hoped that having both the Americium and the Titanium sealed in a vacuum would have acted more like a capacitor. That it would have allwed the source and the plate to gather large voltages, at microscopic amperage. My goal was to communicate the gathered charges outside the vessel to drive a device similar to the Oxford Electric Bell. Because the current draw of such a device is 2 or 3 nanoamps, I hoped that the device could last for centuries, like the Oxford Bell already has. I have built a replica of the Oxford Bell that works great, so doing some work with nanoamps is not new to me. (view it here http://youtu.be/nZDxzaNsI7Q) I was hoping the Americium would generate a voltage of around 1-2 thousand volts (or even higher), at 2-3 nanoamps. That would meet my needs perfectly. From your answer, what I was considering would fail totally. 5 volts is far too low. I need 1KV or more at nanoamps. Again, Thank you for your answer. I will have to figure out a better way.
- Paul (age 54)
Orlando, fl

Well, maybe you shouldn't give up yet. 1kV, as opposed to 50 kV,  is within the range that can be handled by some commercial spacers that aren't all that big. For example, I remember having a 900V battery hooked up by some standard commercial connectors with only a few mm spacing. I believe those used teflon spacers. We did have to be careful to keep the surfaces very clean. (Also, the reason I remember the spacing between the leads quite well is that I touched both of them simultaneously with the tip of a single finger, leaving what looked like the bite of a small snake.)

What I'm not sure of is how well any of those insulating materials (epoxy is also used, and might be good since you need vacuum seals) stand up to radiation. A clever design might work, but here you really should talk with someone experienced with radiation damage etc., perhaps a nuclear engineer.

Here's a thought. The gathering electrode could be a spherical shell of something sturdy and inert- platinum would be great, but expensive. A hole in it, perhaps about 1 cm diameter would be needed. Through it there would go an insulating rod, sealed to the shell to maintain vacuum. At the end of the rod, in the middle of the shell, the radioactive source would sit. There would have to be a wire from it to the outside through the middle of the insulating rod. Probably the rod would need some sort of guard ring on it inside the shell to protect the region of the seal, where electrical breakdown is most likely, from radiation.

Using the shell would both provide protection and gather the maximum amount of current, allowing use of minimal amounts of radioactivity.

Then if all worked well you might have about 1kV (maybe more?) between that wire and the conducting shell. If you can ground either side, the shell is obviously the part to ground, leaving the wire as the hot lead.

It doesn't sound easy, but I wouldn't be shocked (unlike that time with the 900V) if some high-tech group could get it to work. Maybe the insulating rod would have to be thicker if you want higher voltage.

Of course, there also can be safety issues working with radioactive sources and with high voltages. So please consult someone with more detailed knowledge. But this is fun to think about.

Mike W.

p.s. Here's a maybe more basic issue. A typical smoke detector has about 0.3 μg of ²⁴¹Am. That's about 10^-9 moles or 10¹⁵ atoms. With a decay time of about 10¹⁰ sec, you only get a current of around 10⁵ alphas/sec. With each alpha having a charge of about 3*10^-19 C, you only get a current around 3*10^-14 A. You wanted about 10⁵ times that. You'd need a pretty big, and unsafe, amount of ²⁴¹Am to power this.

(published on 11/12/2012)