Post History

According to [this site](https://physics.bu.edu/~duffy/sc546_notes09/alpha_decay.html), the kinetic energy of the two alpha particles emitted is 4.273 MeV.

As you said, these aren't neutral atoms yet with balancing electrons. So, they are a charged plasma.

Imagining, instead of a shielded lump, that you had your sample crushed fine on a dish, (so that the helium has an exit path) contained within a conducting tube to both provide a vent, and collect the electricity by [Lenz's Law](http://www.waynesthisandthat.com/Lenz's%20Law%20Magnet%20Falling%20Through%20a%20Copper%20Tube%20or%20Pipe.html)

Let's try it.

$$ B = {{\mu_0} \over {4 \pi}} {{q v} \over {r^2}} $$

- $\mu_0$ is the permeability of free space at $ 4 \pi \times 10^{-7}$$ $${{N s^2} \over {C}}$
- q is the charge, in the case of helium $ \approx 3.204 \times 10^{-19}$ C
- m is the weight of each atom of the helium plasma ~ 4 atomic mass units (u) ($1.66 \times 10^{-27}$ ${ {kg} \over {u} }$ ) $ \approx 6.64 \times 10^{-27}$ kg
- E is the energy of the plasma. Assuming the 4.273 MeV ($1.602 \times 10^{-13}$ $J \over {MeV}$) is split evenly between the two atoms emitted for each decay reaction, this is $ \approx 3.422 \times 10^{-13} $ J
- c is the speed of light, approx $3 \times 10^8$ ${ {m} \over {s} }$
- v is the velocity of the helium plasma. $ v = \sqrt{{2(E - mc^2)} \over {m}}$ $ \approx 1 \times 10^7 $ $ { {m} \over {s} }$, if I've done the math correctly
- r is the average distance to our conductor; this is arbitrary, but let's try half an inch (1 centimeter = 0.01 m)

So, B from each particle of hot helium plasma is approximately $3.204 \times 10^{-17}$ Teslas

Per Lenz's law, the conducting tube will react to the change in magnetic flux with a flux of it's own

This should [cause the production of a current](http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magloo.html) in our tube.  For [our arbitrary 1 centimeter radius tube](http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c1), the voltage induced should be $V = N B A \div t \approx 1 \times 10^{-20}$ Volts per hydrogen atom, but you can scale this up by turning N into a coiled wire with thousands of turns.

So how often are we getting decay events? 

U-238 has a half-life of 4 billion years ($1.26 \times 10^{17}$ seconds) and a density of 19 grams per cubic centimeter. In our dust layer one centimeter in radius, and one centimeter thick, then, we expect 59.6 grams of material. According to [this resource](https://www.webqc.org/molecular-weight-of-U238.html), the molar mass of U-238 is 56,650 grams; so this layer contains ${ {59.6} \over {56,650} } = 0.001$ moles, or $6.022 \times 10^{20}$ atoms of U-238.

So, we expect $3.022 \times 10^{20}$ decay events (half the sample) in $1.26 \times 10^{17}$ seconds. If I've done all that math correctly, about 2,398 decay events per second. 

For two atoms of helium per event, that'd be $4,796 \times 10^{-20}$ volts per second ($4 \times 10^{-17}$ volts). And a typical induction coil contains a thousand windings, so you would be able to get 1,000 times that ($4 \times 10^{-14}$ volts).

This scales with the cube of fissile material radius (assuming your filling the dish with more U-238), and the inverse square of inductor radius.

Scaling up then, to find out what it would take to get a useful output --

If you had a floor about 20 meters wide (10 m in radius) of fissile material, and a bunch of our 1 centimeter induction tubes, cut down in size to 0.1 millimeters, hanging over the floor... this system would produce about one-tenth of a volt and 16 amps passively for about 4 billion years.