Electricity from nuclear decay
Can electricity be generated directly from nuclear decay?
It occurred to me that in alpha decay two protons fly off rapidly, leaving their associated electrons behind. If the alpha particle was captured, perhaps by a conducting shell around a lump of the decaying element then a usable current might flow between the shell and the lump.
The current is clearly going to be minute, but the potential difference might be high enough to provide useful energy anyway.
1 answer
According to this site, 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
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 in our tube. For our arbitrary 1 centimeter radius tube, 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, 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 and magnetically isolated from one another with some sort of diamagnetic, maybe mu metal. This system would produce about 0.1 volt and 16 amps passively for about 4 billion years.
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