Building a Betavoltaic battery

  • Thread starter wil3
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Main Question or Discussion Point

Hello, I would like to start by saying that I know very little about nuclear decay, so please do not flame me if this is a silly question:

I am wondering if it is possible to generate enough potential to light an LED indefinitely using pure beta decay. Let's say the circuit consists of a beta source, such as potassium chloride, crushed up and mixed with a continuous piece of aluminum foil in such a way as to maximize the surface area that the potassium chloride touches.

Let's say that I then attached the aluminum foil to an LED, and connected the other end to ground. Is it theoretically possible that the electrons from the beta decay process could light the LED? About how much potassium chloride (such as salt substitute) would be necessary? I am not looking for efficiency, just a demonstration.
 

Answers and Replies

  • #2
It's possible to light an LED with a betavoltaic battery.

I don't think its possible to make a betavoltaic battery with potassium chloride and aluminum foil. Potassium chloride is not much of a beta source.
 
  • #3
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I was thinking about that- are there any other easily obtainable beta sources?
 
  • #4
I don't know, I guess you'll have to do some research on wikipedia.
 
  • #5
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Wiki fails me on this front. Thank you very much for your reply.
 
  • #6
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A typical LED operates at about 2 V forward voltage and 20 mA forward current.
That's 40 mW of electrical power.

The beta decay energy of K-40 is 1.31 MeV.
(Let's assume that the antineutrinos don't exist and the electron carries off 100% of the energy.)

If we make an extra-conservative assumption that we can capture and use 100% of the beta energy, we need 190 GBq of K-40.

This corresponds to 716 kg of pure K-40, or 6000 tonnes of natural potassium.

:)
 
  • #7
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Haha. Assuming minerva is right...that's awesome.
 
  • #8
QuantumPion
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A typical LED operates at about 2 V forward voltage and 20 mA forward current.
That's 40 mW of electrical power.

The beta decay energy of K-40 is 1.31 MeV.
(Let's assume that the antineutrinos don't exist and the electron carries off 100% of the energy.)

If we make an extra-conservative assumption that we can capture and use 100% of the beta energy, we need 190 GBq of K-40.

This corresponds to 716 kg of pure K-40, or 6000 tonnes of natural potassium.

:)
Yeah but if you took that 716 kg of K-40 and mixed it with an olympic-sized swimming pool of phosphorous you could light up a city with the radioluminescence. I think that just goes to show you how inefficient it is to try to capture betas for electrical current.
 
  • #9
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For a more realistic example, we might consider strontium-90 as the beta source.
Making similar extra-conservative spherical-frictionless-cow assumptions as per the above, to put 40 mW through the LED would require 12 millicuries (87 micrograms) of Sr-90, which is a much more plausibly sized radioactive source.
 
  • #10
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For a more realistic example, we might consider strontium-90 as the beta source.
Making similar extra-conservative spherical-frictionless-cow assumptions as per the above, to put 40 mW through the LED would require 12 millicuries (87 micrograms) of Sr-90, which is a much more plausibly sized radioactive source.
That's with the secondary decay of Y-90, right? I came up with 174 μg with total specific power:
Pd (decay power) for Sr-90/Y-90 = (0.546 MeV/decay + 2.28 MeV/decay)
hl = 28.2 yr
N(t) = N₀ e^(-ln2 * t/hl)
N₀ = 1 decay/90 amu
Rd (decay rate) = [as t->0] (N₀ - N(t))/t = 5.014 TBq / g
Ps (specific power) = Pd * Rd = 2.31 W/g (0.446 W/g[Sr-90] + 1.864 W/g[Y-90])

Yeah but if you took that 716 kg of K-40 and mixed it with an olympic-sized swimming pool of phosphorous you could light up a city with the radioluminescence.
I don't think so. For the same calculations above, K-40 has a specific power of:
Pd (decay power) for K-40 = (1.31 MeV / decay)
hl = 1.2483 Gyr
N(t) = N₀ e^(-ln2 * t/hl)
N₀ = 1 decay/40 amu
Rd (decay rate) = [as t->0] (N₀ - N(t))/t = 271.61 kBq / g
Ps (specific power) = Pd * Rd = 57 nW/g
Po (output power) = Ps * m = 57 nW/g * 716 kg = 40 mW

40 mW would not light up a city (and is not surprising, as 40mW was the target to which the 716 kg was aimed) - and that's even assuming that the phosphor used (which, ironically, would not likely contain any phosphorous) is 100% efficient in converting induced energy into light.

The unfortunate thing about betavoltaics is that the higher their power density desired, the more radioactive the isotope you have to use to get it, and the more damage you are going to do to your semiconductor matrix.

As such, most of the research in this area is in induced beta decay of low energy isotopes (e.g., if you could make K-40 decay quickly, rather than over billions of years, you'd get some nice power density, and you'd have some control over dissipation), and in making beta-damage-resistant semiconductors (so that the cells will last at least a half-life of the fuel).

Anyway, I love them conceptually; a Lithium-Ion battery gets about 0.34 W/g, so, even with low efficiency and mass sacrificed to semiconductors, you could basically exploit the laws of physics to have a battery you'd have to replace once in 20-30 years. Not only that, but it provides a valuable destination for one of the more concerning components of spent nuclear fuel.

Neat, right? Totally worth the research money.
 

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