Hi Metatron,
Your post sounded very interesting to me at first (as virtually all of them do), but after I read the linked website (BetaVoltaic) some red flags went up. Early in the page-1 description the author said this:
"The limitation that this cell had was that only very low powered isotopes could be used. This is due to the types of semiconductors that would accept a P and N type dopant was limited to silicone and germanium for the most part."
Being a EE who deals with silicon every day, it struck me that anyone who can manage to mispell this word who claims actually to be working with silicon devices such as pn junctions cannot possibly have real working knowledge of electron devices.
After reading some more, I came upon this statement about powering electronic devices via beta-decay of the isotope beryllium-10:
"A microgram of this naturaly abundant isotope would power an electronic device for its entire expected lifetime."
This is nonsense, and I will show the calculations which prove it. Basically, I calculate the number of electrons produced per second by Be-10 decay, and convert this to an equivalent electrical current.
First we calculate the time constant of the exponential radioactive decay. With a half-life for Be-10 of 1.5 million years, this works out to be tau = 6.829e13 seconds. I.e., if you plug this number into the equation: Quantity = exp( - t / tau ) with t = 1.5 million years worth of seconds (1.5e6 * 365.25 * 24 * 60 * 60), you get Quantity = 1/2, which is the definition of "half-life".
Next we calculate the fraction of material lost in one second due to this decay: Fraction lost = exp( -1 / tau) = 1.464e-14 per second.
Next we calculate the number of atoms in a microgram of Be-10: A mole of Be-10 contains 6.022e23 (Avogadro's number) atoms, a mole of Be-10 weighs about 10 grams, so: Number of atoms = (1e-6 gm / 10 gm) * 6.022e23 = 6.022e16 atoms.
As each atom decays, it emits one beta-electron, which according to BetaVoltaic, forms a sort of electrical current when harnessed properly. So the number of electrons available per second is equal to the number of atoms in one microgram of Be-10 multiplied by the fraction lost per second: 6.022e16 * 1.464e-14 = 880 electrons per second.
Finally we divide the number of electrons emitted per second by the number of electrons in one coulomb of charge, to get a current (since current in Amperes is by definition 1 coulomb per second): 880 electrons per second / 6.24e18 electrons per coulomb = 1.4e-16 coulombs per second = 1.4e-16 Amperes. This is a very small current.
Now, the BetaVoltaic website claims that they know how to speed up the decay of Be-10 to get a half-life of only a few years. Let's say they can speed it up to get a half-life of only 1.5 years. Then the rate of electrons produced per second would go up by a factor of a million, and the current available from the decay would go up by a factor of a million. Thus we might theoretically have available 1e6 * 1.4e-16 = 1.4e-10 Amperes. In EE language this is 0.14 nanoamps -- still an extremely small current.
Now, I know of no electron devices that operate at this level of current. In fact, currents this small tend to be leakage currents (unwanted stray currents). Leakage currents in typical CMOS transistors tend to be at least 1 nanoamp and can easily be 100 or more times larger, and 10 to 100 times higher still if temperature is high. That's why a typical Intel microprocesser has a leakage current of up to 10 Amps when it's completely shut off! Which means that for each of the 50-some-odd million transistors on the Pentium-4, each one has a leakage of some 200 nanoamps.
So it appears that BetaVoltaic's ideas are complete nonsense. I have no doubt that these people have never even tried to produce a working demonstration. This whole idea sounds to me like trying to make a perpetual motion machine.
Other problems: how does one manage to gather all the Be-10 necessary to power things on the vast scale that would be necessary? Do they have the slightest idea how difficult it is to collect and purify the stuff? I think not. Do they have any idea how difficult it is to handle radioactive material? I don't think so.
Sorry to rain on your parade, Metatron, but I don't want to see you go down the garden path with this nonsense.
AlanF