IMAGINARY PHYSICS FROM THE BETAVOLTAIC FOLKS

Eric Krieg and Shawn Bishop

On the Betavoltaic home-page , we are told:

The isotope Potassium-40 or K40 if you will ejects a 1.311 MeV beta-electron and has a half-life of 1.7 billion years. There is enough power in 1 kilo of K40 to equal 35,000 gallons of gasoline. Also because it is so long lived it has a low specific activity. Making it pretty safe and easy to shield against exposure from its low level of natural radiation. You probably consumed some small amount of K40 in the last banana you had for breakfast. Because it releases its radiation so slowly and over such a long time period it is mostly harmless to us in small naturally occurring amounts.

The Aluminum in a gum wrapper will shield the beta-electrons ejected from this isotope.

The half-life of 40K is 1.28 billion years, not 1.7 billion. But this is a minor inconsistency. The more serious issue is the false statement in the final sentence regarding the gum wrapper. NIST has a great website (the ESTAR program) where one can produce calculated stopping powers of electrons through various materials.

I've used it to generate the included figure of the stopping power of electrons through aluminum.

A straightforward calculation, using the masses of the nuclei involved in the decay of 40K --> e + 40Ca, yields an average kinetic energy for the electron (E) of 0.8 MeV. (average because the decay is 3-body involving a neutrino, thus allowing the electron's kinetic energy to take on a spectrum of values)

The graph shows that the areals stopping power (to be thought of as the kinetic energy loss through a mono-layer of the material being traversed) for an electron at this energy in Al is around 1.5 MeV-cm2/g. We can convert this into more "household" units by multiplying the areal stopping power by the density of aluminum to obtain the stopping power per unit length travelled by the electron. The density of Al = 2.7g/cm3.

Multiplying the areal stopping power by this number gives us an electron stopping power per unit length of about 4 MeV/cm. That is, for each 1 cm of travel in Al, the electron will lose about 4 MeV of kinetic energy. Note also that we're working in the area of the minimum of the curve, and so this rough calculation is actually going to be reasonably insensitive to the slope of the stopping power function.

We can use this number to work out the approximate thickness, D, of Al required to stop a 0.8 MeV electron: D = 0.8 / 4 = 0.2 cm, or 2 mm of Al.

Has anyone ever seen bubble-gum wrap that is a full 2 mm thick? That's thicker than the walls of a beer can; in fact it's about ten times thicker.

Browsing the PlasmaVolt/Betavoltaic web site, we find this statement on enhancing the radioactive decay rate of a nucleus:

This goes against the conventional wisdom but has never been given a fair chance to be proven or disapproved. We intend to make this one of our first priorities upon completion of initial funding of our company. To date his theory has only received 2 tests and both confirmed his theory. We have contacted the Monolithic technology team at ORNL who have agreed in principle to provide a set of proofs of his theory after we have obtained initial funding.

Besides Dr. Santilli's breakthrough work on accelerated decay of an isotope there are some other peer reviewed scientific papers of note that cover some of the theory of accelerated decay of an isotope. These are sited below:

Friar,J. L., and Reiss, Howard R., Modification of Nuclear Beta Decay by Intense Low-Frequency Electromagnetic Waves, Physical Review C, Vol. 36, No. 1, July 1987, p. 283

We've looked up that paper. It is a purely theoretical paper in which the authors explore the effects of high intensity, low frequency, electromagnetic waves on the beta decay rate of a nucleus. The conclusion the authors reach upon concluding their calculations, is summarized in the abstract of the paper thusly:

...integration of the electron spectrum for the nonrelativistic (electron) case demonstrates that there is no field-dependent enhancement of the total rate in the long-wavelength limit, which agrees with some previous conclusions. A classical argument is presented which demonstrates in all of our cases that there is no enhancement of the total rate in that limit.

There is no theoretical enhancement of the decay rate!

It is misrepresentative for the proponents of the betavoltaic device to tout this paper as giving scientific weight to their claims. The paper, in fact, flatly contradicts those claims; the theoretical calculations presented show there is no enhancement whatever of the beta decay rate as the result of exposing nuclei to EM fields. Further, there is as yet not one shred of experimental evidence to indicate that these rates can be changed. Even more bizarre, Betavoltaics does not even propose any methods by which they intend to approach the problem of enhancing beta decay rates, making the citation to this paper all the more a "red herring".

Another page on the site displays "an animation of the Plasma Vortex Fusion device in action." We are told that fusion between a 24Mg nuclei and 9Be nuclei is taking place, resulting in the production of 40K.

The second link bespeaks their ignorance of what they are dealing with. They state: "This fusion device produces no neutron flux or any other harmful radiation." Quite right on the first account of neutrons. Quite wrong on the second account of "other harmful radiation". That device will be producing copious quantities of X-rays; neither of us would willingly choose to stand close to it during operation. Skeptical about there being X-rays? Then perhaps you should look at the story of Roentgen's simple experiment that led to his discovery of X-rays and a subsequent Nobel Prize.

More non sequiter from their homepage:

Potassium-40 is produced in our lab using the plasmavolt transmutation device at a high rate by causing a low energy transmutation of Magnesium, and Beryllium-9, into Potassium-40 and hydrogen. These elements are caused to fuse at a high efficiency within the hydrogen plasma vortex of the device.

To recognize the nonsense of this statement, one only need know simple addition. There are 3 stable isotopes of Mg: 24Mg, 25Mg, and 26Mg, with natural abundances of 79%, 10% and 11% respectively. The only stable isotope of beryllium is, indeed, 9Be.

Let's do the math: 24Mg/25Mg/26Mg + 9Be --> 33S/34S/35S, which is 7,6, or 5 nucleons short of the forty required to reach 40K, to say nothing of the immense Coulomb barrier one would need to overcome just to get this reaction to go. One of us (Bishop) has done nuclear reactions at a national laboratory, fusing a proton with a 21Na nucleus. We needed a 21Na beam with about 4.5 MeV of energy to observe any yield, and the yield observed was around six 22Mg fusion nuclei for every trillion 21Na nuclei fired into our hydrogen gas target. The Coulomb barrier for the 21Na reaction was about 3.5 MeV. Bishop's reaction, it must be stressed, took place at centre of mass energies approximately 10 times lower than this Coulomb barrier! His reaction relied completely on the phenomenon of quantum tunnelling. By comparison, the Coulomb barrier for the 24Mg + 9Be --> 33S reaction is about 12 MeV.

The probability of a reaction occurring, from quantum mechanical considerations, is proportional to the exponential of the quantity: -0.046 x Z1 Z2 c/v, where Z1 and Z2 are the nuclear charges of each reacting particle, c is the speed of light and v is the relative velocity between the two reacting particles in the centre of mass frame. For Bishop's reaction, this quantity was -1.17. Now, let's consider the case of 24Mg + 9Be --> 33S with optimistic considerations. Let us allow them the full benefit of the doubt. Since their Coulomb barrier is 12 MeV, let us consider the case where the 24Mg and 9Be collide head-on with a centre of mass energy of 12 MeV. This is the most optimistic case for two reasons:

What is the exponential factor in this most optimistic of cases? My calculation gives it as -21.1. Thus, the relative probability of a reaction occurring in their bottle to a reaction occurring in Bishop's experiment is proportional to:

exp(-21.1)/exp(-1.17) = 2.2 E-9

Let us think about this result. It is telling us, roughly, that for every chance there is for a reaction to occur in Bishop's experiment, there is 2 chances in a billion of a reaction occurring in Beltavoltaics bottle! And the reason for this is that the Coulomb barrier in the 24Mg + 9Be --> 33S reaction greatly diminishes the probability of a reaction occurring! Considering that Bishop's experiment had about 6 fusion reactions for every trillion 21Na nuclei fired at his hydrogen target, this numerical result implies that Betavoltaic would require a billion-trillion reaction events just to produce just one sulfur-33 nucleus! To go on from here to obtain 40K requires yet more fusion reactions through even larger Coulomb barriers than the preceeding one! Betavoltaic's statement of producing 40K with "high efficiency" is absurd to say the least! And this is the most optimistic estimate, to say nothing of how their table-top glass bottle would be able to hold 10's of millions of volts to accelerate the particles to the necessary energies!

National nuclear and particle physics laboratories around the wold have spent hundreds of millions of dollars to equip themselves with the necessary accelerators to provide the energies sufficient for studying nuclear reactions. These machines can be anywhere from 5 meters in length, to kilometers in length. We're supposed to believe that a glass bottle with a solenoid in its neck, that is capable of generating impressive-looking sparks, is capable of nuclear fusion on a scale vastly in excess of what the national laboratories can do? Vastly in excess of what Supernovae and Novae can do; some of the most powerful explosions in the cosmos?

The claims made of the device are untenable and are replete with errors in scientific understanding of what it is they desire to achieve and errors in understanding the science of what they claim to be doing.

Note: A person from Betavolt says that all the lines on their original page that we comment about are not even on the web site

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