WHY EXPECT NEUTRONS?
NEN Editorial by Hal Fox
The story of cold fusion includes the hypothesis of the dead graduate student. Graduate students are assigned to monitor tests. If cold fusion worked as expected, then there would be lots of neutrons released and those who closely monitored the tests (most probably a graduate student) would be severely harmed by neutron radiation. Following the classical gas plasma branching ratio for deuterium plus deuterium fusion, there is about a fifty percent probability that a d+d fusion will produce neutrons. However, in the over 600 papers reporting on successful replication of the Pons-Fleischmann cold fusion (or variations and improvements), very small amounts of neutrons were observed. These experimental results coupled with the strong belief that metal lattice fusion must be like gas-plasma fusion has lead to considerable criticism of cold fusion experiments.
Now we are finding that there is a similar criticism of low-energy nuclear reactions, such as for the latest invention by Stan Gleeson of the Cincinnati Group. The purpose of this article is to investigate the scientific reasoning behind the perceived need for there to be neutrons emitted. Because the current Gleeson demonstrations make use of naturally radioactive thorium, consider the natural radioactive decay process for thorium:
Thorium Nitrate, Th(NO3)4, is used because it is very soluble in water. Thorium, in nature, consists of about 100% 90Th232 which is mildly radioactive with a half life of 1.4 x 1010 years. The entire chain of radioactive thorium decay is:
90Th232 -> 88Ra228 -> 89Ac228 -> 90Th228 -> 88Ra224 ->
86Rn220 -> 84Po216 -> 82Pb212 -> 83Bi212 -> 84Po212 ->
or 81Tl208 -> 82Pb208.
If the thorium nitrate were absolutely 100% pure and freshly constituted, one could assume that there would be no daughter products that had time to accumulate. However, the radioactive decay products are continuously produced by the radioactive decay of thorium. The amount of each daughter product in a prepared sample is a function of previous purity of the thorium and the shelf life of the product. Table 1. provides information on half-lives, decay modes, and energies of emitted particles.
[Table 1 shows these Elements, Half-Liives, Alpha (Mev), Beta (Mev), and Gamma (Mev) Particle Energies.]
Note that 5 of the 11 radioactive elements produce beta particles (high speed electrons), 7 of the 11 produce energetic alpha particles (helium nuclei), and all but one produce gamma emissions (however the Thorium-232 and the Radium-228 gamma emissions are weak, low energy levels). None of these radioactive daughter products produce neutrons. However, there are two neutrons and two protons in each alpha particle.
It is the current working hypothesis, subject to further experimental verification, that during the processing of the thorium, the Gleeson invention injects protons (one or more) into the thorium nuclei; the thorium nuclei becomes immediately unstable; spontaneous fission occurs; and the thorium plus proton combination breaks into two parts to produce stable elements.
Conservation of Baryon Number
One of the several conservation rules that must be obeyed (according to classical physics) is the conservation of baryon number. A baryon is an elementary particle of the nuclei and any nuclear reaction must conserve (same number before and after) the baryon number during any nuclear reaction. The major baryons are protons and neutrons. There are other relatively scarce baryons that occur when breaking up nuclei under high energy particle accelerators. For our discussion, we must have a nuclear reaction (a kind of an equation) in which the number of neutrons plus protons are the same on both sides of the nuclear reaction equation.
90Th232 + 1H1 xAy + uBv + Q (energy emitted or absorbed)
To preserve baryon numbers 90 + 1 must equal x + u, and 232 + 1 must equal y + v. The energy Q is computed from the conservation of energy and involves the balancing of mass and energy on both sides of the equation by use of E = mc2 formula to equate loss (or gain) of mass with a gain (or loss) of energy. To be precise, we should include the possibility of a neutron or an alpha particle being emitted on the right side of the above equation. The point of this article is to determine if we must have neutrons to balance such an equation.
If we look at a table of Radio-Nuclides, a table of all of the possible stable and unstable isotopes of all of the elements, we would find that there are many thorium plus proton reactions that are possible. By 'possible', it is meant that the reactions obey all six of the conservation rules (baryon number, energy, spin, iso-spin, charge, and parity.) However, nature does not work just on possible reactions. Nature has a way of creating the most probable reactions. At the present time, we lack a detailed understanding of the nature of the most probable reactions and will require substantial experimental evidence to determine if any of the nuclear models (such as the nucleon cluster model or the alpha-particle model, or the liquid drop model, etc.) are correct.
The latest experimental evidence (based on the new Gleeson process) indicates that most of the thorium is reduced to Cu-65 and Titanium. Begin with the left hand side of the nuclear reaction equation as given above and substitute copper and titanium for the right hand side and we have:
90Th232 + 1H1 -> 29Cu65 + 22Tiv + pXq
Note: due to experimental evidence that Cu-65 is predominant in the measured elements created, we have replaced y with 65 in the above reaction equation.
There are five stable isotopes of titanium so that v can be 46, 47, 48, 49, or 50. The equation does not balance with baryon numbers so an unknown, pXq, has been added. The next step is to determine the values for p and q to balance the equation: p must equal 90 + 1 -29 - 22 or 40. q must equal 232 + 1 - 65 - (46, 47, 48, 49, or 50) or q must range from 122 to 118. Looking at the table of radio-nuclides, Zr (p = 40) has stable elements of Zr-92, Zr-94, & Zr-96. However, q should be in the range of 118 to 122. Therefore, for this reaction to occur we would have to account for at least 118 - 96 or 22 neutrons. This is an example of why many nuclear scientists would immediately ask about the neutrons from these reactions, and about a possible dead graduate student.
If the experimental evidence indicates that there are few, if any neutrons, and if the evidence indicates that there is indeed Cu-65 produced, how can the lack of neutrons be explained? Here is one possibility: there is a nuclear reaction labeled as §-, or beta decay. Under beta decay (§-) the original nucleus loses an electron and creates a proton. Therefore, one missing neutron from the above reaction is accounted for. Looking at Cu-65, what is the possibility of a chain of §- decays? A possible chain would be Ni-65 (half-life of 2.517 hours), Co-65 (half-life of 1.25 seconds), Fe-65 (extremely short half-life), Mn-65 (extremely short half-life). However, this chain of beta-decays would account for a missing neutron at each step.
Next, examine the table of nuclides for possible beta-decay processes for producing Titanium. Ti-49 could have a beta-decay chain of Sc-49 (57.3 mins), Ca-49 (8.72 min), K-49 (1.26 sec). Ti-50 could have a beta-decay chain of Sc-50 (1.71 min), Ca-50 (14 sec), and K-50 (472 msec). Each of these steps account for a missing neutron.
It is important to know that the table of nuclides has been experimentally established mainly with the use of high-energy particle bombardment of isotopes. It may be that we do not, as yet, fully understand all of the variations that can occur in this new type of nuclear reactions. However, it would be surprising if baryon number were not conserved. It is more likely that there are more extended chains of §- processes that we have not, as yet, included in our table of nuclides. Alternatively (as suggested by Dr. S.X. Jin), there could be more complex chains of reactions which do not produce neutrons, specifically, proton bombardment can produce an alpha particle and another element.
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August 19, 1997.