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Hot Fusion
Certain small atomic nuclei can combine to make a larger nucleus, releasing energy. For example \[ {}^2\text{H} + {}^2\text{H} \rightarrow {}^4\text{He} + 23.8\, \text{MeV}. \] Alternatively they can transfer a neutron or proton from one to the other to make two new nuclei, \[ \begin{aligned} {}^2\text{H} + {}^2\text{H} & \rightarrow {}^3\text{H} + {}^1\text{H} + 4.0\, \text{MeV} & \text{(neutron transfer)} \\ {}^2\text{H} + {}^2\text{H} & \rightarrow {}^3\text{H} + \text{n} + 3.3\, \text{MeV}. & \text{(proton transfer)} \end{aligned} \] These reactions conserve neutrons, protons, and electrons. They do not occur at low temperature because of the coulomb barrier that inhibits charged nuclei from approaching close enough to react. But they do occur at high temperature where thermal agitation thrusts the reactants closer together. These reactions are termed “hot fusion” reactions.
Cold Fusion
Fusion reactions can have more than two reactants and more than two products, and need not conserve neutrons, protons, or electrons provided the total number of nucleons and the total electric charge are conserved. An example is \[ {}^2\text{H} + {}^2\text{H} + {}^2\text{H} + {}^2\text{H} \rightarrow {}^1\text{H} + {}^1\text{H} + {}^1\text{H} + {}^1\text{H} + {}^4\text{He} + 21.0\,\text{MeV}. \] In more compact notation this becomes \[ 4\, ({}^2\text{H}) \rightarrow 4\, ({}^1\text{H}) + {}^4\text{He} + 21.0 \, \text{MeV}. \qquad\qquad\text{(cold fusion)} \]
Hypotheses:
- This reaction can occur at low temperatures in a series of catalytic reactions and beta decays for which there are no coulomb barriers.
- Neutron isotopes are the catalysts of these reactions.
The remainder of this page is organized as follows: First the catalyst is introduced and then the sequence of reactions that produce cold fusion are described.
Neutron Isotopes (Polyneutrons)
The chemical properties of an element are determined by the number of protons in each of its atomic nuclei. Each element has isotopes that differ by the number of neutrons per nucleus. Hydrogen has three isotopes \({}^1\text{H}\), \({}^2\text{H}\), \({}^3\text{H}\) containing one proton and respectively 0, 1, 2 neutrons. Isotope \({}^3\text{H}\) is unstable against beta decay to \({}^3\text{He}\).
The neutron is not charged and does not participate in ordinary chemistry. But it does participate significantly in nuclear chemistry (e.g. in nuclear fission reactions), and it can in principle have isotopes \({}^2\text{n}\), \({}^3\text{n}\), \({}^4\text{n}\), …, \({}^A\text{n}\), … analogous to the hydrogen isotopes. At present only the isotope \({}^1\text{n}\) (the neutron) has been confirmed to exist. Experiment has shown that \({}^2\text{n}\) and \({}^3\text{n}\) are not bound. Evidence for \({}^4\text{n}\) has been published but has not been confirmed.
Hypotheses:
- Larger neutron isotopes are bound and that they are stable except for beta decay.
- I call these stable neutron isotopes polyneutrons.
- The polyneutrons involved in cold fusion reactions have neutron numbers \(\text{A}\) larger than about 50.
- Their masses can be approximated by a liquid-drop model having a binding energy proportional to \(\text{A}\) and a surface energy proportional to \(\text{A}^{2/3}\).
Neutron Transfer Reactions
Polyneutron reactions are uninhibited by a coulomb barrier, and like single neutrons polyneutrons are highly reactive. Consider the reaction of a polyneutron with deuterium, \[ {}^\text{A}\text{n} + {}^2\text{H} \rightarrow {}^{\text{A}+1}\text{n} + {}^1\text{H}. \] I assume the reaction is exothermic (i.e. it proceeds with the emission of energy), which puts a constraint on the allowable parameters of the liquid-drop model. Physically the transferred neutron must be more strongly attached to the polyneutron than to the deuterium nucleus from which it came.
The \({}^{\text{A}+1}\text{n}\) can interact with another \({}^2\text{H}\), \[ {}^{\text{A}+1}\text{n} + {}^2\text{H} \rightarrow {}^{\text{A}+2}\text{n} + {}^1\text{H}, \] and so on. After four such reactions we have overall \[ {}^\text{A}\text{n} + {}^2\text{H} + {}^2\text{H} + {}^2\text{H} + {}^2\text{H} \rightarrow {}^{\text{A}+4}\text{n} + {}^1\text{H} + {}^1\text{H} + {}^1\text{H} + {}^1\text{H}. \] In more compact notation this becomes \[ {}^\text{A}\text{n} + 4\,({}^2\text{H}) \rightarrow {}^{\text{A}+4}\text{n} + 4\, ({}^1\text{H}) . \qquad\qquad \text{(neutron transfer)} \]
Polyneutron Beta Decay
Free neutrons are unstable against beta decay—in which a neutron changes into a proton plus an electron plus an antineutrino. This can be written \[ \text{n} \rightarrow {}^1\text{H} + 0.78\, \text{MeV}. \] The antineutrino has near-zero mass and is not written down. The released \(0.78\, \text{MeV}\) energy is shared among the ionization of the \({}^1\text{H}\) atom, the kinetic energies of the proton and the electron, and energy of the antineutrino.
A polyneutron undergoes beta decay when one of its constituent neutrons changes into a proton. I write this reaction \[ {}^\text{A}\text{n} \rightarrow {}^\text{A}\text{H} + 0.78\, \text{MeV}. \qquad\qquad\text{(polyneutron beta decay)} \] The proton is not released as a free proton, which would require considerable energy, but it remains as a bound nucleon interacting in the same wave function with the same nuclear symmetry and nuclear interaction energy as the neutron it replaced. I refer to \({}^\text{A}\text{H}\) as polyhydrogen.
Polyhydrogen Beta Decay
The polyhydrogen nucleus \({}^\text{A}\text{H}\) contains \(\text{A}-1\) neutrons, any one of which can undergo beta decay to generate polyhelium \[ {}^\text{A}\text{H} \rightarrow {}^\text{A}\text{He} + 0.78\, \text{MeV} - \text{E}_c \qquad\qquad\text{(polyhydrogen beta decay)} \] where both protons in \({}^\text{A}\text{He}\) share the nuclear symmetry and contribute to the nuclear portion of the energy the same as though they were neutrons. Now however their mutual coulomb energy \(\text{E}_c\) increases the internal energy of the nucleus and reduces the energy that is liberated in the decay. I assume that \(\text{E}_c < 0.78\, \text{MeV}\) and in consequence the reaction is exothermic.
Polyhelium Alpha Decay
Polyhelium \({}^\text{A}\text{He}\) is a massive helium isotope with two protons and \(\text{A}-2\) neutrons. It can undergo alpha decay in which the two protons join with two neutrons to form a nucleus of ordinary helium, \({}^4\text{He}\), that splits off from the \({}^\text{A}\text{He}\) leaving behind \({}^{\text{A}-4}\text{n}\), \[ {}^\text{A}\text{He} \rightarrow {}^4\text{He} + {}^{\text{A}-4}\text{n} + \text{E}_\alpha. \qquad\qquad\text{(polyhelium alpha decay)} \] I assume that this reaction is exothermic with positive \(\text{E}_\alpha\) and that polyhelium is unstable with respect to alpha decay.
Steady State Reactions: Cold Fusion
Consider the following sequence of reactions: \[ \begin{aligned} {}^\text{A}\text{n} + 4\,({}^2\text{H}) &\rightarrow {}^{\text{A}+4}\text{n} + 4\, ({}^1\text{H}) & \text{(neutron transfer)} \\ {}^{\text{A}+4}\text{n} &\rightarrow {}^{\text{A}+4}\text{H} & \text{(polyneutron beta decay)} \\ {}^{\text{A}+4}\text{H} &\rightarrow {}^{\text{A}+4}\text{He} & \text{(polyhydrogen beta decay)} \\ {}^{\text{A}+4}\text{He} &\rightarrow {}^4\text{He} + {}^\text{A}\text{n}. & \text{(polyhelium alpha decay)} \end{aligned} \] Overall they accomplish \[ 4\,({}^2\text{H}) \rightarrow 4\,({}^1\text{H}) + {}^4\text{He} + 21.0 \, \text{MeV} \qquad\qquad\text{(cold fusion)} \] which is just the cold fusion reaction stated at the beginning.
Notice that the initial polyneutron in the neutron transfer reaction is restored in the polyhelium alpha decay so that there is no net change in polyneutron size. The polyneutron has served as a catalyst.
Although we do not have accurate values for the energies (i.e. masses) of polyneutrons, polyhydrogen, or polyhelium, these energies do not affect the overall reaction energy, which depends only on the known energies of \({}^2\text{H}\), \({}^1\text{H}\), and \({}^4\text{He}\). We can conclude that in any steady state reaction (or more generally in any reaction where polyneutron sizes and numbers are the same at the beginning of measurement as at the end of measurement) the energy released in cold fusion of deuterium is 21.0 MeV per helium atom produced. (Because the energy of the two emitted antineutrinos is not measurable, the observed energy per helium atom is expected to be about 20 MeV.)