Deflation fusion theory provides a potential solution to the riddle
of why the radioactive byproducts 59CU29, 61Cu29 and 62Cu29 to the Ni
+ p reactions do not appear in Rossi's byproducts. This solution of
the specific problem byproducts is manifest if the following rules
are obeyed by the environment, except in extremely improbable instances:
1. The initial wavefunction collapse involves the Ni nucleus
plus two p*
2. As with all LENR, radioactive byproducts are energetically
disallowed.
Here p* represents a deflated hydrogen atom, consisting of a proton
and electron in a magnetically bound orbital, and v represents a
neutrino.
The above two rules result in the following energetically feasible
reactions:
58Ni28 + 2 p* --> 60Ni28 + 2 v + 18.822 MeV [-0.085]
60Ni28 + 2 p* --> 62Ni28 + 2 v + 16.852 MeV [-1.842]
60Ni28 + 2 p* --> 58Ni28 + 4He2 + 7.909 MeV [-10.786]
60Ni28 + 2 p* --> 61Ni28 + 1H1 + v + 7.038 MeV [-11.657]
61Ni28 + 2 p* --> 62Ni28 + 1H1 + v + 9.814 MeV [-8.777]
62Ni28 + 2 p* --> 64Ni28 + 2 v + 14.931 Mev [-3.560]
62Ni28 + 2 p* --> 64Zn30 + 13.835 MeV [-4.656]
62Ni28 + 2 p* --> 60Ni28 + 4He2 + 9.879 MeV [-8.612]
62Ni28 + 2 p* --> 63Cu29 + 1H1 + 6.122 MeV [-12.369]
62Ni28 + 2 p* --> 59Co27 + 4He2 + 1H1 + 00.346 MeV [-18.145]
64Ni28 + 2 p* --> 66Zn30 + 16.378 MeV [-1.918]
64Ni28 + 2 p* --> 62Ni28 + 4He2 + 11.800 MeV [-6.497]
64Ni28 + 2 p* --> 65Cu29 + 1H1 + 7.453 MeV [-10.843]
Ni28 + 2 p* ---> 2 1H1 + 0 MeV
Note that in the case where the second p* is rejected and results in
1H1, ultimately a hydrogen atom, that the electron and proton are not
ejected at the same time. The large positive nuclear charge ejects
the proton immediately with approximately 6 MeV kinetic energy.
This kind of zero point energy fueled proton ejection should result
in detectible brehmstrahlung. This energy is in addition to the mass
change energy listed above. The approximately 6 MeV free energy so
gained is made up from the zero point field via uncertainty pressure
expanding any remaining trapped electron's wavefunction. Such energy
may also be obtained from the direct magnetic attraction of a pair of
deflated protons, without the aid of a lattice nucleus. This is of
the form:
p* + P* --> 2 1H1
However, the repulsion of a proton from a proton is far less than
from a large nucleus, and the electrons in this case are not trapped
when the protons separate. However, some EuV radiation can be
expected from the ensemble breakup. A very very small rate of pep
reactions may occur:
p + p* --> D + e+ v + 0.42 MeV
These are followed immediately by:
e- + e+ --> 2 gamma + 0.59 MeV
and this gamma producing reaction was not observed above background
in the Rossi E-cats.
The following represent energetically feasible initial strong
reactions based on deflation fusion theory:
Compare to 18.822 MeV:
58Ni28 + p* --> 59Cu29 * + 3.419 MeV [-4.867 MeV]
58Ni28 + 2 p* --> 56Ni28 * + 4He2 + 5.829 MeV [-10.650 MeV]
58Ni28 + 2 p* --> 60Zn30 * + 8.538 MeV [-7.941 MeV]
Compare to: 16.852 MeV:
60Ni28 + p* --> 61Cu29 * + 4.801 MeV [-3.394 MeV]
60Ni28 + 2 p* --> 58Ni28 + 4He2 + 7.909 MeV [-8.391 MeV]
60Ni28 + 2 p* --> 62Zn30 * + 11.277 MeV [-5.022 MeV]
Compare to: 9.814 MeV
61Ni28 + p* --> 58Co27 * + 4He2 + 00.489 MeV [-7.661 MeV]
61Ni28 + p* --> 62Cu29 * + 5.866 MeV [-2.284 MeV]
61Ni28 + 2 p* --> 59Ni28 * + 4He2 + 9.088 MeV [-7.125 MeV]
61Ni28 + 2 p* --> 62Cu29 * + 1H1 + 5.866 MeV [-10.347 MeV]
61Ni28 + 2 p* --> 63Zn30 * + 12.570 MeV [-3.643 MeV]
Compare to: 14.931 Mev
62Ni28 + p* --> 59Co27 + 4He2 + 00.346 MeV [-7.760 MeV]
62Ni28 + p* --> 63Cu29 + 6.122 MeV [-1.984 MeV]
62Ni28 + 2 p* --> 64Zn30 + 13.835 MeV [-2.293 MeV]
Compare to: 16.378 MeV
64Ni28 + p* --> 65Cu29 + 7.453 MeV [-0.569 MeV]
64Ni28 + 2 p* --> 66Zn30 + 16.378 MeV [00.415 MeV]
In all cases the net reaction energies of the proposed reactions
exceed those the net energies from reactions that produce radioactive
isotopes. This makes rule 2 reasonable and understandable on an
energy only basis. The mechanism that enforces the rule is more
difficult to understand. Understanding the mechanism requires
understanding the initial energy deficit due to the trapped electron.
This deficit is shown in brackets above. This deficit provides a
limit to how far an energetically ejected electron can travel out of
the coulomb well before being pulled back. If an electron is in the
nucleus at the site of the initial reaction, then a large part of the
energy that normally goes into ejecting a gamma goes into ejecting
the trapped electron. However, given that this energy is
insufficient, the electron has numerous delayed passes through the
nucleus in which to effect a weak reaction. The electron, when
outside the nucleus and accelerating, is free to radiate large
numbers of gammas in much smaller than normal energies. It is also
notable that the electron energy deficits noted are only initial
lower limits. The actual energy deficit can be much higher,
depending on the radius of the deflated proton or deflated quark
involved.
For background on deflation fusion theory see:
http://www.mail-archive.com/vortex-l@eskimo.com/msg59132.html
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
