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
60Ni28 + 2 p* --> 62Ni28 + 2 v + 16.852 MeV
60Ni28 + 2 p* --> 58Ni28 + 4He2 + 7.909 MeV
60Ni28 + 2 p* --> 61Ni28 + 1H1 + v + 7.038 MeV
61Ni28 + 2 p* --> 62Ni28 + 1H1 + v + 9.814 MeV
62Ni28 + 2 p* --> 64Ni28 + 2 v + 14.931 Mev
62Ni28 + 2 p* --> 64Zn30 + 13.835 MeV
62Ni28 + 2 p* --> 60Ni28 + 4He2 + 9.879 MeV
62Ni28 + 2 p* --> 63Cu29 + 1H1 + 6.122 MeV
62Ni28 + 2 p* --> 59Co27 + 4He2 + 1H1 + 00.346 MeV
64Ni28 + 2 p* --> 66Zn30 + 16.378 MeV
64Ni28 + 2 p* --> 62Ni28 + 4He2 + 11.800 MeV
64Ni28 + 2 p* --> 65Cu29 + 1H1 + 7.453 MeV
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
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.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
