Lou,
 
If W&L works and you capture a neutron to form Ni63 from Ni62 you would have a 
half life of 98.7 years before the beta decay.   There would be a very small 
quantity of Cu63 as a result.

I have made numerous calculations which demonstrate that you get the same 
amount of energy release regardless of the path between two of these points as 
long as you start and end at the same isotopes.  As example, Ni62 + neutron = 
Ni63 plus energy.  Next Ni63 beta decays to Cu63 and releases energy.  The sum 
of these energy releases is exactly the same as if you take Ni62 and add a 
proton and electron to arrive at Cu63.  You must subtract the energy of 
formation for the neutron from the first path to make the balance.

In this beta decay:Ni63 to Cu63 the neutrino has 49.52 keV while the beta 
particle has 17.425 keV.  Most of the energy shows up in the neutrino which 
seems typical of the beta plus decays as well.  The decay energy in this case 
is much less than in the other more typical examples.  The beta plus decay of 
Cu59 into Ni59 releases 3.7773 MeV typically.  Only 1.7226 MeV is given to the 
positron.

Dave




-----Original Message-----
From: pagnucco <pagnu...@htdconnect.com>
To: vortex-l <vortex-l@eskimo.com>
Sent: Tue, Jun 5, 2012 11:54 pm
Subject: Re: [Vo]:about Triumph Management (and LENR)


David,
According to my rough calculations the Ni-to-Cu transmutations do release
ignificant energy, as you claimed - see my reply to ny.min.  I am not
ure how much energy is siphoned off by the neutrino, though.
I assumed a neutron capture by the Ni nucleus, so no there would be no
eed to consider the coulomb barrier.  Excess energy is released, if we
ssume the conversion of electron+proton--to--neutron takes approximately
80 Kev.  So W-L theory, or hydrinos, or some kind of screening, could
xplain the transmutations, if they are real.
Lou Pagnucco
David Roberson wrote:
 I took a table of nuclides and performed a Energy difference between the
 different nickel isotopes, plus associated proton and electron, and the
 daughter copper isotopes and compared the net released energy to the
 energy required to overcome the coulomb barrier.  The most significant
 energy release which exceeds the barrier occurs when the Ni62 or Ni64
 isotopes are converted.  The delayed beta plus decay which is present for
 all of the other transformations looses a large amount of energy to a
 neutrino which promptly escapes the device.  I will demonstrate the
 numbers below.

 These components are required to build Cu63 from Ni62
 1 u = 931.494 MeV  Disregard the slight rounding errors, excel chart
 source of data

 Ni62  Mass=61.92835 (u)  Energy=57685.88 MeV
 Proton Mass=1.007276 (u)  Energy=938.2716 MeV
 Electron Mass=.000549 (u) Energy=.510999 MeV

 Coulomb Barrier Energy ~5.6 MeV according to Rossi in his paper

 Cu63 Mass=62.9296 (u) Energy=58618.54 MeV

 Mass of components of Cu63;Ni62 + Proton + Electron=62.936175 (u)
 Energy=58624.66 MeV

 Mass decrease that must be released as energy=62.936175 - 62.9296 =
 .006575 (u) Energy=  6.12457 MeV - 5.6 MeV Barrier = .52457 MeV;Same
 Calculation for Cu65 yields 1.8532 MeV
 In these reactions there are no Beta Plus Decay radiation losses due to
 neutrino release and no 511 keV gammas.

 Please note that I also calculated the expected energy release due to W&L
 process on the isotopes such as Ni60 and had perfect energy correlation
 when the energy required to make a neutron from a proton and electron is
 included.

 Exactly the same energy is seen in both paths (Rossi and W&L)  when the
 starting point is a nickel isotope with a proton and an electron, and the
 final point is the next higher isotope of nickel.

 I am working very hard to get a clear understanding of the coulomb barrier
 energy behavior.  I can show that the alpha process within stars stops
 once iron has been synthesized, but this is only true if the barrier
 energy is trapped within the nucleus in the form of mass.  I am
 approaching the problem from different directions to prove whether or not
 this hypothesis is accurate.

 Dave


 -----Original Message-----
 From: pagnucco <pagnu...@htdconnect.com>
 To: vortex-l <vortex-l@eskimo.com>
 Sent: Mon, Jun 4, 2012 11:49 pm
 Subject: Re: [Vo]:about Triumph Management (and LENR)


 David,
 Can you explain your conclusion.
  can't see how any energy is released in these Ni --> Cu transmutations.
 Lou Pagnucco
 David Roberson wrote:
  I may have been a bit to fast in pointing out the possible endothermic
  nature of the Ni62 and Ni64 reactions.  They actually are the best two
  isotopes to use if you were not to rely upon the beta plus decay for a
  substantial portion of the energy release.  They further are not
  susceptible to having the 511 keV gammas that would no doubt be released
  by the reactions involving the other nickel isotopes since copper 63 and
  65 are stable and do not decay into nickel by that process.

  So, if Rossi is actually able to overcome the coulomb barrier by some
  mechanism and his device only uses the Ni62 and Ni64 isotopes then it
  could be functional.  The energy released per atom for these two isotopes
  is only 1 or 2  MeV after satisfying the coulomb barrier, but that is a
  lot more than any chemical reaction can deliver.  I wonder if the
  relatively modest amount of energy release also can be more safely
  directed toward useful forms such as vibrational coupling into the
  surrounding structure.

  All of my estimates and calculations assume the reaction path that has
  been suggested by Rossi instead of the W&L process which would be much
  more energetic.  Each of these proposed mechanisms has it's own
 particular
  problems to overcome.

  Dave



  -----Original Message-----
  From: David Roberson <dlrober...@aol.com>
  To: vortex-l <vortex-l@eskimo.com>
  Sent: Mon, Jun 4, 2012 10:30 am
  Subject: Re: [Vo]:about Triumph Management (and LENR)


  I just wanted to point out that if Ni62 and Ni64 are the only isotopes
  that work then the addition of a proton to either results in the
  production of a stable isotope of copper which does not undergo beta plus
  decay.  Much less energy is released per atom if the beta plus decay is
  avoided.  My calculations suggest that these two reactions might actually
  be endothermic due to the large coulomb barrier.

  Perhaps this is a bit of misdirection?

  Dave



  -----Original Message-----
  From: integral.property.service <integral.property.serv...@gmail.com>
  To: vortex-l <vortex-l@eskimo.com>
  Sent: Mon, Jun 4, 2012 9:30 am
  Subject: Re: [Vo]:about Triumph Management (and LENR)


  A.R. from Florida with love,
  "Andrea Rossi






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