On Feb 25, 2011, at 12:25 PM, mix...@bigpond.com wrote:
In reply to Jed Rothwell's message of Fri, 25 Feb 2011 09:17:40
-0500:
Hi,
[snip]
By a person, not Google translate:
http://www.nyteknik.se/nyheter/energi_miljo/energi/article3111124.ece
- Jed
Ni has roughly the following isotopes/percentages:-
Ni-58 68%
Ni-60 26%
Ni-61 1%
Ni-62 4%
Ni-64 1%
If 30% of the Ni is converted to copper over the long term, then
it's possible
that it's just the heavier isotopes that are reacting and that
Ni-58 is not
involved.
It does not look to me to be credible that this happens, unless maybe
very large hydrogen clusters are involved - and that is not credible
for a mechanism that is so effective it can consume 94% of the
available consumable isotopes and convert all that to copper.
The main problem lies with 60Ni28. Looking at every strong force
reaction energetically feasible involving 4 or fewer protons:
60Ni28 + 2 p* --> 32S16 + 30Si14 + 00.554 MeV [-16.327 MeV] (B_Ni:2)
60Ni28 + 2 p* --> 34S16 + 28Si14 + 1.530 MeV [-15.351 MeV] (B_Ni:3)
60Ni28 + 2 p* --> 50Cr24 + 12C6 + 00.365 MeV [-16.516 MeV] (B_Ni:4)
60Ni28 + 2 p* --> 58Ni28 + 4He2 + 7.909 MeV [-8.973 MeV] (B_Ni:5)
60Ni28 + 3 p* --> 32S16 + 31P15 + 7.851 MeV [-18.205 MeV] (B_Ni:6)
60Ni28 + 3 p* --> 35Cl17 + 28Si14 + 7.901 MeV [-18.155 MeV] (B_Ni:7)
60Ni28 + 3 p* --> 36Ar18 + 27Al13 + 4.823 MeV [-21.233 MeV] (B_Ni:8)
60Ni28 + 3 p* --> 39K19 + 24Mg12 + 5.135 MeV [-20.921 MeV] (B_Ni:9)
60Ni28 + 3 p* --> 40Ca20 + 23Na11 + 1.771 MeV [-24.285 MeV] (B_Ni:10)
60Ni28 + 4 p* --> 32S16 + 32S16 + 16.715 MeV [-18.997 MeV] (B_Ni:11)
60Ni28 + 4 p* --> 36Ar18 + 28Si14 + 16.408 MeV [-19.304 MeV] (B_Ni:12)
60Ni28 + 4 p* --> 40Ca20 + 24Mg12 + 13.464 MeV [-22.248 MeV] (B_Ni:13)
there appears to be no energetically feasible reaction that can
produce copper from 60Ni without creating radioactive nuclei. Same
is true considering weak reactions, which take an extra 782.353 kEv
away. If radioactive nuclei are created then some will remain in the
leftover material, but it was denied that there was any such
radioactive "ash".
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
