On Feb 24, 2011, at 11:50 AM, [email protected] wrote:
In reply to Horace Heffner's message of Thu, 24 Feb 2011 10:56:08
-0900:
Hi,
The reaction
Ni-60 + 4 H (cluster) => Ca-40 + Mg-24 + 13.5 MeV
readily produces Calcium.
Good point! I ignored cluster reactions because I considered them
too unlikely for production of readily measurable amounts. Also, I
was looking for prospect that create both K and Ca. If you include
cluster possibilities up to 4 hydrogens then the following reactions
are feasible producers of K and Ca:
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* --> 40Ca20 + 24Mg12 + 13.464 MeV [-22.248 MeV] (B_Ni:13)
61Ni28 + 3 p* --> 39K19 + 25Mg12 + 4.646 MeV [-21.274 MeV] (B_Ni:19)
61Ni28 + 3 p* --> 40K19 + 24Mg12 + 5.115 MeV [-20.805 MeV] (B_Ni:20)
61Ni28 + 4 p* --> 40Ca20 + 25Mg12 + 12.974 MeV [-22.554 MeV] (B_Ni:26)
62Ni28 + 3 p* --> 39K19 + 26Mg12 + 5.142 MeV [-20.644 MeV] (B_Ni:39)
62Ni28 + 3 p* --> 40K19 + 25Mg12 + 1.849 MeV [-23.938 MeV] (B_Ni:40)
62Ni28 + 3 p* --> 41K19 + 24Mg12 + 4.613 MeV [-21.173 MeV] (B_Ni:41)
62Ni28 + 3 p* --> 42Ca20 + 23Na11 + 3.198 MeV [-22.589 MeV] (B_Ni:42)
62Ni28 + 4 p* --> 39K19 + 27Al13 + 13.413 MeV [-21.934 MeV] (B_Ni:54)
62Ni28 + 4 p* --> 40Ca20 + 26Mg12 + 13.471 MeV [-21.877 MeV] (B_Ni:55)
62Ni28 + 4 p* --> 42Ca20 + 24Mg12 + 14.890 MeV [-20.457 MeV] (B_Ni:56)
64Ni28 + 3 p* --> 41K19 + 26Mg12 + 6.541 MeV [-18.986 MeV] (B_Ni:70)
64Ni28 + 3 p* --> 44Ca20 + 23Na11 + 5.766 MeV [-19.761 MeV] (B_Ni:71)
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
