On Dec 7, 2009, at 2:16 PM, Terry Blanton wrote:
Maybe someone can explain this comment in the paper:
"In the carbon breakup reaction, a
metastable 13C shatters into three α particles and the
residuals of the reaction can be viewed in the CR-39
detector as a three-prong star where each prong represents
each charged particle that occurs in the decay (Antolković
and Dolenec 1975)."
Is this saying that the reaction happens only with the 13C isotope,
typically 1% of the content?
Terry
No. When a two particle nuclear reaction occurs it typically
involves creating an "excited nucleus" of combined mass. For
example, ordinary fusion is not just a one step process. Excited
helium is created as a middle step, denoted:
D + D --> He*
The unstable He* can decay in one of three ways, releasing a gamma, a
T and p, or an He3 and an n. I give an extended description of this
process on pages 4-7 of:
http://www.mtaonline.net/~hheffner/CFnuclearReactions.pdf
here are the reactions:
D(D,p)T 4.03 MeV
D(D,n)He3 3.27 MeV
D(D,gamma)He4 23.9 MeV
The He* in hot fusion has 23.9 MeV of "fusion energy". If it
fissions then those fissions require energy to break up the He*. The
bond breaking energy to enable the D(D,p)T reaction is 23.9 MeV -
4.03 MeV = 19.87 MeV. Similarly the bond breaking energy to enable
the D(D,n)He3 reaction is 23.9 MeV - 3.27 MeV = 20.63 MeV.
Similarly, the neutron reaction with 12C creates 13C*:
n + 12C --> 13C*
which then fissions:
13C* --> 3 4He
if the 13C* has enough energy.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
