On Sep 13, 2011, at 3:10 AM, [email protected] wrote:
Hi,
Could it be that Rossi uses a Cobalt60 gamma source as catalyzer?
Cobalt 60 decays to Nickel60 and emits gamma rays. The gamma
spectrum could be just the right spectrum and energy to excite the
Nickel nucleus.
Maybe it is mainly the Cobalt60 that needs screening and not the
reactor? This would explain the thickness of the lead screening.
Focardi has calculated the lead screening. So he must know about
the gamma rays and probably knows about the catalyzer.
Once he said "I dont know and dont want to know"... Now I would
believe he doesnt know, but I can hardly believe he doesnt want to
know ;-)
Peter
It is notable that if a gamma plus high energy beta source were used
for stimulation it could be kept in a container that isolated it from
the nickel, and thus it would not be seen in post experiment analysis
of the fuel. There are numerous reports of the effectiveness of
radiation stimulation of LENR. The problem is that shielding merely
attenuates gammas. Some always gets through. This could be detected
externally.
Almost all (99.88%) C60 decay is 0.32 MeV beta followed by 1.12 MeV
gamma, followed by 1.33 MeV gamma. About 0.12% is 1.48 MeV beta
followed by 1.3325 MeV gamma.
I think the lead shielding was stated by Rossi to be 2 cm thick, but
don't have a reference handy.
Very roughly, the mass attenuation coefficient in lead for 1.12 MeV
gammas is about 0.062 cm^2/gm and for 1.33 gammas about 0.057 cm^2/
gm. The linear attenuation coefficient mu for 1.12 MeV gammas is
(0.062 cm^2/gm) * (11.4 gm/cm^3) = 0.707/cm, and for 1.33 MeV gammas
is (0.057 cm^2/gm) * (11.4 gm/cm^3) = 0.65/cm.
The gamma attenuation, from internal intensity I0 to external
intensity I at distance x is given by:
I = I0 * exp(-mu * x)
so for 1.12 MeV gammas we have:
I = I0 * exp(-(0.707/cm)*(2 cm)) = I0 * exp(-1.414)
I = I0 * 0.243
for 1.33 MeV gammas we have:
I = I0 * exp(-(0.065/cm)*(2 cm)) = I0 * exp(-1.3)
I = I0 * 0.88
There is in effect (assuming no calc. errors on my part), with
regards to either safety or detection, no practical attenuation
offered for cobalt 60 gammas by 2 cm of lead.
However, cobalt 60 is not the only possibility given the radioactive
source is (and must be to avoid post experiment detection) physically
isolated from the nickel. There is no requirement for it to decay
into Ni, or Cu, or Zn, which were found in the used fuel. Thus, it
is possible to choose an alpha or beta source which does not produce
large gamma signatures through 2 cm of lead.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
