In reply to Eric Walker's message of Fri, 22 Mar 2013 01:34:47 -0700: Hi, [snip] >Lou, > >If LENR neutrons are indeed generated as proposed by W-L, almost all will >> be in the thermal range - quite a low momentum by fusion standards. >> > >They speak about "ultra low momentum neutrons," which I think is >significantly lower than thermal energies. These would then collide with >nickel substrate atoms in inelastic and elastic collisions as well as be >absorbed. The highest absorption cross sections in the graphs you point to >for nickel are ~1000 for 63Ni and ~10000 for 59Ni. 63Ni is only synthetic, >and 59Ni exists only in trace quantities, so in general the absorption >cross section for unenriched nickel will be lower than these. According to >the charts, the cross section for 58Ni, the most common isotope (68 >percent), is ~100 barns, and that for 60Ni (26 percent) is ~50 barns. So I >think you would take the weighted average of these to get an upper bound on >the absorption cross section of a block of normal nickel; e.g., 100 * .68 + >50 * .26 = 81 barns. That would be the upper bound, I think, neglecting >other isotopes that exist in small amounts. > >I looked, and it is difficult to pin down exactly how to calculate the half >value layer (the amount of material needed to decrease the intensity of an >incident neutron beam by half) starting from the microscopic total cross >section. Here we have the absorption cross section rather than the total >cross section. The other two relevant cross sections -- elastic and >inelastic -- are going to bounce our neutrons around and then out of the >system, so I wonder if they can be neglected. It seems that shielding >thickness is something that is experimentally determined and not calculated >analytically so much, although perhaps Robin or someone else can help us >out with a calculation.
The mean free path (mfp) should be (roughly) the atomic volume divided by the cross section. For a cross section of 81 barns and Ni, this works out to 1.351 mm. The fraction that gets transmitted is e^(-d/mfp) where d is any given distance. In this case e^(-d/1.351mm). So for a thickness of 10 mm, you get a transmission fraction of 6.1E-4 = 0.061%. For 1 mm you get 47.7%. Note however that the absorption cross section increases as the speed of the neutrons decreases, hence WL's emphasis on "ultra cold". (See e.g. http://atom.kaeri.re.kr/cgi-bin/endfplot.pl?j=f&d=mcnp&f=mcnp/Ni-58) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
