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. An absorption cross section of 81 is not perfect. It is not hard to imagine that some neutrons would get through. To get a sense of how many neutrons we're talking about, consider the number needed to produce by 1 W of power production through absorption into nickel. The upper bound on the amount of energy that will be provided by a single Ni(n,*) reaction will be around 10 MeV, if I've done my calculation right. For 1W power, 1J of energy is produced during one 1s. To get 1J energy, at you need 6.24150974E12 MeV / 10 MeV = 6.24E11 neutron captures per second. The trick is to figure out how efficient 81 barns is at stopping that kind of flux. If even 0.1 percent of the neutrons escape, that's 624 million neutrons escaping from the system per second. I assume that is a lot, and that that would set off a GM counter. If this is correct, the question becomes whether 81 barns is going to stop a lot more than 99.9 percent of the neutrons being generated and captured. Also bear in mind that there is a saturation that occurs, where the nickel cannot be further activated, after which it starts to transmit neutrons. At that point I think they would need to be absorbed by other isotopes that have evolved lest they escape in large numbers. Eric
