Question 5.11: Suppose that energy groups 142, 143, and 144 in an ultrafine......
Suppose that energy groups 142, 143, and 144 in an ultrafine cross-section library have scattering cross sections of 2.5, 2.8, and 3.1 barns, respectively. If the energy groups have the same width and they are exposed to a neutron flux where the number of neutrons in each energy group is 1.0 × 10^8 neutrons/cm²/s, 0.9 × 10^8 neutrons/cm²/s, and 0.8 × 10^8 neutrons/cm²/s, respectively, what is the effective scattering cross section when all three of these groups are collapsed into a single energy group?
The effective scattering cross section in this case can be found from the equation σ_{sg} = Σ_{g′} σ_{sg′} ϕ_{g′} ΔE_{g′}/E_{g′} ϕ_{g′} ΔE_{g′}, where g′ = 142, 143, and 144. Since all three groups have the same width, the values ofΔE_{g′} cancel from the numerator and the denominator, and we are left with σ_s = (2.5 × 1.0 + 2.8 × 0.9 + 3.1 × 0.8)/(1.0 + 0.9 + 0.8) = 7.5/2.7 = 2.77 barns as the “effective” scattering cross section for the collapsed group.