A steady fission source of strength operates inside a large tank of light water

PHYSICS · Multi-part · Fermi University Physics · Nuclear Physics · Nuclear reactions · Nuclear reactions

The Nucleus Resolved, Ch Neutrons: the reagent with no barrier

Problem

A steady fission source of strength S=2.0×105 cm−3 s−1S = 2.0\times10^{5}\,\mathrm{cm^{-3}\,s^{-1}} operates inside a large tank of light water, treated as an infinite, non-absorbing moderator with ξΣs=1.35 cm−1\xi\Sigma_s = 1.35\,\mathrm{cm^{-1}} (uniform, no spatial dependence). ((a)) Starting from a balance of the slowing-down density in the lethargy variable u=ln⁡(E0/E)u = \ln(E_0/E), derive the flux spectrum ϕ(E)\phi(E) in the region below the source energy, and evaluate it at E=5.0 eVE = 5.0\,\mathrm{eV}. ((b)) Deep in the same tank the slowed neutrons thermalise into a Maxwellian population at temperature TT. A thin gold foil, whose neutron-capture cross section follows the 1/v1/v law with σ0=98.8 b\sigma_0 = 98.8\,\mathrm{b} at v0=2200 m s−1v_0 = 2200\,\mathrm{m\,s^{-1}}, is immersed in a thermal neutron density n=1.0×1013 cm−3n = 1.0\times10^{13}\,\mathrm{cm^{-3}}. Starting from the rate integral ∫σ(v) v n(v) dv\int \sigma(v)\,v\,n(v)\,\mathrm{d}v, derive the capture rate per gold atom and show that its value does not depend on whether the water is at T=293.6 KT=293.6\,\mathrm{K} or T=600 KT=600\,\mathrm{K}. Give the numerical value.

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