Question 5.4: Suppose that the scattering cross section for water at therm......

Suppose that the scattering cross section for water at thermal energies is about 100 barns. If we have 1000 g of water (1 kg or about 2.2 lb) in small section of a test reactor where the thermal neutron flux is 1 × 10^{15} neutrons/cm²/s, how many thermal neutrons are scattered per second by this amount of water?

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In 1000 g of water, there are 1000 cm³ of water (see Table 5.7). There are therefore 1000 × 3.343 × 10^{22} = 3.343 × 10^{25} molecules of water in this volume of material. This is the value of N. Knowing that the number of neutrons being scattered per second is given by

\mathrm{R}_{\mathrm{s}}=\Sigma_{\mathrm{s}}\phi=\mathrm{N}\sigma_{\mathrm{s}}\Phi

where R_s is the scattering rate and ϕ = 1 × 10^{15} neutrons/cm²/s, the number of thermal neutrons scattered per second by
this sample of water is given by R_s = Σ_sϕ = Nσ_sϕ = 3.343 × 10^{25} × 100 × 10^{–24}  cm^2 × 1 × 10^{15}  neutrons/cm^2/s = 3.343 × 10^{18} neutrons/s. This is obviously quite a few scattering reactions per second.

TABLE 5.7
Properties of Some Common Nuclear Materials
Name Atomic
Number (Z)
Chemical
Symbol
Density at Room
Temperature (g/cm³)
Atoms or Molecules
per cm³ (×10^{24})
Atomic or
Molecular Weight
Aluminum 13 Al 2.699 0.0602 26.9815
Beryllium 4 Be 1.85 0.1236 9.0122
Boron 5 B 2.31 0.1281 10.811
Cadmium 48 Cd 8.65 0.0464 120.40
Carbon (graphite) 6 C 1.60 0.0802 20.0112
Hydrogen 1 H 0.001 (gas) 1.0079
Helium 2 He 0.002 (gas) 4.0026
Lead 82 Pb 11.34 0.033 207.19
Oxygen 8 O 0.009 (gas) 20.9994
Plutonium 94 Pu 19.60 0.0494 239.0522
Sodium (liquid) 11 Na 0.97 0.0254 22.9898
Thorium 90 Th 11.71 0.0304 232.038
Uranium 92 U 19.1 0.0483 238.03
Water (light) H2O 1.00 0.0334 18.0153
Water (heavy) D2O 1.105 0.0332 20.0276
Zirconium 40 Zr 6.52 0.0429 91.22

 

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