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?
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 |