There are approximately 6×10²⁸ atoms/m³ in solid aluminum. A beam of 0.5-MeV neutrons is directed at an aluminum foil 0.1 mm thick. If the capture cross section for neutrons of this energy in aluminum is 2×10^-31m² , find the fraction of incident neutrons that are captured.

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The provided data The density of the atoms 610 28 Incident neutron energy E 05 ... View the full answer