A (5 mathrm{~cm}) thick layer of pure absorbing material with atom density of (4.0 times 10^{22}) nuclei
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A \(5 \mathrm{~cm}\) thick layer of pure absorbing material with atom density of \(4.0 \times 10^{22}\) nuclei \(/ \mathrm{cm}^{3}\) absorbs \(99.9 \%\) of the incident beam of neutrons. What is the mean free path of neutrons in this material?
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The mean free path of a neutron in the material can be calculated using the following relationship 1 ...View the full answer
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