Question: 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

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 full answer

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