Question

Calculate the thickness of Indium foil which will absorb 1% of neutrons incident at the resonance energy for Indium (1.44 eV) where σ = 28, 000 barns. At. Wt of Indium = 114.7 amu, density of Indium = 7.3 g/cm³.

Answer 1

If 1% of neutrons are absorbed then 99% are transmitted. The transmitted number I are related to the incident number by 

I = I₀ exp(−μx) (1) 

where μ is the absorption coefficient and x is the thickness of the foil. 

μ = Σ = σ N = σ N₀ρ/A 

where σ is the microscopic cross-section, N₀ is the Avagardro’s number, ρ the density and A the atomic weight.