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A narrow beam of thermal neutrons passes through a plate of iron whose absorption and scattering effective cross-sections are equal to `sigma_(a)= 2.5 b` and `sigma_(s)=11 b` respectively. Find the fraction of neutrons quitting the beam due to scattering if the thickness of the plate is `d= 0.50 cm`.

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In traversing a distance `d` the fraction which is either scattered or absorbed is clearly
`1-e^(-n(sigma_(s)+sigma_(a))d)`
by the usual definition of the attenuation factor. Of this, the fraction scattered is (by defination of scattering and absorption cross section)
`w={1-e^(-n(sigma_(s)+sigma_(a))d)} (sigma_(s))/(sigma_(s)+sigma_(a))`
In iron `n=(rhoxxN_(A))/(M)=8.3xx10^(22)per c c`
Substitution gives `w=0.352`

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