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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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Answer» Solution :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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