Question 3.4: A neutron scatters off of a Carbon-12 nucleus, which is freq......
A neutron scatters off of a Carbon-12 nucleus, which is frequently used as a moderating material in thermal graphite reactors. If the scattering reaction is an elastic one, what is the average scattering angle of the neutron? What percentage of the initial kinetic energy of the neutron is lost in a single elastic collision of this type?
The average scattering of the neutron is given by <θ> = cos^{–1}(2/3A). Since A = 12, <q> = cos^{–1}(2/36) = 86.5°. According to Equation 3.67, α′ = {(1 + α)/2 + (1 − α)/2 ⋅ cos θ} = 0.86 + 0.14 cos θ. The fractional energy loss is therefore ΔE = 0.14 (1 − cos θ) = 0.14 (1 − cos θ) = 0.14 × (1 − 0.06) = 0.13 = 13%.
2\mathrm{EM}_{\mathrm{n}}=2\mathrm{E}^{\prime}\mathrm{M}_{\mathrm{n}}+2\mathrm{E}_{\mathrm{a}}^{\prime}\mathrm{M}_{\mathrm{a}}-2\sqrt{\left(2\mathrm{E}^{\prime}\mathrm{M}_{\mathrm{n}}\right)}\sqrt{\left(2\mathrm{E}_{\mathrm{a}}^{\prime}\mathrm{M}_{\mathrm{a}}\right)}\cos\theta (3.67)