Question 3.3: In Eq.(3.31) suppose that the neutron slowing down past Eo i......
In Eq.(3.31) suppose that the neutron slowing down past E_o is due entirely from elastic scattering from a single nuclide with A > 1, and with no absorption for E > E_o . Show that s(E) then takes the form
s(E) = \Bigg \{\begin {matrix}\frac{1 }{ (1 – α) ζ} \bigg (\frac{1}{E_o} – \frac{α}{E} \bigg ) , \ \ αE_o < E < E_o \\ 0, \ \ \ \ \ \ \ \ \ \ E < αE_o \end {matrix}
First write Eq. (3.32) as :s(E)q_{o}=\int_{{E}_{o}}^{\infty}\ \Sigma_{s}(E^{\prime})p(E^{\prime}\to E)\varphi(E^{\prime})d E^{\prime} . For a pure scatter we my simplify Eq. (3.29) to \varphi(E)=\;\frac{q_{o}}{ξ \Sigma_{s}(E)E}\,. Thus s(E)=\int_{{E}_{0}}^{{\infty}}\ p(E^{\prime}\to E)\frac{1}{{ }{\xi } \ E^{\prime}}d E^{\prime} We may rewrite the scattering probability of Eq. (2.47) as p(E^{\prime}\rightarrow E)={\frac{1}{1-\alpha}}\cdot{\frac{1}{E^{\prime}}} for E < E^{\prime} < E / α Thus
s(E)=\int_{E_{o}}^{E/\alpha}~\frac{1}{1-\alpha}\frac{1}{\xi E^{\prime 2}}d E^{\prime}=\frac{1}{1-\alpha}\frac{1}{\xi}\bigg [-\frac{1}{E^{\prime}}\bigg ]_{E_o}^{E /\alpha}=\frac{1}{1-\alpha}\frac{1}{\xi}\biggl[\frac{1}{E_{o}}-\frac{\alpha}{E}\biggr]\,. Below α E_o
the source must vanish, since a neutron cannot scatter elastically form E > E_o to E < α E_o . Note that you can show that this expression for S(E) satisfies the normalization condition of problem [3.2].