Question 2.18: Show that for elastic scattering a. E - E′ ≡ ∫ (E - E′)P(E →......

a. Using Eq. (2.47) we have

{\overline{{E-E^{\prime}}}}\equiv\int_{\alpha E}^{E}~(E-E^{\prime}){\frac{1}{(1-\alpha)E}}\,d E^{\prime}={\frac{1}{(1-\alpha)}}\int_{\alpha E}^{E}~d E^{\prime}-{\frac{1}{(1-\alpha)E}}\int_{\alpha E}^{E}~E^{\prime}d E^{\prime}

Performing the integrals:

\overline{{{E-E^{\prime}}}}=\frac{1}{(1-\alpha)}(E-\alpha E)-\frac{1}{(1-\alpha)E}\frac{1}{2}(E^{2}-\alpha^{2}E^{2})=\left(1-\frac{1}{2}\frac{1-\alpha^{2}}{1-\alpha}\right)E

But since 1-\alpha^{2}=(1+\alpha)(1-\alpha) We have {\overline{{E-E^{\prime}}}}={\frac{1-\alpha}{2}}E

b For water, we must combine Eqs. (2.47), (2.53) and Assuming the cross sections are energy independent we have

\overline{{{E-E^{\prime}}}}=\frac1{\sum_{s}}\sum_{i}\Sigma_{s i}\int~~(E-E^{\prime})p_{i}(E\to E^{\prime})d E^{\prime}.

{\overline{{E-E}}}^{\prime}={\frac{1}{\sum_{s}}}\sum_{i}\Sigma_{s_{i}}{\int_{\alpha_{i}E}^E}\ \ (E-E^{\prime}){\frac{1}{(1-\alpha_{i})E}}d E^{\prime}.

But from part a. we know that the integrals are just equal to {\frac{1-\alpha_{i}}{2}}E\,. Hence

{\overline{{E-E^{\prime}}}}={\frac{1}{\sum_{s}}}\sum_{i}\Sigma_{s i}{\frac{1}{2}}(1-\alpha_{i})E

Since we may write the cross section for water as \Sigma_{s}=N(2\sigma_{s}^{H}+\sigma_{s}^{o})\,,\,\ \Sigma_{s}^{H}=N2\sigma_{s}^{H} and \Sigma_{s}^{o}=N\sigma_{s}^{o} we may reduce this expression to

{\overline{E-E^{\prime}}}=\frac{1}{2\sigma_{s}^{H}+\sigma_{s}^{o}}\bigl[2\sigma_{s}^{H}(1-\alpha_{H})+\sigma_{s}^{o}(1-\alpha_{O})\bigr]\frac{1}{2}E

We know that \alpha_{H}=0\mathrm{~and~}\alpha_{0}=(15\,/\,17)^{2}=0.779 Thus

\overline{{{E-E^{\prime}}}}=\frac{2{\sigma}_{s}^{H}+0.221{\sigma}_{s}^{O}}{2{\sigma}_{s}^{H}+{\sigma}_{s}^{O}}\frac{1}{2}E Taking \sigma_{s}^{H}=20b\,\,\,\mathrm{and}\,\,\,\sigma_{s}^{O}=3.8b\,,

{\overline{E-E^{\prime}}}={\frac{2\cdot20+0.221\cdot3.8}{2\cdot20+3.8}}{\frac{1}{2}}E=0.466E Which is fairly close to the value of 0.5E for pure hydrogen.