Question 3.4: For thermal neutrons calculate η as a function of uranium en......

We have \overline{{{\eta}}}=\frac{(\nu N\sigma_{f})^{25}}{(N\sigma_{a})^{25}+(N\sigma_{a})^{28}} =\frac{\nu^{25}\sigma_{f}^{\ 25}}{\sigma_{a}^{\ 25}}\frac{1}{1+(N^{28}/N^{25})\sigma_{a}^{28}/\sigma_{a}^{\ 25}}

Since the enrichment is \tilde{e}=N^{25}/(N^{25}+N^{28}) we have

\overline{{{\eta}}}=\frac{\nu^{25}{\sigma_{f}}^{ 25}}{{\sigma_{a}}^{ 25}}\frac{1}{{{1+(\hat{e}^{-1}-1){\sigma_{\!a}^{ 28}}\,/\,{\sigma_{\!a}^{\ 25}}}}}\, = 2.08{\frac{1}{1+0.00409\cdot({\hat{e}}^{-1}-1)}}