Question 3.6: In the wide resonance approximation (Also called narrow reso......
In the wide resonance approximation (Also called narrow resonance infinite mass approximation because the fuel is assumed to have an infinite mass) A^ƒ → ∞ and thus α^ƒ → 1 in the first integral on the left of Eq. (3.28) , while the remaining approximations are the same as in narrow resonance approximation. Determine φ(E) through the resonance. How does it differ from Eq. (3.29)? In which case is there more energy self shielding?
Since the domain of integration between E and E / α^ƒ in the first integral of Eq. (3.28) becomes infinitely narrow, we may approximate \Sigma_{s}^{f}(E^{\prime})\varphi(E^{\prime})/E^{\prime}\to\Sigma_{s}^{f}(E)\varphi(E)/E within this interval. We thus have
\operatorname*{lim}_{\alpha^{f}\to1}\int_{E}^{{{E / \alpha^f}}}\ \frac{1}{(1-\alpha^{{f}})E^{\prime}}\Sigma_{s}^{f}(E^{\prime})\varphi(E^{\prime})d E^{\prime} = \operatorname*{lim}_{\alpha^{f}\rightarrow1}{\frac{\Sigma_{s}^{f}(E)\varphi(E)}{E}}\int_{{E}}^{{E / \alpha^{f}}}\;{\frac{1}{(1-\alpha^{f})}}d E^{\prime}
={\frac{\Sigma_{s}^{f}(E)\varphi(E)}{E}}\operatorname*{lim}_{\alpha^{f}\rightarrow1}{\frac{1}{(1-\alpha^{f})}}\Big(E/\alpha^{f}-E\Big)
=\Sigma_{s}^{f}(E)\varphi(E)\operatorname*{lim}_{\alpha^{f}\rightarrow1}\frac{1}{(1-\alpha^{f})}\Big(1/\alpha^{f}-1\Big) = \Sigma_{s}^{f}(E)\varphi(E)\operatorname*{lim}_{\alpha^{f}\to1}{\frac{1}{\alpha^{f}}}= \Sigma_{s}^{f}\left(E\right)\varphi(E)
Substituting this limiting case for the first integral and Eq. (3.26) for the second integral reduces Eq. (3.28) to
\Sigma_{t}(E)\varphi(E)=\Sigma_{s}^{f}(E)\varphi(E)+{\frac{1}{\xi E}}q or
\varphi(E)={\frac{1}{\xi\left[\Sigma_{t}(E)-\Sigma_{s}^{f}(E)\right]E}}q
Since \Sigma_{t}(E) -\Sigma_{s}^{f}(E) is smaller that \Sigma_{t}(E) appearing in the denominator of Eq. (3.29), the flux is larger, and therefore there is less self-shielding in this approximation