Question 3.5: Suppose a new isotope is discovered with a ‘1/E’ absorption ......
Suppose a new isotope is discovered with a ‘1/E’ absorption cross section given by
Σ_α(E) = (E_o / E) Σ_α (E_0). Determine the energy-averaged cross section if the isotope is placed the thermal flux distribution given by Eq. (3.35??).
The Maxwell Boltzmann distribution may be written, from Eq. (3.35??) to be
\varphi_{M}(E)=A E\exp(-E/\,k T) where A=\frac{2\pi n^{\prime \prime \prime}}{\left(\pi k T\right)^{3/2}}\left(\frac{2}{m}\right)^{1/2}
Thus \overline {\Sigma}_{a}=\frac{{\int}_0^{\infty}(E_{o}/E)\Sigma_{a}(E_{o})A E\exp(-E/k T)d E}{{\int_{o}}^{\infty}{A E\exp(-E/k T)d E}}= E_o \Sigma_{\alpha} (E_0) \frac{\int_0^\infty exp(-E / kT)dE}{\int_0^\infty E\exp(-E / kT)dE}
Change variables: x = E / kT:
{\int_0^{\infty}}\ \ E\exp(-E/k T)d E = (k T)^2 \int_0^{\infty}x\exp(-x)d x=(k T)^{2}
{\int_0^{\infty}} exp(-E /kT)dE = kT {\int_0^{\infty}} exp(-x)dx = kT
Hence: \overline{{{\Sigma}}}_{a}=\left(\frac{E_{o}}{k T}\right)\Sigma_{a}(E_{o})