Question 2.14: Neutrons scatter elastically at 1.0 MeV. After one scatterin......
Neutrons scatter elastically at 1.0 MeV. After one scattering collision, determine fraction of the neutrons will have energy of less than 0.5 MeV if they scatters from
a. hydrogen,
b. deuterium,
c. carbon-12,
d. uranium-238.
Let E=1.0~M e V\qquad E_{1}=0.9~M e V\quad{\frac{E_{1}}{E}}=0.9
f r = \Bigg \{\begin {matrix} \int_{ \alpha E}^{E_1} \frac{1}{(1-\alpha)E}d E^{\prime}=\frac{E_{1}-\alpha E}{1-\alpha}=\frac{1}{1-\alpha}\bigg(\frac{E_{1}}{E}-\alpha\bigg)\ \ \mathrm{i}f\ \ \ \ \ \alpha\lt \frac{E_{1}}{E} \\ 0\;\;\;\;\ if \quad \alpha\geq\frac{E_{1}}{E} \end {matrix}
\alpha=\left({\frac{A-1}{A+1}}\right)^{2}
a. hydrogen \alpha=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;fr=\frac{1}{1}(0.9-0.0)=0.9
b. deuterium \alpha=0.111\;\;\;\;\;\;\;\;\;\;\;\;\;\;fr=\frac{1}{0.889}(0.9-0.111)=0.886
c. carbon-12 \alpha=0.716\;\;\;\;\;\;\;\;\;\;\;\;\;\;fr=\frac{1}{0.284}(0.9-0.716)=0.648
d. uraniium-238 \alpha=0.983\;\;\;\;\;\;\;\;\;\;\;\;\;\;fr = 0 \text{since} \alpha > E_1 / E