Slowing down of neutrons: In a nuclear reactor,a neutron of high speed (typically $10^{7} \; m s^{-1}$) must be slowed to $10^{3} \; m s^{-1}$ so that it can have a high probability of interacting with isotope $^{235}_{92}U$ and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nucleus like deuterium or carbon,which has a mass of only a few times the neutron mass. The material making up the light nuclei,usually heavy water $(D_{2}O)$ or graphite,is called a moderator.

(N/A) The initial kinetic energy of the neutron is $K_{1i} = \frac{1}{2} m_{1} v_{1i}^{2}$.
The final kinetic energy of the neutron after an elastic collision is $K_{1f} = \frac{1}{2} m_{1} v_{1f}^{2} = \frac{1}{2} m_{1} \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)^{2} v_{1i}^{2}$.
The fractional kinetic energy remaining is $f_{1} = \frac{K_{1f}}{K_{1i}} = \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)^{2}$.
The fractional kinetic energy gained by the moderating nucleus is $f_{2} = 1 - f_{1} = \frac{4 m_{1} m_{2}}{(m_{1} + m_{2})^{2}}$.
For deuterium,$m_{2} = 2m_{1}$. Substituting this,we get $f_{1} = \left( \frac{m_{1} - 2m_{1}}{m_{1} + 2m_{1}} \right)^{2} = \left( \frac{-m_{1}}{3m_{1}} \right)^{2} = \frac{1}{9} \approx 11.1\%$. Thus,$f_{2} = 1 - \frac{1}{9} = \frac{8}{9} \approx 88.9\%$. Almost $89\%$ of the neutron's energy is transferred to deuterium.
For carbon,$m_{2} \approx 12m_{1}$. Substituting this,$f_{1} = \left( \frac{m_{1} - 12m_{1}}{m_{1} + 12m_{1}} \right)^{2} = \left( \frac{-11}{13} \right)^{2} = \frac{121}{169} \approx 71.6\%$. Thus,$f_{2} = 1 - 0.716 = 0.284 = 28.4\%$. In practice,these values represent the maximum energy transfer occurring in head-on collisions.