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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 nuclel like deuterlum or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclel, usually heavy water $\left( D _{2} O \right)$ or graphite, is called a moderator.
Solution
The initial kinetic energy of the neutron is
$K_{1 i}=\frac{1}{2} m_{1} v_{1 i}^{2}$
while its final kinetic energy from Eq. $(6.27)$
$K_{1 f}=\frac{1}{2} m_{1} v_{1 f}^{2}=\frac{1}{2} m_{1}\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)^{2} v_{1 t}^{2}$
The fractional kinetic energy lost is
$f_{1}=\frac{K_{1 f}}{K_{1 i}}=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)^{2}$
while the fractional kinetic energy gained by the moderating nucle $K_{2 f} / K_{1 i}$ is $f_{2}=1-f_{1}$ (elastic collision)
$=\frac{4 m_{1} m_{2}}{\left(m_{1}+m_{2}\right)^{2}}$
One can also verify this result by substituting from Equation
For deuterium $m_{2}=2 m_{1}$ and we obtain $f_{1}=1 / 9$ while $f_{2}=8 / 9 .$ Almost $90 \%$ of the neutron's energy is transferred to deuterlum. For carbon $f_{1}=71.6 \%$ and $f_{2}=28.4 \% .$ In practice however, this number is smaller since head-on collisions are rare.
