In a nuclear reactor, a neutro of high speed `(-10^(7)ms^(-1))` has to be slowed down to `10^(3)ms^(-1)` so that it can have a high probability of interacting with the isotope `""_(95)^(235)U` and cause it to fission. Show that fractional KE lost is about `10%` when a neutron has an elastic collision with a high nuclei of deuterim. (The material making up the light nucli, usually heavy water `(D_(2)O)` or graphite, is called nuclei moderatior).

Explanation

The initial kinetic energy of the neutron is
`K_(1i)=(1)/(2)m_(1)v_(1i)^(2)rArr K_(1f)=(1)/(2)m_(1)v_(1f)^(2)=(1)/(2)m_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)v_(1i)^(2)`
The fractional kinetic energy lost is
`f_(1)=(K_(1f))/(K_(1i))=((m_(1)-m_(2))/(m_(1)+m_(2)))`
While the fractional kinetic energy gained by the moderating nuclei `K_(2f)//K_(1i)` is
`f_(2)=1-f_(1)` (elastic collision) `=(4m_(1)m_(2))/((m_(1)+m_(2))^(2))`
For deutrium `m_(2)=2m_(1)` and we obtain `f_(1)=1//9 = 11 %` while `f_(1)=8//9=89%`. Almost 90% of the neutrons energy is transfered to the duterium.
For carbon `f_(1)=71.6%` and `f_(2)=28.4%`.
In practice, however, this numner is smaller since head - on collisions are rare.