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

Initial KE of the neutron is `K_(i)=(1)/(2)m_(1)v_(1)^(2)`
final KE `K_(1f)` using equation (48) is
`K_(1f)=(1)/(2)m_(1)v_(1f)^(2)=(1)/(2)m_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)v_(1)^(2)`
Fractional Ke retained is
`f_(1)=(K_(1f))/(K_(1i))=((1)/(2)m_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)v_(1i)^(2))/((1)/(2)m_(1)v_(1i)^(2))=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)`
Fractional KE gained by moderating nuclei `(K_(2f))/(K_(1i))` is
`f_(2)=1-f_(1)" (elastic collision)"`
`=1-((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)`
`=((m_(1)+m_(2))^(2)-(m_(1)-m_(2))^(2))/((m_(1)+m_(2))^(2))`
`=((4m_(1)m_(2)))/((m_(1)+m_(2))^(2))`
You can verify this result by putting from eqquation (49)
For deuterium, `m_(2)=2m_(1)`
`f_(1)=((m_(1)-2m_(1))/(m_(1)+2m_(1)))^(2)=(1)/(9)`
and `f_(2)=(8)/(9)`
`f_(1)` expressed as present `=(1)/(2)xx100=11.1%`
or`" "` almost `90%`of the neutron.s energy is transferred to deuterium.