In a nuclear reactor, a neutron of high speed `(~~10^(7)ms^(-1))` must be slowed down to `10^(3)ms^(-1)` so that it can have a high probality of interacting with isotipe `_92U^(235)` and causing it to fission. Show that a neutron can lose most of its K.E. in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a fewe times the neutron mass. The material making up the light nuclei usually heavy water `(D_(2)O)` or graphite is called modertaor.

Explanation

Initial KE of the neutron is `K_(i)=1/2m_(1)v_(1)^(2)`
Final KE `K_(1f)` using equation (48) is
`K_(1f)=1/2m_(1)v_(1f)^(2)=1/2m_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)+v_(1f)^(2)`
Fractional KE retained is
`f_(1)=(K_(1f))/(K_(1i))=(1/2m_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)v_(1i)^(2))/(1/2m_(1)v_(1i)^(2))=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)`
Fractional KE gained by moderating nuclei `(K_(2t))/(K_(1i))` is
`f_(2)=1-f_(1)` (elastic collision)
`=1-((m_(1)-m_(2))/(m_(1)+2m_(1)))^(2)=1/9and f_(2)=8/9`
`f_(1)` expressed as percent `=1/9xx100-11.1%`
or almost `90%` of the neutron's energy is t ransferred to deuterium