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

The initial kinetic energy of the neutron is
`k_(li) = 1/2m_(1)v_(lt)^(2)`
While its final kinetic energy from Eq.
`K_(1)f = 1/2 m_(1)v_(1)^(2)` = `1/2m_(1)((m_(1)-m_(2))/(m_(1)+m_(2))^(2)`
The fractional kinetic energy lost is
`f_(1) = (K_(1)f)/(K_(1i)) = (m_(1)-m_(2))/(m_(1)+m_(2))^()2)`
while the fractional kinetic energy gained by hte moderating nuclei `K_(2f)//K_(lt)` is
`f_(2)=1-f_(1)` (Eleastic collison)
`=(4m_(1)m_(2))/(m_(1)+m_(2))^(2)`
One can also verify this result by substituing from eq. For deuterium `m_(2) = 2m_(1)` and we obtain `f_(1)=1//9` while `f_(2)=8//9`. Almost `90%` of the neutrons energy is transferred to deuterium. For carbon `f_(1)= 71.6%` and `f_(2) = 28.4%`. In practice. however, this number is smaller since head-on colisons are rare.