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

Let Mass of neutron `m_(1) = m `
Mass of terget `m_(2) = M `
Intial speed of neutron , `u_(1) = u `
Intial speed of target `u_(3) = 0 `
We know that final speed of mass M is given by
`v_(2) = (2m_(1))/(m(1)+m_(2)).u_(1) +(m_(2)-m_(1))/(m_(1)+m_(2)).u_(2)`
` = (2m)/(m+M . u+0 = (2"mu")/(M+m)`
Intial Kinectic energy of neutron is
`K_(1)=1/2 "mu"^(2)`
Final K.E of mass M
`K_(2) = 1/2 Mv_(2)^(2) = 1/2 M ((2"mu")/(M+m))^(2)`
` (2Mm^(2)u^(2))/((M+m)^(2))`
Fraction of kinetic energy of neutron transferred to the target is
`f = (K_(2))/(K_(1)) = (2Mm^(2)u^(2))/((m+M)^(2))xx 2/("mu"^(2)) = (4mM)/((m+M)^(2))`
If teh target nucleous is deuterium ,
M = 2m
` :. " " f = (4m xx 2m)/((m+2m)^(2)) = 8/9 = 0.888`
About 88.8 % of the neutron.s energy is transferred to the deutrium .
If the target nucleus is carbon ,
M = 12 m
` :. " " f = (4m xx 12m)/((m+12m)^(2)) = 48/169 = 0.284 `
About 28.4 % of neutron energy is tranferred to carbon .