In a nuclear reactor a neutron of high speed (typically `10^(7) ms^(-1)`) must be slowed to of interacting with isotope `._(92)^(235)U` and causing it to fission . Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass . the material making up the light nuclei , usually heavy water `(D_(2)O)` or graphite is called a moderator .

Explanation

Suppose the mass of neutron is `m_(1)` and its initial velocity is `v_(1i)` .
` :. ` Its initial kinetic energy is
`K_(li) =1/2 m_(1)v_(1)._(f)^(2)`
`=1/2 m_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))^(2).v_(1)^(2) " "....(2)`
`[ "where " v_(1)f =((m_(1)-m_(2))/(m_(1)+m_(2)))vli] `
` :. `The frictional kinetic lost is ,
`f_(1) =(K_(1f))/(K_(1f))`
`f_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)` [ From equation (1) and (2) ]
Final kinetic energy of moderating nuclei
`K_(2f) =1/2 m_(2)v_(2f)^(2)`
`=1/2 m_(2)xx(4m_(1)^(2))/((m_(1)+m_(2)).v_(li)^(2) " "...(3)`
` :. ` The frictional kinetic energy gained has the moderating nuclei ,
`f_(2) =(K_(2f))/(K_(li))`
` :. f_(2) =(4m_(1)m_(2))/((m_(1)+m_(2))^(2))` [ from result (3) and (4) ]
For deuterium `m_(2) =2m_(1)`
`:. f_(1) = ((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)`
`=((m_(1)-2m_(1))/(m_(1)+2m_(1)))^(2)`
`=((-m_(1))/(3m_(1)))^(2)`
`=1/9`
and `f_(2) = (4m_(1)m_(2))/((m_(1)+m_(2))^(2)) `
`=(4m_(1)xx2m_(1))/((m_(1)+2m_(1))^(2))`
` = (8m_(1)^(2))/((3m_(1))^(2)=(8m_(1)^(2))/(9m_(1)^(2))`
`= 8/9`
For deuterium ,
`f_(1) = 1/9 and f_(2) = 8/9 `
For carbon
`f_(1) = 71.6 % and f_(2) = 28.4 % `