In a nuclear reactor a neutron of high speed("typically" 10^(7) ms^(-1)) must be slowed to 10^(3)ms^(-1) so that it canhave ahigh probality of interacting with isotope ""_(92)""^(235)Uand 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.

In a nuclear reactor a neutron of high speed("typically" 10^(7) ms^(-1

1.	In a nuclear reactor a neutron of high speed("typically" 10^(7) ms^(-1)) must be slowed to 10^(3)ms^(-1) so that it canhave ahigh probality of interacting with isotope ""_(92)""^(235)Uand 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.
Answer» Solution :The initial kinetic ENERGY of the NEUTRON is `K_(1i)=(1)/(2)m_(1)v_(1i)^(2)rArrK_(1f)(1)/(2)m_(1)v_(1f)^(2)=(1)/(2)m_(1)((m_(1)-m_(2))/(m_(1) + m_(2)))^(2)v_(1i)^(2)` The fractional kinetic energy lost is `f_(1)=(K_(1f))/(K_(1i))=((m_(1)-m_(2))/(m_(1)+m_(2)))` While the fractional kinetic energy gained by the moderating nuclei `K_(2f)//K_(1i)` is `f_(2) = 1 - f_(1)` (elastic collision) = `(4m_(1)m_(2))/((m_(1) + m_(2))^(2))` For deutrium `m_(2) = 2m_(1)` and we obtain `f_(1)=1//9=11% "while"f_(1)=8//9=89%.` Almost90% of the neutrons energy is transfered to the deuterium. For CARBON `f_(1) = 71.6% and f_(2) = 28.4 %` In practice, however, this number is smaller since head - on collisions are rare.