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 can have a high probability of interacting with istope ""_(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 euterium 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 can have a high probability of interacting with istope ""_(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 euterium 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)rArr K_(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%`. Almost 90% of the neutrons energy is transfered to the duterium. For carbon `f_(1)=71.6%` and `f_(2)=28.4%`. In practice, however, this numner is SMALLER SINCE head - on collisions are rare.