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 isotope {:(235),(92):}U and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elasticcollision with a light nuclet like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclet, usually heavy water (D_(2)O) or graphite, is called a moderator.

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 isotope {:(235),(92):}U and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elasticcollision with a light nuclet like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclet, usually heavy water (D_(2)O) or graphite, is called a moderator.
Answer» Solution :The initial kinetic energy of the NEUTRON is `K_(1i)= (1)/(2)m_(1)v_(1i)^(2)` While its final kinetic energy from Eq. (6.27) `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))^(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)` One can also verify this RESULT by substituting from Eq. (6. 28). For deuterium `m_(2)= 2m_(1)` and we obtain `f_(1)= 1"/"9" while "f_(2)= 8"/"9`. Almost 90% of the neutron.s energy is transferred to deuterium. For carbon `f_(1)= 71.6%" and "f_(2)= 28.4%`. In practice, howerver, this number is smaller since head-on-collisions are rare.