In a nuclear reactor a neutron of high speed (typically 10^(7) ms^(-1)) must be slowed to of interactingwith isotope._(92)^(235)U and causing it to fission . Show that a neutroncan lose mostof its kineticenergy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutronmass . the material makingup the light nuclei , 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 of interactingwith isotope._(92)^(235)U and causing it to fission . Show that a neutroncan lose mostof its kineticenergy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutronmass . the material makingup the light nuclei , 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.