In a nuclear reactor, a neutron of high speed `(~~10^(7)ms^(-1))` must be slowed down to `10^(3)ms^(-1)` so that it can have a high probality of interacting with isotipe `_92U^(235)` and causing it to fission. Show that a neutron can lose most of its K.E. in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a fewe times the neutron mass. The material making up the light nuclei usually heavy water `(D_(2)O)` or graphite is called modertaor.

Explanation

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.