Question

In a nuclear reactor a neutron of high speed (typically 107 m s–1) must be slowed to 103 m s–1 so that it can have a high probability of interacting with isotope 92235U 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 deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water (D2O) or graphite, is called a moderator.

Answer 1

At the following we shall assume that the nuclei is are stationary. We can use momentum conservation

$$m\vec v = m\vec v' + M\vec V$$

where $m$ - mass of the neutron, ${\vec v}$ - initial velocity of the neutron, $\vec v'$ - velocity of the neutron after collision, $M$ - mass of the nucleus, $\vec V'$ - velocity of the nucleus after collision, ${\vec V}$ - initial velocity of the nucleus (assumed to be 0 in the laboratory system). We shall also use dimensionless value $\lambda = {M \over m}$. Let's go to the center-of-mass system. The center of mass is moving with the velocity

$${{\vec u}_{cm}} = {{m\vec v} \over {m + M}} = {{\vec v} \over {1 + \lambda }}$$

In the center of mass the initial velocities are

$${{\vec v}_c} = \vec v - {{\vec u}_{cm}} = \vec v(1 - {1 \over {1 + \lambda }}) = \vec v{\lambda \over {\lambda + 1}}$$ $${{\vec V}_c} = \vec 0 - {{\vec u}_{cm}} = - {{\vec v} \over {1 + \lambda }}$$

Because in the center of mass system the sum of momentum are equal to zero, the magnitudes of velocities (speeds) do not changes, thus

$$\left| {{{\vec v}_c}'} \right| = {\lambda \over {\lambda + 1}}\left| {\vec v} \right|$$ $$\left| {{{\vec V}_c}'} \right| = {1 \over {1 + \lambda }}\left| {\vec v} \right|$$

Now use the vectors ${{\vec v}_c}',\vec v',{{\vec u}_{cm}}$ (we use that $\left| {{{\vec v}_c}'} \right| = \left| {{{\vec v}_c}} \right|$ )and apply cosine theorem to get

$${\left| {\vec v'} \right|^2} = {\left| {{{\vec v}_c}'} \right|^2} + {\left| {{{\vec u}_{cm}}} \right|^2} - 2\cos (\pi - {\theta _c}) \cdot \left| {{{\vec u}_{cm}}} \right|\left| {{{\vec v}_c}'} \right|$$

(where ${{\theta _c}}$ is the scattering angle in the center of mass system, it can be connected to the scattering angle in the laboratoty system through $\tan \theta = {{\sin {\theta _c}} \over {\cos {\theta _c} + {\lambda ^{ - 1}}}}$ ).

Now we can return to our goal - we need to calculate the ration of kinetic energies after and before collision

$$k = {{K'} \over K} = {{{{m{{\left| {\vec v'} \right|}^2}} \over 2}} \over {{{m{{\left| {\vec v} \right|}^2}} \over 2}}}$$

Thus

$$k = {{{{\left| {{v_c}'} \right|}^2} + {{\left| {{{\vec u}_{cm}}} \right|}^2} + 2\cos {\theta _c} \cdot \left| {{{\vec u}_{cm}}} \right|\left| {{v_c}'} \right|} \over {\left| {\vec v} \right|}}$$ $$k = {{{\lambda ^2} + 2\lambda \cos {\theta _c} + 1} \over {{{(\lambda + 1)}^2}}}$$

Now we can see, that the energy loss depend on the angle ${{\theta _c}}$ . In the simpliest case - the central head-on collision (${\theta _c} = \pi$), we get

$${k_{Head - on}} = {({{\lambda - 1} \over {\lambda + 1}})^2}$$

Let's calculate the relative energy loss as

$$\left| {{{\Delta K} \over K}} \right| = 1 - k$$

Then in collision with deiterium $\lambda = 2$ the maximum energy loss (in the central head-on collision)

$$\left| {{{\Delta K} \over K}} \right| = 1 - {k_{Head - on}} = {{4\lambda } \over {{{(\lambda + 1)}^2}}} = {8 \over 9} \approx 89\%$$

Now, in general case, we should average the $k$ by averaging the cosine function

$${1 \over \pi }\int\limits_0^\pi {\cos {\theta _c}d{\theta _c}} = 0$$

We can see that in this case $k = {{{\lambda ^2} + 1} \over {{{(\lambda + 1)}^2}}}$ and

$$\left\langle {\left| {{{\Delta K} \over K}} \right|} \right\rangle = {{2\lambda } \over {{{(\lambda + 1)}^2}}}$$