Q.Slowing down of neutrons: In a nuclear reactor a neutron of high speed (typically $10^{7}\ \text{m s}^{-1}$) must be slowed to $10^{3}\ \text{m s}^{-1}$ so that it can have a high probability of interacting with isotope $^{235}_{92}\text{U}$ 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 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 ($\text{D}_2\text{O}$) or graphite, is called a moderator.

The key idea is Conservation of Momentum and Conservation of Kinetic Energy in a one-dimensional elastic collision. The neutron transfers a large fraction of its energy to a stationary target nucleus when the masses are comparable.

Step 1: Set up the elastic collision.

Let a neutron of mass $m$ and initial speed $u$ collide head-on with a stationary nucleus of mass $M$. After collision, let the neutron have speed $v$ and the nucleus have speed $V$. Both move along the same line.

Step 2: Apply conservation laws.

Momentum: $m u = m v + M V$

Kinetic energy: $\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + \frac{1}{2} M V^2$

Step 3: Solve for the neutron's final speed.

From the standard result for a 1D elastic collision:

The kinetic energy of the neutron after collision is:

Step 4: Interpret for a light moderator. …

The key insight here is Conservation of Momentum combined with Conservation of Kinetic Energy — the hallmark of an elastic collision. When a fast neutron hits a stationary nucleus, the two bounce off each other without any energy being lost to heat or deformation. The neutron slows down, and the nucleus recoils, carrying away the energy.

Why does a light target work best? Think of a ball hitting a wall versus hitting a pillow. A heavy wall barely moves — the ball bounces back with nearly the same speed. A light pillow flies backward, taking most of the ball’s energy with it. In nuclear terms, a light nucleus (mass close to the neutron’s) can “absorb” a large share of the kinetic energy, while a heavy nucleus (like uranium) barely budges, leaving the neutron still fast.

Let’s prove this quantitatively.

1. Set up the collision in one dimension (head-on, maximum energy transfer). Let the neutron have mass $m$ and initial speed $v_i$. The target nucleus has mass $M$ and is initially at rest. After the collision, the neutron moves with speed $v_f$ and the nucleus with speed $V$. By conservation of momentum:

By conservation of kinetic energy (elastic):

2. Solve for the final neutron speed $v_f$. From the momentum equation: $V = \frac{m}{M}(v_i - v_f)$. Substitute into the energy equation:

Simplify:

Divide through by $m$:

This is a quadratic in $v_f$. Expand and rearrange:

Multiply through by $M$:

Bring terms together:

Rearranging into standard quadratic form:

3. Solve the quadratic for $v_f$. Using the quadratic formula:

Notice that $(M+m)(m-M) = m^2 - M^2$, so the discriminant becomes:

Hence:

The plus sign gives $v_f = v_i$ (no collision — the neutron passes through), which is physically trivial. The minus sign gives the real result:

For a head-on elastic collision, the final speed of the lighter particle (neutron) is reduced by the factor $\frac{|m-M|}{m+M}$. If $M > m$, the neutron reverses direction (negative $v_f$), but the magnitude is what matters for energy.

4. Find the fraction of kinetic energy retained by the neutron. Initial kinetic energy: $K_i = \frac{1}{2} m v_i^2$ Final kinetic energy: $K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m \left( \frac{m - M}{m + M} \right)^2 v_i^2$ So the fraction retained is:

The fraction lost to the nucleus is:

5. Apply to real moderators. Neutron mass $m \approx 1$ u (atomic mass unit).
   • Deuterium (in heavy water): $M \approx 2$ u $\frac{K_f}{K_i} = \left( \frac{1 - 2}{1 + 2} \right)^2 = \left( \frac{-1}{3} \right)^2 = \frac{1}{9} \approx 0.111$ …

Concept: Elastic Collision — Fraction of Kinetic Energy Transferred

Step 1: Set up momentum and kinetic-energy conservation for a 1D elastic collision.

Neutron (mass $m$, initial speed $u$) hits a stationary nucleus (mass $M$):

Step 2: Solve the standard result for the neutron's final speed.

Step 3: Express the fraction of kinetic energy the neutron retains.

Step 4: Evaluate for deuterium ($M \approx 2m$) and carbon ($M \approx 12m$).

Deuterium: $\left(\frac{1-2}{1+2}\right)^2 = \frac{1}{9} \approx 0.111$ retained $\Rightarrow$ 89% lost. …