Kvant Physics Problem 199

The physical system consists of a neutron entering a layer of material.

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Problem

A neutron passes easily through a layer of lead, but is stopped in an equivalent layer of paraffin, water, or another hydrogen-containing compound. Explain why.

P. L. Kapitsa

Setup and Assumptions

The physical system consists of a neutron entering a layer of material. The materials to be compared are lead and a hydrogen-containing substance such as paraffin or water.

The quantity of interest is the ability of the material to stop the neutron. Stopping a neutron means reducing its kinetic energy through collisions until it becomes a slow, thermal neutron that is readily absorbed or no longer penetrates deeply into the material.

Let the neutron mass be $m_n$. Let the mass of a nucleus with which the neutron collides be $M$. The collisions are treated as elastic, since neutron energies relevant to moderation are usually reduced mainly through elastic scattering. The nuclei are initially at rest in the laboratory frame. Effects of absorption are neglected during the first stage of the analysis, because the central question is why the neutron loses energy efficiently in hydrogen-containing substances.

For hydrogen nuclei,

$$M \approx m_n.$$

For lead nuclei,

$$M \approx 207,m_n.$$

The problem reduces to determining how much kinetic energy a neutron loses in a typical elastic collision with a light nucleus and with a heavy nucleus.

Physical Principles

The analysis rests on conservation of momentum and conservation of kinetic energy during an elastic collision.

For a head-on elastic collision between a particle of mass $m_n$ and a stationary target of mass $M$, the final neutron velocity is

$$v_n'=\frac{m_n-M}{m_n+M},v_n,$$

where $v_n$ is the initial neutron velocity.

The corresponding neutron kinetic energies before and after the collision are

$$E=\frac12 m_n v_n^2,$$

and

$$E'=\frac12 m_n {v_n'}^2.$$

Combining these relations gives the fraction of kinetic energy retained by the neutron after a head-on collision:

$$\frac{E'}{E} = \left( \frac{m_n-M}{m_n+M} \right)^2.$$

The energy transferred to the nucleus is

$$\Delta E = E-E'.$$

Efficient neutron slowing requires a large value of $\Delta E$ in each collision.

Derivation

Consider first a collision with a hydrogen nucleus. Since the hydrogen nucleus is a proton,

$$M \approx m_n.$$

Substituting into the energy-retention formula gives

$$\frac{E'}{E} = \left( \frac{m_n-m_n}{m_n+m_n} \right)^2 = 0.$$

For a perfectly head-on collision, the neutron can transfer essentially all of its kinetic energy to the proton and come to rest. Even for collisions that are not exactly head-on, a large fraction of the neutron's energy is typically transferred because the masses are nearly equal.

Now consider a collision with a lead nucleus. The nucleus is much heavier than the neutron:

$$M \approx 207,m_n.$$

Substituting into the same expression,

$$\frac{E'}{E} = \left( \frac{1-207}{1+207} \right)^2 = \left( \frac{-206}{208} \right)^2 = \frac{42436}{43264} \approx 0.981.$$

The fraction of energy lost in such a collision is

$$\frac{\Delta E}{E} = 1-0.981 \approx 0.019.$$

Thus only about

$$1.9%$$

of the neutron's kinetic energy is transferred in a typical maximum-energy-transfer collision with lead.

A neutron entering a hydrogen-containing substance undergoes repeated collisions with nuclei whose mass is nearly equal to its own. Each collision can remove a substantial fraction of the neutron's energy. After only a modest number of collisions, the neutron becomes very slow.

In lead, each collision removes only a tiny fraction of the neutron's energy because the lead nucleus is much heavier than the neutron. A very large number of collisions would be required to achieve the same reduction in energy. Consequently the neutron penetrates much more deeply through lead than through an equivalent thickness of paraffin or water.

Result

The fraction of neutron kinetic energy remaining after a head-on elastic collision with a stationary nucleus of mass $M$ is

$$\frac{E'}{E} = \left( \frac{m_n-M}{m_n+M} \right)^2.$$

For hydrogen,

$$M \approx m_n,$$

so

$$\frac{E'}{E}\approx 0, \qquad \frac{\Delta E}{E}\approx 1.$$

For lead,

$$M \approx 207,m_n,$$

so

$$\frac{E'}{E}\approx 0.981, \qquad \frac{\Delta E}{E}\approx 0.019.$$

The neutron loses nearly all of its energy in a collision with hydrogen, but only about $1.9%$ of its energy in a collision with lead.

$$\boxed{\text{Hydrogen-containing materials are excellent neutron moderators because } M \approx m_n, \text{ allowing very efficient energy transfer.}}$$

$$\boxed{\text{Lead nuclei are much heavier than neutrons, so each collision removes only a small fraction of the neutron's energy.}}$$

Sanity Checks

The expression

$$\left( \frac{m_n-M}{m_n+M} \right)^2$$

is dimensionless, as required for a ratio of energies.

In the limiting case $M=m_n$, the retained energy becomes zero for a head-on collision. This matches the familiar behavior of two equal masses in an elastic collision, where the incoming particle can transfer all of its kinetic energy to the target.

In the opposite limit $M \gg m_n$,

$$\frac{E'}{E} \to 1,$$

which means the neutron keeps almost all of its energy after the collision. This agrees with physical intuition, since a very light particle cannot significantly accelerate a much heavier object.

The most common source of error is confusing the fraction of energy retained,

$$\frac{E'}{E},$$

with the fraction of energy lost,

$$1-\frac{E'}{E}.$$

Using the retained-energy fraction in place of the energy-loss fraction would reverse the physical conclusion. The calculation shows that hydrogen removes energy from neutrons efficiently, whereas lead does not, explaining why paraffin, water, and other hydrogen-containing substances stop neutrons much more effectively than an equivalent thickness of lead.