# How does neutron reflection work?

What I mean here isn't what is it used for, but what's actually going on with the reflector. There are a lot of sub questions to this.

1. how much energy is lost in the act of reflection.

2. not everything is reflected, I know beryllium absorbs a lot of neutrons, do any pass through? How much of each occurs?

3. whats the new direction of the neutron?

4. I know this must change with energy values, because the whole idea of moderation is to lower the energy to something more likely to hit the target, but how does it change? Is it consistent that lower energy neutrons are always more likely to interact with a given material, or could a reflector be essentially transparent to lower energy neutrons, the way air blocks some above visible EM frequencies then goes transparent again.

Equations and further reading is appreciated for all of this.

• This would be rather improved if you mentioned what kind of system you are thinking of. A power reactor? A test beam? An explosive device? As it stands you've asked about "the reflector" as if there is only one such object. Not terribly helpful for someone who isn't thinking of the same problem you are. – dmckee --- ex-moderator kitten Sep 23 '15 at 21:29

You're mistaken in your question #2. For thermal neutrons beryllium has capture cross section $\sigma_\text{capture} = 7.6\,\rm millibarn$ and scattering cross section $\sigma_\text{scatter} = 7.6\,\rm barn$, so a thermal neutron in beryllium will on average scatter 1000 times before capturing. For some other nuclei it's the case that scattering cross section is roughly independent of the neutron's kinetic energy while capture cross section is proportional to $1/\sqrt E$, so for faster neutrons the difference is even more pronounced. Beryllium is used as a neutron reflector because it scatters neutrons without capturing them. The mean free path between scatters is $\ell = 1/n\sigma$, where $n$ is the number density of beryllium nuclei; I haven't computed it today but the length scale is centimeters.
The scattered neutron's wavefunction is predominantly $s$-wave. (You may remember the "Born approximation" for scattering, in which you assume that the scattered particle's wavefunction is entirely $s$-wave; this works because the $p$-wave and higher states don't have much overlap with the nucleus.) The $s$-wave wavefunctions are spherically symmetric: in the center-of-mass frame for the incident neutron and the beryllium scatterer, the scattered neutron is equally likely to be found in any direction. Since beryllium is an order of magnitude heavier than the neutron, it only takes a couple of scatters before any information about the neutron's incident direction is completely lost. The scattered neutrons within the beryllium go in all directions.
What this means is that, at a distance $\sim\ell$ from the the surface of the beryllium, the incident neutron has a 50-50 chance of having turned around and being headed back out.