# Reflection of Gamma Rays

I have an experiment with a radioactive Cs$$^{137}$$ source which beta decays to Ba$$^{137}$$. Some of the barium are in an excited nuclear state which decay to the ground state and emit a $$662$$ keV $$\gamma$$-ray.

To measure the attenuation of aluminum, these gamma rays are passed through blocks of aluminum, and then the counts are recorded with an NaI crystal/PMT.

A gamma ray may enter the aluminum block at such an angle that it won't reach the NaI crystal. However, would it be possible for a gamma ray like this to be "totally internally reflected" in one of the aluminum blocks, and thus have its trajectory altered so that it does reach the NaI crystal?

• A side note, be careful with that Cs-137 source. A lot of people have received fatal radiation dose from mishandling those. And it is even suspected that Marie Curie, the same person who invented radioactivity, ultimately died because of radiation exposure to cesium-137 that she discovered. Oct 18, 2022 at 22:20
• Caution with radioactive sources is always wise, but...This device spectrumtechniques.com/products/sources/isotope-generator-kit or its predecessor from Union Carbide has been around in post-secondary education for decades Oct 19, 2022 at 4:03

Yes. Kind of.

Total internal reflection is not a thing at these energies, the wavelengths of these gammas are way too small. The dominant interaction at these energies is via Compton scattering - that is also how the gammas get attenuated in your aluminum block in the first place: they Compton scatter and thus loose energy.

Compton scattering occurs at all kinds of angles, and so those gammas that do interact in the aluminum effectively be deflected from their straight path. In effect, your detector will be hit by gammas that do not just come from the direct line of sight, but also from the sides. Add a "collimator" in front of it if you want to reduce this effect, i.e. some lead shielding with a hole that points straight at your source.

would it be possible for a gamma ray like this to be "totally internally reflected" in one of the aluminum blocks, and thus have its trajectory altered so that it does reach the NaI crystal?

Gamma rays are much "harder" than optical frequency light. For most purposes, we can treat gamma rays as so high energy that the value of the "optical constant" $$n$$ is effectively equal to $$1$$ for most materials.

The condition for "total internal reflection" is: $$\theta_c = \arcsin(n_1/n_2) \approx \arcsin(1/1) = \pi/2\;,$$ so we do not expect any "total internal reflection."

Similarly, the reflection coefficient for gamma rays is likely very small because their energy is high.

The so-called "optical constants" like $$n$$, $$\epsilon$$, reflectivity, etc are actually not constant, but change with photon energy. Gammas are very energetic photons so you can think of them as "hard" corpuscles that just zoom right on through material with very little reflection. (Of course, gammas do interact with atoms, etc, which causes absorption, scattering, reflection, and whatnot, but if you ignore absorption "edges" and some other phenomena, the behavior of gammas is in some sense simpler than optical light.)