Suppose a sample of strontium-90 is stored in a lead container with lead walls. It is know that X-ray radiation may be detected outside the lead container. After some discussion with my peers, it seems that we have differing theories on how the X-ray radiation is formed.

1. Beta particles emitted by the decay of strontium-90 collide with the walls of the container, and in the process emit photons. These photons, when energetic enough, have high penetration power and can penetrate the walls of the container.
2. Similarly, the beta emission by the decay of strontium-90 produces photons, but the photons instead tunnel through the walls.

The first theory seems to show a picture of the emitted x-rays actually passing through the empty space between lead particles in the wall of the container. Under this theory, the term "penetration" is not well defined enough.

Under the second theory, it seems that quantum tunnelling is not directly applicable because there is no potential barrier involved.

Which would be the correct explanation, or is there a better explanation?

I don't know what "penetration power" is or why quantum tunneling needs to be invoked.

Sr-90 decays entirely via beta emission with up to $0.546\ \mathrm{MeV}$ given to the electron, and its daughter isotope similarly decays with up to $2.28\ \mathrm{MeV}$ given to the electron.

These energy ranges are right around the $1.71\ \mathrm{MeV}$ of P-32, whose beta emissions are known to induce significant bremsstrahlung in lead. Bremsstrahlung can easily produce photons of energies similar to the incident energy of the charged particle.

Any photon will have some attenuation length dependent on the frequency and the material it is passing through. Here is the NIST chart for lead to stop photons. As you can see, the value of $\mu/\rho$ is about $0.5\ \mathrm{cm^2/g}$ for photon energies of $2\ \mathrm{MeV}$. At a density of $11\ \mathrm{g/cm^3}$, this means the attenuation length is about $5.7\ \mathrm{cm}$. Even a $10\ \mathrm{cm}$ thick lead wall will only stop about 80% of such photons.

• "he value of $\mu/\rho$ is about $0.5 \,\mathrm{cm^2/g}$ [...]" They use goofy units at NIST. Or is this common in shielding applications outside of the experimental world? Most particle experimenters think in terms of the energy loss which is roughly $2 \,\mathrm{MeV/(g\,cm^2)}$ at minimum inonization or the radiation length. I had to read that a couple of times to parse it correctly I agree that worrying about showering is always the main issue with photons. – dmckee --- ex-moderator kitten Oct 2 '14 at 0:24