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

Any photon will have some [attenuation length][1] dependent on the frequency and the material it is passing through. [Sr-90 decays][2] 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][3], 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.

[Here][4] 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.

  [1]: http://en.wikipedia.org/wiki/Mass_attenuation_coefficient
  [2]: http://en.wikipedia.org/wiki/Strontium-90
  [3]: http://en.wikipedia.org/wiki/Phosphorus#Isotopes
  [4]: http://physics.nist.gov/PhysRefData/XrayMassCoef/ElemTab/z82.html