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.

Please do follow Jon Custer's answer and properly line the storage container.

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.

Please do follow Jon Custer's answer and properly line the storage container.

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

Any photon will have some attenuation length dependent on the frequency and the material it is passing through. 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.

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.

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.

Please do follow Jon Custer's answer and properly line the storage container.