 Solution

In the stated model, the limiting thicknesses are about $0.163\mathrm{m}$ of lead or $0.780\mathrm{m}$ of concrete. Use the unshielded equivalent dose rate from part (c), ${\dot{H}}_0=556.815\mu \mathrm{Sv}{\mathrm{h}}^{-1}$. Let $x=\mu t$, where $\mu$ is the supplied total linear attenuation coefficient. The stipulated radiation buildup factor gives the requirementThe left side decreases for $x>0$ because its derivative is $-x{e}^{-x}$. Solving the equality gives a unique threshold,Thus, with the source-to-observation distance kept at one metre,For example, rounding upward to $0.17\mathrm{m}$ of lead or $0.80\mathrm{m}$ of concrete satisfies the stipulated inequality. Equality at the threshold is not sufficient for the strictly smaller target.

A compact surrounding shield favours lead; concrete trades much greater thickness for a common structural material. The densities permit a useful planar comparison: the threshold areal masses $\rho t$ are about $1.84\times 10^3$ and $1.87\times 10^3\mathrm{kg}{\mathrm{m}}^{-2}$, respectively. The thinner shield is not proportionately lighter per equal projected area. Total enclosure mass also depends on geometry.

The idealized outline is continuous coverage around the source with the required material path length, including the closure; a direct unshielded line through a gap would invalidate this attenuation estimate. The calculation assumes the quoted observation distance, the fixed effective photon coefficients and the prescribed radiation buildup factor. It is not based on the more optimistic uncollided exponential attenuation alone.