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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.

A factor accounting for scattered radiation that contributes to a specified response beyond the uncollided exponential attenuation. It depends on material, geometry, spectrum and the response being counted; $B(x)=1+x$ is a stipulated simple model, not a universal law.

Consider unit energy of incident radiation travelling downward into a conservatively scattering slab. Let $t$ be its unscattered transmission fraction. Of the fraction $1-t$ that scatters, suppose a fraction $a$ continues exactly forward and the remaining fraction $1-a$ has an isotropic outgoing distribution. Approximate every scattered contribution as escaping without a second interaction.

The isotropic part splits equally between upward and downward directions. The escape fractions are thereforeThey sum to one, as required by conservation of energy without electromagnetic absorption. For example, $t=0.6$ and $a=0.5$ give $f_{\uparrow }=0.1$ and $f_{\downarrow }=0.9$. Exactly forward scattering does not remove downward energy in this model; treating all scattering as backward loss would give a different result.

For a specified direct path, exponential attenuation gives $t=e^{-{\tau }_s}$ with scattering optical depth ${\tau }_s$. The single-escape assumption becomes unreliable when repeated scattering is important. A thick slab generally requires the angular scattering phase function and the full radiative transfer equation, rather than this two-branch bookkeeping.