Each equation is stored thermal energy changing at the rate “input minus output”. Define the three heat capacities $C_j=M_j{c}_j$, and introduce the internal heat-transfer ratesIn the fuel balance, $C_F{\dot{\overline{T}}}_F$ is accumulation, $P$ is thermal power generation and $Q_{FM}$ is transfer to the neutron moderator. In the moderator balance, $Q_{FM}$ enters, $Q_{MC}$ leaves and $C_M{\dot{\overline{T}}}_M$ is accumulation. In the coolant balance, $Q_{MC}$ enters, $C_C{\dot{\overline{T}}}_C$ is accumulation and $2W_C{c}_C({\overline{T}}_C-T_i)$ is net energy carried away by flowing coolant.

The factor two follows the model's arithmetic-mean approximation ${\overline{T}}_C=(T_i+T_o)/2$, since the advective energy removal is $W_C{c}_C(T_o-T_i)$. The parameters $R_1$ and $R_2$ are effective thermal resistances, in $\mathrm{K}{\mathrm{W}}^{-1}$, for fuel-to-moderator and moderator-to-coolant transfer. They summarize the relevant geometry and heat-transfer mechanisms; they are not heat capacities.

The minimum-volume aspect ratio is $H/R\simeq 1.847$. Keep the material parameter $B_m^2=(\eta -1){\Sigma }_a/D$ fixed and positive. Write $a=2.405$ and $x=H/R$. The constraint from part (c) givesThe cylinder volume is consequentlyIts logarithmic derivative isThe denominator is positive. The derivative changes from negative to positive exactly once, and $V$ diverges as $x\to 0$ or $x\to \infty$. Hence this stationary point is the global minimum:Equivalently $H/(2R)\simeq 0.924$. This is an aspect ratio in the supplied idealized boundary value problem; it does not determine an absolute size without a specified material parameter.

The one-group particle balance with a prescribed external volume source $S$ can be writtenHere $\varphi$ is scalar neutron flux, $D$ is the neutron diffusion coefficient, and ${\Sigma }_a,{\Sigma }_f$ are macroscopic cross sections. The terms describe accumulation, current divergence, absorption, nuclear fission production and external input.

For a steady state, set ${\partial }_t\varphi =0$; source-free means $S=0$. In a homogeneous medium $D$ is spatially constant, so substitution givesThe supplied notation defines $\eta$ as the effective production per absorption, $\eta =\nu {\Sigma }_f/{\Sigma }_a$. ThereforeThis identification uses the absorption accounting for the homogeneous model; a fuel-only reproduction factor with other absorbers excluded would not automatically equal this $\eta$. The derivation assumes one prescribed energy group, constant effective coefficients and the diffusion approximation; all neutron production included here is represented by the steady nuclear fission term.

One energy group and a nearly isotropic local neutron distribution are the main approximations. The one-group neutron diffusion equation replaces the full energy-dependent transport problem by one scalar neutron flux and averaged coefficients. A single group cannot resolve spectral changes, resonance absorption or differences between rapidly moving and thermal neutrons; its effective parameters depend on the spectrum used for averaging.

The diffusion approximation also replaces angular transport by a current proportional to the negative neutron flux gradient. It works best where repeated scattering makes the angular distribution nearly isotropic over distances comparable to the field's variation length. It is less reliable close to a sharply localized source, a vacuum boundary, a strong absorber or an abrupt material interface, and in weakly scattering regions. Approximate boundary extrapolation does not restore the lost angular information.

Thus an apparently smooth solution of the one-group neutron diffusion equation need not accurately represent a transport field outside the assumptions that produced its coefficients and boundary conditions.

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.

The idealized external photon estimate is approximately $0.557\mathrm{mSv}{\mathrm{h}}^{-1}$ at one metre. First apply radioactive decay to the certified radioactive activity:Treat the source as isotropic and small compared with the observation distance. Neglect photon self-absorption, capsule attenuation and attenuation in air. The metal and capsule stop the primary beta particles, so the direct external contribution is taken to be the two gamma rays; secondary bremsstrahlung is neglected. Their total energy per transformation isThe incident photon energy flow per area at distance $r$ is thereforeInterpret the given $3{\mathrm{m}}^{-1}$ as the linear energy-absorption coefficient ${\mu }_{\mathrm{en}}$, rather than a total photon-removal coefficient. For an absorbing medium of mass density $\rho$, a thin volume $Sdx$ absorbs energy per time ${\Psi }_{E}S{\mu }_{\mathrm{en}}dx$ and has mass $\rho Sdx$. Its absorbed dose rate is consequentlyFor the estimate assume a water-like $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$ and local deposition of the secondary charged-particle energy. At $r=1\mathrm{m}$,For gamma rays, the radiation weighting factor is one, giving $\dot{H}=0.556815\mathrm{mSv}{\mathrm{h}}^{-1}$ under these assumptions.^[ref]

The phrase “equivalent absorbed dose” mixes two quantity names: absorbed dose is measured in gray, while equivalent dose is measured in sievert. Their rates are numerically equal here because the weighting factor is one. The estimate is a physical energy-deposition calculation; the assumed mass density is essential, and the stated inverse-length coefficient must not be used as if it already had area-per-mass units.

The cobalt-60 mass is approximately $5.73\times 10^{-8}\mathrm{kg}$. The nuclear decay constant follows from the half-life, using a $365.25$-day year:Since radioactive activity is $A=\lambda N$, the parent population at certification isConvert the supplied atomic mass into molar mass, $M=0.05993382\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, and use the Avogadro constant:This is about $57.3\mu \mathrm{g}$ of the specified isotope. It is not the total mass of the pellets or capsule, which can also contain other material.

Reduce time, increase distance and use radiation-appropriate shielding. At fixed conditions, accumulated absorbed dose is proportional to exposure time. For an approximately isotropic point source, the inverse-square law reduces the uncollided radiation field as distance increases. Confinement of radioactive material prevents contamination; it addresses a different pathway from attenuation of an external field.

For alpha particles, short stopping ranges make light barriers sufficient against external particles, but confinement remains important. For beta particles, a low-atomic number shield such as plastic limits both penetration and secondary bremsstrahlung; high-atomic number material can then attenuate residual photons. For penetrating gamma rays, dense material such as lead, or a thicker bulk shield such as concrete, provides attenuation. One shield material is not equally appropriate for every radiation type.^[ref][ref]