Past exam of the MPhil in nuclear energy course of the University of Cambridge / 2021 / Module NE1 / 3

Cadmium acts as an energy-selective neutron absorber in the stated comparison. Its large thermal capture collision cross-section preferentially reduces the thermal neutron field reaching the source material, while its much lower fast-neutron capture probability allows more of the emitted fast neutrons to escape. This limits thermal neutron-induced changes in the source while retaining its qualitative role as a neutron emitter.

The point is selectivity between incoming thermal and outgoing fast components, rather than a claim that cadmium is transparent to every neutron or that capture produces no secondary radiation. This explains the stated material choice at a conceptual level; no source construction, cladding dimensions or irradiation specification follows from it.

For the prescribed homogeneous one-group model, number balance isUse steady state, $\partial n/\partial t=0$, and the diffusion approximation $\mathbf{J}=-D\nabla \varphi$. A spatially constant $D$ givesFor a spherically symmetric scalar field,Dividing by $D$ reproduces the requested radial equation. The assumptions are a steady source, spherical symmetry, homogeneous effective coefficients, a single energy-group description and a diffusion closure. In particular, replacing the general production term by the supplied $(\eta -1){\Sigma }_a\varphi$ is a model assumption, not a general identity for heterogeneous reactor materials.

This part is a radial differential-equation identity. Expanding the spherical derivative givesPut $\psi =r\varphi$. Since ${\psi }^{\prime \prime }=r{\varphi }^{\prime \prime }+2{\varphi }^{\prime }$, multiplication of the equation by $r$ yields ${\psi }^{\prime \prime }+B^2\psi =0$. Its real solution isFor a regular field at the centre, the cosine-over-radius contribution must vanish. The sine-over-radius term is regular because $\sin (Br)/r\to B$ as $r\to 0$.

Write $R_e=R^{+}$ and apply the stated constant-coefficient boundary value problem. A constant particular solution is $-S/(D{B}^2)$. Regularity at the origin removes $\cos (Br)/r$, soThe outer boundary condition gives $A=S{R}_e/[D{B}^2\sin (B{R}_e)]$, henceAt $r=0$, use the limit $\sin (Br)/r\to B$ rather than substitute into the apparent $0/0$. The result satisfies both the differential equation and $\varphi (R_e)=0$ provided $\sin (B{R}_e)\ne 0$. This is the ideal extrapolated-boundary model specified in the question; a separately resolved outer transport layer would need its own matching conditions.

At the stated threshold the idealized system admits a self-sustaining neutron mode. Production and losses balance without the external source. A nonzero continuing source then makes the assumed linear, time-independent response singular rather than selecting a finite equilibrium amplitude.

An apparent infinite steady flux is therefore a failure of that steady linear approximation at the threshold, not a physical prediction that an actual neutron population instantly becomes infinite. Time dependence and feedback must be considered to discuss subsequent behaviour.