Past exam of the MPhil in nuclear energy course of the University of Cambridge / 2021 / Module NE1 (Reactor physics)

All four printed alternatives are represented. Questions occupy physical pages 44–47. The solutions use the stipulated idealized equations and state their assumptions.

The lighter product receives the fraction $1/(1+m/M)$ of the two massive products' kinetic energy. Neglecting photon momentum, conservation of momentum gives equal and opposite daughter momenta, ${\mathbf{p}}_m=-{\mathbf{p}}_M$. In the nonrelativistic approximation,Hence $K_{\mathrm{products}}=K_m(1+m/M)$ andThis partitions kinetic energy after any stipulated photon energy has been removed from the available Q value. It does not assign the complete mass-defect energy to the light particle.

The stipulated decay branch gives $K_{\alpha }\simeq 4.51\mathrm{MeV}$. Using the supplied atomic masses, the mass defect isso the total Q value is $Q\simeq 0.00580(931.494)=5.4027\mathrm{MeV}$. The photon carries away the stated $0.802\mathrm{MeV}$, leavingApplying the recoil partition,The daughter recoil takes approximately $0.0877\mathrm{MeV}$. Atomic electron counts balance in the neutral-atom mass difference; small electronic binding corrections and photon recoil are neglected consistently with the problem. These values follow the branch specified in the examination, rather than an assumed average over every physical decay branch.

Over $N$ half-lives, the duration is $T=N{t}_{1/2}=N\ln 2/\lambda$. The time-averaged thermal power isSince $\lambda T=N\ln 2$ and $e^{-\lambda T}=2^{-N}$,This is a time average, not the arithmetic mean of the initial and final powers. It follows directly from the assumed constant nuclear decay constant and constant recovered energy per decay.

After three half-lives, $P(T)=P_0/8$. Therefore the minimum initial thermal power isEquality gives the stated terminal power exactly in the idealized exponential decay model. The associated duration is $415.2$ days. No thermal-to-electrical efficiency is needed because both specified powers are thermal.

The idealized isotope mass is $1.63\times 10^{-2}\mathrm{kg}$, about $16.3\mathrm{g}$. Both the alpha particle and daughter recoil energy are recovered in the stipulated heat balance. Thus the recovered energy per transformation isand the nuclear decay constant isWith $N_0=(m/M_{\mathrm{mol}})N_A$, thermal power is $P_0=\lambda N_0{E}_h$. ConsequentlyThe conversion of $210\mathrm{kg}\mathrm{kmo}{\mathrm{l}}^{-1}$ to $0.210\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ is essential. This is an isotope-only energy-accounting result under the paper's prescribed branch and perfect heat-recovery assumptions; it is not the mass or design of a complete generator.

Alpha decay releases energy in charged particles with short stopping ranges in matter, so much of their kinetic energy becomes local heat. An appropriate radioisotope thermoelectric generator can use a predictable radioactive decay heat source without requiring sunlight or a sustained nuclear chain reaction. Low penetrating photon and neutron output is desirable, but being an alpha emitter does not by itself establish the complete radiation spectrum.

The isotope's half-life must suit the mission. A short half-life gives high initial specific activity but a rapid decline in useful thermal power; a much longer one gives a steadier output but less activity per atom. Selection also depends on recoverable decay energy, specific power, daughter emissions, chemical and thermal stability, compatibility with the power-conversion system, availability and cost. Practical containment and radiological acceptability remain separate system-level requirements.

A thermoelectric generator converts a temperature difference into electrical output through the Seebeck effect; not all decay heat becomes electricity. NASA's operational account describes this heat-to-electricity distinction and the need to match source lifetime to long space missions.^[ref]

For the assumed three-half-life mission, the thermal power falls by a factor of eight. That large decline is the important disadvantage of this particular half-life even though the initial specific thermal power is high.

Delayed neutrons are emitted after precursor radioactive transformations, rather than promptly at fission. Some neutron-rich fission products undergo beta decay into excited daughter states that can emit a neutron. The waiting time is governed chiefly by the precursor's radioactive lifetime, not by a neutron moving unusually slowly after it has already been emitted.

Although their fraction is small, delayed neutrons introduce a much slower population timescale into the nuclear chain reaction. In the delayed-neutron-controlled regime, changes can therefore develop over times comparable with precursor lifetimes rather than only the prompt neutron generation time. That timescale separation is central to qualitative reactor controllability. It does not imply that every positive-reactivity transient is slow.

The point kinetics model replaces the spatially and spectrally distributed neutron field by a single amplitude $n(t)$, effectively assuming a fixed or sufficiently slowly varying flux shape. The source amplitude is $s$. A single effective delayed neutron precursor population $c(t)$, delayed neutron fraction $\beta$ and decay rate $\lambda$ stand in for many precursor species with different lifetimes. $\Lambda$ is the prompt neutron generation time and $\rho$ the specified reactor reactivity.

The displayed equations omit explicit thermal-hydraulic, fuel-temperature and isotope-composition feedback. For the stated constant-coefficient transients, $\rho ,\beta ,\Lambda ,\lambda$ are prescribed rather than being solved as evolving fields. The model is useful for an elementary timescale calculation; it does not establish the spatial stability or operating limits of an actual core.

At equilibrium, precursor production balances precursor decay:ThereforeThe delayed neutron precursor population can exceed the instantaneous neutron population even when the delayed neutron fraction is small: the delayed neutron precursor reservoir persists much longer than an individual prompt-neutron generation.

Multiply the supplied inhour equation by $\omega +\lambda$ and collect powers of the inverse period:With the stipulated constants this becomeswith roots ${\omega }_{+}=0.0165064{\mathrm{s}}^{-1}$ and ${\omega }_{-}=-12.1165{\mathrm{s}}^{-1}$. The negative root describes a rapidly decaying component. The positive root governs the late-time growth, givingThe growing-mode e-folding time is not the precursor mean lifetime, and the negative root is a legitimate transient mode rather than an unphysical negative time.

After the short prompt adjustment, the prompt jump approximation takes the neutron amplitude to remain nearly in balance with the slowly changing delayed neutron precursor reservoir. For the source-free model,Substitute this quasi-steady relation into the precursor equation:For constant coefficients,The approximation is meaningful in the stated delayed-neutron regime with $\rho <\beta$ and a separated fast transient. The delayed neutron precursor reservoir remains continuous across an idealized instantaneous reactivity step; the approximation does not resolve the short prompt adjustment itself.

Differentiating the supplied relation givesThus the tangent approximation is $\rho \simeq \omega (\Lambda +\beta /\lambda )$ andIt is about $16.4\%$ longer than $60.58\mathrm{s}$ and therefore predicts slower growth. For this positive step it is not conservative as a growth-rate estimate. The local expansion needs $|\omega |\ll \lambda$; here the actual ratio is approximately $0.165$, so its error is appreciable.

The precursor exponential from part (e) has inverse period $\rho \lambda /(\beta -\rho )$. HenceThis is approximately $0.96\%$ shorter than the exact dominant period, so it slightly overestimates growth for these stipulated constants. Neglecting the small $\Lambda$ contribution is more accurate here than replacing the complete frequency dependence by its tangent at the origin.

With no delayed neutrons, set $\beta =0$ and omit the precursor equation. The source-free model reduces to $\dot{n}=(\rho /\Lambda )n$, givingThe delayed-neutron model's dominant e-folding time is about $121$ times larger. This comparison illustrates why a small delayed neutron fraction strongly alters the kinetics. It is a comparison of the supplied idealized models at the same imposed positive reactivity, not a control prescription or a claim that every transient has the same timescale.

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.

The first balance treats the fertile parent as being removed by neutron absorption. The second has a production contribution associated with capture in the parent and a removal contribution from absorption in the fissile population. Capture and total absorption are not interchangeable: in general absorption includes capture and fission, and a capture channel is the one that feeds the specified transmutation chain.

The lumped description assumes spatially averaged populations and flux, fixed effective reaction coefficients, no transport of material into or out of the system, and a shortened chain in which intermediate decay delays are neglected. It omits explicit precursor accumulation, competing reaction chains and the changing neutron spectrum. It also prescribes flux rather than solving a coupled power-and-composition feedback problem. Neglecting intermediate lifetimes requires them to be short relative to the variation being modelled; that assumption can fail during a rapid change in irradiation.

The first equation is a prescribed first-order removal law. For a constant absorption coefficient,Integrating from the initial condition givesand henceThe neutron fluence $\Phi$ is the time integral of scalar neutron flux; it has inverse-area units, making $\sigma \Phi$ dimensionless. Constant flux is not needed for this integration, but a constant effective coefficient is.

Use neutron fluence $\Phi$ as the independent variable and abbreviate $a={\sigma }_{a\mathrm{Th}}$, $b={\sigma }_{a\mathrm{U}}$ and $c={\sigma }_{c\mathrm{Th}}$. Substituting the parent solution into the second linear differential equation givesThe integrating factor is $e^{b\Phi }$. Integrating from the initial state yieldsThe first term is survival of the initial population; the second accumulates production at earlier exposures, weighted by subsequent survival. For $b=a$, the continuous limit is $(N_{U0}+c{N}_{Th0}\Phi )e^{-a\Phi }$. The same formula applies to a time-varying prescribed flux through $\Phi (t)$; intervals of zero flux leave both populations unchanged.

The production rate is ${\sigma }_{c\mathrm{Th}}\varphi N_{\mathrm{Th}}$, while depletion includes both capture and fission of the fissile population, ${\sigma }_{a\mathrm{U}}\varphi N_{\mathrm{U}}$. Thus the conversion ratio in this stipulated model isAt zero instantaneous flux this expression describes the limiting ratio of the two reaction rates. The total absorption coefficients are $a=3.08+0.02=3.10\mathrm{b}$ and $b=7.10+57.10=64.20\mathrm{b}$, soIt is below one: fissile nuclei are initially removed slightly faster than this simplified chain replaces them. Consequently this model is not initially self-replacing in fissile inventory. Inventory replacement is distinct from the instantaneous neutron balance that determines criticality.

At this point $N_{\mathrm{Th}}/N_{Th0}=0.95$, so $a\Phi =-\ln 0.95$ and $e^{-b\Phi }=0.95^{b/a}=0.345671$. The solution from part (c) givesThereforeThe fissile population has declined by about $4.51\%$ from its initial value, while the conversion ratio has decreased slightly from $0.9595$. The replacement shortfall persists over this interval. Maintaining a specified long-term fuel inventory would require replenishment or a different balance of reactions; the prescribed-flux equations alone do not establish a constant-power operating state. Intermediate decay delays and the changing neutron spectrum were excluded by the original lumped model.