Original solutions to all questions and alternatives. ^[ref]

Original source: Released examination PDF, physical pages 347–350. All three note alternatives and both longer questions are treated.

Original paper: Public original PDF, physical pages 299–302; cover on page 299 and questions on pages 300–302.

Original paper: Public original PDF, physical pages 254–257; cover on page 254 and questions on pages 255–257.

Original public paper: Physical PDF pages 826–828. All three questions and all alternatives in question 2 are answered. Optical amplitudes and quadratures use the normalization printed in the paper; factors of two depend on that convention.

Original paper: 22 April 2015, physical PDF pages 751–754.

Original paper: 23 April 2014, physical PDF pages 678–680.

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.

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.

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 $r\to 0$.

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.

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.

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 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.

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 approximate population peaks about $11.3\mathrm{h}$ after flux removal. Differentiate the preceding difference of exponentials. The stationary condition isso for distinct nuclear decay constants,Using hours consistently,and thereforeThe approximate population is positive for $t>0$, starts at zero, tends back to zero and has only this one stationary point, so it is a maximum.

The delayed maximum corresponds to a delayed maximum in the magnitude of the absorber's negative contribution to reactor reactivity, within the fixed model. It does not determine a permissible restart time, which would require information absent here. The estimate also inherits the high-flux initial-population approximation of part (ii); the exact solution can peak at a different time when that approximation is unsuitable.

The general solution contains both decay of the initial xenon and continued feeding from the initial iodine. Use $t=0$ for removal of the flux, and retain $\varphi$ as the pre-shutdown value in the initial conditions. Neglect post-shutdown nuclear fission and neutron capture. ThenMultiply this first-order linear differential equation by the integrating factor $e^{{\lambda }_{X}t}$ and integrate:With the preceding steady initial conditions, ${\lambda }_I{I}_0=G={\gamma }_I{\Sigma }_f\varphi$ and $X_0=G/({\lambda }_X+\varphi {\sigma }_{aX})$, this isIf ${\lambda }_X+\varphi {\sigma }_{aX}\gg |{\lambda }_X-{\lambda }_I|$, the coefficient of the first exponential term is small compared with the coefficients of the feeding term, yieldingThis approximation neglects the finite initial xenon population and is consequently not a small relative error at $t=0$, where its value is zero while the exact $X_0$ is positive. The exact expression supplies the initial value and satisfies $\dot{X}(0)=G-{\lambda }_X{X}_0$. For equal nuclear decay constants, the continuous limit of the exact solution is $X(t)=(X_0+{\lambda }_I{I}_{0}t)e^{-{\lambda }_{I}t}$.

Let $G={\gamma }_I{\Sigma }_f\varphi$ be the constant precursor production rate per volume. Here ${\gamma }_I$ is the precursor yield per nuclear fission, and $I$ and $X$ are number densities. At a steady state,For fixed homogeneous coefficients and spatial shape, the reactivity contribution of an added absorption coefficient ${\Sigma }_{aX}={\sigma }_{aX}{X}_0$ is $-{\Sigma }_{aX}/(\nu {\Sigma }_f)$. Thus the model givesWhen capture removal dominates xenon radioactive decay, $\varphi {\sigma }_{aX}\gg {\lambda }_X$, and henceThe limit is independent of the xenon neutron absorption cross section: at high neutron flux, nearly every xenon atom supplied by the precursor ultimately consumes a neutron. The assumptions include constant pre-shutdown neutron flux, sufficient time for precursor and xenon equilibrium, the supplied single-precursor model with no direct xenon fission source, and fixed parameters for translating absorption into reactor reactivity.