Past exam of the MPhil in nuclear energy course of the University of Cambridge / 2021 / Module NE1 / 1

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.