Question 5 of 7: Energetics and Radioactivity of Thermal-Neutron Fission

Question 5: Energetics and Radioactivity of Thermal-Neutron Fission (16 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The thermal-neutron fission equation $n+{}^{235}\text{U}\rightarrow{}^{236}\text{U}^*\rightarrow FF_1+FF_2+3n+\gamma$; the binding-energy-per-nucleon curve, which peaks near $A\approx56$ (iron/nickel) and is lower at both $A\approx235$ and at the mid-mass fragment values ($A\approx90$–140).

Find. (a) why the reaction releases energy; (b) the approximate percentage split of that energy among fragments, neutrons and gamma rays; (c)(i) why fission fragments are radioactive; (c)(ii) which decay mode dominates, and why it stabilizes the fragment.

Approach. Compare binding energy per nucleon before and after fission via the mass-defect argument; use standard illustrative energy-partition figures (Krane) for part (b); use the parent's fixed $N/Z$ ratio, inherited by the fragments, to explain (c).

1. Part (a) — why energy is released. ${}^{235}$U (mass number 235) sits on the low, heavy-mass side of the binding-energy-per-nucleon curve (BE/A $\approx7.6$ MeV/nucleon), while the two mid-mass fragments $FF_1,FF_2$ land much closer to the curve's peak (BE/A $\approx8.5$ MeV/nucleon in the $A\approx90$–140 region). Splitting the nucleus therefore increases total binding energy (more binding energy means a more negative, i.e. lower, total mass-energy), so the products' total rest mass is measurably less than the initial $n+{}^{235}$U rest mass. By $E=\Delta mc^2$, that mass deficit reappears as kinetic energy of the fragments, neutrons, and as gamma/beta/neutrino energy from subsequent decay — roughly 200 MeV per fission, an enormous energy release for a single nuclear event.
2. Part (b) — energy distribution (illustrative percentages). Using standard textbook energy-partition figures for ${}^{235}$U thermal fission (total ∼207 MeV): $$\boxed{\text{Fission fragments (kinetic energy)}: 168\ \text{MeV}\approx81\%}$$ $$\boxed{\text{Prompt neutrons (kinetic energy)}: 5\ \text{MeV}\approx2\%}$$ $$\boxed{\text{Prompt gamma rays}: 7\ \text{MeV}\approx3\%}$$ The remaining ∼14% (beta particles ∼8 MeV, decay gammas ∼7 MeV, antineutrinos ∼12 MeV) is released later, from the radioactive decay of the fragments described in part (c), rather than by the immediate reaction products shown in the given equation — antineutrino energy in particular escapes the reactor entirely and is never recovered as heat. The overwhelming majority of the prompt energy stays with the fragments themselves, because they carry almost all of the momentum shared between only two heavy bodies (versus three light neutrons and a massless photon).
3. Part (c)(i) — why fragments are radioactive. Stable nuclei need a progressively higher neutron-to-proton ratio as mass number increases, but that ratio never needs to be as high as ${}^{235}$U's own ($N/Z\approx1.55$). When ${}^{235}$U splits, each fragment inherits essentially the same high $N/Z$ ratio as the parent, but a stable nucleus at the fragment's much lower mass number ($A\approx90$–140) needs a substantially lower $N/Z$ (∼1.3–1.4). Every fragment is therefore born well above the stability line — too neutron-rich for its mass — which is precisely why essentially all fission fragments are radioactive from the moment they are created.
4. Part (c)(ii) — dominant decay mode and why it stabilizes. Being neutron-rich (not proton-rich, and not simply "too heavy"), fragments decay overwhelmingly by $\beta^-$ emission ($n\rightarrow p+\beta^-+\bar\nu$), not by $\alpha$ decay (which is instead characteristic of very heavy, proton-and-neutron-rich nuclei near the top of the periodic table, not mid-mass fragments). Each $\beta^-$ decay converts one excess neutron into a proton within the same nucleus (mass number $A$ unchanged, atomic number $Z$ increases by one), moving the fragment one step along its isobaric chain toward the stability valley — directly reducing the neutron excess that made it unstable in the first place, and generally increasing its binding energy per nucleon as it approaches the bottom of that mass-parabola.