Revise Fission for IB Physics (E.4)

Einstein's mass–energy equivalence states that mass and energy are interchangeable, related by $E = mc^2$. Because the speed of light $c \approx 3.0 \times 10^8$ m s⁻¹ is so large, $c^2$ is about $9 \times 10^{16}$, so even a minute change in mass corresponds to a very large energy. This is why nuclear processes, which convert a tiny fraction of a percent of the mass, release millions of times more energy per event than chemical reactions.

If you weigh a nucleus carefully, it is always slightly lighter than the sum of the masses of its separate protons and neutrons. This difference is the mass defect, $\Delta m$. It is not that mass has vanished: forming the nucleus released energy, and by $E=mc^2$ that released energy shows up as reduced mass. Masses at this scale are usually quoted in unified atomic mass units (u), where $1\text{ u} = 1.661 \times 10^{-27}$ kg, and the handy conversion $1\text{ u} = 931.5$ MeV$c^{-2}$ lets you turn a mass defect straight into an energy.

The binding energy of a nucleus is the energy that would be needed to pull it completely apart into free protons and neutrons — equivalently, the energy released when those nucleons come together. It is simply the mass defect expressed as energy: $E_\text{binding} = \Delta m\,c^2$. A larger binding energy means a more tightly held nucleus, but total binding energy is not itself a fair measure of stability, because bigger nuclei naturally have more of it just by having more nucleons.

The fair measure is the binding energy per nucleon: the total binding energy divided by the nucleon number $A$. This tells you how tightly, on average, each individual nucleon is bound, and it is the single most important quantity in this topic. The rule to memorise: the higher the binding energy per nucleon, the more stable the nucleus.

Plot binding energy per nucleon against nucleon number and you get the single most useful graph in nuclear physics. It rises steeply for the lightest nuclei, climbs to a broad maximum around iron ($^{56}\text{Fe}$, at roughly 8.8 MeV per nucleon), then declines slowly for the heaviest nuclei such as uranium (around 7.6 MeV per nucleon). The peak marks the most tightly bound, most stable nuclei of all — which is why iron and nickel are so abundant as the end products of stellar nuclear burning.

Certain heavy nuclei can be made to split. For uranium-235, the fuel in most reactors, induced fission happens when the nucleus absorbs a slow-moving thermal neutron. The neutron's capture forms a highly excited uranium-236 nucleus that deforms, oscillates, and splits into two lighter fragments, releasing typically two or three further neutrons and a large amount of energy. The fragments are not fixed — many different pairs can form — and they emerge with enormous kinetic energy, which is ultimately the heat we harness.

A representative fission reaction is: $^{1}_{0}n + {}^{235}_{92}\text{U} \rightarrow [{}^{236}_{92}\text{U}]^* \rightarrow {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3\,{}^{1}_{0}n + \text{energy}$

Notice how the reaction obeys two conservation rules that let you balance any fission equation. The top numbers (nucleon number $A$) must balance: $1 + 235 = 141 + 92 + 3(1) = 236$. The bottom numbers (proton number $Z$) must balance: $0 + 92 = 56 + 36 + 0 = 92$. The energy released comes not from the neutron but from the increase in binding energy: the fragments sit higher up the binding-energy curve than uranium, so overall a small mass is lost and converted to energy through $E = \Delta m c^2$.

The reason fission can power a reactor is that each event releases more neutrons, which can trigger further fissions — a chain reaction. Whether it grows, holds steady or fades depends on the average number of neutrons per fission that go on to cause another fission.

To sustain a chain reaction there must be enough fissile material packed closely enough that neutrons find another nucleus before escaping through the surface. The minimum amount is the critical mass. Below it, too many neutrons leak out and the reaction cannot sustain itself; the critical mass depends on the material's purity, density and shape (a sphere, with the least surface area for its volume, needs the least mass).

A reactor is a machine for holding a chain reaction exactly at critical ($k=1$) so it produces heat at a steady rate. Two components make this possible. The moderator (water or graphite) slows the fast fission neutrons to thermal speeds, because U-235 captures slow neutrons far more readily than fast ones — the moderator does not absorb neutrons, it only slows them. The control rods (boron or cadmium) absorb neutrons, and operators raise or lower them to tune the fission rate precisely: pushed in, they soak up neutrons and slow the reaction; withdrawn, they let it speed up.

Fission's appeal is its energy density: a few kilograms of uranium can release as much energy as thousands of tonnes of coal, with essentially no carbon dioxide emitted during operation, and it delivers steady large-scale output regardless of weather. Against that stand real risks: the fission fragments are intensely radioactive and some remain dangerous for thousands of years, demanding secure long-term storage; a serious accident can release radioactive material over a wide area; and the same technology and materials overlap with nuclear weapons, raising security and proliferation concerns. Weighing these trade-offs is exactly the kind of evaluation an exam question may ask you to make.

In Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an answer mark (A) — and error-carried-forward (ECF) means a wrong number early on does not have to cost you the marks that follow. But that protection only exists if your method is written down. Study how each mark below is earned by a specific line in a mass-defect calculation, where the substitution, the joules answer and the MeV conversion are each marked separately.

The single idea behind this whole topic — that stability is measured by binding energy per nucleon, and that moving up that curve releases energy — carries directly into fusion, the process that powers the Sun and the stars, where light nuclei on the left of the curve join to release even more energy per nucleon than fission. The mass–energy equivalence you used here reappears throughout modern physics, from particle production to the energy of the Sun. Master the habit — find the mass defect, convert with $E = mc^2$, compare stability per nucleon, balance every equation — and both fission and fusion become variations on a method you already own.