Fission

Nuclear physics rests on one revolutionary idea from Einstein: mass and energy are two forms of the same thing, linked by

$E = \Delta m\,c^{2}$

where $c = 3.0\times10^{8}\,\text{m s}^{-1}$ is the speed of light. Because $c$ is enormous and it is squared, a minute change in mass corresponds to a huge amount of energy. This is why nuclear reactions release millions of times more energy per kilogram than chemical (burning) reactions.

The mass defect — the surprising fact. If you could weigh a nucleus and separately weigh all the protons and neutrons (nucleons) that make it up, the nucleus is always lighter than the sum of its parts. The difference is the mass defect:

$\Delta m = \left(Z\,m_{p} + N\,m_{n}\right) - m_{\text{nucleus}}$

where $Z$ is the number of protons (each of mass $m_p$), $N$ the number of neutrons (each of mass $m_n$), and $m_{\text{nucleus}}$ the measured mass of the assembled nucleus.

Where did the mass go? When separate nucleons come together to form a nucleus, the strong nuclear force does work and energy is released. By $E = \Delta m c^2$, releasing that energy means the system loses mass — so the finished nucleus is lighter. That lost mass, converted to energy, is exactly the binding energy (next section).

Deriving the energy released — a first-principles walkthrough. Suppose a reaction has a mass defect $\Delta m$ (the total mass decreases by $\Delta m$). The energy released is:

$E = \Delta m\,c^{2}$

There are two equally valid ways to put numbers in:

1. Work in SI units. Convert $\Delta m$ to kilograms, multiply by $c^2 = (3.0\times10^{8})^2 = 9.0\times10^{16}\,\text{m}^2\text{s}^{-2}$, and the answer comes out in joules.
2. Use the shortcut for atomic mass units. Physicists pre-computed the energy equivalent of one unified atomic mass unit: $1\,\text{u} \;\rightarrow\; (1.66\times10^{-27})(3.0\times10^{8})^2 = 1.49\times10^{-10}\,\text{J} \approx 931.5\,\text{MeV}.$ So if $\Delta m$ is given in u, the energy in MeV is simply $E\,(\text{MeV}) = \Delta m\,(\text{u}) \times 931.5.$

That single line — multiply the mass defect in u by 931.5 to get MeV — is the workhorse of every fission energy calculation.

The units you must know cold:

• Unified atomic mass unit: $1\,\text{u} = 1.66\times10^{-27}\,\text{kg} = 931.5\,\text{MeV}/c^{2}$ (roughly the mass of one nucleon).
• Megaelectronvolt: $1\,\text{MeV} = 1.60\times10^{-13}\,\text{J}$ (a convenient energy unit for the nuclear world).

Binding energy is the energy that would be needed to completely separate a nucleus into its individual nucleons — equivalently, it is the energy released when those nucleons come together to form the nucleus. It is simply the energy equivalent of the mass defect:

$E_{\text{binding}} = \Delta m\,c^{2}.$

A larger binding energy means the nucleus is more tightly held together and harder to break apart.

Why "per nucleon"? A big nucleus has a large total binding energy just because it has more nucleons — that does not by itself tell you how stable it is. To compare fairly, we use the binding energy per nucleon:

$\frac{E_{\text{binding}}}{A} \quad\text{(measured in MeV per nucleon)},$

where $A$ is the nucleon (mass) number. The higher the binding energy per nucleon, the more stable the nucleus.

The curve every IB student must be able to sketch. Plotting binding energy per nucleon against nucleon number $A$ gives a famous shape:

• It rises steeply for the lightest nuclei (hydrogen, helium…).
• It reaches a broad peak at around iron-56 (Fe-56), at roughly 8.8 MeV per nucleon — the most stable nuclei.
• It then falls slowly for heavier nuclei, so uranium (A ≈ 235) sits at about 7.6 MeV per nucleon, below the peak.

Iron-56 sitting at the top is the reason it is the "end of the road" for energy release: you cannot get energy by fusing or splitting nuclei that are already at the peak.

Nuclear fission is the splitting of a heavy nucleus (such as uranium-235) into two lighter nuclei ("daughter" fragments), together with a few neutrons and a release of energy.

The energy comes from the binding-energy-per-nucleon curve. Follow the logic carefully — this is the heart of E.4:

1. Uranium-235 lies on the right-hand, downhill part of the curve, at about 7.6 MeV per nucleon.
2. When it splits, the two fragments are medium-mass nuclei ($A \approx 90$–$145$) that sit closer to the iron-56 peak, at about 8.5 MeV per nucleon.
3. So after fission, each nucleon is more tightly bound — the binding energy per nucleon has increased.
4. An increase in binding energy means energy has been released (the products are more stable, at lower energy). By $E = \Delta m c^2$, this shows up as a mass defect: the total mass of the products is slightly less than that of the reactants.

Estimating the energy released from the curve. If all $A \approx 236$ nucleons each become more tightly bound by roughly $(8.5 - 7.6) = 0.9\,\text{MeV}$:

$E \approx 236 \times 0.9 \approx 210\,\text{MeV per fission}.$

That ~200 MeV is millions of times the few eV released when a single atom takes part in a chemical (burning) reaction — the reason nuclear fuel is so energy-dense.

Fission vs fusion — the same principle, opposite sides of the peak. Both processes move nuclei towards the iron-56 peak and both release energy:

• Fission: heavy nuclei split → move left/up towards the peak.
• Fusion: light nuclei join → move right/up towards the peak.

You cannot release energy by splitting a light nucleus or fusing a heavy one — that would move away from the peak and absorb energy.

Most fission in a reactor is induced (neutron-induced) fission: a uranium-235 nucleus absorbs a slow-moving (thermal) neutron, becomes highly unstable, and splits. A typical reaction is:

$^{235}_{\;92}\text{U} + \,^{1}_{0}\text{n} \;\rightarrow\; ^{141}_{\;56}\text{Ba} + \,^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n} + \text{energy}$

Check the conservation rules — always do this:

• Nucleon number $A$: $235 + 1 = 236$ on the left; $141 + 92 + 3(1) = 236$ on the right. ✓
• Proton number $Z$: $92 + 0 = 92$ on the left; $56 + 36 + 3(0) = 92$ on the right. ✓

The exact fragments vary from event to event, but 2–3 neutrons are typically released each time, and $A$ and $Z$ must always balance.

The chain reaction. Here is the crucial feedback loop: each fission is triggered by one neutron but releases 2–3 neutrons. Those free neutrons can go on to be absorbed by other U-235 nuclei, causing further fissions, which release yet more neutrons — a self-sustaining chain reaction.

Critical mass. Whether the chain reaction sustains itself depends on how many neutrons escape the fuel without causing a fission. The critical mass is the minimum mass of fissile material needed for a self-sustaining chain reaction — enough that, on average, at least one neutron from each fission goes on to trigger another. Below the critical mass too many neutrons leak out and the reaction dies; at (or above) it, the reaction is sustained. In a power reactor the aim is a steady, controlled rate (exactly one further fission per fission on average), not a runaway increase.

A nuclear reactor is engineered to keep a controlled chain reaction running and to turn the released energy into electricity. Learn the five components and their roles — examiners ask for these directly, and love to test the moderator-vs-control-rod distinction.

Component	What it is	Its role
Fuel	Enriched uranium (raised proportion of U-235)	Provides the fissile nuclei that split and release energy.
Moderator	Water or graphite (surrounds the fuel)	Slows down the fast neutrons released by fission (through collisions) so they are slow enough to be absorbed by U-235 and cause further fission.
Control rods	Boron or cadmium (raised/lowered)	Absorb neutrons to control the rate of the chain reaction. Lower them in → fewer neutrons → slower reaction; raise them → faster.
Coolant / heat exchanger	A fluid (water, CO₂, …) pumped through the core	Carries the thermal energy away from the core to a heat exchanger, where it boils water to make steam that drives a turbine and generator.
Shielding	Thick concrete and/or lead around the core	Absorbs the radiation (neutrons and gamma rays) so it does not escape and harm people or the environment.

The key contrast (examiners love this): the moderator SLOWS neutrons (so they can be absorbed to keep fission going), whereas the control rods ABSORB neutrons (to remove them and control the rate). One keeps the reaction going; the other reins it in.

Radioactive waste (qualitative). The fission fragments (the daughter nuclei) are themselves radioactive, some with very long half-lives. This spent fuel / nuclear waste must be handled carefully: stored initially in cooling ponds, then sealed in shielded containers and stored securely (often deep underground) for a very long time until its activity has fallen to safe levels. Managing this long-lived waste is one of the main drawbacks of nuclear power, set against its advantage of generating large amounts of energy without emitting greenhouse gases.

Almost every numerical fission question in Paper 2 asks you to calculate the energy released. There are two standard routes — pick whichever the data suggests.

Route 1 — from a mass defect (given masses).

1. Add up the total mass of the reactants (parent nucleus + the absorbed neutron).
2. Add up the total mass of the products (both fragments + all the released neutrons).
3. Find the mass defect: $\Delta m = m_{\text{before}} - m_{\text{after}}$ (this is positive because mass is lost).
4. Convert to energy:
   • if $\Delta m$ is in u, use the shortcut $E(\text{MeV}) = \Delta m \times 931.5$;
   • if you need joules, convert $\Delta m$ to kg ($\times 1.66\times10^{-27}$) and use $E = \Delta m c^{2}$, or convert the MeV answer with $1\,\text{MeV} = 1.60\times10^{-13}\,\text{J}$.

Route 2 — from a binding-energy-per-nucleon comparison (given the curve). $E \approx A\left[\left(\tfrac{E_b}{A}\right)_{\text{after}} - \left(\tfrac{E_b}{A}\right)_{\text{before}}\right],$ where $A$ (≈ 236 for U-235 fission) is the number of nucleons. Read the two binding-energy-per-nucleon values off the graph, take the increase, and multiply by the number of nucleons.

A worked mini-example (Route 1). For $^{235}\text{U} + \text{n} \rightarrow \,^{141}\text{Ba} + \,^{92}\text{Kr} + 3\text{n}$ with masses (in u): U-235 = 235.0439, n = 1.0087, Ba-141 = 140.9144, Kr-92 = 91.9262: $m_{\text{before}} = 235.0439 + 1.0087 = 236.0526\,\text{u}$ $m_{\text{after}} = 140.9144 + 91.9262 + 3(1.0087) = 235.8667\,\text{u}$ $\Delta m = 0.1859\,\text{u} \;\Rightarrow\; E = 0.1859 \times 931.5 = 173\,\text{MeV}.$

Scaling up to a reactor. Once you know the energy per fission in joules ($173\,\text{MeV} \approx 2.8\times10^{-11}\,\text{J}$), a power output $P$ (in watts = joules per second) tells you the number of fissions per second: $N = \frac{P}{E_{\text{per fission}}}.$ This is how a lump of uranium the size of a fingertip can run a power station: even at ~10⁻¹¹ J per fission, ~10¹⁹ fissions happen every second.

Units discipline (where marks are won and lost): decide before you start whether you want the answer in MeV or J, keep the two conversion factors straight ($931.5\,\text{MeV per u}$ and $1.60\times10^{-13}\,\text{J per MeV}$), and always quote the unit with the final number.