Nuclear Fission and Fusion

Take a nucleus apart into its separate protons and neutrons and, remarkably, the separate nucleons together are more massive than the nucleus was. The difference is the mass defect:

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

where $Z$ is the number of protons, $N$ the number of neutrons. The bound nucleus is always the lighter object, so $\Delta m$ is positive.

Where did the mass go? By mass–energy equivalence it was released as energy when the nucleus formed:

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

This energy $\Delta E$ is the binding energy — the energy you would have to supply to pull the nucleus completely apart again into free nucleons. Equivalently, it is the energy released when the free nucleons come together. Because $c^2 = 9.00 \times 10^{16}$ is huge, even a mass defect of a tiny fraction of a u corresponds to millions of electronvolts.

Units — get these straight before every calculation.

• Masses of nuclei are usually quoted in the atomic mass unit (u), where $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$ (one-twelfth the mass of a carbon-12 atom).
• To use $\Delta E = c^2\Delta m$ for an answer in joules, first convert $\Delta m$ to kilograms.
• To convert joules to MeV, divide by $1.60 \times 10^{-13}$ (since $1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}$).
• Shortcut: $1\ \text{u} \equiv 931.5\ \text{MeV}/c^2$, so a mass defect in u × $931.5$ gives the energy directly in MeV. Use this as your default for MeV answers and as a cross-check for joule answers.

A nucleus with 235 nucleons naturally has a bigger total binding energy than one with 4 — simply because there are more nucleons to bind. So total binding energy is a poor measure of stability. Instead we use the binding energy per nucleon:

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

This is the average energy holding each nucleon in place. The higher the binding energy per nucleon, the more tightly bound and the more stable the nucleus.

Plotting binding energy per nucleon against nucleon number $A$ for all the elements gives the single most important graph in nuclear physics:

• It rises steeply for the lightest nuclei (hydrogen, helium, ...).
• It peaks near iron-56 at about $8.8\ \text{MeV}$ per nucleon — iron-56 is the most stable nucleus.
• It then falls slowly across the heavier elements up to uranium.

Why the shape matters. Any reaction that moves nucleons to a position of higher binding energy per nucleon makes them more tightly bound. More binding means less mass, so a mass defect appears and energy is released ($\Delta E = c^2\Delta m$). Because iron-56 sits at the top, iron-56 can release energy by neither fission nor fusion — it is the end of the line.

Nuclear fission is the splitting of a large (heavy) nucleus into two smaller nuclei, releasing energy. A uranium-235 nucleus absorbs a slow neutron, becomes unstable, and splits into two lighter 'daughter' nuclei plus two or three fast neutrons and a large amount of energy (about 200 MeV per fission). A typical example:

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

Balancing any nuclear equation uses two conservation rules, applied separately:

• Nucleon number $A$ (the top number) is conserved: $235 + 1 = 141 + 92 + 3(1) = 236$. ✓
• Proton number $Z$ (the bottom number) is conserved: $92 + 0 = 56 + 36 + 0 = 92$. ✓

A free neutron is written $^{1}_{0}\text{n}$ — it adds 1 to the nucleon total and 0 to the proton total.

The chain reaction. Each fission releases 2–3 neutrons, and each of those can be absorbed by another U-235 nucleus, triggering further fissions — a self-sustaining chain reaction.

Controlling the reaction in a reactor. Left unchecked the chain reaction runs away, so a reactor uses two components (you should know their jobs qualitatively):

• Moderator (e.g. water, graphite): fission neutrons are too fast to be absorbed efficiently by U-235. The moderator slows the neutrons down through collisions so that they are readily absorbed and the reaction is sustained.
• Control rods (e.g. boron, cadmium): these absorb neutrons. Lowering them into the core removes neutrons and slows the reaction; raising them speeds it up. They keep the reaction steady (or shut it down).

The energy released mostly appears as kinetic energy of the daughter nuclei and neutrons, which heats a coolant to raise steam and drive turbines.

Nuclear fusion is the joining of two small (light) nuclei to form a single larger nucleus, releasing energy. A key reaction in fusion research is deuterium–tritium fusion:

$^{2}_{1}\text{H} + ^{3}_{1}\text{H} \rightarrow ^{4}_{2}\text{He} + ^{1}_{0}\text{n}$

Check the balance: $A$: $2 + 3 = 4 + 1$ ✓; $Z$: $1 + 1 = 2 + 0$ ✓. This single event releases about 17.6 MeV. Fusion is the energy source of the stars: in the Sun, hydrogen nuclei fuse in stages to form helium.

Why fusion needs extreme temperatures. Both nuclei are positively charged, so as they approach they feel an increasingly strong electrostatic (Coulomb) repulsion. To fuse, they must get close enough (about $10^{-15}\ \text{m}$) for the very short-range strong nuclear force — attractive at that range — to take over and bind them.

To overcome that repulsion, the nuclei must collide at very high speed, i.e. with very high kinetic energy. Temperature is a measure of mean kinetic energy, so a very high temperature (of order $10^{7}\ \text{K}$ in the Sun's core) is required. High pressure/density is also needed so that collisions occur often enough for a useful rate.

Stars versus Earth. A star's enormous mass produces immense gravitational pressure that compresses and heats its core to these conditions and confines the plasma for billions of years — so fusion is self-sustaining. On Earth we cannot use gravity to confine the plasma, so reaching $\sim 10^{7}\ \text{K}$ and holding a hot, dense plasma together long enough for net energy gain (using magnetic confinement, since no material container can touch the plasma) remains a major engineering challenge.

For any fission or fusion reaction the recipe is the same:

1. Add the masses of the reactants (left-hand side).
2. Add the masses of the products (right-hand side).
3. Mass difference $\Delta m = m_{\text{before}} - m_{\text{after}}$. In an energy-releasing reaction the products are lighter, so $\Delta m > 0$.
4. Convert to energy. For MeV: $\Delta E = \Delta m\,(\text{in u}) \times 931.5$. For joules: convert $\Delta m$ to kg and use $\Delta E = c^2\Delta m$ (then ÷ $1.60 \times 10^{-13}$ if you also want MeV).

Worked pattern (D–T fusion). With $^{2}_{1}\text{H} = 2.01410\ \text{u}$, $^{3}_{1}\text{H} = 3.01605\ \text{u}$, $^{4}_{2}\text{He} = 4.00260\ \text{u}$, $^{1}_{0}\text{n} = 1.00867\ \text{u}$: $m_{\text{before}} = 2.01410 + 3.01605 = 5.03015\ \text{u}$ $m_{\text{after}} = 4.00260 + 1.00867 = 5.01127\ \text{u}$ $\Delta m = 0.01888\ \text{u} \;\Rightarrow\; \Delta E = 0.01888 \times 931.5 = 17.6\ \text{MeV}$

Precision warning. Reactant and product masses are almost equal, so the whole answer lives in the last few decimal places. Keep all the quoted decimal places when subtracting, and only round the final energy to 3 s.f. Rounding the masses first can wipe out $\Delta m$ entirely.

Nuclear versus chemical energy. A fission event releases about $200\ \text{MeV}$; a chemical reaction (rearranging outer electrons) releases only a few eV per event. So per unit mass, nuclear fuel releases roughly a million times more energy than chemical fuel — because nuclear reactions tap the strong nuclear force between nucleons, whereas chemistry only shuffles electrons.