Account for the energy released in nuclear fission and fusion in terms of mass defect and binding energy, using E = mc^2 and the binding energy curve

What this dot point is asking

NESA wants you to define mass defect and binding energy, calculate them from atomic mass data using $E = m c^2$ (with $1 \text{ u} \to 931.5$ MeV), describe the shape of the binding-energy-per-nucleon curve, and use it to explain why both nuclear fission of heavy nuclei and nuclear fusion of light nuclei release energy.

The answer

Mass defect and binding energy

The mass of a nucleus is always slightly less than the sum of the masses of its constituent protons and neutrons. The difference is the mass defect:

By $E = m c^2$, this "missing" mass corresponds to the binding energy of the nucleus:

This is the energy required to separate the nucleus into free protons and neutrons. Equivalently, it is the energy released when free nucleons assemble into the bound nucleus.

A useful unit conversion: $1 \text{ u} \cdot c^2 = 931.5$ MeV, so a mass defect in atomic mass units converts directly to a binding energy in MeV.

Binding energy per nucleon

The bound state of a nucleus is more meaningful when normalised by the number of nucleons:

This is the average energy needed to remove one nucleon. Plotted against mass number $A$, it produces the famous binding-energy-per-nucleon curve:

• IMATH_10 (hydrogen-1): 0 (single proton, no binding).
• IMATH_11 (deuterium): 1.11 MeV per nucleon.
• IMATH_12 (helium-3, tritium): 2.6-2.8 MeV per nucleon.
• IMATH_13 (helium-4): 7.07 MeV per nucleon (a local peak, very tightly bound).
• IMATH_14 (carbon-12): 7.7 MeV per nucleon.
• IMATH_15 (iron-56): 8.79 MeV per nucleon. Near the maximum.
• IMATH_16 (uranium-235): 7.6 MeV per nucleon. Declining.

The curve rises steeply for light nuclei, peaks near $A = 56$, and falls slowly for heavy nuclei.

Rule for releasing energy

A nuclear process releases energy if the products have higher binding energy per nucleon than the reactants. Geometrically, this means the reaction moves the nucleons "uphill" on the binding-energy-per-nucleon curve, toward the iron peak.

• Light side of the peak. Combining light nuclei (fusion) moves up toward iron: energy is released.
• Heavy side of the peak. Splitting heavy nuclei (fission) moves up toward iron: energy is released.
• At the peak (iron). Neither fission nor fusion releases energy. Iron is the end point of stellar nucleosynthesis (see the related dot point on stars).

Nuclear fission

A heavy unstable nucleus splits into two medium-mass fragments and a few neutrons. Example, induced fission of uranium-235 by a thermal neutron:

The barium and krypton fragments have $E_B / A \approx 8.5$ MeV per nucleon, while uranium has 7.6 MeV per nucleon. The increase of about 0.9 MeV per nucleon over 235 nucleons gives about 200 MeV released per fission event, carried away as kinetic energy of the fragments and neutrons plus prompt and delayed gamma radiation.

A chain reaction can occur if the released neutrons trigger further fissions. In a controlled reactor, a moderator (water, graphite) slows neutrons to thermal energies for efficient capture by $^{235}$U, and control rods absorb excess neutrons to keep the reaction critical (one fission triggers exactly one more). In a fission weapon, no control is used.

Nuclear fusion

Two light nuclei combine into a heavier one. Example, deuterium-tritium fusion (the easiest practical fusion reaction):

Deuterium has 1.11 MeV per nucleon and tritium 2.83 MeV per nucleon. Helium-4 has 7.07 MeV per nucleon. The energy released is $\Delta m \, c^2 = 17.6$ MeV per reaction. Even though only 5 nucleons rearrange, the change per nucleon is much larger than for fission, so fusion is more energy-dense per kilogram of fuel.

Fusion requires temperatures of $\sim 10^8$ K to overcome the Coulomb repulsion between the positively charged nuclei, plus high densities and confinement times. In stars (Sun and similar), the proton-proton chain fuses four protons into a helium-4 nucleus, releasing 26.7 MeV per cycle, the energy source that sustains stellar luminosity.

On Earth, achieving net-energy-positive fusion is the long-running goal of magnetic-confinement (tokamak) and inertial-confinement (laser-driven) research.

Worked example: deuterium-tritium fusion

Calculate the energy released. Atomic masses: $m(^2$H$) = 2.01410$ u, $m(^3$H$) = 3.01605$ u, $m(^4$He$) = 4.00260$ u, $m(n) = 1.00867$ u.

Reactants: $2.01410 + 3.01605 = 5.03015$ u.

Products: $4.00260 + 1.00867 = 5.01127$ u.

Mass defect: $\Delta m = 5.03015 - 5.01127 = 0.01888$ u.

Energy: $E = \Delta m \times 931.5 = 0.01888 \times 931.5 = 17.6$ MeV.

Most of this energy is carried by the neutron (about 14.1 MeV) because the helium-4 nucleus is heavier and recoils more slowly. In a power reactor the neutron would deposit its energy in a blanket of lithium to breed more tritium and heat a working fluid.

Try it: Mass-energy calculator to convert between mass defects (in u or kg) and energy releases (in MeV or J) for any nuclear reaction.

Worked example: uranium-235 fission

Approximate energy per fission. Take the change in $E_B / A$ as 0.9 MeV per nucleon and $A = 235$ nucleons.

$E \approx 235 \times 0.9 = 210$ MeV. (Typical experimental value 200 MeV.)

For comparison, burning a single carbon atom in a chemical reaction releases only a few eV. Fission releases about $10^8$ times more energy per atom, which is why nuclear processes can power cities and weapons with kilograms of fuel.

Common traps

Forgetting that nuclear processes are governed by binding energy per nucleon, not total binding energy. A heavy nucleus has a larger total binding energy than a light one (more nucleons), but the energetic question is the energy per nucleon, the depth of the well each nucleon sits in.

Treating mass defect as a loss of nucleons. All nucleons are still present after the reaction; the mass loss is in the bound system as a whole, converted to kinetic energy and radiation.

Mixing up fission and fusion. Fission = heavy nucleus splits; fusion = light nuclei combine. Both move nucleons toward the iron peak and both release energy.

Using non-relativistic kinetic energy at nuclear scales. Fission fragments and fusion products often have several MeV of kinetic energy, well within the non-relativistic regime for the heavy fragments, but the relativistic energy budget $E = m c^2$ is the consistent accounting tool.

Saying fusion takes "no fuel". Fusion fuel (deuterium, tritium, lithium) is needed; tritium in particular must be bred in the reactor blanket from neutron capture on lithium.

In one sentence

Both fission and fusion release energy because the products have higher binding energy per nucleon than the reactants; the mass defect $\Delta m$ between reactants and products converts to energy via $E = \Delta m \, c^2$, with about 200 MeV per fission event and 17.6 MeV per D-T fusion event.