Binding energy per nucleon

From mass defect to binding energy. A nucleus of mass number $A$ and atomic number $Z$ is built from $Z$ protons and $(A-Z)$ neutrons. Measure the mass of the free nucleons and compare it to the measured mass of the nucleus, and the nucleus is always lighter. This shortfall is the mass defect $\Delta m$, and by mass-energy equivalence it corresponds to the total binding energy:

where: $\Delta m$ is the mass defect (kg or, more conveniently, in atomic mass units u, where $1\ \text{u} = 931.5\ \text{MeV}/c^2$) and $c$ is the speed of light. $BE$ is the energy that would have to be supplied to break the nucleus apart into free, separated nucleons — equivalently, the energy released when those free nucleons come together to form the nucleus.

From binding energy to binding energy per nucleon. Total binding energy by itself is not a fair way to compare nuclei of different sizes, since a bigger nucleus naturally has more total binding energy just because it has more nucleons. Dividing by $A$ fixes this and gives the quantity that actually measures how tightly each nucleon is held:

where: $\overline{BE}$ is the binding energy per nucleon (typically quoted in MeV), $BE$ is total binding energy, and $A$ is the mass number. A larger $\overline{BE}$ means the nucleus is, on average, more stable and harder to pull apart.

Shape of the curve. Plotting $\overline{BE}$ against $A$ for every stable nucleus produces one of the most information-dense graphs in physics. It rises sharply for the lightest nuclei — from about $1.1\ \text{MeV}$ for deuterium ($^2_1\text{H}$) up through helium, carbon and oxygen — reaches a broad, flat maximum of roughly $8.7$–$8.8\ \text{MeV}$ per nucleon in the region $A\approx56$–$62$ around iron and nickel, and then decreases slowly for heavier nuclei, falling to about $7.6\ \text{MeV}$ per nucleon by uranium ($A\approx238$).

Why the curve explains fission and fusion. Nuclei near $A\approx56$ sit at the peak — the most tightly bound, most stable nuclei that exist. Any nuclear process that moves nucleons toward that peak releases energy, because the products end up more tightly bound than the reactants. In fission, a heavy nucleus (low $\overline{BE}$, right side of the curve) splits into two medium-mass fragments that sit higher on the curve. In fusion, two very light nuclei (low $\overline{BE}$, left side of the curve) combine into a heavier product that again sits higher on the curve. Iron itself is essentially "nuclear ash": sitting at the peak, no ordinary fission or fusion reaction starting from iron can release energy, since there is nowhere higher on the curve to move to.

Given: the binding energy of $^{56}_{26}\text{Fe}$ is $492.26\ \text{MeV}$. Calculate the binding energy per nucleon and compare it with deuterium ($\overline{BE}\approx1.11\ \text{MeV}$). Then estimate the energy released when a uranium-235 nucleus ($\overline{BE}\approx7.6\ \text{MeV}$) undergoes fission into two fragments of mass number about 117–118 with $\overline{BE}\approx8.5\ \text{MeV}$ each.

Step 1. Iron-56 has mass number $A=56$, so

Step 2. Compare with deuterium: $8.79\ \text{MeV}$ versus $1.11\ \text{MeV}$ shows iron nucleons are far more tightly bound — roughly eight times more energy per nucleon would be needed to break iron apart than to break deuterium apart, consistent with iron sitting near the peak of the curve.

Step 3. For the uranium fission estimate, find the total binding energy before splitting, using $A=235$ throughout since the total number of nucleons is conserved: $BE_{\text{before}} = 235\times7.6 = 1786\ \text{MeV}$.

Step 4. Find the total binding energy after splitting, using the fragments' higher average $\overline{BE}\approx8.5\ \text{MeV}$ over the same 235 nucleons: $BE_{\text{after}} = 235\times8.5 = 1997.5\ \text{MeV}$.

Step 5. The energy released equals the increase in total binding energy, since the nucleons end up more tightly bound and the surplus binding energy appears as kinetic energy and radiation: $\Delta E = BE_{\text{after}} - BE_{\text{before}} = 1997.5 - 1786 = 211.5\ \text{MeV}$.

Answer: $\overline{BE}_{\text{Fe}} = 8.79\ \text{MeV per nucleon}$, about $8$ times that of deuterium, and the estimated fission energy release is $\Delta E \approx 211.5\ \text{MeV}$ — matching the commonly quoted figure of roughly $200\ \text{MeV}$ per uranium-235 fission event.

• Confusing total binding energy (which generally increases with $A$, since more nucleons means more total binding) with binding energy per nucleon (which peaks near $A\approx56$ and is the quantity that actually measures stability).
• Assuming the curve keeps rising for all heavy elements — it falls slowly past the iron peak, which is exactly why heavy nuclei like uranium are less stable and can undergo fission.
• Thinking iron itself can release energy by fission or fusion — it sits at the maximum of the curve, so neither splitting nor combining it releases energy.
• Forgetting that the energy released in fission or fusion equals the difference in total binding energy between products and reactants, not simply the binding energy of the products alone.

Ask the tutor: Want to work through another fission or fusion energy-release calculation, or see how the binding-energy curve connects to nuclear reactors and stars?