Mass Defect and Binding Energy

Spec mapping: OCR H556 Module 6.4 — Nuclear and Particle Physics (mass defect $\Delta m$ as the difference between summed nucleon masses and the assembled nuclear mass; binding energy $E_{B} = \Delta m \cdot c^{2}$ ; binding energy per nucleon $E_{B}/A$ and its variation with $A$ ; the curve peaks near iron-56; both fission of heavy nuclei and fusion of light nuclei release energy by moving toward the peak). Refer to the official OCR H556 specification document for exact wording.

In the last lesson we met Einstein's relation $E = mc^{2}$ and used it to compute the energy released in individual nuclear decays. The same idea, applied to entire nuclei, leads directly to the concept of binding energy: the amount of energy that holds a nucleus together. Graphing the binding energy per nucleon against mass number produces one of the most famous diagrams in physics, and explains in a single picture why iron is so abundant in the universe, why heavy nuclei can release energy by fission, and why light nuclei can release energy by fusion.

This lesson covers Module 6.4 of the OCR A-Level Physics A specification (H556) and lays the groundwork for the next lesson on fission and fusion.

The Mass Defect

If you measure the mass of a helium-4 nucleus very precisely, you find $m({}^{4}\text{He nucleus}) \approx 4.00150\,\text{u}$ . Now add up the masses of its constituents — two protons and two neutrons:

$2 m_{p} + 2 m_{n} = 2(1.00728) + 2(1.00867) = 4.03190\,\text{u}.$

The bound nucleus is lighter than the sum of its separate parts. The difference is

$\Delta m = 4.03190 - 4.00150 = 0.03040\,\text{u},$

and is called the mass defect. It is a genuine, measurable mass difference: the helium nucleus weighs less on a mass spectrometer than the four particles it is made of.

Definition. The mass defect of a nucleus is the difference between the total mass of its separated constituent nucleons and the mass of the assembled nucleus:

$\boxed{\; \Delta m = Z\, m_{p} + (A - Z)\, m_{n} - m_{\text{nucleus}} \;}$

Every stable bound nucleus has $\Delta m > 0$ . The bound state is the lower-energy configuration; forming the nucleus releases $\Delta m \cdot c^{2}$ of energy, and dismantling it requires putting that energy back in.

The Binding Energy

By $E = mc^{2}$ , the mass defect corresponds to an energy:

$\boxed{\; E_{B} = \Delta m \cdot c^{2} \;}$

This is the binding energy of the nucleus. It is the amount of energy you would need to supply to pull the nucleus apart into its constituent nucleons, or equivalently, the amount of energy released when those free nucleons assemble themselves into the nucleus.

For helium-4:

$E_{B} = 0.03040\,\text{u} \times 931.5\,\text{MeV/u} \approx 28.3\,\text{MeV}.$

That is the total binding energy of $^{4}\text{He}$ . It takes 28.3 MeV to dismantle a helium nucleus into two free protons and two free neutrons — an enormous amount of energy compared to chemical binding (which is of order a few eV per atom). The strong nuclear force is about a million times stronger than the electromagnetic force at the femtometre distances involved inside a nucleus.

Binding Energy per Nucleon

The total binding energy grows roughly with nucleon number $A$ : bigger nuclei have more nucleons and hence more bonds. The quantity that matters physically — and that is plotted in every textbook — is the binding energy per nucleon, $E_{B}/A$ . This tells you, on average, how tightly each nucleon is bound in that nucleus. A high value means tightly bound, stable, hard to dismantle. A low value means loosely bound, potentially ready to rearrange into a more tightly bound configuration (either by fission or by fusion).

For helium-4: $E_{B}/A = 28.3/4 \approx 7.1\,\text{MeV per nucleon}$ .

Worked Example 1 — Binding energy of oxygen-16

Oxygen-16 has a nuclear mass of $15.99052\,\text{u}$ . Find its total binding energy and binding energy per nucleon.

Sum of nucleons: $8(1.00728) + 8(1.00867) = 16.12760\,\text{u}$ .

$\Delta m = 16.12760 - 15.99052 = 0.13708\,\text{u},$

$E_{B} = 0.13708 \times 931.5 \approx 127.7\,\text{MeV},\qquad \frac{E_{B}}{A} = \frac{127.7}{16} \approx 7.98\,\text{MeV per nucleon}.$

Oxygen-16 is one of the most tightly bound light nuclei — a consequence of its magic-number structure (8 protons and 8 neutrons, both filled nuclear shells).

Worked Example 2 — Binding energy of iron-56

Iron-56 has a nuclear mass of $55.92068\,\text{u}$ . Find its binding energy per nucleon.

Sum of nucleons: $26(1.00728) + 30(1.00867) = 56.44938\,\text{u}$ .

$\Delta m = 56.44938 - 55.92068 = 0.52870\,\text{u},$

$E_{B} = 0.52870 \times 931.5 \approx 492.5\,\text{MeV},\qquad \frac{E_{B}}{A} \approx \frac{492.5}{56} \approx 8.79\,\text{MeV per nucleon}.$

Iron-56 has the highest binding energy per nucleon of any nucleus. It sits at the peak of the binding-energy curve and is the most stable nucleus in existence. In the cores of massive stars, nuclear burning proceeds through successive fusion stages until it reaches iron — and then stops, because no more energy can be extracted by further fusion. The next supernova explodes the resulting iron core back out into the galaxy, where it becomes the seed for second-generation rocky planets like Earth.

The Binding-Energy-per-Nucleon Curve

If you plot $E_{B}/A$ against $A$ for all the stable nuclei, you get a curve that rises steeply from hydrogen, passes through helium, carbon and oxygen, peaks near iron ( $A \approx 56$ ), and slopes gently downward toward uranium. The peak value is about $8.8\,\text{MeV per nucleon}$ ; typical values across the middle of the periodic table are around $8.5\,\text{MeV per nucleon}$ ; uranium-238 sits at about $7.6\,\text{MeV per nucleon}$ .

The qualitative shape is far more important than any individual value. You should be able to sketch this curve from memory, labelling:

Hydrogen ( $A = 1$ ) at $E_{B}/A = 0$
Deuterium ( $A = 2$ ) very low ( $\approx 1.1\,\text{MeV/nucleon}$ )
Helium-4 as an unusually bound "spike" near $A = 4$ ( $\approx 7.1\,\text{MeV/nucleon}$ )
The rising slope from $A \approx 4$ to $A \approx 60$
Iron-56 at the peak ( $E_{B}/A \approx 8.8\,\text{MeV per nucleon}$ )
A gentle decline to uranium-238 ( $\approx 7.6\,\text{MeV/nucleon}$ )

From this single curve you can read off two profoundly important facts:

Fusion of light nuclei releases energy. Moving right along the steeply rising slope (say, fusing two hydrogens to make helium) increases $E_{B}/A$ , so the products are more tightly bound than the reactants. The extra binding appears as kinetic energy of the products.
Fission of heavy nuclei releases energy. Moving left along the gently falling slope (say, splitting uranium into two middle-weight fragments) also increases $E_{B}/A$ . Again, the extra binding appears as kinetic energy.

Both fission and fusion exploit the same principle: they move the nucleons towards the peak of the binding-energy curve, releasing the difference as energy. This is the most counter-intuitive feature of nuclear energetics and the most heavily examined: do not put the peak at uranium.

Why Does the Curve Have This Shape?

The shape of the binding-energy curve can be understood from two competing effects:

The strong nuclear force is short-range. A nucleon in the interior of a nucleus is pulled equally from all sides by its immediate neighbours; it feels essentially no net force from distant nucleons on the other side of the nucleus. So the contribution to the binding energy scales roughly with the number of nucleons — $E_{B}$ is roughly proportional to $A$ , and $E_{B}/A$ is roughly constant — apart from a correction for surface nucleons, which have fewer neighbours.
The electromagnetic (Coulomb) force is long-range. Every proton repels every other proton, no matter how far apart. The total Coulomb energy is proportional to roughly $Z(Z-1)/R \propto Z^{2}/A^{1/3}$ .

For small nuclei the Coulomb effect is negligible and $E_{B}/A$ rises as more nucleons are added (because surface effects decrease: interior nucleons have more neighbours than surface ones). For larger nuclei the growing Coulomb repulsion starts to erode the binding, and $E_{B}/A$ slowly decreases. The peak, where these effects balance, is at iron. This intuition is formalised in the semi-empirical mass formula (Weizsäcker), discussed in "Going further".

There are also shell effects: nuclei with magic numbers (2, 8, 20, 28, 50, 82, 126) of protons or neutrons are unusually tightly bound. Helium-4 (2 protons, 2 neutrons), oxygen-16 (8, 8) and lead-208 (82, 126) are all local "spikes" above the smooth trend.

Worked Example 3 — Energy release in U-235 fission (binding-energy route)

A single uranium-235 nucleus undergoes fission into (approximately) two equal fragments. Take $E_{B}/A$ for U-235 as $7.6\,\text{MeV}$ and for the fragments (nuclei with $A \approx 117$ each) as $8.5\,\text{MeV}$ . How much energy is released per fission?

Total binding energy before fission: $E_{B}(\text{U-235}) = 235 \times 7.6 \approx 1786\,\text{MeV}$ .

Total binding energy after fission (the same 235 nucleons, now rearranged): $E_{B}(\text{fragments}) \approx 235 \times 8.5 \approx 1998\,\text{MeV}$ .

The "extra" binding (i.e. the additional energy released when the more-tightly-bound fragment configuration forms) is

$Q \approx 1998 - 1786 \approx 212\,\text{MeV}.$

About 200 MeV per fission is released — a figure you should commit to memory, because it is the standard energy yield of a single U-235 fission event.

Worked Example 4 — Energy release in D-D fusion

Compute the energy released when two deuterons ( $^{2}\text{H}$ ) fuse to a helium-4 nucleus. Take $E_{B}/A$ for $^{2}\text{H}$ as $1.1\,\text{MeV}$ and for $^{4}\text{He}$ as $7.1\,\text{MeV}$ .

E_{B}(2 \times {}^{2}\text{H}) &= 2 \times (2 \times 1.1) = 4.4\,\text{MeV}, \\ E_{B}({}^{4}\text{He}) &= 4 \times 7.1 = 28.4\,\text{MeV}, \\ Q &= 28.4 - 4.4 = 24.0\,\text{MeV}. \end{aligned}$$ About $24\,\text{MeV}$ per fusion event in this simplified two-deuteron → alpha picture. The actual D–D reaction proceeds via a tritium or helium-3 intermediate and releases less (about $3.3$–$4.0\,\text{MeV}$); the larger figure here uses an unrealistic simplification. Per unit mass, fusion releases several times more energy than fission — but it requires much higher temperatures to initiate, because the reacting nuclei must overcome Coulomb repulsion. --- ## Macroscopic Energy Yield Binding energies come out in MeV when you use atomic mass units and multiply by $931.5$. They come out in joules when you use kilograms and multiply by $c^{2} = 9 \times 10^{16}\,\text{m}^{2}\text{s}^{-2}$. For macroscopic energies, convert MeV per event to joules and multiply by the number of nuclei. Complete fission of 1 kg of U-235 would release: $$N = \frac{10^{3}\,\text{g}}{235\,\text{g mol}^{-1}} \times N_{A} \approx 2.56 \times 10^{24}\,\text{nuclei},$$ $$E = N \times 200\,\text{MeV} \times 1.60 \times 10^{-13}\,\text{J/MeV} \approx 8.2 \times 10^{13}\,\text{J}.$$ That is roughly $2.3 \times 10^{7}\,\text{kWh}$, or about 20 kilotonnes of TNT equivalent. A kilogram of uranium, fully fissioned, releases the same energy as about 20 000 tonnes of TNT — the figure that haunted the twentieth century. --- ## Synoptic Links

Mass Defect and Binding Energy

Mass Defect and Binding Energy

The Mass Defect

The Binding Energy

Binding Energy per Nucleon

Worked Example 1 — Binding energy of oxygen-16

Worked Example 2 — Binding energy of iron-56

The Binding-Energy-per-Nucleon Curve

Why Does the Curve Have This Shape?

Worked Example 3 — Energy release in U-235 fission (binding-energy route)

Worked Example 4 — Energy release in D-D fusion

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