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Why are some nuclei more stable than others, and what holds a nucleus together?

Explain nuclear binding energy and the stability of nuclei using the binding-energy curve

A focused answer to the WACE Year 12 Physics Unit 4 content point on nuclear stability. The strong nuclear force, mass defect and binding energy, binding energy per nucleon, the shape of the binding-energy curve, and why fission and fusion release energy.

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

WACE wants you to explain what holds the nucleus together, define binding energy and binding energy per nucleon, and use the binding-energy curve to explain why fission and fusion release energy. This ties mass-energy equivalence to nuclear stability.

The strong nuclear force

Protons in a nucleus repel each other electrostatically, yet nuclei hold together. The strong nuclear force, attractive and far stronger than the electrostatic force but acting only over very short ranges (about the size of a nucleus), binds protons and neutrons (nucleons) together. Neutrons add this binding without adding repulsion, which is why heavier stable nuclei need more neutrons than protons.

Mass defect and binding energy

The mass of a nucleus is less than the total mass of its separate nucleons. This mass defect Δm\Delta m corresponds to the binding energy, the energy that would be needed to pull the nucleus apart into free nucleons:

Eb=Δmc2.E_b=\Delta m\,c^2.

Equivalently, this energy was released when the nucleus formed. A larger binding energy means a more tightly bound nucleus.

Binding energy per nucleon

Total binding energy grows with nucleus size, so to compare stability fairly we use the binding energy per nucleon,

EbA,\frac{E_b}{A},

where AA is the number of nucleons. This is the better stability measure: the higher it is, the more energy holds each nucleon and the more stable the nucleus.

The binding-energy curve

Plotting binding energy per nucleon against mass number gives a curve that rises steeply for light nuclei, peaks around iron and nickel (mass number near 5656), then declines slowly for heavy nuclei. Iron-region nuclei are the most stable. Any reaction that moves nuclei toward this peak releases energy, because the products are more tightly bound than the reactants.

Linking to fission and fusion

Heavy nuclei such as uranium lie to the right of the peak, so splitting them (fission) moves the fragments up toward the peak and releases energy. Light nuclei such as hydrogen isotopes lie to the left, so joining them (fusion) also moves toward the peak and releases energy. Both processes increase binding energy per nucleon, which is the unifying reason they are energy sources.

Calculating a mass defect from nucleon masses

The most common quantitative task is to build the mass defect from scratch. Count the protons (ZZ, the proton number) and neutrons (AZA-Z) in the nucleus, multiply by their individual masses, and add to get the total mass of the separate nucleons. Subtract the measured mass of the bound nucleus to get the mass defect Δm\Delta m. Then convert with either Eb=Δmc2E_b=\Delta mc^2 (if Δm\Delta m is in kilograms) or, more conveniently, by multiplying Δm\Delta m in atomic mass units by 931.5 MeV per u931.5\ \text{MeV per u}. Keep all five decimal places of the nucleon masses through the subtraction, because the mass defect is a small difference of large numbers and rounding early destroys the answer.

Comparing nuclei

To judge which of two nuclei is more stable, compare binding energy per nucleon, not total binding energy. A large nucleus can have a big total binding energy yet a lower per-nucleon value, making it less stable than a small, tightly bound nucleus.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksA nucleus of lithium-7 (37Li^7_3\text{Li}) has a measured mass of 7.01435 u7.01435\ \text{u}. A proton has mass 1.00728 u1.00728\ \text{u} and a neutron 1.00867 u1.00867\ \text{u}. (a) Calculate the mass defect of the nucleus. (b) Calculate its binding energy in MeV. (c) Calculate its binding energy per nucleon.
Show worked answer →

A 7 mark calculation rewards the mass defect, the binding energy and the per-nucleon value.

(a) Mass defect
Lithium-7 has 33 protons and 44 neutrons. Total mass of separate nucleons =3(1.00728)+4(1.00867)=3.02184+4.03468=7.05652 u=3(1.00728)+4(1.00867)=3.02184+4.03468=7.05652\ \text{u}. Mass defect Δm=7.056527.01435=0.04217 u\Delta m=7.05652-7.01435=0.04217\ \text{u}.
(b) Binding energy
Using 1 u931.5 MeV1\ \text{u}\to931.5\ \text{MeV}: Eb=0.04217×931.5=39.3 MeVE_b=0.04217\times931.5=39.3\ \text{MeV}.
(c) Per nucleon
EbA=39.37=5.6 MeV per nucleon\dfrac{E_b}{A}=\dfrac{39.3}{7}=5.6\ \text{MeV per nucleon}.

Markers reward summing the nucleon masses, the mass defect 0.04217 u0.04217\ \text{u}, Eb=39.3 MeVE_b=39.3\ \text{MeV} and the per-nucleon value of 5.6 MeV5.6\ \text{MeV}.

WACE 20215 marksExplain what binding energy is, and explain why iron-56 is one of the most stable nuclei using the shape of the binding-energy-per-nucleon curve.
Show worked answer →

A 5 mark explanation needs the definition and the curve-peak argument.

Binding energy. Binding energy is the energy that would be needed to separate a nucleus completely into its individual protons and neutrons. It equals the mass defect (the difference between the mass of the separate nucleons and the mass of the bound nucleus) multiplied by c2c^2, Eb=Δmc2E_b=\Delta mc^2. It was also the energy released when the nucleus formed.

Why iron-56 is stable. Plotting binding energy per nucleon against mass number gives a curve that rises for light nuclei, peaks near mass number 5656 (iron and nickel), then falls gradually for heavy nuclei. Iron-56 sits at the peak, so each of its nucleons is held by the maximum binding energy. Any reaction moving away from the peak would have to supply energy, so iron-56 is exceptionally stable and is the end point of fusion in massive stars.

Markers reward the energy-to-separate definition with Eb=Δmc2E_b=\Delta mc^2 and iron sitting at the peak of the per-nucleon curve.

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