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What does Einstein's equation E=mc2E = mc^2 tell us about the relationship between mass and energy?

Apply mass-energy equivalence to rest energy and to the mass defect and energy release in nuclear processes.

How E=mc2E = mc^2 expresses mass and energy as interchangeable, the meaning of rest energy, and how mass defect explains the energy released in nuclear reactions, with worked examples.

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

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  1. What this dot point is asking
  2. Mass and energy are the same thing
  3. Mass defect and binding energy
  4. Fission, fusion and the Sun
  5. How SACE assesses this
  6. Energy can also create mass

What this dot point is asking

You need to apply E=mc2E = mc^2 to calculate rest energy, and to relate the mass defect in a nuclear process to the energy released.

Mass and energy are the same thing

Einstein's special relativity shows that mass is a highly concentrated form of energy. Even an object at rest possesses energy purely because it has mass.

This does not mean ordinary objects easily release this energy. In chemical reactions the mass change is far too small to detect. It is only in nuclear reactions that a measurable fraction of mass is converted.

Mass defect and binding energy

The mass of a nucleus is slightly less than the total mass of its individual protons and neutrons. This difference is the mass defect, and the equivalent energy is the binding energy that holds the nucleus together.

In any nuclear reaction (fission, fusion or radioactive decay), if the products are more tightly bound than the reactants, the surplus mass is released as kinetic energy of the products and radiation.

Fission, fusion and the Sun

  • Fission splits a heavy nucleus (like uranium-235) into lighter fragments that are more tightly bound; the mass defect appears as energy. This powers nuclear reactors.
  • Fusion joins light nuclei (like hydrogen isotopes) into a more tightly bound nucleus (helium). The Sun fuses hydrogen into helium, converting about 4 million tonnes of mass into energy every second.

Both release energy because the products sit higher in binding energy per nucleon than the reactants. The binding-energy-per-nucleon curve peaks near iron, so nuclei lighter than iron release energy by fusing toward it, while nuclei heavier than iron release energy by splitting toward it. This single curve explains why fusion powers stars but fission powers reactors: each process moves nuclei toward the more tightly bound middle of the curve, releasing the surplus mass as energy.

How SACE assesses this

SACE Stage 2 questions almost always supply rest masses (often in kilograms) and ask for the energy released by annihilation, the energy from a stated mass defect, or the frequency of an emitted photon. Quote E=mc2E = mc^2 before substituting, keep the factor c2=9.00×1016c^2 = 9.00\times10^{16} explicit, and when two photons are produced remember to halve the total energy before applying E=hfE = hf. Finishing with a sentence in context, such as identifying the radiation region, secures the communication marks.

Energy can also create mass

The equivalence works both ways. In particle accelerators, kinetic energy can be converted into new particles with real mass, and in pair production a photon can create an electron-positron pair. Mass-energy together is conserved, even though neither alone always is. Annihilation, as in the PET-scan question above, is the reverse process: mass becomes pure photon energy.

Exam-style practice questions

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

SACE 20254 marksIn a PET scan, a positron and an electron annihilate to produce two identical photons. The electron and positron each have rest mass 9.11×1031 kg9.11\times10^{-31}\ \text{kg}. Calculate the frequency of one of the photons. Use c=3.00×108 m s1c = 3.00\times10^8\ \text{m s}^{-1} and h=6.63×1034 J sh = 6.63\times10^{-34}\ \text{J s}.
Show worked answer →

First the total energy from the combined rest mass via E=mc2E = mc^2:

Etotal=(2×9.11×1031)(3.00×108)2=(1.822×1030)(9.00×1016)=1.64×1013 J.E_\text{total} = (2\times9.11\times10^{-31})(3.00\times10^8)^2 = (1.822\times10^{-30})(9.00\times10^{16}) = 1.64\times10^{-13}\ \text{J}.

Two identical photons share this equally:

Ephoton=12(1.64×1013)=8.20×1014 J.E_\text{photon} = \tfrac{1}{2}(1.64\times10^{-13}) = 8.20\times10^{-14}\ \text{J}.

Use E=hfE = hf:

f=Eh=8.20×10146.63×1034=1.24×1020 Hz.f = \dfrac{E}{h} = \dfrac{8.20\times10^{-14}}{6.63\times10^{-34}} = 1.24\times10^{20}\ \text{Hz}.

1 mark for total energy, 1 mark for halving between two photons, 1 mark for using E=hfE = hf, 1 mark for the answer of about 1.24×1020 Hz1.24\times10^{20}\ \text{Hz}.

SACE 20243 marksA proton and an anti-proton annihilate, each of rest mass 1.67×1027 kg1.67\times10^{-27}\ \text{kg}. Calculate the total energy released in this annihilation. Use c=3.00×108 m s1c = 3.00\times10^8\ \text{m s}^{-1}.
Show worked answer →

All of the combined rest mass is converted to energy via E=mc2E = mc^2.

Combined mass: m=2×1.67×1027=3.34×1027 kgm = 2\times1.67\times10^{-27} = 3.34\times10^{-27}\ \text{kg}.

E=mc2=(3.34×1027)(3.00×108)2=(3.34×1027)(9.00×1016)=3.01×1010 J.E = mc^2 = (3.34\times10^{-27})(3.00\times10^8)^2 = (3.34\times10^{-27})(9.00\times10^{16}) = 3.01\times10^{-10}\ \text{J}.

1 mark for the combined rest mass, 1 mark for substitution into E=mc2E = mc^2, 1 mark for the answer of about 3.0×1010 J3.0\times10^{-10}\ \text{J}.

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