What does Einstein's equation E = 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 = 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.
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What this dot point is asking
You need to apply 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.
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.
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.
2023 SACE Stage 22 marksAn electron and a positron annihilated, producing two gamma photons. Determine the total energy released in the annihilation.Show worked answer →
In annihilation, the rest mass of both particles is converted entirely into photon energy via E = m c squared.
The electron and positron each have a rest mass of 9.11 x 10^-31 kg, so the total mass converted is 2 x 9.11 x 10^-31 = 1.822 x 10^-30 kg.
E = m c squared = (1.822 x 10^-30)(3.00 x 10^8) squared = (1.822 x 10^-30)(9.00 x 10^16) = 1.64 x 10^-13 J.
1 mark for using the total mass of both particles, 1 mark for the answer of about 1.64 x 10^-13 J (equivalent to roughly 1.02 MeV).
2024 SACE Stage 23 marksA proton and an anti-proton annihilate. Calculate the energy released in this annihilation.Show worked answer →
All of the rest mass of the proton and anti-proton is converted into energy using E = m c squared.
A proton has a rest mass of 1.67 x 10^-27 kg, and the anti-proton has the same mass, so the total mass converted is 2 x 1.67 x 10^-27 = 3.34 x 10^-27 kg.
E = m c squared = (3.34 x 10^-27)(3.00 x 10^8) squared = (3.34 x 10^-27)(9.00 x 10^16) = 3.01 x 10^-10 J.
1 mark for the combined rest mass, 1 mark for substitution into E = m c squared, 1 mark for the answer of about 3.0 x 10^-10 J.
2025 SACE Stage 24 marksIn a PET scan, a positron and an electron annihilate to produce two photons. Calculate the frequency of one of the photons that is produced in the annihilation.Show worked answer →
First find the total energy released using E = m c squared with the combined rest mass of the electron and positron, each 9.11 x 10^-31 kg.
E total = (2 x 9.11 x 10^-31)(3.00 x 10^8) squared = (1.822 x 10^-30)(9.00 x 10^16) = 1.64 x 10^-13 J.
Two identical photons are produced, so each carries half: E photon = 1.64 x 10^-13 / 2 = 8.20 x 10^-14 J.
Use E = h f to find the frequency: f = E / h = (8.20 x 10^-14) / (6.63 x 10^-34) = 1.24 x 10^20 Hz.
1 mark for total energy, 1 mark for halving it between two photons, 1 mark for using E = h f, 1 mark for the answer of about 1.24 x 10^20 Hz.