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Mass-energy equivalence (SACE Stage 2 Physics)

Syllabus dot point

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

9 min read

Quick answer

Mass and energy are equivalent: a mass mm has a rest energy E=mc2E = mc^2. Because c2c^2 is enormous, a tiny amount of mass corresponds to a huge amount of energy. In nuclear reactions the products have slightly less mass than the reactants (the mass defect Δm\Delta m); the missing mass is released as energy E=Δm c2E = \Delta m\, c^2.

Jump to a section
  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
  7. Exam-style questions

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.

Energy from a mass defect

A nuclear reaction loses 0.0030 u of mass. How much energy is released?

The atomic mass unit 1 u=1.66×10−27 kg1\ \text{u} = 1.66\times10^{-27}\ \text{kg}.

Mass lost:

Δm=0.0030×1.66×10−27=4.98×10−30 kg.\Delta m = 0.0030 \times 1.66\times10^{-27} = 4.98\times10^{-30}\ \text{kg}.

Energy released:

E=Δm c2=(4.98×10−30)(3.00×108)2=(4.98×10−30)(9.00×1016)=4.5×10−13 J.E = \Delta m\, c^2 = (4.98\times10^{-30})(3.00\times10^8)^2 = (4.98\times10^{-30})(9.00\times10^{16}) = 4.5\times10^{-13}\ \text{J}.

This is per reaction. Multiplied over the vast number of nuclei in even a small sample, it accounts for the enormous energy of nuclear power and weapons.

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.

Rest energy

E=mc2E = mc^2

EE is the rest energy (J), mm the mass (kg), and c=3.00×108 m s−1c = 3.00\times10^8\ \text{m s}^{-1} the speed of light. Because c2≈9×1016c^2 \approx 9\times10^{16}, a single gram of matter holds about 9×1013 J9\times10^{13}\ \text{J} of energy.

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.

Mass defect and binding energy

The mass defect Δm\Delta m is the difference between the combined mass of the separate nucleons and the actual mass of the nucleus. The binding energy is E=Δm c2E = \Delta m\, c^2 - the energy needed to pull the nucleus apart into individual nucleons, and equally the energy released when it forms.

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.

Common errors
Forgetting to square cc
The factor is c2≈9×1016c^2 \approx 9\times10^{16}, not cc. Missing the square is an enormous error.
Unit slips with mass defect
Convert atomic mass units to kilograms (1 u=1.66×10−27 kg1\ \text{u} = 1.66\times10^{-27}\ \text{kg}) before using E=Δm c2E = \Delta m\,c^2, unless working in MeV with a conversion factor.
Thinking mass is destroyed
Mass is converted to energy; the total mass-energy is conserved. The "missing" mass reappears as energy.
Applying it to chemical reactions in practice
The principle holds, but the mass change in chemical reactions is immeasurably small. The conversion only becomes significant in nuclear processes.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2025-style4 marks
In a PET scan, a positron and an electron annihilate to produce two identical photons. The electron and positron each have rest mass 9.11×10−31 kg9.11\times10^{-31}\ \text{kg}. Calculate the frequency of one of the photons. Use c=3.00×108 m s−1c = 3.00\times10^8\ \text{m s}^{-1} and h=6.63×10−34 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=(2×9.11×10−31)(3.00×108)2=(1.822×10−30)(9.00×1016)=1.64×10−13 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(1.64×10−13)=8.20×10−14 J.E_\text{photon} = \tfrac{1}{2}(1.64\times10^{-13}) = 8.20\times10^{-14}\ \text{J}.

Use E=hfE = hf:

f=Eh=8.20×10−146.63×10−34=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 2024-style3 marks
A proton and an anti-proton annihilate, each of rest mass 1.67×10−27 kg1.67\times10^{-27}\ \text{kg}. Calculate the total energy released in this annihilation. Use c=3.00×108 m s−1c = 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=2×1.67×10−27=3.34×10−27 kgm = 2\times1.67\times10^{-27} = 3.34\times10^{-27}\ \text{kg}.

E=mc2=(3.34×10−27)(3.00×108)2=(3.34×10−27)(9.00×1016)=3.01×10−10 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 3.0×10−10 J3.0\times10^{-10}\ \text{J}.

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

ExamExplained