Mass-Energy Equivalence (E = mc²)
Einstein's most famous equation: mass and energy are equivalent, every mass corresponds to an enormous rest energy.
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Formula
E = m \cdot c^2Variables & units – Mass-Energy Equivalence (E = mc²)
| Symbol | Meaning | Unit |
|---|---|---|
| E | Rest energy | J |
| m | Mass (rest mass) | kg |
| c | Speed of light in vacuum (3×10⁸ m/s) | m/s |
Derivation & background – Mass-Energy Equivalence (E = mc²)
Albert Einstein derived the equivalence in 1905 from special relativity. Because c² is huge (9×10¹⁶ m²/s²), tiny masses contain enormous energy. In nuclear fission and fusion the mass defect is released: the products are lighter than the initial nuclei, and the difference appears as energy. The Sun loses about 4 million tonnes of mass per second this way.
Exam blueprint
Validity range
E = mc² gives the rest energy of a mass. For moving particles the total energy is E = γmc²; the kinetic energy is the difference E_kin = (γ−1)mc².
Derivation steps
From special relativity it follows that energy and inertial mass are the same property in different units.
- 1Einstein 1905: if a body emits energy ΔE, its mass decreases by Δm = ΔE/c².
- 2Conversely every mass corresponds to the rest energy E = mc².
Rearrangements
Mass defect from the released energy
Because c² = 9×10¹⁶ m²/s², the mass changes are tiny.
Energy in electron volts
Common in nuclear and particle physics: MeV instead of joules.
Task variant
The Sun radiates 3.8×10²⁶ W. How much mass does it lose per second?
Δm = E/c² = 3.8×10²⁶ / 9×10¹⁶ ≈ 4.2×10⁹ kg, about 4 million tonnes per second.
Compute the rest energy of an electron (m = 9.11×10⁻³¹ kg).
E = 9.11×10⁻³¹ × 9×10¹⁶ ≈ 8.2×10⁻¹⁴ J ≈ 0.511 MeV.
Common mistakes
Not squaring c or computing with 3×10⁸ instead of 9×10¹⁶.
c² = (3×10⁸)² = 9×10¹⁶ m²/s²; the exponent doubles.
Believing the entire mass disappears in nuclear fission.
Only the mass defect (below 0.1%) is converted into energy.
Interpreting E = mc² as kinetic energy.
It is the rest energy; kinetic energy is added via the γ factor.
Exam context
- Typical: converting the mass defect of nuclear reactions to MeV, the energy balance of the Sun, pair annihilation in PET scanners.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Relativistic energy
Connects nuclear physics (mass defect) with energy conservation.
Worked example
Complete conversion of m = 1 g = 0.001 kg: E = 0.001 × (3×10⁸)² = 9×10¹³ J = 25 GWh, the annual electricity consumption of about 7,000 households.
Applications
Nuclear power plants, nuclear fusion (Sun, ITER), PET diagnostics (pair annihilation), particle physics
Quanta exam set
Curated exam set for "Mass-Energy Equivalence (E = mc²)":
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Which formula describes Mass-Energy Equivalence (E = mc²)?
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How do you rearrange E = mc² for Mass defect from the released energy?
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Which common mistake happens with Mass-Energy Equivalence (E = mc²)?
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Scientific sources
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Frequently asked questions about Mass-Energy Equivalence (E = mc²)
How do you calculate with E = mc²?+
Multiply the mass in kilograms by the square of the speed of light: c² = (3×10⁸ m/s)² = 9×10¹⁶ m²/s². Example: one gram of matter (0.001 kg) corresponds to E = 0.001 × 9×10¹⁶ = 9×10¹³ J, equal to 25 GWh, the annual electricity consumption of about 7,000 households. The most common mistake is forgetting the square and multiplying only by 3×10⁸, missing a factor of 300 million. Conversely, m = E/c² gives the mass change for an energy release. In nuclear and particle physics one often converts to electron volts: 1 eV = 1.602×10⁻¹⁹ J.
What is the mass defect in nuclear reactions?+
If you weigh the building blocks of an atomic nucleus individually and compare with the finished nucleus, mass is missing: the bound nucleus is lighter than the sum of its protons and neutrons. This difference is the mass defect, and it was released as binding energy during assembly, according to ΔE = Δm·c². In the fission of uranium-235 the defect is about 0.1% of the initial mass, in the fusion of hydrogen to helium as much as 0.7%, which is why fusion yields more per kilogram of fuel. Sample calculation: a defect of 0.2 u ≈ 3.32×10⁻²⁸ kg corresponds to E = 3.32×10⁻²⁸ × 9×10¹⁶ ≈ 3×10⁻¹¹ J ≈ 186 MeV, the typical scale of a single fission event.
Does the Sun really lose mass by shining?+
Yes, and massively so: the Sun radiates a power of about 3.8×10²⁶ W. By Δm = E/c² it loses Δm = 3.8×10²⁶ / 9×10¹⁶ ≈ 4.2×10⁹ kg per second, a good four million tonnes every second. The energy comes from nuclear fusion in the solar core: four hydrogen nuclei fuse in several steps into one helium nucleus that is 0.7% lighter than the initial particles; exactly this mass defect is released as radiation. Despite the huge numbers the loss is insignificant for the Sun: at a mass of 2×10³⁰ kg it loses only about 0.007% of its mass to radiation even in a billion years.
Does E = mc² mean mass can be completely converted into energy?+
In principle yes, in practice almost never. Complete conversion happens only in annihilation: when a particle meets its antiparticle, for instance an electron meets a positron, the entire rest mass becomes radiation energy, used in PET diagnostics, where the two 511 keV photons reveal the annihilation site. Fission and fusion, by contrast, convert only the tiny mass defect (0.1 to 0.7%), and chemical reactions a trillion times less still. The rest of the mass persists as particles because conservation laws (such as baryon number) forbid the complete conversion of ordinary matter. E = mc² is thus a statement of equivalence; it does not say the conversion is technically feasible at will.
Does mass also change in chemical reactions or on heating?+
Yes. Every energy change of a system also changes its mass by Δm = ΔE/c², only the effect is immeasurably small outside nuclear physics. Combustion example: one kilogram of petrol releases about 4.3×10⁷ J; the products are thus lighter by Δm = 4.3×10⁷/9×10¹⁶ ≈ 4.8×10⁻¹⁰ kg, half a microgram, far below any balance precision. A heated body or a stretched spring is likewise minimally heavier than in its low-energy state. That chemists may work with Lavoisier conservation of mass is due only to the tininess of the effect. Only in nuclear processes, where millions of times more energy is converted per particle, does the mass difference become measurable and technically usable.
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How do you calculate with Mass-Energy Equivalence (E = mc²)?
Here is how to work through a typical Mass-Energy Equivalence (E = mc²) (E = mc²) task step by step:
- 1
Task
The Sun radiates 3.8×10²⁶ W. How much mass does it lose per second?
Solution path
Δm = E/c² = 3.8×10²⁶ / 9×10¹⁶ ≈ 4.2×10⁹ kg, about 4 million tonnes per second.
- 2
Task
Compute the rest energy of an electron (m = 9.11×10⁻³¹ kg).
Solution path
E = 9.11×10⁻³¹ × 9×10¹⁶ ≈ 8.2×10⁻¹⁴ J ≈ 0.511 MeV.