Physics/Lessons/Nuclear & Particle Physics/Mass–Energy Equivalence (E = mc²)
Physics · Nuclear Physics · A2 Level

Mass–Energy Equivalence (E = mc²)

Mass and energy aren't two separate things linked by a formula — they're the same physical quantity, measured in different units. This lesson makes that idea quantitative: how to convert between them with $E=mc^2$, why particle physicists routinely measure mass in MeV/c², and how pair production and annihilation obey this relationship exactly.

10 min read

Mass and energy as the same quantity

Einstein's mass–energy equivalence relation states that mass and energy are interconvertible, related by:

$E = mc^2$

where $E$ is energy in joules, $m$ is mass in kilograms, and $c$ is the speed of light in a vacuum, $c = 3.00\times10^{8}\,\text{m s}^{-1}$. Because $c^2$ is such an enormous number ($9.00\times10^{16}\,\text{m}^2\text{s}^{-2}$), even a very small mass corresponds to a very large amount of energy — the fact underlying both nuclear power and the energy released in particle collisions.

The relation works both ways. A system that loses energy — by emitting radiation or releasing heat — loses a corresponding mass, $\Delta m = \dfrac{\Delta E}{c^2}$, usually far too small to detect by weighing. A system that gains energy gains mass by the same relationship. This isn't confined to nuclear or particle physics — it applies universally, just at a scale too small to notice outside a dedicated laboratory.

Mass defect and binding energy — a preview

When separate nucleons come together to form a nucleus, the nucleus has slightly less mass than the sum of its parts — the missing mass is the mass defect, and $E=mc^2$ says it corresponds exactly to the energy released binding the nucleus together (the binding energy), covered properly in the next topic. For now: any time energy is released or absorbed in a nuclear or particle process, a calculable, real change in total mass goes with it.

The electron-volt and MeV/c² as a mass unit

The joule is inconvenient at particle scales — particle energies and masses are so much smaller than one joule that the numbers become unwieldy. Particle physicists instead work in electron-volts (eV), where $1\,\text{eV} = 1.60\times10^{-19}\,\text{J}$ is the energy gained by a single electron accelerated through a potential difference of 1 volt, and its larger multiples: keV ($10^3$ eV), MeV ($10^6$ eV) and GeV ($10^9$ eV).

Because mass and energy are interconvertible via $E=mc^2$, it's standard practice to express a particle's rest mass directly in energy units, as $\text{MeV}/c^2$ or $\text{GeV}/c^2$, rather than kilograms — shorthand for "the energy equivalent of this mass, divided by $c^2$", though in practice masses are simply quoted and compared directly in these units.

DerivationThe electron's rest mass in MeV/c²
1
The electron's rest mass in SI units is $m_e = 9.11\times10^{-31}\,\text{kg}$.
2
Its rest energy is $E = m_ec^2 = 9.11\times10^{-31}\times(3.00\times10^{8})^2 = 8.20\times10^{-14}\,\text{J}$.
3
Converting to eV: $E = \dfrac{8.20\times10^{-14}}{1.60\times10^{-19}} = 5.12\times10^{5}\,\text{eV} = 0.512\,\text{MeV}$.
4
So the electron's rest mass is quoted as $m_ec^2 \approx 0.511\,\text{MeV}$, or equivalently $m_e \approx 0.511\,\text{MeV}/c^2$ — a figure worth simply remembering, since it recurs constantly in pair-production and annihilation calculations.
Small rounding differences come from the precision of $m_e$ and $c$ used; the accepted value is $0.511\,\text{MeV}/c^2$.
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Pair production, revisited quantitatively

A sufficiently energetic photon, passing close to a nucleus (needed to conserve momentum), can convert directly into a particle–antiparticle pair — most commonly an electron and a positron:

$\gamma \rightarrow e^{-} + e^{+}$

Mass–energy equivalence makes the threshold for this process exact. Since electron and positron each have rest mass $0.511\,\text{MeV}/c^2$, the photon must supply at least enough energy to create both rest masses:

Worked exampleThreshold photon energy for electron–positron pair production
1
Minimum energy needed $= $ rest energy of electron $+$ rest energy of positron $= 0.511 + 0.511 = 1.022\,\text{MeV}$.
At the threshold, both particles are created essentially at rest, with no leftover kinetic energy — any photon energy above this threshold appears as shared kinetic energy of the pair instead.
2
A photon with energy below $1.022\,\text{MeV}$ cannot produce an electron–positron pair under any circumstances — there simply isn't enough energy present to create the rest mass of both particles, regardless of where or how the photon interacts.
3
Photon energy is related to frequency by $E=hf$, so this threshold energy corresponds to a minimum photon frequency of $f = \dfrac{E}{h} = \dfrac{1.022\times1.60\times10^{-13}}{6.63\times10^{-34}} \approx 2.47\times10^{20}\,\text{Hz}$ — deep in the gamma-ray part of the electromagnetic spectrum, consistent with pair production only being observed for high-energy gamma photons.
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Annihilation, revisited quantitatively

Running the process in reverse, a particle and its antiparticle meeting can annihilate, converting their combined mass entirely into energy, typically released as two gamma-ray photons (two, rather than one, so that momentum as well as energy can be conserved):

$e^{-} + e^{+} \rightarrow \gamma + \gamma$
Worked exampleEnergy released when an electron and positron annihilate at rest
1
If the electron and positron are (approximately) at rest before annihilating, the only energy available is their combined rest energy: $E_{\text{total}} = 0.511 + 0.511 = 1.022\,\text{MeV}$.
2
This energy is shared equally between the two emitted photons (by symmetry, since momentum must also be conserved and the initial total momentum is zero), so each photon carries $\dfrac{1.022}{2} = 0.511\,\text{MeV}$.
3
Converting to joules: $0.511\times1.60\times10^{-13} = 8.18\times10^{-14}\,\text{J}$ per photon — matching, as it should, the rest energy of a single electron found earlier, since mass–energy equivalence guarantees the numbers come out consistent either way round.
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A general worked example: mass defect to energy released

Worked exampleConverting a mass defect directly to energy
1
Suppose a nuclear process releases a mass defect of $\Delta m = 3.20\times10^{-28}\,\text{kg}$ (a typical order of magnitude for a single nuclear reaction).
2
Energy released: $\Delta E = \Delta m\,c^2 = 3.20\times10^{-28}\times(3.00\times10^8)^2 = 2.88\times10^{-11}\,\text{J}$.
3
Converting to MeV: $\dfrac{2.88\times10^{-11}}{1.60\times10^{-13}} \approx 180\,\text{MeV}$ — a realistic value for the energy released in a single fission event, which is exactly where this method is applied in the next topic.
Note the units: since 1 MeV $=1.60\times10^{-13}\,\text{J}$ (not $10^{-19}$ — that's the eV), always convert carefully between eV, MeV and joules using the correct power of ten.
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Key takeaways

  • $E=mc^2$: mass and energy are interconvertible, with $c^2 \approx 9.00\times10^{16}\,\text{m}^2\text{s}^{-2}$ meaning even a tiny mass corresponds to a large energy.
  • Particle physicists quote mass in $\text{MeV}/c^2$ (or $\text{GeV}/c^2$) rather than kilograms, since it's the energy-equivalent value that's usually needed; the electron's rest mass is $0.511\,\text{MeV}/c^2$, worth memorising.
  • Pair production has an exact minimum (threshold) photon energy equal to the combined rest energy of the particle–antiparticle pair produced — $1.022\,\text{MeV}$ for an electron–positron pair.
  • Annihilation converts the entire rest mass of a particle–antiparticle pair into energy, typically as two photons (for momentum conservation), each carrying half the total energy released when annihilation occurs at rest.
  • A mass defect $\Delta m$ converts to energy released via $\Delta E = \Delta m\,c^2$ — the same equation underlying nuclear binding energy, fission and fusion, covered quantitatively in the next topic.
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