Derive and apply the mass-energy equivalence E = mc^2, including the calculation of mass defect and binding energy in nuclear reactions

What this dot point is asking

NESA wants you to state and apply $E = m c^2$, link it to the rest energy of a particle and the total relativistic energy, calculate mass defect and binding energy for simple nuclear reactions, and use the unified atomic mass unit conversion $1$ u $= 931.5$ MeV/$c^2$.

The answer

The famous equation

In Einstein's 1905 special relativity, the total energy of a free particle of rest mass $m$ moving at speed $v$ is:

where $\gamma = 1 / \sqrt{1 - v^2 / c^2}$. When the particle is at rest ($v = 0$, $\gamma = 1$), this reduces to the rest energy:

This is the famous "mass-energy equivalence": mass is a form of energy, and the conversion factor is $c^2 \approx 9.0 \times 10^{16}$ m$^2$ s$^{-2}$ - an enormous number. Even a few grams of mass converted to energy yields a colossal output.

The kinetic energy in special relativity

Splitting the total energy gives kinetic energy as:

In the non-relativistic limit $v \ll c$, a Taylor expansion gives $KE \approx \tfrac{1}{2} m v^2$, recovering classical mechanics. Near $c$, $KE$ diverges, which is why no massive object can reach $c$ (infinite energy would be required).

Unit conventions

For atomic and nuclear calculations, the unified atomic mass unit is convenient:

In energy units ($E = m c^2$):

Particle masses are often quoted in MeV/$c^2$: $m_e c^2 = 0.511$ MeV, $m_p c^2 = 938.3$ MeV, $m_n c^2 = 939.6$ MeV. Energy and mass are interconvertible currencies.

Mass defect and binding energy

The mass of a bound nucleus is less than the sum of the masses of its free constituents (protons and neutrons). The difference is the mass defect:

The corresponding energy:

is the binding energy, the energy that was released when the nucleus formed (or equivalently, the energy that must be supplied to dissociate it back into free nucleons).

Dividing by the number of nucleons gives the binding energy per nucleon, which peaks near iron-56 at about $8.8$ MeV per nucleon. Light nuclei (below iron) can release energy by fusion (smaller systems combine into more strongly bound systems). Heavy nuclei (above iron) can release energy by fission (large systems split into more strongly bound systems).

Three worked examples

Deuteron binding energy. $\Delta m = (m_p + m_n) - m_d = (1.00728 + 1.00866) - 2.01355 = 2.39 \times 10^{-3}$ u. $E_b = 2.39 \times 10^{-3} \times 931.5 = 2.23$ MeV. Per nucleon: $1.11$ MeV.

Iron-56 binding energy. $\Delta m \approx 0.528$ u. $E_b \approx 492$ MeV. Per nucleon: $\approx 8.79$ MeV - the peak of the binding-energy curve.

Fission energy release. When U-235 captures a neutron and fissions into Ba-141 + Kr-92 + 3 neutrons, the mass defect is approximately $0.215$ u, giving about $200$ MeV per fission event. A reactor running at $1$ GW thermal fissions about $3 \times 10^{19}$ U-235 atoms per second.

Pair production and annihilation

The cleanest demonstrations of $E = m c^2$ are at particle level:

• A gamma photon of at least $1.022$ MeV ($= 2 m_e c^2$) can convert in the field of a nucleus into an electron-positron pair, creating mass out of pure radiation energy.
• The reverse: an electron and a positron annihilate to produce two $0.511$-MeV gamma photons (or three for a parallel-spin state).

These processes are routine in particle physics; they conserve energy, momentum and charge while showing mass-energy conversion in both directions.

Energy in a chemical bond

For comparison, chemical-bond energies are of order eV per molecule, six orders of magnitude smaller than nuclear binding energies. That is why nuclear reactions release millions of times more energy per atom than chemical reactions.

Common traps

Quoting $E = m c^2$ when the particle is moving. That formula is only the rest energy. The total energy of a moving particle is $E = \gamma m c^2$.

Using kilograms with $c$ in km/s, or u with $c$ in m/s. Stick to SI throughout, or use the $1$ u $= 931.5$ MeV shortcut.

Confusing mass defect with the masses themselves. $\Delta m$ is a small difference, often $10^{-3}$ u or less; the binding energy in MeV is the headline result.

Saying mass is destroyed in a nuclear reaction. Mass and energy are equivalent forms; what is conserved is total relativistic energy (which includes rest energy). It is more accurate to say "rest mass is converted to other forms of energy".

Treating chemical and nuclear energy releases as the same scale. Chemical reactions release of order eV per atom; nuclear reactions release of order MeV per nucleon - a factor of one million larger.

In one sentence

Mass and energy are equivalent through $E = m c^2$ for the rest energy and $E = \gamma m c^2$ for the total relativistic energy, so the mass defect $\Delta m$ between a bound nucleus and its free constituents corresponds to the binding energy $E_b = \Delta m c^2$, with $1$ u $\leftrightarrow 931.5$ MeV.