Mass–Energy Equivalence in the Nucleus

The equation on this page is not a museum piece; it is read out every day in three very different places. In hospitals, positron-emission tomography (PET) injects a positron-emitting tracer, waits for each positron to annihilate with an electron, and images the body from the back-to-back $0.511\ \text{MeV}$ photons — a direct, clinical use of $E = mc^2$ turning mass into detectable light. In power stations and stars, the energy comes from mass: a working reactor or a fusing stellar core is very slightly lighter after releasing its energy than before, and the Sun's four-million-tonnes-a-second mass loss is the fuel bill for sunlight. And in the deepest reading of all, particle physics has found that most of the mass of ordinary matter is itself binding energy — only about one percent of a proton's mass is the rest mass of its quarks; the rest is the energy of the gluon field that binds them, mass in the sense of $E = mc^2$. When you step on a scale, you are mostly weighing energy. The full four-momentum framework that makes equation (4) precise — where energy and momentum join into a single relativistic object — is the university seam named below; here we have the piece of it, $E_0 = mc^2$, that the nucleus runs on.

In Newtonian mechanics mass is the amount of "stuff" in a body and energy is something the body has — kinetic, potential, thermal. The two are related through motion but are fundamentally different bookkeeping quantities, and each is separately conserved. Einstein's 1905 result collapsed that distinction. Mass, he showed, is simply the energy a body carries when it is sitting still, divided by $c^2$. Heat a brick and it gets heavier; wind a spring and it gains mass; the effect is far too small to weigh for anything on a human scale, but it is real, and inside the nucleus it is enormous compared with the masses involved. The reason the nucleus is where this stops being a curiosity is the sheer strength of the nuclear binding. Pulling the four nucleons of a helium nucleus apart takes about $28\ \text{MeV}$ of energy — and by $E = mc^2$ that energy has a mass, about $0.03\ \text{u}$, which is exactly the amount by which the assembled nucleus is lighter than its separated parts. The "missing mass" is not missing at all; it left as binding energy when the nucleus formed. This single idea — that a bound system weighs less than its ingredients by (binding energy)$/c^2$ — is the engine behind radioactivity, fission, fusion and the light of every star. Here we build only the conversion tool; the binding-energy accounting itself is the next article's job.

The barcode of energy scales in the figure is the whole reason the nucleus needs its own unit. Rearranging the electrons of an atom — making and breaking chemical bonds, absorbing and emitting visible light — costs a few electronvolts. Rearranging the nucleons of a nucleus costs millions of electronvolts. That factor of a million is not an accident of chemistry; it is the ratio of nuclear binding to atomic binding, and $E = mc^2$ is the ruler that measures both.

Two units do all the work in this chapter, and both are chosen to make nuclear numbers convenient. The first is the unit of mass. Nuclear masses are far too small to quote in kilograms without a swarm of powers of ten, so physicists define the unified atomic mass unit, symbol u (older books write "amu"), by a single anchor: one atom of carbon-12 has a mass of exactly 12 u. That fixes the unit at

where: the value comes from dividing the measured mass of a carbon-12 atom by 12. On this scale a proton is $1.007276\ \text{u}$, a neutron $1.008665\ \text{u}$, and a neutral hydrogen atom $1.007825\ \text{u}$ — every nucleon is close to $1\ \text{u}$, which is the whole point of the definition. The physical reading is simple: u is a convenient unit of mass, roughly the mass of one nucleon, and a nucleus of mass number $A$ has a mass of about $A$ atomic mass units (not exactly, and that small difference is where the physics hides). The second unit is the unit of energy. The joule is hopelessly large for single particles, so we use the electronvolt: the energy one elementary charge gains falling through a potential difference of one volt.

where: $e = 1.60\times10^{-19}\ \text{C}$ is the elementary charge. Atomic and chemical energies come out as a few eV; nuclear energies come out as a few MeV — a million times larger, exactly the gap in the ladder above. Keeping the two prefixes straight (eV for the electron cloud, MeV for the nucleus) is half the battle in this chapter.

The centrepiece is Einstein's relation between the mass of a body and the energy it holds when at rest:

where: $E_0$ is the rest energy — the energy a body has purely by virtue of its mass, with no motion — $m$ is the (rest) mass, and $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ is the speed of light. The physical reading is that mass is frozen energy: even a motionless object is an energy store of $mc^2$, and if that mass could be entirely converted, that is the energy you would get out. Because $c^2 = 9.0\times10^{16}\ \mathrm{m^2\,s^{-2}}$ is so vast, the rest energy of even a gram of matter is staggering — we compute it below. Equation (3) is the rest case of a more general relativistic law, which we state but do not derive (its proof needs the machinery of special relativity, beyond this article's algebra):

where: $p$ is the particle's momentum. Setting $p = 0$ (a particle at rest) recovers $E = mc^2$; setting $m = 0$ (a photon, which has no rest mass) gives $E = pc$, the energy-momentum relation for light. Between those limits, a moving massive particle carries $E = \gamma mc^2 = mc^2 + \tfrac12 mv^2 + \dots$, so its rest energy $mc^2$ is the floor and its kinetic energy is the extra on top; for everyday speeds ($v \ll c$) the extra reduces to the familiar $\tfrac12 mv^2$. We only quote these; the honest place they are derived is the university relativity seam named at the end. What matters here is the rest term: every mass is a parcel of energy $mc^2$. A short but crucial disclaimer on rigour: a full derivation of $E = mc^2$ or of equation (4) from the Lorentz transformations — and any use of the integral or differential calculus that relativity leans on — is deliberately outside this article. We take $E = mc^2$ as a given experimental and theoretical fact (confirmed in the laboratory by Cockcroft and Walton in 1932) and put it to work as a units converter. The cleanest demonstration that rest mass is energy is annihilation: bring an electron and its antiparticle the positron together at rest, and they vanish entirely, their whole rest energy reappearing as two gamma photons flying apart back-to-back.

With the conversion $1\ \text{u} = 931.5\ \text{MeV}$ in hand, every particle mass becomes an energy by one multiplication. These three rest energies are worth memorising, because they anchor every calculation to come.

where: the electron mass is $m_e = 0.000549\ \text{u}$, the proton $m_p = 1.007276\ \text{u}$, and the neutron $m_n = 1.008665\ \text{u}$. Three physical readings fall straight out. First, the electron's rest energy $0.511\ \text{MeV}$ is tiny beside a nucleon's — an electron weighs about $1/1836$ of a proton — which is why we may often neglect the electron masses when we do nuclear energy bookkeeping using neutral-atom masses (the electron masses on the two sides of a reaction cancel; more on this in the next article). Second, the neutron is very slightly heavier than the proton, by $939.6 - 938.3 = 1.3\ \text{MeV}$ — a small difference with a large consequence, since it is what lets a free neutron beta-decay into a proton. Third, notice that $1\ \text{u} = 931.5\ \text{MeV}$ is a little less than the proton's $938.3\ \text{MeV}$: the atomic mass unit is not exactly a nucleon mass, precisely because carbon-12, which defines it, is a bound system whose binding lowers its mass.

Board level — electron-positron annihilation and the 0.511 MeV photon

An electron meets a positron (its antiparticle, of equal mass $m_e = 0.000549\ \text{u}$), both essentially at rest, and they annihilate into two identical gamma-ray photons. Find (a) the total energy released and (b) the energy of each photon. Take $1\ \text{u} = 931.5\ \text{MeV}$.

Step 1. Both particles are at rest, so the only energy available is their combined rest energy. Each has $m_e c^2 = 0.511\ \text{MeV}$, so the total released is $E = 2\times 0.511 = \boxed{1.022\ \text{MeV}}$. In joules, $1.022\times10^{6}\times1.602\times10^{-19} = 1.64\times10^{-13}\ \text{J}$.

Step 2. Two photons are emitted. Because the initial momentum was zero, the two photons must fly off back-to-back with equal energy, so each carries half the total: $\boxed{0.511\ \text{MeV}}$ per photon. This is the single most important number in nuclear medicine — every PET scan detects exactly these back-to-back $0.511\ \text{MeV}$ gammas. Here rest mass has been turned entirely into radiation: a clean, complete demonstration that mass is energy.

Board level — the mass defect of the deuteron as binding energy

A deuteron is a bound proton-plus-neutron. Using neutral-atom masses $M(^1\text{H}) = 1.007825\ \text{u}$, $m_n = 1.008665\ \text{u}$ and $M(^2\text{H}) = 2.014102\ \text{u}$, find how much lighter the deuteron is than its parts, and express that mass difference as an energy in MeV.

Step 1. Add the parts and subtract the whole. Working with neutral-atom masses lets the one electron on each side cancel, so we may use hydrogen-atom masses directly: $\Delta m = M(^1\text{H}) + m_n - M(^2\text{H}) = 1.007825 + 1.008665 - 2.014102 = 0.002388\ \text{u}$. The assembled deuteron is lighter than its ingredients by this much.

Step 2. Convert the missing mass to energy: $\Delta m\,c^2 = 0.002388\ \text{u}\times 931.5\ \text{MeV/u} = \boxed{2.224\ \text{MeV}}$. That is the deuteron's binding energy — the energy that left when it formed, and the energy you must supply to pull it apart. Notice we never touched a kilogram or a joule: the $931.5$ conversion did the whole job. (The full binding-energy story is the next article; this is a first taste.)

Competitive level — the energy locked in one gram, and the Sun's mass loss

Estimate (a) the energy that would be released if one gram of matter were completely converted to energy, comparing it to TNT, and (b) the rate at which the Sun loses mass, given it radiates $3.8\times10^{26}\ \text{W}$. Take $c = 3.0\times10^{8}\ \mathrm{m\,s^{-1}}$ and $1\ \text{tonne of TNT} = 4.18\times10^{9}\ \text{J}$.

Step 1. One gram is $m = 1.0\times10^{-3}\ \text{kg}$. Its rest energy is $E = mc^2 = (1.0\times10^{-3})(9.0\times10^{16}) = \boxed{9.0\times10^{13}\ \text{J}}$. Dividing by the TNT figure, $9.0\times10^{13}/4.18\times10^{9} = 2.15\times10^{4}$ tonnes, i.e. about $\boxed{21.5\ \text{kilotonnes}}$ of TNT — more than the Hiroshima bomb, from one gram if fully converted. (Real bombs convert only a small fraction, which is why kilograms of fuel yield kilotonnes, not megatonnes, of TNT-equivalent.)

Step 2. The Sun pours out energy, and every joule it radiates carries mass $E/c^2$. So it loses mass at $\dfrac{dm}{dt} = \dfrac{P}{c^2} = \dfrac{3.8\times10^{26}}{9.0\times10^{16}} = \boxed{4.2\times10^{9}\ \mathrm{kg\,s^{-1}}}$ — roughly four million tonnes every second, turned from mass into sunlight, and it has been doing so for over four billion years without appreciable change. Timing note: about 90 seconds. Both parts are one application of $E = mc^2$; the only traps are unit slips (grams to kilograms, watts as joules per second). Commit $c^2 = 9\times10^{16}$ to memory so part (a) is a one-line multiply, and recognise that "mass-loss rate" is just power over $c^2$.

Competitive level — the threshold for pair production

The reverse of annihilation is pair production: a gamma photon, passing near a nucleus, can turn into an electron-positron pair. Find (a) the minimum photon energy needed, and (b) the total kinetic energy shared by the pair if the photon has energy $1.50\ \text{MeV}$. Take $m_e c^2 = 0.511\ \text{MeV}$.

Step 1. To create the pair the photon must supply at least the two particles' rest energies. The threshold is $E_\text{min} = 2 m_e c^2 = 2\times0.511 = \boxed{1.022\ \text{MeV}}$ — exactly the annihilation energy run backwards. A photon below $1.022\ \text{MeV}$ simply cannot make an electron-positron pair, however you arrange it. (A nearby nucleus is needed to absorb the recoil momentum; it takes essentially none of the energy.)

Step 2. A $1.50\ \text{MeV}$ photon has $1.50 - 1.022 = \boxed{0.478\ \text{MeV}}$ left over after paying the mass bill, and this appears as kinetic energy shared between the electron and positron. So the same $E = mc^2$ that released $1.022\ \text{MeV}$ when the pair annihilated is the price of admission to create it — mass-energy equivalence read in both directions. Timing note: under a minute once you see the structure — threshold is "buy the two rest masses", leftover is "photon minus threshold goes into motion". The reusable idea for the whole chapter is that any reaction's energy release, or energy cost, is (mass of reactants minus mass of products) times $c^2$.

Mixing up eV, keV and MeV. The single commonest error in the whole nuclear chapter. Atomic and photon energies are in eV or keV; nuclear energies are in MeV, a million times larger. Writing an ionisation energy in MeV, or a nuclear binding energy in eV, is a factor-of-$10^6$ blunder that examiners on CBSE, NEET and JEE plant deliberately. Anchor yourself: $m_e c^2 = 0.511\ \text{MeV}$, nuclear binding $\approx 8\ \text{MeV}$ per nucleon.

Treating mass as destroyed. Do not write "mass disappears and becomes energy." Mass is energy; the total is conserved. In a reaction the products are lighter than the reactants by (energy released)$/c^2$ — nothing is annihilated except the distinction between the two ledgers. Marks are lost for language that implies conservation is broken.

Forgetting to use $1\ \text{u} = 931.5\ \text{MeV}$ and grinding through SI. In an exam you almost never convert to kilograms and joules. Keep masses in u and multiply the mass difference by $931.5$ to get MeV directly. Students who reach for $c^2 = 9\times10^{16}$ in a binding-energy question waste minutes and invite arithmetic slips.

Confusing rest energy with kinetic energy. $E_0 = mc^2$ is the energy of a particle at rest; a moving particle has more, $E = mc^2 + \text{KE}$. In annihilation-at-rest and threshold problems the kinetic energies start at zero, so the whole budget is rest energy — but say so, do not assume it silently.

Board scope. Quantitative mass-energy equivalence — $E = mc^2$, $1\ \text{u} = 931.5\ \text{MeV}$, rest energies and mass defects — is examined on CBSE, ICSE (as ISC), JEE Main, JEE Advanced, NEET, AP Physics 2, IB HL and A-Level 9702. It is not in the IB Standard Level or IGCSE 0625 core: IGCSE treats nuclei only qualitatively (isotopes, the three radiations, half-life, fission and fusion as uses) with no $931.5\ \text{MeV}$ conversion, no Q-values and no exponential decay law. If you are on IGCSE or IB SL, know the idea that mass and energy are equivalent, but you will not be asked to compute in MeV.

Texts. NCERT Physics Class 12, Chapter 13 ("Nuclei"), the mass-energy and atomic-mass-unit sections · H. C. Verma, Concepts of Physics Vol. 2, "The Nucleus" · Halliday, Resnick and Walker, Fundamentals of Physics, the "Relativity" and "Nuclear Physics" chapters · Krane, Introductory Nuclear Physics, for the mass-energy conventions used throughout.

Provenance. Standard constants (CODATA / LANE_SPEC). Recompute: $1\ \text{u} = 1.6605\times10^{-27}\ \text{kg}$; $c^2 = 9.0\times10^{16}\ \mathrm{m^2\,s^{-2}}$ (rounded), $8.988\times10^{16}$ (precise); $1\ \text{u}\to E$: $mc^2 = 1.4924\times10^{-10}\ \text{J} = 9.315\times10^{8}\ \text{eV} = 931.5\ \text{MeV}$; $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$, $1\ \text{MeV} = 1.602\times10^{-13}\ \text{J}$; rest energies $m_e c^2 = 0.511\ \text{MeV}$, $m_p c^2 = 938.3\ \text{MeV}$, $m_n c^2 = 939.6\ \text{MeV}$, $m_n - m_p = 1.3\ \text{MeV}$; proton mass $1.673\times10^{-27}\ \text{kg}$; annihilation $2\times0.511 = 1.022\ \text{MeV}$, $1.64\times10^{-13}\ \text{J}$; deuteron $\Delta m = 0.002388\ \text{u} \to 2.224\ \text{MeV}$; one gram $\to 9.0\times10^{13}\ \text{J} \approx 21.5\ \text{kt}$ TNT; Sun mass loss $4.2\times10^{9}\ \mathrm{kg\,s^{-1}}$; pair-production threshold $1.022\ \text{MeV}$, leftover from $1.50\ \text{MeV}$ is $0.478\ \text{MeV}$.

Prerequisites. Inside the Nucleus: Protons, Neutrons and Isotopes — for $Z$, $N$, $A$ and the nuclide notation these masses attach to. Leads to. Mass Defect and Nuclear Binding Energy, which cashes $1\ \text{u} = 931.5\ \text{MeV}$ into the binding energy of real nuclei. For enrichment, the university seams Mass-Energy Equivalence in Special Relativity — where $E = mc^2$ is derived rather than quoted — and Four-Momentum, Mass and Energy, which builds the full relation $E^2 = (pc)^2 + (mc^2)^2$ that equation (4) states (go-deeper only, never an exam dependency here).