Mass Defect and Binding Energy

At the Bhabha Atomic Research Centre in Trombay, Mumbai, sits a vacuum chamber called a mass spectrometer, and if you put one helium-4 atom into it and measure its mass — very carefully, to about ten decimal places — you get the number 4.002602 atomic mass units (u). Now do the same for its building blocks. Two protons: 2 \times 1.007276 = 2.014552 u. Two neutrons: 2 \times 1.008665 = 2.017330 u. Two electrons: 2 \times 0.000549 = 0.001098 u. Add them: 4.032980 u.

The helium atom is lighter than the parts it is made of. 4.032980 - 4.002602 = 0.030378 u is missing — about 0.75% of the total. And "missing" here does not mean "we haven't found it" or "the measurement is off". The measurement is correct. The mass is really, physically, not there.

Where did it go? Einstein's 1905 paper on mass-energy equivalence, E = mc^2, gives the answer. The missing mass was converted, in the act of assembling the helium nucleus from free protons and neutrons, into binding energy — the energy released as gamma rays, neutrinos, and kinetic energy during the formation. To tear the helium nucleus apart again, you would have to put that energy back in. The missing mass is the signature of the binding.

This is not a small effect. 0.030378 u times c^2 is 28.3 MeV — nearly seven million times the energy holding together a water molecule. It is the energy you would have to supply to pry the two protons and two neutrons of a helium nucleus apart. It is also the energy released every time four hydrogen nuclei, deep inside the sun, fuse to make one helium nucleus — the primary energy source of our nearest star and of every star you see in the night sky. It is the energy that, split four different ways, lights up the cities of Tamil Nadu from the six reactors at Kudankulam.

This article is about where that missing mass goes, how it is measured, why it matters, and how the plot of binding energy per nucleon against mass number — a simple graph you can sketch on the back of a notebook — explains both why the sun shines and why the first atomic bomb used uranium rather than iron.

What "binding" means, in terms you can hold

Before getting into nuclei, hold on to a simpler picture. Take a hydrogen atom: one proton, one electron, bound together. The electron is bound to the proton with 13.6 eV — meaning it takes 13.6 eV of energy to tear them apart (set the electron free at infinity with zero kinetic energy). Now weigh the hydrogen atom and compare with free proton + free electron. The atom is lighter by 13.6\ \text{eV}/c^2 = 2.4\times 10^{-8} u — a mass defect you cannot detect with any normal balance, but it is there.

The same logic applies to nuclei, but with one crucial difference: the binding is a million times stronger. Nuclear binding energies are in MeV (million eV), not eV. The mass defect per nucleon is correspondingly larger — about one part in a hundred, easily detectable.

The nuclear force (the residual strong interaction) is what holds nucleons together. It is attractive, short-ranged, charge-independent (protons and neutrons feel it about equally), and strong enough to overwhelm the Coulomb repulsion between the positively charged protons. Every pair of nucleons that sits close enough to feel the nuclear attraction contributes to lowering the total energy — and therefore, via E = mc^2, lowering the total mass.

The mass defect, formally

For a nucleus with Z protons and N = A - Z neutrons, total mass number A:

Masses are conventionally quoted in atomic mass units (u) where 1\,\text{u} = 1.66054\times 10^{-27} kg. The defining convention is that a neutral carbon-12 atom has mass exactly 12\,\text{u}.

Reference masses:

• Proton: m_p = 1.007276 u
• Neutron: m_n = 1.008665 u
• Electron: m_e = 0.000549 u

When mass tables in textbooks quote an "atomic mass" (like m(^4\text{He}) = 4.002602 u), they mean the neutral atom — nucleus plus electrons. This is the form you get from mass spectrometry. To get the nuclear mass alone, subtract Z\cdot m_e. But because Z\cdot m_p + Z\cdot m_e = Z\cdot m_H to extremely good approximation (the hydrogen-atom binding energy is a negligible 13.6 eV — about 1.5\times 10^{-8} u), you can equivalently write:

Using neutral-atom masses on both sides is the standard convention — it cancels out the electron contributions without any further book-keeping. The tiny chemical binding-energy corrections are below the precision of tabulated masses.

From mass defect to binding energy — Einstein's equation

Every joule of energy has an associated mass of E/c^2. Equivalently, every kilogram of mass has an associated energy of mc^2. So the missing \Delta m corresponds to an energy

The standard shortcut for nuclear physics: 1\,\text{u}\cdot c^2 = 931.5 MeV. You don't need to plug in c = 3\times 10^8 m/s every time — if \Delta m is in u, then B in MeV is

Worked derivation: helium-4

\Delta m = 2m_H + 2m_n - m(^4\text{He})

= 2(1.007825) + 2(1.008665) - 4.002602

= 2.015650 + 2.017330 - 4.002602 = 0.030378 u.

B = 0.030378 \times 931.5 = 28.30 MeV.

Why: the helium nucleus is bound by 28.3 MeV. Four nucleons share that binding, so the binding energy per nucleon is 28.30/4 = 7.07 MeV. This is already high for a light nucleus — helium-4 is a famously stable combination, sometimes called a "magic" nucleus, because it is doubly closed-shell (2 protons closing a shell, 2 neutrons closing a shell).

The binding energy per nucleon curve

The informative quantity is not B itself but B/A — the average binding energy per nucleon. If every nucleon in every nucleus were bound by the same amount, B/A would be a horizontal line. It is not. Plotting B/A against A gives one of the most important figures in physics.

The shape of this curve is the single most important fact in nuclear physics. Everything that follows — fusion, fission, why certain nuclei are stable, what powers stars, why atomic bombs work — is a consequence.

Explore the curve

Drag the slider below to move through mass numbers A and read off the approximate B/A for stable nuclei in that range. The total binding energy B = A\cdot(B/A) grows with A, but the per-nucleon value peaks and then falls.

Fusion and fission — movement toward the peak

Any reaction where the total B/A increases releases energy. That follows directly from the mass-energy balance.

Fusion — light nuclei going up

Take four hydrogen nuclei (four protons). Each is a free nucleon, B/A = 0. They fuse, through a series of steps including the weak-interaction p + p \to d + e^+ + \nu_e, into one helium-4 nucleus with B/A = 7.07 MeV. The total binding increased from zero (free nucleons) to 4 \times 7.07 = 28.28 MeV (the helium nucleus). That energy — 28.28 MeV per helium nucleus formed, or 7.07 MeV per nucleon — is released. Most of it ends up as kinetic energy of the fusion products and gamma photons; a bit is carried away by neutrinos.

This is the proton-proton chain that powers the sun. The sun consumes about 6\times 10^{11} kg of hydrogen every second and converts it to helium, releasing energy at the rate of 3.8\times 10^{26} W.

For comparison, the Indian National Fusion Programme at the Institute for Plasma Research (IPR) in Gandhinagar operates a tokamak (Aditya-U) trying to achieve the same fusion reactions on Earth — not with hydrogen but with deuterium and tritium, which fuse more easily:

17.6 MeV per reaction — about 3.5 MeV per nucleon — is the energy released because the helium-4 product sits much higher on the B/A curve than the deuterium-tritium reactants.

Fission — heavy nuclei going up

Take one uranium-235 nucleus with B/A = 7.59 MeV. Break it into two fragments of roughly equal mass — say ^{139}\text{Ba} (B/A = 8.39 MeV) and ^{94}\text{Kr} (B/A = 8.65 MeV) — plus a couple of free neutrons. The total binding went from 235 \times 7.59 = 1784 MeV (uranium) to 139 \times 8.39 + 94 \times 8.65 \approx 1166 + 813 = 1979 MeV (fragments) — an increase of \sim 195 MeV per fissioning uranium nucleus.

That number — about 200 MeV per fission — is the foundation of every nuclear reactor on Earth. At Tarapur (the first Indian nuclear power plant, commissioned in 1969) and Kudankulam (the largest, built with Russian collaboration), ^{235}\text{U} undergoes induced fission when it absorbs a slow neutron:

The released neutrons induce more fissions, producing a chain reaction. The ~200 MeV per event becomes heat, which boils water, which drives a turbine, which generates electricity. One kilogram of fully fissioned uranium-235 produces as much energy as three million kilograms of coal.

Why iron is the end of the road

In a massive star, fusion proceeds: hydrogen to helium, helium to carbon, carbon to oxygen, oxygen to silicon, silicon to... iron. When the star's core is iron, fusion can go no further — iron is at the peak of the B/A curve, so fusing iron with anything produces a less tightly bound product. Any further fusion would absorb energy rather than release it. The star has no more nuclear fuel. The core collapses under gravity and, in a supernova, produces the heavier elements (up to uranium) through explosive neutron-capture processes. Every atom of iron in your blood (the centre of the haemoglobin molecule) was made at the last stable stage of some dead star. Everything heavier was forged in the supernova that killed that star.

The semi-empirical mass formula

The B/A curve is not just "a fact" — its shape is understood quantitatively, term by term, via the semi-empirical mass formula (also called the Bethe-Weizsäcker formula). It models the nucleus as a drop of a liquid — a good first approximation — and gives:

The coefficients are fitted to measured nuclear masses. Typical values:

• a_V \approx 15.5 MeV (volume)
• a_S \approx 17.8 MeV (surface)
• a_C \approx 0.71 MeV (Coulomb)
• a_A \approx 23.2 MeV (asymmetry)
• \delta is a pairing term (about \pm 12/\sqrt{A} MeV for even-even / odd-odd nuclei)

Each term captures one aspect of the physics:

Volume term: +a_V A

The nuclear force is short-ranged. Each nucleon interacts only with its nearest neighbours — like molecules in a drop of water. So the total binding is proportional to the number of nucleons, i.e., to the nuclear volume. Since A \propto V (nuclear density is roughly constant), this term scales as A. This is the single biggest effect and the reason B/A is roughly constant across the periodic table at about 8 MeV.

Surface term: -a_S A^{2/3}

Nucleons on the surface of the nucleus have fewer neighbours than nucleons in the bulk — they are bound less tightly. The fraction of surface nucleons scales as the surface area divided by the volume, which for a nucleus of radius R \propto A^{1/3} goes as R^2/R^3 = 1/R \sim A^{-1/3}. Multiplying by A gives a correction proportional to A^{2/3}, subtracted from the volume contribution. This is why light nuclei have lower B/A than heavy ones — the surface "tax" is a larger fraction of the total.

Coulomb term: -a_C Z(Z-1)/A^{1/3}

Every pair of protons repels. There are Z(Z-1)/2 such pairs; the average distance between them scales as R \propto A^{1/3}. Coulomb self-energy of a uniform sphere scales as Z^2/R \sim Z^2/A^{1/3}, which destabilises heavy nuclei. This is what kills B/A beyond iron: as Z grows (roughly half of A for stable nuclei), the Coulomb term grows faster than the surface term, and more than compensates the volume term's growth.

Asymmetry term: -a_A (A - 2Z)^2/A

The Pauli exclusion principle (same principle as in atomic orbitals) applies to nucleons too. Protons and neutrons each fill their own energy levels. If you have more neutrons than protons, the extras must sit in higher-energy states (because the lower states are full), lowering the binding. The penalty is minimised when Z = N (so A = 2Z). For heavy nuclei, this competes with the Coulomb term: too many protons push up the Coulomb energy, so stable heavy nuclei have N > Z (uranium-238 has 92 protons and 146 neutrons).

Pairing term: \delta

Nucleons like to pair up with opposite spins (like electrons in an orbital — see quantum numbers). A nucleus with an even number of protons and an even number of neutrons gets an extra bonus (+\delta); with an odd count of either it is roughly average; with odd-odd it is penalised (-\delta). This pairing is why all but a handful of stable nuclei have even A.

Put them together and you get the smooth curve you drew above, with systematic wiggles superimposed (for magic numbers, pairing, etc.). Stability against beta decay picks out a narrow valley in the (Z, N) plane along which the semi-empirical formula is minimised for each A — the "valley of stability" that atomic nuclei live in.

Worked examples

Common confusions

• "Mass is never created or destroyed — conservation of mass." In pre-relativistic mechanics, yes. Once you include Einstein's equation, mass and energy are interconvertible, and only their sum E = mc^2 + KE is conserved. In chemical reactions the mass change is tiny (\sim 10^{-10} fraction) because binding energies are in eV. In nuclear reactions it is \sim 10^{-3} and readily measurable. Conservation of mass alone is a non-relativistic approximation that fails in nuclear physics.

• "E = mc^2 applies only to converting matter into energy (like annihilation)." No — it applies to every form of energy. A hot object has more mass than the same object cold, by E_\text{thermal}/c^2. A compressed spring has more mass than a relaxed one. A bound nucleus has less mass than its separated constituents. These are all tiny fractions in everyday life, but each is exactly \Delta E/c^2. Binding energy is just a negative \Delta E — and so produces negative \Delta m, a mass defect.

• "Mass defect is a measurement error." The opposite — it is so precisely known that modern atomic mass evaluations quote masses to 10 decimal places. The sensitivity of modern mass spectrometry (Penning traps) is better than 10^{-11}, easily sufficient to detect the ~1% mass defects of nuclei.

• "^{56}\text{Fe} is the most tightly bound nucleus." Technically, ^{62}\text{Ni} has a slightly higher B/A (about 8.7945 MeV) than ^{56}\text{Fe} (8.7903 MeV) — but the difference is tiny and the iron value is close enough that every textbook rounds to "iron is at the peak". For JEE and school physics, "peak at iron" is the correct answer.

• "Fission products are more massive than the original nucleus." No — they are less massive, by exactly Q/c^2 where Q is the energy released. The two fragments plus free neutrons always sum to less than the parent nucleus because they are more tightly bound and have a larger total B (smaller total mass).

• "Fusion combines nuclei, so the product should be more massive." The product nucleus has slightly less mass than the sum of the reactants, because binding energy (= released energy) appears as the mass defect. It is this mass loss that becomes the kinetic energy of the products and the photons they emit.

• "Binding energy and Q-value are the same thing." Related but different. Binding energy is the energy to tear a bound nucleus apart into free nucleons. Q-value is the energy released in a specific reaction. For a general reaction A + B → C + D, Q = (m_A + m_B - m_C - m_D)c^2 = B_C + B_D - B_A - B_B (the difference in total binding energies). A reaction is exothermic (Q > 0) when the products are more tightly bound than the reactants.

Where this leads next

• The nucleus — composition and size — what a nucleus is made of and how large it is, the starting point for any binding-energy calculation.
• Radioactivity — types of decay — how unstable nuclei shed energy to move toward the valley of stability.
• Nuclear fission — the energy release when heavy nuclei split toward the iron peak.
• Nuclear fusion — the energy release when light nuclei merge toward the iron peak.
• Photons and wave-particle duality of light — the photons that carry away the binding energy during nuclear reactions.
• Energy levels and quantum numbers — the analogous quantum-number structure for electrons (which also gives rise to binding energies and Pauli-driven shell structures, the nuclear version of which explains the magic numbers).