Mass Defect & Nuclear Binding Energy

Try a piece of nuclear book-keeping that should be trivial. Take two protons and two neutrons, weigh them separately on the world's best balance, add the four numbers, and then weigh a helium-4 nucleus made of exactly those four particles. The two totals ought to match — matter is conserved, nothing has been thrown away. They do not match. The assembled nucleus is lighter than its own ingredients, by about three parts in four hundred. Some mass has gone missing in the act of sticking the pieces together. That missing mass is not an error and it is not lost. By Einstein's $E = mc^2$ — the subject of the previous article — mass and energy are the same currency, and the mass shortfall was paid out as energy the instant the nucleus formed. Run the film backwards and the same number tells you the price of demolition: to tear a helium nucleus back into two free protons and two free neutrons you must supply exactly that much energy. It is the glue bill. Physicists call the missing mass the mass defect and the energy it represents the binding energy — the single most important number in all of nuclear physics, because it decides which nuclei are stable, which ones decay, and why both the Sun and the bomb release the energy they do. This article shows how to compute it from a table of masses with nothing but arithmetic, why we quietly use atomic masses to make the electrons cancel, and why the truly telling quantity is not the binding energy itself but the binding energy per nucleon. Every number here was weighed into existence by Francis Aston's mass spectrograph a century ago.

Key takeaways

Mass defect $\Delta m$ : a bound nucleus weighs less than the free protons and neutrons that make it — the difference is $\Delta m = Z m_p + N m_n - M_{\text{nucleus}}$ .
Binding energy $BE = \Delta m\, c^2$ : the energy released when the nucleus is assembled, equal to the energy you must supply to pull it completely apart. For $^4$ He, $\Delta m = 0.030377\ \text{u}$ gives $BE = 28.3\ \text{MeV}$ .
Use atomic masses. Tabulated masses include the electrons; if you use the neutral-atom mass of hydrogen $M(^1\text{H})$ for each proton and the atomic mass $M$ of the nuclide, the $Z$ electron masses cancel exactly: $BE = [\,Z\,M(^1\text{H}) + N m_n - M(^A_Z X)\,]\,c^2$ .
Binding energy per nucleon $BE/A$ is the real stability meter — the average glue holding each nucleon. Deuterium sits at a feeble $1.11\ \text{MeV}$ ; helium-4 at a robust $7.07\ \text{MeV}$ .
The conversion factor is $1\ \text{u} \to 931.5\ \text{MeV}$ , so a mass defect of one thousandth of a u already means about $0.93\ \text{MeV}$ — MeV energies, a million times the eV scale of chemistry.
Key idea: a nucleus is bound precisely because it is lighter than its parts; the missing mass, converted by $E=mc^2$ , is the binding energy, and its per-nucleon value $BE/A$ measures how tightly the nucleus is held together.

The core insight

A bound nucleus is lighter than the separate protons and neutrons that compose it by the mass defect $\Delta m$ ; multiplied by $c^2$ this becomes the binding energy $BE = \Delta m\,c^2$ , the energy released on assembly and required for complete disassembly, and dividing by the nucleon number gives $BE/A$ , the intensive measure of nuclear stability.

Where this lives today

The mass defect is not a museum piece; it is the running total behind the world's energy. Every fusion reaction in the Sun and every fission event in a power reactor is the cashing-in of a difference in binding energy, computed exactly as on this page. Assembling helium from its nucleons releases $28.3\ \text{MeV}$ per nucleus — scale that up and forming just one gram of helium this way would liberate about $6.8\times10^{11}\ \text{J}$ , roughly the energy in twenty tonnes of oil. That colossal density is why nuclear processes dominate the energy budgets of stars and the ambitions of reactor engineers. Aston's descendants — modern mass spectrometers — now weigh atoms to parts per billion, and those measurements feed the binding-energy tables used to predict which exotic, short-lived nuclei can exist at all, guiding the search for new elements at facilities like GSI and RIKEN. The same idea underlies the "atomic mass evaluation" that metrologists maintain, and even the redefinition of physical constants. Whenever you see an energy quoted in MeV next to a nucleus, a mass defect is doing the work.

The physical picture

Picture the four nucleons of a helium nucleus — two protons, two neutrons — held far apart from one another, each at rest, not yet a nucleus. In this pulled-apart state each has its own free-particle mass, and the system's total mass is simply the sum of the four. Now let the strong nuclear force draw them together into a single bound nucleus. As they fall into the attractive well they speed up and radiate away energy — as a gamma photon, as the recoil kinetic energy of the products — and the system settles into its low-energy bound state. Because it has shed energy, and because energy has mass, the bound nucleus weighs less than the four free particles did. The system paid for its own binding by giving up some of its mass-energy. This is exactly analogous to a ball rolling into a valley: to get it out again you must supply the energy it released rolling in. The binding energy is the depth of the nuclear valley. It is positive by definition for any bound system — it is the energy you would have to invest to climb back out to the separated state. A nucleus with a large binding energy sits in a deep valley and is hard to break apart; one with a small binding energy sits in a shallow dip and is fragile.

Figure

Weighing helium-4 against its parts: two hydrogen atoms plus two neutrons total 4.032980 u but the neutral helium-4 atom is only 4.002603 u, the missing 0.030377 u appearing as 28.3 MeV of binding energy

Two readings of the same number sit at the heart of everything below. The binding energy is the energy released when the free nucleons are assembled into the nucleus, and it is the energy required to disassemble the nucleus back into free nucleons. They are equal because the two processes are the same journey run in opposite directions. When a student is asked "how much energy is needed to break this nucleus into its constituent protons and neutrons?", the answer is simply its binding energy — no new calculation.

Mass defect and binding energy

Take a nucleus $^A_Z X$ with $Z$ protons and $N = A - Z$ neutrons. If it were merely a loose bag of those particles its mass would be $Z m_p + N m_n$ . The actual bound nucleus has a smaller measured mass $M_{\text{nucleus}}$ . The shortfall is the mass defect:

\Delta m = Z m_p + N m_n - M_{\text{nucleus}} \qquad (1)

where: $m_p = 1.007276\ \text{u}$ is the proton mass, $m_n = 1.008665\ \text{u}$ the neutron mass, and $Z, N$ the numbers of each. The physical reading is that $\Delta m$ is the mass that "disappeared" when the nucleus formed — always positive for a bound nucleus, because binding always lowers the energy and hence the mass. Convert that mass to energy with Einstein's relation, and you have the binding energy:

BE = \Delta m \, c^2 = \big(Z m_p + N m_n - M_{\text{nucleus}}\big)\,c^2 \qquad (2)

where: $c = 3.00\times10^{8}\ \text{m s}^{-1}$ . The physical reading of (2) is the whole point of the article: the binding energy is the mass defect expressed in energy units. In practice you never multiply by $c^2$ in SI; you use the nuclear conversion factor $1\ \text{u} = 931.5\ \text{MeV}/c^2$ , so a mass defect quoted in u becomes an energy in MeV just by multiplying by $931.5$ . A mass defect of a single thousandth of a u is already $0.93\ \text{MeV}$ — which is why nuclear energies live at the MeV scale, a million times the electronvolt scale of chemical bonds. Chemistry rearranges electrons; nuclear physics rearranges nucleons, and the mass changes are a million times larger.

Using atomic masses — why the electrons cancel

There is a subtlety that trips up nearly everyone the first time. The masses in every data table are atomic masses — the mass of the neutral atom, nucleus plus its $Z$ electrons — not bare nuclear masses. If you carelessly mix a bare proton mass with an atomic nuclide mass, you leave $Z$ electron masses unbalanced and get the wrong answer. The clean fix is to use atomic masses everywhere, and let the electrons cancel on their own. The trick is to represent each proton by the neutral hydrogen atom $^1$ H, whose mass $M(^1\text{H}) = 1.007825\ \text{u}$ is the proton plus one electron. Then $Z$ hydrogen atoms carry $Z$ protons and $Z$ electrons — exactly the electron count of the neutral atom $^A_Z X$ on the other side. The electrons match up and drop out:

BE = \big[\,Z\,M(^1\text{H}) + N m_n - M(^A_Z X)\,\big]\,c^2 \qquad (3)

where: $M(^1\text{H}) = 1.007825\ \text{u}$ is the mass of a neutral hydrogen atom, $m_n = 1.008665\ \text{u}$ the neutron mass (a neutron has no electron to worry about), and $M(^A_Z X)$ the atomic mass of the nuclide. The physical reading: equation (3) is numerically identical to (2) but uses only the masses you can look up, and the tiny electron binding energies (a few eV) that this ignores are utterly negligible against MeV nuclear energies. From here on we always use atomic masses and equation (3). This is the formula every board expects you to write.

Derivation AFrom mass defect to binding energy, with the electron cancellation

The goal is to derive the working formula $BE = [\,Z\,M(^1\text{H}) + N m_n - M(^A_Z X)\,]\,c^2$ from the bare-nucleus definition $BE = (Z m_p + N m_n - M_{\text{nucleus}})c^2$ , showing explicitly that the $Z$ electron masses cancel, and then to apply it to the simplest bound nucleus of all — the deuteron — to get its binding energy. We use only the definitions above and school algebra; the relation $E = mc^2$ is quoted from the previous article, not re-derived.

Step 1. Start from the bare-nucleus binding energy, equation (2): $BE = (Z m_p + N m_n - M_{\text{nucleus}})\,c^2$ . Everything here is a nuclear mass; we now trade each nuclear mass for an atomic one.

Step 2. A neutral hydrogen atom is one proton plus one electron, so $M(^1\text{H}) = m_p + m_e$ , giving $m_p = M(^1\text{H}) - m_e$ . Likewise the neutral atom $^A_Z X$ is its nucleus plus $Z$ electrons, so $M(^A_Z X) = M_{\text{nucleus}} + Z m_e$ , giving $M_{\text{nucleus}} = M(^A_Z X) - Z m_e$ . (Both statements drop the eV-scale electron binding energies, negligible here.)

Step 3. Substitute both into (2). The proton term becomes $Z m_p = Z\,M(^1\text{H}) - Z m_e$ , and the nuclear-mass term becomes $-M_{\text{nucleus}} = -M(^A_Z X) + Z m_e$ . Adding:

BE = \big[\,Z\,M(^1\text{H}) - Z m_e + N m_n - M(^A_Z X) + Z m_e\,\big]\,c^2.

The two $Z m_e$ terms are equal and opposite — the electrons cancel exactly — leaving equation (3): $BE = [\,Z\,M(^1\text{H}) + N m_n - M(^A_Z X)\,]\,c^2$ .

Step 4. Apply it to the deuteron, $^2_1$ H (one proton, one neutron), whose atomic mass is $M(^2\text{H}) = 2.014102\ \text{u}$ . Here $Z=1$ , $N=1$ : $\Delta m = M(^1\text{H}) + m_n - M(^2\text{H}) = 1.007825 + 1.008665 - 2.014102 = 0.002388\ \text{u}$ . Convert: $BE = 0.002388 \times 931.5 = 2.224\ \text{MeV}$ . The deuteron is bound by a mere $2.224\ \text{MeV}$ — the most weakly bound stable nucleus there is, which is why a modest gamma photon can split it apart. Verified: $\Delta m(^2\text{H}) = 0.002388\ \text{u}$ and $BE = 2.224\ \text{MeV}$ , with the electron cancellation checked symbolically — recomputed in verify_hs2-mass-defect-binding-energy.py.

Binding energy per nucleon — the real stability meter

The total binding energy grows steadily with size — a uranium nucleus, with its 235 nucleons, is bound by about $1800\ \text{MeV}$ , far more than helium's $28\ \text{MeV}$ — but that does not make uranium more stable. A bigger nucleus has more bonds simply because it has more nucleons, just as a longer chain has more links. To compare how tightly different nuclei are held, you must divide out the size and ask for the binding energy shared per nucleon:

\frac{BE}{A} = \frac{\big[\,Z\,M(^1\text{H}) + N m_n - M(^A_Z X)\,\big]\times 931.5}{A}\ \ \text{MeV per nucleon} \qquad (4)

where: $A = Z + N$ is the total nucleon number. The physical reading: $BE/A$ is the average energy needed to remove one nucleon, an intensive measure that does not automatically grow with size — so it can genuinely rank nuclei by stability. The deuteron's $BE/A$ is only $2.224/2 = 1.11\ \text{MeV}$ , whereas helium-4's is $28.3/4 = 7.07\ \text{MeV}$ : each nucleon in the alpha particle is held more than six times as firmly as each nucleon in the deuteron. That single comparison already explains why the alpha particle is such a rugged, favoured building block of the nuclear world.

Figure

Binding energy per nucleon for four nuclides: deuterium 1.11, helium-4 7.07, iron-56 8.79 at the peak, uranium-238 about 7.6 MeV per nucleon, showing tightly bound medium nuclei flanked by looser light and heavy ones

This per-nucleon view is the doorway to everything that follows in the chapter. Line the nuclides up by mass number and plot $BE/A$ , and you get the celebrated binding-energy curve: it climbs steeply through the light nuclei, peaks in a broad hump around iron-56 at about $8.8\ \text{MeV}$ per nucleon, and eases gently back down toward the heaviest elements. Fusing light nuclei climbs up that curve toward the peak, and splitting heavy ones climbs up the other side — both release energy. That is the whole logic of stars and reactors, and it is the job of the next article. Here we only build the tool: $BE/A$ , computed from a table of masses.

Go deeper · Derivation B · 1 minHelium-4: the binding energy and binding energy per nucleon

Derivation BHelium-4: the binding energy and binding energy per nucleon

The goal is to compute, from atomic masses alone, the total binding energy and the binding energy per nucleon of the alpha particle $^4_2$ He — the anchor value of nuclear physics — and to express the same result three ways (in u, in MeV, and as a mass in kilograms) so the $E=mc^2$ conversion is transparent. We use equation (3), atomic masses, and the factor $1\ \text{u} = 931.5\ \text{MeV}$ .

Step 1. Count the constituents. Helium-4 has $Z=2$ protons and $N = A - Z = 4 - 2 = 2$ neutrons. Using atomic masses, the free constituents weigh $2\,M(^1\text{H}) + 2\,m_n = 2(1.007825) + 2(1.008665) = 2.015650 + 2.017330 = 4.032980\ \text{u}$ .

Step 2. Subtract the measured atomic mass $M(^4\text{He}) = 4.002603\ \text{u}$ to get the mass defect: $\Delta m = 4.032980 - 4.002603 = 0.030377\ \text{u}$ . As a fraction of the whole, that is $0.030377/4.032980 = 0.75\%$ — three parts in four hundred have turned into binding energy.

Step 3. Convert to energy: $BE = 0.030377 \times 931.5 = 28.3\ \text{MeV}$ . Cross-check the conversion through SI: $\Delta m = 0.030377 \times 1.66\times10^{-27} = 5.04\times10^{-29}\ \text{kg}$ , and $BE = \Delta m\,c^2 = 5.04\times10^{-29}\times(3.00\times10^{8})^2 = 4.54\times10^{-12}\ \text{J}$ , which divided by $1.602\times10^{-13}\ \text{J/MeV}$ is $28.3\ \text{MeV}$ — the same number by two roads.

Step 4. Divide by the four nucleons: $BE/A = 28.3/4 = 7.07\ \text{MeV}$ per nucleon. Compare the deuteron's $1.11\ \text{MeV}$ : helium-4 is far more tightly bound per nucleon, which is exactly why alpha particles are ejected whole in alpha decay and why helium is the ash of stellar fusion. Verified: constituents $4.032980\ \text{u}$ , $\Delta m = 0.030377\ \text{u}$ , $BE = 28.3\ \text{MeV}$ (both by $\times931.5$ and by the SI $\Delta m c^2$ route), $BE/A = 7.07\ \text{MeV}$ — recomputed in verify_hs2-mass-defect-binding-energy.py.

Limiting cases

case	result	meaning
a single free nucleon ( $A=1$ )	$\Delta m = 0$ , $BE = 0$	nothing to bind — a lone proton or neutron has no binding energy
the deuteron $^2$ H	$BE = 2.224\ \text{MeV}$ , $BE/A = 1.11$	the most weakly bound stable nucleus; a $2.22\ \text{MeV}$ photon splits it
the alpha particle $^4$ He	$BE = 28.3\ \text{MeV}$ , $BE/A = 7.07$	a local peak — unusually tightly bound for a light nucleus
medium nuclei near iron-56	$BE/A \approx 8.8\ \text{MeV}$	the summit of stability — most tightly bound of all
very heavy nuclei ( $^{238}$ U)	$BE/A \approx 7.6\ \text{MeV}$	past the peak — loosely enough bound to fission
$\Delta m \to 0$ (unbound)	$BE \le 0$	no mass defect means no binding — the "nucleus" would fly apart

Common trapThe mass is not "lost" and the nucleons do not shrink

Two twin misconceptions haunt this topic. The first is that the missing mass has simply vanished — a violation of conservation. It has not: it was carried off as energy (a photon, kinetic energy of products) when the nucleus formed, and mass-plus-energy is conserved throughout. The mass defect is bookkeeping, not disappearance. The second, subtler error is to imagine that a proton inside a nucleus is a lighter object than a free proton — that binding physically compresses each particle. It does not. The individual nucleons keep their identities and their free masses; what is lighter is the bound system as a whole, because the system has shed energy. Always attribute $\Delta m$ to the system, never to the individual nucleons. And never forget the sign: for a bound nucleus $\Delta m$ and $BE$ are positive — if your arithmetic gives a negative binding energy, you have subtracted the wrong way round.

Worked examples

Board level — the binding energy of helium-4 (CBSE and ICSE pattern)

Given the atomic masses $M(^1\text{H}) = 1.007825\ \text{u}$ , $m_n = 1.008665\ \text{u}$ and $M(^4\text{He}) = 4.002603\ \text{u}$ , and $1\ \text{u} = 931.5\ \text{MeV}$ , find (a) the mass defect, (b) the total binding energy, and (c) the binding energy per nucleon of the alpha particle.

Step 1. Identify $Z=2$ , $N=2$ , $A=4$ . State the atomic-mass formula (3) so the electrons cancel: $\Delta m = 2\,M(^1\text{H}) + 2\,m_n - M(^4\text{He})$ .

Step 2. Constituent mass: $2(1.007825) + 2(1.008665) = 4.032980\ \text{u}$ . Mass defect: $\Delta m = 4.032980 - 4.002603 = \boxed{0.030377\ \text{u}}$ .

Step 3. Binding energy: $BE = 0.030377 \times 931.5 = \boxed{28.3\ \text{MeV}}$ .

Step 4. Per nucleon: $BE/A = 28.3/4 = \boxed{7.07\ \text{MeV per nucleon}}$ . State the meaning in one line: it takes on average $7.07\ \text{MeV}$ to prise one nucleon out of an alpha particle.

Competitive level — iron-56 and the peak of stability (JEE and NEET pattern)

The atomic mass of iron-56 is $M(^{56}_{26}\text{Fe}) = 55.934937\ \text{u}$ . Using $M(^1\text{H}) = 1.007825\ \text{u}$ , $m_n = 1.008665\ \text{u}$ and $1\ \text{u} = 931.5\ \text{MeV}$ , find its binding energy per nucleon, and comment on the result.

Step 1. Here $Z=26$ , $N = 56 - 26 = 30$ . Constituent mass $= 26\,M(^1\text{H}) + 30\,m_n = 26(1.007825) + 30(1.008665) = 26.203450 + 30.259950 = 56.463400\ \text{u}$ .

Step 2. Mass defect: $\Delta m = 56.463400 - 55.934937 = 0.528463\ \text{u}$ .

Step 3. Total binding energy: $BE = 0.528463 \times 931.5 = 492.3\ \text{MeV}$ .

Step 4. Per nucleon: $BE/A = 492.3/56 = \boxed{8.79\ \text{MeV per nucleon}}$ . This is essentially the highest $BE/A$ of any nuclide — iron-56 sits at the top of the binding-energy curve, which is why iron is the end-point of fusion in massive stars and the most stable nucleus in nature. Nothing above or below it is bound quite as tightly per nucleon. Timing note: about two minutes by hand. The one place to lose marks is the neutron count — with $A=56$ and $Z=26$ it is $N=30$ , not $26$ . Keep four decimal places in the constituent sum (a rounding slip in the fourth place shifts $BE$ by several MeV), and only round at the very end. A useful sanity check: $BE/A$ for a mid-mass nucleus should land near $8.8\ \text{MeV}$ ; if you get $88$ or $0.88$ , you have mislaid a factor of $10$ .

Competitive level — energy released when two deuterons make helium-4 (JEE Advanced pattern)

Two deuterons combine to form a helium-4 nucleus. Given $BE(^2\text{H}) = 2.224\ \text{MeV}$ and $BE(^4\text{He}) = 28.3\ \text{MeV}$ , find the energy released, without looking up any masses.

Step 1. The energy released in a nuclear rearrangement is the increase in total binding energy — the products are more tightly bound than the reactants, and the difference comes out as released energy. This lets you skip the mass table entirely and work with binding energies directly.

Step 2. Reactant binding energy: two deuterons carry $2 \times BE(^2\text{H}) = 2 \times 2.224 = 4.448\ \text{MeV}$ of binding between them.

Step 3. Product binding energy: the single helium-4 nucleus carries $BE(^4\text{He}) = 28.3\ \text{MeV}$ .

Step 4. Energy released $= BE(\text{products}) - BE(\text{reactants}) = 28.3 - 4.448 = \boxed{23.85\ \text{MeV}}$ . The helium is far more tightly bound than the two loose deuterons, and that difference is liberated — this is a fusion reaction, and the same climb up the binding-energy curve powers the Sun. Timing note: under a minute once you see it is a binding-energy difference, not a mass calculation. The trap is to compute the released energy as $BE(\text{reactants}) - BE(\text{products})$ , which gives the wrong sign; energy is released when binding increases, so it is always products minus reactants. If the number comes out negative, the reaction would absorb energy, not release it.

Exam traps

Mixing bare and atomic masses. The single most common error. Tables list atomic masses. Use $M(^1\text{H})$ for each proton (not the bare $m_p$ ) and the atomic mass of the nuclide, and the electrons cancel automatically. Mixing a bare proton mass with an atomic nuclide mass leaves $Z$ electron masses unbalanced and corrupts the answer.

Getting the neutron number wrong. $N = A - Z$ , not $A$ . For iron-56, $N = 56 - 26 = 30$ . A surprising number of scripts multiply the neutron mass by $A$ or by $Z$ . Write $Z$ and $N$ explicitly before you start.

Rounding too early. Mass defects are small differences of large, nearly equal numbers, so they are exquisitely sensitive to rounding. Carry the atomic masses to at least four decimal places (six is safer) and round only the final MeV answer. Truncating $1.007825$ to $1.01$ can throw the binding energy off by tens of MeV.

Confusing $BE$ with $BE/A$ . Total binding energy grows with size, so a bigger nucleus almost always has a larger $BE$ but not a larger $BE/A$ . Stability comparisons must use $BE/A$ . Uranium's $BE$ ( $\approx 1800\ \text{MeV}$ ) dwarfs helium's ( $28\ \text{MeV}$ ), yet helium is more tightly bound per nucleon.

Sign of the mass defect. For any bound nucleus, constituents minus nucleus is positive. If you write nucleus minus constituents you get a negative number and may wrongly conclude the nucleus is unbound. Binding energy is positive by definition.

Board scope. The quantitative mass defect and binding energy — the $931.5\ \text{MeV}$ conversion, $BE = [\,Z M(^1\text{H}) + N m_n - M\,]c^2$ , and $BE/A$ — are core on CBSE, ICSE (ISC), JEE Main, JEE Advanced, NEET, AP Physics 2, IB SL, IB HL and A-Level 9702. IGCSE 0625 carries the qualitative nuclear atom (isotopes, the strong force) but not the $931.5\ \text{MeV}$ arithmetic or mass-defect calculations — so an IGCSE student meets the idea that nuclei are bound, but is never asked to compute $BE$ . Do not spend IGCSE revision time on these sums.

Origins · 1 minFrancis William Aston and how this was discovered

Francis William Aston

1877–1945 · Birmingham and Cambridge, England

Francis William Aston was the English chemist-physicist whose mass spectrograph, built at Cambridge's Cavendish Laboratory in 1919, first measured atomic masses precisely enough to make this whole article possible. Working from J. J. Thomson's positive-ray apparatus, Aston separated ions by mass and discovered that most elements are mixtures of isotopes with nearly whole-number masses — his "whole-number rule". But his most consequential finding was the tiny departures from whole numbers: he charted the "packing fraction", the fractional deviation of each atomic mass from an integer, and showed it varies smoothly across the periodic table. Those deviations are precisely the mass defects of this article, and his packing-fraction curve was the direct empirical ancestor of the binding-energy-per-nucleon curve. Aston received the 1922 Nobel Prize in Chemistry, and foresaw with startling clarity the energy locked in the nucleus.

History

1905

— Albert Einstein's mass-energy relation E = mc² makes mass a form of energy, the principle that turns a mass defect into a binding energy.

1919

— Francis Aston builds the mass spectrograph at the Cavendish.

1920

— Aston states the whole-number rule for isotopes.

1922

— Aston is awarded the Nobel Prize in Chemistry for his mass-spectrograph work on isotopes and atomic masses.

1927

— Aston, using his far more accurate second-generation mass spectrograph, charts the packing fraction, revealing that nuclei weigh slightly less than the sum of their nucleons.

1932

— James Chadwick discovers the neutron, completing the proton–neutron picture from which the mass defect is computed.

Going further

Texts. NCERT Physics Class 12, Chapter 13 ("Nuclei"), the mass-defect and binding-energy sections · H. C. Verma, Concepts of Physics Vol. 2, "The Nucleus" · Halliday, Resnick and Walker, "Nuclear Physics" · Krane, Introductory Nuclear Physics, for the mass-spectrometry route to atomic masses.

Provenance. Standard atomic masses (u). Recompute: deuteron $\Delta m = M(^1\text{H})+m_n-M(^2\text{H}) = 0.002388\ \text{u}$ , $BE = 2.224\ \text{MeV}$ , $BE/A = 1.11\ \text{MeV}$ ; helium-4 constituents $= 4.032980\ \text{u}$ , $\Delta m = 0.030377\ \text{u}$ ( $0.75\%$ of the mass), $BE = 28.3\ \text{MeV}$ (both by $\times931.5$ and by $\Delta m c^2 = 4.54\times10^{-12}\ \text{J}$ ), $BE/A = 7.07\ \text{MeV}$ ; iron-56 constituents $= 56.463400\ \text{u}$ , $\Delta m = 0.528463\ \text{u}$ , $BE = 492.3\ \text{MeV}$ , $BE/A = 8.79\ \text{MeV}$ ; $2\,^2\text{H}\to{}^4\text{He}$ releases $28.3 - 2(2.224) = 23.85\ \text{MeV}$ ; conversion $1\ \text{u} = 931.5\ \text{MeV} = 1.66\times10^{-27}\ \text{kg}$ ; one gram of helium so formed releases $6.8\times10^{11}\ \text{J}$ ; constants $M(^1\text{H})=1.007825$ , $m_n=1.008665$ , $m_p=1.007276$ , $m_e=0.000549\ \text{u}$ ( $=0.511\ \text{MeV}$ ).

Prerequisites. Mass–Energy Equivalence in the Nucleus — where $E=mc^2$ and the $1\ \text{u}=931.5\ \text{MeV}$ conversion this article rests on are established. Leads to. The Binding-Energy Curve, which turns the $BE/A$ values computed here into the master curve that governs fission and fusion. For enrichment, the university seam Binding Energy and Mass Defect in Special Relativity (go-deeper only, never an exam dependency here) sets the same idea in a fully relativistic frame.

What exactly is the mass defect — is mass being destroyed?

No mass is destroyed; it is converted. When free nucleons bind into a nucleus the system releases energy (as a photon or as kinetic energy), and because energy carries mass, the bound system ends up lighter by exactly the energy released divided by $c^2$ . The "missing" mass left the system as energy. The total mass-plus-energy of everything involved is conserved throughout. So the mass defect $\Delta m$ is best read as "the amount of mass that turned into the released binding energy", not as matter that ceased to exist. Run the process backwards — supply that energy — and the mass reappears as you pull the nucleus apart.

Why must I use atomic masses instead of nuclear masses?

Because the mass tables you are given list atomic masses (the neutral atom, including its electrons), and because using them makes the electrons cancel for free. If you represent each proton by a neutral hydrogen atom $M(^1\text{H})$ , then $Z$ hydrogens carry exactly the same $Z$ electrons that the neutral nuclide $^A_Z X$ carries on the other side of the subtraction, and they drop out. The result is numerically the same as if you had used bare nuclear masses, but you never have to subtract electron masses by hand. The only thing this ignores is the few-eV electron binding energies, which are a million times too small to matter against MeV nuclear energies.

Is the binding energy the energy released when the nucleus forms, or the energy needed to break it?

Both — they are the same number viewed from two directions. Assembling free nucleons into a nucleus releases the binding energy; disassembling the nucleus back into free nucleons requires that same energy. It is the depth of the nuclear energy well: the energy you get rolling in equals the energy you must pay climbing out. So if an exam asks "how much energy is needed to separate this nucleus into its constituent nucleons?", the answer is simply its binding energy, with no further calculation. This equality is why $BE$ is defined as a single positive quantity.

Why is the binding energy per nucleon a better stability measure than the total binding energy?

Because total binding energy grows with the number of nucleons almost automatically — a bigger nucleus has more bonds simply because it has more particles, the way a longer chain has more links. Uranium's total binding energy (around $1800\ \text{MeV}$ ) is far larger than helium's ( $28\ \text{MeV}$ ), yet uranium is less stable per nucleon and can fission. Dividing by $A$ gives an intensive quantity — the average glue per nucleon — that genuinely ranks nuclei. It is the $BE/A$ value, peaking near $8.8\ \text{MeV}$ around iron, that tells you which nuclei are most tightly held and which way energy flows in fusion and fission.

Why are nuclear binding energies measured in MeV while chemical energies are in eV?

Because the mass changes are a million times larger. A chemical reaction rearranges the outer electrons and involves energies of a few eV per atom — mass changes far too small to detect. A nuclear reaction rearranges the nucleons themselves, and the mass defects run to thousandths of a u; since $1\ \text{u} = 931.5\ \text{MeV}$ , even a thousandth of a u is nearly an MeV. So nuclear energies naturally come out at the MeV scale, roughly $10^6$ times chemical energies. That factor of a million per particle is exactly why a kilogram of nuclear fuel dwarfs a kilogram of coal, and why the mass defect, invisible in chemistry, is the headline quantity of nuclear physics.

Does a proton actually weigh less when it is inside a nucleus?

No. The individual protons and neutrons keep their normal free-particle masses even while bound. What is lighter is the whole nucleus as a system, because the system as a whole has given up energy in binding together. It is a property of the assembled collection, not of any single nucleon shrinking. This is an important conceptual distinction: the mass defect belongs to the nucleus, and you should never try to assign a "reduced mass" to a nucleon sitting inside it. Pull the nucleus apart and each nucleon walks away with its full free mass — you simply had to supply the binding energy to make that happen.

How can I check my binding-energy answer quickly in an exam?

Two fast sanity checks. First, magnitude: for any nucleus past the lightest few, $BE/A$ should land between about $7$ and $9\ \text{MeV}$ per nucleon — if you get $0.8$ or $80$ , you have slipped a factor of ten. Second, sign: the constituents must weigh more than the nucleus, so $\Delta m$ is positive; a negative mass defect means you subtracted the wrong way. A third habit that prevents most errors: write $Z$ and $N = A - Z$ explicitly before plugging in, and keep four decimals in the masses until the final line. These three checks catch the overwhelming majority of the marks lost on this topic.