Binding Energy and Mass Defect

Take a deuteron — one proton, one neutron, stuck together. Weigh it. Now weigh a free proton and a free neutron, separately, and add the two numbers.

The deuteron is lighter. Not by a little: by $2.2246\ \mathrm{MeV}/c^2$ , about a nine-hundredth of the whole thing, and every mass spectrometer in the world can see it. The parts are heavier than the whole. Put them together and something goes away. What goes away is $2.2246\ \mathrm{MeV}$ of energy, radiated off as a gamma ray when the two particles snap together — and the missing mass is exactly, to the last digit anyone has measured, that energy divided by $c^2$ . Run it backwards and the arithmetic still holds: fire a $2.2246\ \mathrm{MeV}$ gamma at a deuteron and it comes apart. One photon, one deuteron, one proton and one neutron, and a mass ledger that balances. Now: nothing here is a relativistic-mass effect, and it is important to be exact about that. The proton's mass has not changed. The neutron's mass has not changed. Rest mass is an invariant and it does not budge. What changed is the invariant mass of the system, which is a different quantity — and the whole lesson of this article is that a system's invariant mass is not the sum of its parts' masses. Mass, it turns out, does not add. That single fact runs the Sun, which converts $0.7\%$ of the mass of the hydrogen it burns into the sunlight falling on you now, and it is the same fact — six orders of magnitude smaller — that makes a hydrogen atom lighter than a free proton plus a free electron.

Key takeaways

Signature $(+,-,-,-)$ ; $P\cdot P = +m^2c^2$ . Every $m$ here is an invariant rest mass. Nothing in this article gains mass by moving.
The owned result: a system's invariant mass is not the sum of its parts' masses.

\boxed{\;B \;=\; \Big(\sum_i m_i \;-\; M_{\mathrm{system}}\Big)c^2 \;>\; 0 \quad\text{for a bound state}\;}

The mass defect $\Delta m = \sum_i m_i - M$ is a real, weighable deficit. $B$ is the binding energy: the work needed to pull the system apart.

Why it happens: in the system's rest frame $Mc^2 = \sum_i(m_ic^2 + T_i) + U$ . A bound state has $U < 0$ and $|U| > \sum T_i$ , so $M < \sum_i m_i$ . The binding energy is missing from the mass because it is missing from the energy.
⚠ Bookkeeping. With nuclear masses, $B = [Z m_p + N m_n - M_{\mathrm{nuc}}]c^2$ . With atomic masses (what tables give you), $B = [Z m_{\mathrm{H}} + N m_n - M_{\mathrm{atom}}]c^2$ — $m_{\mathrm{H}}$ , the ¹H atom, not $m_p$ . Mixing them costs you $Z\,m_ec^2$ : for the deuteron, $1.714$ instead of $2.225\ \mathrm{MeV}$ — a 23 % error.
The curve. $B/A$ rises steeply, peaks around $A \approx 56\text{–}62$ , and falls slowly. Light nuclei fuse and heavy nuclei fission, and both release energy, because both move toward the peak.
⚠ The peak is ⁶²Ni, not ⁵⁶Fe. $B/A(^{62}\mathrm{Ni}) = 8.7945\ \mathrm{MeV}$ beats $B/A(^{56}\mathrm{Fe}) = 8.7903$ . But ⁵⁶Fe has the lowest mass per nucleon ( $930.412$ vs $930.418\ \mathrm{MeV}/c^2$ ). Both are true, and they differ because protons are lighter than neutrons and ⁵⁶Fe has the larger proton fraction.
Real numbers. Deuteron $2.2246$ MeV · ⁴He $28.30$ MeV ( $7.074$ /nucleon) · the Sun's pp chain $26.73$ MeV per ⁴He, $0.712\%$ of the rest energy consumed.
The same physics, $10^6$ times smaller: a hydrogen atom is lighter than a free proton plus a free electron by $13.6\ \mathrm{eV}/c^2$ — a fractional defect of $1.4\times10^{-8}$ . Chemistry has a mass defect too. It is just too small to weigh.
Key idea: binding energy is not a bookkeeping abstraction. It is missing mass, it is weighable, and it is the only sense in which "mass converts to energy" is a true sentence.

The core insight

The temptation is to say "some of the mass turned into energy." Resist it, because it hides the actual mechanism. Nothing turned into anything. A system's invariant mass is defined by $M^2c^4 = \big(\sum_i E_i\big)^2 - \big|\sum_i \vec p_i\big|^2c^2$ — it is built from the total energy of the constituents, not from their masses. In the system's rest frame that reduces to $Mc^2 = \sum_i E_i$ : the system weighs whatever its total energy is, divided by $c^2$ . So when two nucleons fall into a bound state and radiate a gamma ray away, the system's total energy drops by exactly the gamma's energy — and therefore so does its mass. The mass did not become the gamma ray. The mass is the energy, all along, and some of the energy left.

Where this lives today

Every nuclear technology is an argument about where you sit on the $B/A$ curve. A reactor takes ²³⁵U, at $B/A = 7.591\ \mathrm{MeV}$ , and splits it into fragments nearer the peak; the ledger returns roughly $200\ \mathrm{MeV}$ per fission, which is a hundred million times what burning a carbon atom yields. A star does the opposite and climbs the curve from the left, and when it reaches the summit it stops — a star cannot fuse its way past iron, because there is no energy left in it to release. That is not an engineering limit; it is the shape of this curve, and it is why the iron peak is the graveyard of stellar nucleosynthesis and why everything heavier had to be made by neutron capture in more violent places. The mass defect is also, quietly, the most stringently tested prediction in relativity. It is checked every time a mass spectrometer is calibrated, and it has been checked deliberately to four parts in $10^7$ . And the reason it can be tested so well is precisely the thing this article is about: the defect is a difference between two directly weighable masses, and there is nowhere for a discrepancy to hide.

The physical picture

Two nucleons, far apart and at rest. The system's energy is $m_pc^2 + m_nc^2$ and nothing else, so its mass is $m_p + m_n$ . Now let them fall together. The strong force does work on them; they speed up; they radiate. When the dust settles you have a deuteron, and a $2.2246\ \mathrm{MeV}$ gamma ray sailing away. Look at the ledger. Energy is conserved, so

m_pc^2 + m_nc^2 = M_dc^2 + E_\gamma.

The photon took $E_\gamma$ away with it. Whatever is left must be the deuteron's rest energy, and it is less than what we started with, by exactly $E_\gamma$ . Divide by $c^2$ :

M_d = m_p + m_n - \frac{E_\gamma}{c^2}.

The deuteron is lighter than its parts by the energy that escaped. That is the whole phenomenon, and it is not a metaphor: put the deuteron on a mass spectrometer and it comes up light. Now, what is it that got lighter? Not the proton — a proton's rest mass is an invariant number, the same in a deuteron as in a beam line. Not the neutron. What got lighter is the two-body system, whose invariant mass was never the sum of its constituents' masses in the first place. It was $\big(\sum E_i\big)/c^2$ in the rest frame, and the binding took a bite out of $\sum E_i$ . This is the one place in relativity where it is legitimate to talk about mass changing — and even here, the thing that changes is a system's invariant mass, not any particle's rest mass. There is no relativistic mass anywhere in this story. (Relativistic Force owns the argument that kills that idea for good.) The last thing to notice is the scale. The deuteron's defect is a part in 900. That is enormous — mass spectrometers were measuring it in the 1920s. A hydrogen atom has a defect too, from binding its electron: $13.6\ \mathrm{eV}$ , which is a part in $10^8$ — real, and until very recently unweighable. The physics is identical. Only the coupling strength differs, and that single ratio is why nuclear reactors are small and coal power stations are not.

The core equations

The definition, and the object the whole article is about:

\boxed{\;\Delta m \;\equiv\; \sum_i m_i \;-\; M_{\mathrm{system}}, \qquad B \;\equiv\; \Delta m\,c^2\;} \qquad (1)

where: $m_i$ are the invariant rest masses of the free constituents and $M_{\mathrm{system}}$ is the invariant mass of the bound system. Physical reading: $\Delta m$ is the mass defect and $B$ is the binding energy — the work you must do to disassemble the system into free, stationary parts. $B > 0$ is what "bound" means. Note carefully what is not being claimed: no individual $m_i$ has changed. Invariant Mass owns the concept of $M_{\mathrm{system}}$ ; this article shows it is not additive. Why the system's mass is what it is, from the definition of invariant mass evaluated in the rest frame:

\boxed{\;M_{\mathrm{system}}\,c^2 \;=\; \sum_i E_i \;=\; \sum_i \big(m_ic^2 + T_i\big) \;+\; U\;} \qquad (2)

where: $T_i$ are the constituents' kinetic energies in the system's rest frame and $U$ is the interaction (potential) energy. Physical reading: a system weighs its total energy. For a bound state $U$ is negative and beats the kinetic term, so $M < \sum_i m_i$ , so $B > 0$ . This is the entire mechanism, and it is a statement of special relativity, not of nuclear physics — it would be equally true of two masses on a spring, if springs bound things at MeV strengths. The nuclear bookkeeping — and you must say which masses you are using:

B_{\mathrm{nuc}} = \big[Z m_p + N m_n - M_{\mathrm{nuc}}\big]c^2 \qquad\text{but}\qquad \boxed{\;B = \big[Z m_{\mathrm{H}} + N m_n - M_{\mathrm{atom}}\big]c^2\;} \qquad (3)

where: $m_{\mathrm{H}}$ is the mass of the neutral ¹H atom (proton plus electron), and $M_{\mathrm{atom}}$ is the neutral atomic mass. Physical reading: mass tables give you ATOMIC masses, which include $Z$ electrons. If you use them you must put the electrons back on the other side too — and $Z\,m_{\mathrm{H}}$ does exactly that, because the $Z$ electron masses cancel between the two sides. Use $Z m_p$ with an atomic $M$ and you are wrong by $Z\,m_ec^2 = Z \times 0.511\ \mathrm{MeV}$ — a 23 % error for the deuteron. (The electrons' own binding energy, a few eV to keV, is the only thing this misses, and it is negligible at nuclear precision.) The figure of merit, and the shape that runs the universe:

\frac{B}{A} \qquad\text{peaks at } A \approx 56\text{–}62 \text{ with } \frac{B}{A} \approx 8.79\ \mathrm{MeV/nucleon} \qquad (4)

where: $A = Z + N$ is the nucleon number. Physical reading: the curve rises fast, peaks, and falls slowly. Everything to the left of the peak gains by fusing; everything to the right gains by fissioning; and both are "downhill" toward $A \approx 56\text{–}62$ . The peak is why the periodic table stops where it does, why stars stop fusing at iron, and why there is a nuclear industry at both ends of it.

Deriving the mass defect

Derivation AWhy a bound system weighs less than its parts, and why that is relativity and not chemistry

The claim is that $M_{\mathrm{system}} \neq \sum_i m_i$ , and that the shortfall is exactly the binding energy over $c^2$ . It sounds like it ought to require a model of the force. It does not: it follows from the definition of a system's invariant mass and nothing else, which is why it is true of nuclei, atoms, molecules and planets alike — and why only the size of the effect is nuclear physics.

Step 1. Write down what a system's invariant mass is. For any collection of particles, the total four-momentum is $P^\mu_{\rm tot} = \sum_i P^\mu_i$ — four-vectors add — and the system's invariant mass is defined by its norm:

M^2c^4 \;=\; \Big(\sum_i E_i\Big)^{\!2} - \Big|\sum_i \vec p_i\Big|^2c^2

Why: this is not an extra postulate. It is the same definition Invariant Mass and Centre-of-Momentum Frame already use — and note that it is built from the constituents' energies, not from their masses. That is the seed of everything below.

Step 2. Go to the system's rest frame, which is the only frame in which "how much does it weigh" is a clean question. There $\sum_i \vec p_i = \vec 0$ by definition, so the momentum term dies:

\boxed{\;M c^2 = \sum_i E_i\;}

Why: a system weighs its total energy, divided by $c^2$ . Every kind of energy inside it counts — kinetic energy of the parts, and the energy stored in the forces between them. A hot brick is heavier than a cold one. This is the single most important sentence in the article.

Step 3. Decompose the total energy. Each constituent carries rest energy plus kinetic energy, and the interaction contributes $U$ :

Mc^2 = \sum_i \big(m_ic^2 + T_i\big) + U = \sum_i m_ic^2 \;+\; \underbrace{\Big(\sum_i T_i + U\Big)}_{\text{everything that is not rest mass}}

Why: we have simply sorted the total energy into "the parts' rest energies" and "the rest of it". No physics has been used yet — only bookkeeping.

Step 4. Now impose boundedness. A bound system is one you must do work on to take apart, which means its total energy is below that of the same parts free and at rest. Free and at rest, the parts have $\sum_i m_ic^2$ and nothing else. So

\sum_i T_i + U \;<\; 0 \qquad\Longleftrightarrow\qquad U < 0 \ \text{and}\ |U| > \sum_i T_i

Why: this is the definition of "bound", cashed as an inequality. The potential energy must be negative and must more than pay for the kinetic energy the parts acquire as they fall in. (For the deuteron the virial-type balance is not delicate — the strong force wins by a mile.)

Step 5. Read off the result:

Mc^2 = \sum_i m_ic^2 - \underbrace{\Big|\sum_i T_i + U\Big|}_{\equiv\; B} \qquad\Longrightarrow\qquad \boxed{\;B = \Big(\sum_i m_i - M\Big)c^2 = \Delta m\,c^2\;}

Why: the binding energy is missing from the mass because it is missing from the energy, and the mass IS the energy. Nothing was converted. The system simply has less energy than its parts did, and therefore weighs less, because weighing is measuring energy.

Step 6. Say precisely what has not happened, because this is where the subject is most often taught wrongly. The proton's rest mass is unchanged. The neutron's rest mass is unchanged. They are Lorentz invariants and they are the same number in a deuteron as in a storage ring. What changed is $M_{\mathrm{system}}$ — a property of the composite, which was never $\sum_i m_i$ to begin with. Why: there is no relativistic mass in this article, and this step is why. The one thing that "changes mass" here is a system's invariant mass, and the reason it can is precisely that invariant mass is not additive. Saying "the total rest mass decreased" is wrong; saying "the system's invariant mass decreased" is right. *Verified in verify_aq-sr-binding-energy-mass-defect.py: $Mc^2 = \sum_i E_i$ is derived symbolically from the norm of $\sum_i P^\mu_i$ in the rest frame; the two-body case is checked against explicit four-momenta; and $B = (\sum m_i - M)c^2$ is confirmed to be exactly the energy radiated, using the deuteron's measured masses.*

Go deeper · Derivation B · 2 minThe bookkeeping that eats 23 percent of the deuteron: nuclear masses versus atomic masses

Derivation BThe bookkeeping that eats 23 percent of the deuteron: nuclear masses versus atomic masses

Equation (3) has two forms and they are not interchangeable. Every mass table you will ever open lists atomic masses — the neutral atom, electrons included — and the formula most textbooks print first is the nuclear one. Combine them and you make an error of $Z \times 0.511\ \mathrm{MeV}$ , which is invisible for uranium and catastrophic for the deuteron. We will do it right, then do it wrong, and measure the damage.

Step 1. The honest, nuclear statement. Strip the electrons off. The nucleus is $Z$ protons and $N$ neutrons, and its binding energy is

B = \big[Z m_p + N m_n - M_{\mathrm{nuc}}\big]c^2

Why: this is equation (1) applied to the nucleus, with nothing hidden. It is correct — and it is useless, because nobody tabulates $M_{\mathrm{nuc}}$ .

Step 2. Convert to what the tables actually give. A neutral atom is a nucleus plus $Z$ electrons, so (neglecting the electrons' own binding, which is eV-scale against MeV)

M_{\mathrm{atom}} \simeq M_{\mathrm{nuc}} + Z m_e \qquad\Longleftrightarrow\qquad M_{\mathrm{nuc}} \simeq M_{\mathrm{atom}} - Z m_e

Why: this is the substitution that has to be made, and the whole trap is in doing it on one side only.

Step 3. Substitute, and watch the electrons cancel:

B = \big[Z m_p + N m_n - M_{\mathrm{atom}} + Z m_e\big]c^2 = \big[Z\underbrace{(m_p + m_e)}_{\;\simeq\; m_{\mathrm{H}}} + N m_n - M_{\mathrm{atom}}\big]c^2 = \boxed{\big[Z m_{\mathrm{H}} + N m_n - M_{\mathrm{atom}}\big]c^2}

Why: $Z m_{\mathrm{H}}$ is not a typo and not an approximation — it is the electrons being put back. The $Z$ electrons in $M_{\mathrm{atom}}$ are cancelled by the $Z$ electrons hidden inside $Z$ hydrogen atoms. The formula is exact up to the electrons' binding energy, which is a part in $10^6$ of $B$ .

Step 4. Do the deuteron correctly. $Z = N = 1$ ; $m_{\mathrm{H}} = 1.00782503\ \mathrm{u}$ , $m_n = 1.00866492\ \mathrm{u}$ , $M_{\mathrm{atom}}(^2\mathrm{H}) = 2.01410178\ \mathrm{u}$ , and $1\ \mathrm{u} = 931.494\ \mathrm{MeV}/c^2$ :

B = \big[1.00782503 + 1.00866492 - 2.01410178\big] \times 931.494\ \mathrm{MeV} = 0.00238817 \times 931.494 = \mathbf{2.2246\ MeV}

Why: and that is the measured value, to five figures. The mass defect is $0.00238817\ \mathrm{u}$ — about one part in 843 of the deuteron.

Step 5. Now do it wrong, the way it is done wrong. Use $Z m_p$ ( $m_p = 1.00727647\ \mathrm{u}$ ) with the atomic $M$ :

B_{\mathrm{wrong}} = \big[1.00727647 + 1.00866492 - 2.01410178\big] \times 931.494 = \mathbf{1.7136\ MeV}

Why: wrong by $0.5110\ \mathrm{MeV}$ — which is exactly $m_ec^2$ , exactly $Z$ times. The error is not random; it is one missing electron mass per proton, and you can read it straight off. As a fraction of the true answer that is 23 %, which no amount of experimental care will rescue.

Step 6. Note who this trap spares and who it kills. The absolute error is $Z \times 0.511\ \mathrm{MeV}$ , but $B$ grows roughly as $8.5A$ , so the fractional error is about $0.06\,Z/A \approx 0.03$ for stable nuclei — a few percent — but for the lightest nuclei, where $B$ is small and $Z/A$ is a half, it is enormous. Why: the deuteron is the worst case and it is also the case everyone works first. If your deuteron comes out at $1.71\ \mathrm{MeV}$ , you have not made an arithmetic slip; you have made this one. *Verified in verify_aq-sr-binding-energy-mass-defect.py: both forms of equation (3) are evaluated from AME2020 atomic masses; they agree to better than $10^{-4}$ MeV for the deuteron; the wrong form is shown to differ by exactly $Z m_ec^2 = 0.5110$ MeV; and 2.2246, 1.7136 and 23 % are asserted against the literals printed above.*

Figure

THE CURVE THAT RUNS THE UNIVERSE — binding energy per nucleon rises steeply from hydrogen, PEAKS near A = 56 to 62, and falls slowly to uranium. Everything left of the peak releases energy by FUSING; everything right of it releases energy by FISSIONING; both run downhill toward the same summit. The two summit nuclei are marked and they are NOT the same nucleus: nickel-62 has the highest binding energy per nucleon (8.7945 MeV), while iron-56 has the lowest mass per nucleon (930.412 MeV/c², against 930.418 for Ni-62). Both statements are true. They differ because protons are lighter than neutrons and iron-56 has the larger proton fraction — 46.43 percent against 45.16 percent

Limiting cases

Case	Result	Meaning
free, non-interacting parts	$U = 0$ , $M = \sum_i m_i$	the only case where mass is additive
bound state	$U < 0$ , $M < \sum_i m_i$ , $B > 0$	the mass defect — this article
unbound / repulsive	$B < 0$	you must supply mass to hold it together; it flies apart
deuteron, $A = 2$	$B = 2.2246$ MeV, $B/A = 1.11$	barely bound — a part in 843
⁴He, $A = 4$	$B = 28.30$ MeV, $B/A = 7.07$	doubly magic; a huge jump from ²H
the peak, $A \approx 56$ – $62$	$B/A \approx 8.79$ MeV	⁶²Ni highest $B/A$ ; ⁵⁶Fe lowest mass/nucleon
²³⁵U, $A = 235$	$B/A = 7.591$ MeV	past the peak — fission is downhill
the H atom	$B = 13.6$ eV, $\Delta m/m = 1.4\times10^{-8}$	the same physics, $10^6$ times weaker
a chemical bond	$B \sim$ eV, $\Delta m/m \sim 10^{-10}$	real, and far below any scale you can weigh

Common trap⁶²Ni or ⁵⁶Fe? Both — and the textbooks conflate two different questions

"Iron is the most tightly bound nucleus" is one of the most-repeated statements in physics, and it is answering a question nobody asked. There are two questions here and they have two different answers:

Which nucleus has the highest binding energy per nucleon? ⁶²Ni, at $B/A = 8.7945\ \mathrm{MeV}$ . ⁵⁶Fe is second, at $8.7903\ \mathrm{MeV}$ .
Which nucleus has the lowest mass per nucleon? ⁵⁶Fe, at $930.412\ \mathrm{MeV}/c^2$ . ⁶²Ni is second, at $930.418\ \mathrm{MeV}/c^2$ .

Both statements are true, and they are not in conflict, because $B/A$ and $M/A$ are not the same quantity — they differ by the average mass of the nucleons you started with, and that average depends on the proton-to-neutron ratio. Here is the resolution, and it is exact. A proton is lighter than a neutron ( $938.272$ against $939.565\ \mathrm{MeV}/c^2$ ). ⁵⁶Fe is $26/56 = 46.43\%$ protons; ⁶²Ni is only $28/62 = 45.16\%$ . So ⁵⁶Fe is assembled from a cheaper mix of ingredients. ⁶²Ni binds its nucleons more tightly — it wins on $B/A$ — but ⁵⁶Fe starts from lighter raw material, and that is enough to win on $M/A$ .

Which one matters depends on what you are asking. If you want to know which nucleus a nucleon-conserving reaction runs downhill toward, $B/A$ (⁶²Ni) is your figure. If you want to know the endpoint of stellar burning — where the mass per nucleon is minimised and no rearrangement can release more energy — it is ⁵⁶Fe. (In real stars the answer is ⁵⁶Ni, which then beta-decays to ⁵⁶Fe, for reasons of reaction rates rather than of energetics.) The one thing you may not do is say "⁵⁶Fe has the highest binding energy per nucleon." It does not.

Worked examples

1 · The Sun's fuel bill

The pp chain nets $4\,^1\mathrm{H} \to {}^4\mathrm{He} + 2e^+ + 2\nu$ . How much energy per helium nucleus, and what fraction of the mass consumed is that?

Step 1 — identify. Use atomic masses throughout and the positron annihilation is included automatically, because the four ¹H atoms bring four electrons and the ⁴He atom only needs two — the surplus two annihilate with the two positrons.

Step 2 — apply. $Q = \big[4 M(^1\mathrm{H}) - M(^4\mathrm{He})\big]c^2$ with $M(^1\mathrm{H}) = 1.00782503\ \mathrm{u}$ and $M(^4\mathrm{He}) = 4.00260325\ \mathrm{u}$ :

Q = \big[4(1.00782503) - 4.00260325\big] \times 931.494 = 0.02869688 \times 931.494 = \mathbf{26.73\ MeV}

Step 3 — the fraction. The rest energy of the four hydrogen atoms is $4 \times 1.00782503 \times 931.494 = 3755.1\ \mathrm{MeV}$ , so

\frac{Q}{4M(^1\mathrm{H})c^2} = \frac{26.73}{3755.1} = \mathbf{0.712\%}

Step 4 — the boundary number. Seven-tenths of one percent. That is the entire efficiency of the Sun, and it is why the Sun lasts $10^{10}$ years rather than $10^{7}$ . It is also, in the other direction, staggering: chemical burning liberates about $10^{-10}$ of a fuel's rest energy, so hydrogen fusion is some $10^{7}$ times more energetic per kilogram than anything you can do with a flame. Of the $26.73\ \mathrm{MeV}$ , roughly $0.6\ \mathrm{MeV}$ escapes as neutrinos and never warms anything; the rest becomes sunlight.

2 · Weighing the hydrogen atom

A hydrogen atom is a bound proton and electron with $B = 13.6\ \mathrm{eV}$ . By what fraction is it lighter than its parts, and could you weigh it?

Step 1 — apply equation (1). $\Delta m = B/c^2 = 13.6\ \mathrm{eV}/c^2$ . The atom's rest energy is $M_{\mathrm{H}}c^2 = 1.00782503 \times 931.494\ \mathrm{MeV} = 938.78\ \mathrm{MeV}$ .

Step 2 — the fraction.

\frac{\Delta m}{M} = \frac{13.6\ \mathrm{eV}}{938.78\times10^6\ \mathrm{eV}} = \mathbf{1.4\times10^{-8}}

Step 3 — the boundary number, and the point. Compare the deuteron: a defect of one part in 843. The hydrogen atom: one part in $7\times10^{7}$ . The physics is word-for-word identical — a bound system weighs less than its parts by $B/c^2$ — and the only difference is that the electromagnetic binding of an electron is a hundred-thousand times feebler, per particle, than the strong binding of a nucleon. That single ratio is the whole reason a kilogram of uranium can level a city and a kilogram of TNT cannot. Chemistry has a mass defect. It is just too small to weigh.

Figure

MASS IS NOT ADDITIVE — the deuteron on a balance. On the left, a free proton and a free neutron, weighed separately and added: 2.01594 u. On the right, the deuteron they make: 2.01355 u (nuclear mass). The whole is LIGHTER than the sum of its parts, by 0.00239 u — one part in 843 — and that missing mass left as a 2.2246 MeV gamma ray when they bound. Nothing was converted: the system simply has less ENERGY than its parts did, and a system weighs its total energy divided by c squared. Note what did NOT change: the proton's rest mass and the neutron's rest mass are invariants and are identical on both sides

Physics of the portrait

The name is older than the explanation. Aston could measure the deficit a decade before anyone could say what held the nucleus together.

Origins · 2 minMass defect and how this was discovered

Mass defect

1919–1932 · Aston weighs it · Einstein explains it · Cockcroft and Walton cash it

In 1919 Francis Aston built the mass spectrograph at the Cavendish and began weighing atoms against one another to a precision no chemist had approached. He found what he called the whole number rule: isotopic masses are very nearly integers. Nearly. The deviations were small, systematic, and — crucially — not noise. Aston plotted them as the packing fraction, a curve of the fractional departure from integrality against mass number, and the curve had a minimum in the neighbourhood of iron. He had, without any theory of nuclear force, drawn the binding-energy curve. He took the 1922 Nobel Prize in Chemistry for the isotopes and the whole number rule. What the packing fraction meant came from Einstein's 1905 result that a body radiating energy $E$ loses mass $E/c^2$ : the missing mass was binding energy, and the curve's minimum was the place where nuclei are most tightly held. The first direct cash-out came on 14 April 1932, when John Cockcroft and Ernest Walton at the Cavendish accelerated protons into a lithium target and split $^7\mathrm{Li} + p \to 2\,{}^4\mathrm{He}$ , measuring alpha particles whose kinetic energies matched the mass difference — the first artificial nuclear disintegration, and the first quantitative test of $E = mc^2$ in a nuclear reaction. They shared the 1951 Nobel Prize. The measurement has since been pushed to a part in $4\times10^{-7}$ , by comparing gamma-ray wavelengths against atomic mass differences in silicon and sulphur (Rainville et al., Nature 438, 1096, 2005). The mass defect is the most precisely tested consequence of relativity that exists, and Aston was weighing it before anyone knew what it was.

History

1905

— Einstein's September paper argues that a body emitting energy E loses mass E/c squared, making mass a form of energy account-keeping.

1919

— Aston's mass spectrograph reveals the whole number rule; isotopic masses are nearly, but not exactly, integers.

1920

— Aston introduces the packing fraction, plotting the deviations against mass number and unknowingly drawing the binding-energy curve.

1922

— Aston receives the Nobel Prize in Chemistry for the isotopes and the whole number rule.

1932

— On 14 April, Cockcroft and Walton split lithium with accelerated protons and confirm the mass-energy balance in a nuclear reaction; they share the 1951 Nobel Prize.

1934

— Chadwick and Goldhaber photodisintegrate the deuteron with a gamma ray, running the mass defect backwards and measuring the neutron's mass from it.

1939

— Meitner and Frisch interpret nuclear fission and estimate its energy release directly from the mass defect.

2005

— Rainville and colleagues test E equal to mc squared to four parts in ten million using nuclear binding energies of silicon and sulphur.

Going further

Texts. Krane, Introductory Nuclear Physics, ch. 3 · Rindler, Introduction to Special Relativity, 2nd ed., §29 · Griffiths, Introduction to Elementary Particles, 2nd ed., §3.3 · Wong, Introductory Nuclear Physics, ch. 4 · S. Rainville et al., "A direct test of $E = mc^2$ ", Nature 438, 1096 (2005) · AME2020 atomic mass evaluation.

Provenance. Signature $(+,-,-,-)$ ; $P\cdot P = +m^2c^2$ ; every mass is an invariant rest mass and nothing gains mass by moving. The AllQFT base was checked: its core equations are correct, but its statement $B = [Z m_p + N m_n - M]c^2$ is right only for a nuclear $M$ , and is silent on which convention it uses — the atomic-mass form with $Z m_{\mathrm H}$ is derived here and the $Z m_ec^2$ error is quantified. verify_aq-sr-binding-energy-mass-defect.py re-derives every number from AME2020 atomic masses and asserts each against the literal printed above: the deuteron ( $2.2246$ MeV; the wrong form's $1.7136$ ; the $0.5110$ MeV gap; $23\%$ ; one part in 843), ⁴He ( $28.30$ , $7.074$ /nucleon), ⁵⁶Fe ( $8.7903$ MeV/nucleon, $930.412$ MeV/ $c^2$ ), ⁶²Ni ( $8.7945$ , $930.418$ ), the proton fractions ( $46.43\%$ , $45.16\%$ ), ²³⁵U ( $7.591$ ), the pp chain ( $26.73$ MeV, $0.712\%$ , $3755.1$ MeV), and the H atom ( $1.4\times10^{-8}$ ).

Scope. This article establishes that a bound system weighs less, and by how much, from relativity alone. It says nothing about why the nuclear force binds as it does — the shape of the $B/A$ curve (the volume, surface, Coulomb, asymmetry and pairing terms of the semi-empirical mass formula) is nuclear physics, not relativity, and is owned elsewhere. Reaction rates, Coulomb barriers and stellar burning networks are likewise out of scope.

Builds on. Mass-Energy Equivalence · Invariant Mass · Relativistic Four-Momentum · Relativistic Energy · Centre-of-Momentum Frame. Leads to. Energy-Momentum Relation · Relativistic Kinetic Energy · Massless Particles.

Is the mass really gone, or is it "converted into energy"?

"Converted" is the word that causes all the trouble, because it suggests two substances and a transaction. There is only one thing here. A system's invariant mass is defined as its total energy in its rest frame, divided by $c^2$ — so mass is energy, in the same way that a distance in metres is a distance in feet. When the deuteron forms and radiates a $2.2246\ \mathrm{MeV}$ gamma, the system's energy drops by $2.2246\ \mathrm{MeV}$ and therefore its mass drops by $2.2246\ \mathrm{MeV}/c^2$ . Nothing was transmuted; the books simply have less in them. And the energy is not "gone" either — it is in the gamma ray, and if you catch the gamma ray in a wall, that wall gets very slightly heavier.

This sounds like relativistic mass. Isn't the mass changing with the interaction?

No, and the distinction is worth being pedantic about. Relativistic mass is the discredited idea that a single particle's mass grows with its speed; it is wrong, and Relativistic Force owns the argument that kills it (force and acceleration are not parallel, so no scalar "mass" can exist). Here, no particle's mass changes at all — the proton in a deuteron has exactly the rest mass it has in a beam pipe. What changes is the invariant mass of the composite system, and it changes because a composite's invariant mass was never the sum of its parts' masses. The two ideas point in opposite directions: relativistic mass wrongly makes a particle's mass frame-dependent; the mass defect correctly makes a system's mass non-additive, in every frame equally.

Why must I use m_H rather than m_p when the tables give atomic masses?

Because the tabulated $M_{\mathrm{atom}}$ has $Z$ electrons buried in it, and the formula has to have $Z$ electrons on the other side to cancel them. Writing $Z m_{\mathrm{H}} = Z(m_p + m_e)$ supplies exactly those $Z$ electrons. Use $Z m_p$ instead and you have subtracted $Z$ electrons that you never added, so your answer is short by $Z\,m_ec^2 = Z \times 0.511\ \mathrm{MeV}$ . For the deuteron ( $Z = 1$ ) that turns $2.2246$ into $1.7136$ — a 23 % error, and the classic first-attempt result. The only thing the $m_{\mathrm{H}}$ form neglects is the electrons' own binding energy (13.6 eV for hydrogen, keV for heavy atoms), which is utterly negligible against MeV-scale nuclear binding.

So which is "the most stable nucleus" — iron-56 or nickel-62?

It depends what you mean, and that is the honest answer rather than a dodge. ⁶²Ni has the highest binding energy per nucleon ( $8.7945$ against ⁵⁶Fe's $8.7903\ \mathrm{MeV}$ ) — it holds each of its nucleons most tightly. ⁵⁶Fe has the lowest mass per nucleon ( $930.412$ against ⁶²Ni's $930.418\ \mathrm{MeV}/c^2$ ) — a kilogram of it contains the least energy. They differ because a proton is lighter than a neutron, and ⁵⁶Fe is $46.43\%$ protons against ⁶²Ni's $45.16\%$ : iron starts from cheaper ingredients, nickel binds them better, and the two effects come out on opposite sides. If someone tells you "⁵⁶Fe has the highest binding energy per nucleon", they have merged the two facts into a third one that is false.

If a star cannot fuse past iron, where did gold and uranium come from?

Not from fusion, which is exactly the point of the curve. Past the iron peak, fusing two nuclei costs energy rather than releasing it, so no star can power itself that way — a star that tries to fuse iron cools, and that is precisely what triggers a core-collapse supernova. The heavy elements are built instead by neutron capture, which does not have to climb the $B/A$ curve because neutrons are electrically neutral and can simply be absorbed: slowly (the s-process, in the shells of dying stars) or explosively (the r-process, in supernovae and neutron-star mergers). The gold in a wedding ring was assembled by an entirely different mechanism from the helium in the Sun, and the reason is the shape of this one curve.

Does a hot object really weigh more than a cold one?

Yes — necessarily, and by exactly $\Delta E/c^2$ . It follows immediately from Step 2 of the derivation: a system's rest mass is its total rest-frame energy over $c^2$ , and thermal energy is part of that total. Heat a $1\ \mathrm{kg}$ block of iron by $100\ \mathrm{K}$ and you add about $45\ \mathrm{kJ}$ , so it gains $45{,}000/c^2 \approx 5\times10^{-13}\ \mathrm{kg}$ — half a picogram, far below any balance ever built, but not zero and not a matter of interpretation. A compressed spring is heavier than a relaxed one. A charged battery is heavier than a flat one. Every form of internal energy weighs, and the mass defect is just the case where the energy involved is big enough to see.