
Bounded modern-physics survey · Interactive lesson
Mass defect and binding energy per nucleon
This is where E = mc² stops being a slogan and becomes a ledger. Turn atomic-mass tables into binding energies, choose masses so the electrons cancel instead of hoping they do, and read the binding-energy-per-nucleon curve for the one thing it reliably gives: which way a rearrangement of nucleons runs downhill.
UPIII-11.02 · Lesson 2 of 9 in this unit. Position is not a completion score.Bounded modern-physics survey. This modern-physics material is an introductory survey, not a complete course in relativity, quantum, atomic, nuclear or particle physics. Your university determines the required depth and assessment.
Build the model
Bound nucleons weigh less than free ones; the B/A curve says which way energy flows.
The key idea
The binding energy B of a nucleus is the energy needed to separate it into free protons and neutrons at rest; it equals the mass deficit Δm, the free nucleons' mass minus the nucleus's, times c², and B/A is its average over the A nucleons.
Example. For ⁴He, two hydrogen atoms and two neutrons total 4.032980 u, while the helium atom is 4.002603 u. The deficit of 0.030377 u times 931.494 MeV u⁻¹ gives B = 28.30 MeV, or B/A = 7.07 MeV per nucleon.
The complete physical model and assumptions
A nucleus is lighter than the free protons and neutrons it is made of, by close to 1% of its mass. Nothing has gone missing. Taking the nucleus apart into nucleons at rest costs an energy B, energy conservation puts that energy into the rest energy of the separated parts, and so the deficit is Δm = B/c². Mass tables list neutral atoms, measured to better than a part in 10⁸, so the deficit is computed with hydrogen atoms standing in for protons, B = [Z M(¹H) + N mₙ − M(ᴬX)]c², and the Z electrons cancel to within their atomic binding, a few parts in 10⁴ of B at worst. Divided by A, the result is a measured curve: B/A climbs steeply through the light nuclei, flattens into a broad maximum of 8.8 MeV at A ≈ 56 to 62, and sinks slowly to 7.6 MeV at uranium. Whenever nucleons are regrouped with Z and N each conserved, every constituent mass cancels and the energy released is the rise in total binding, so fusing light nuclei and splitting heavy ones both run toward the peak. That is where the model stops. The curve is data, read here and not derived. It tells you whether energy is available, not whether a reaction happens or how fast, and its smooth trend hides the shell and pairing effects that lift ⁴He, at 7.07 MeV per nucleon, far above ⁶Li at 5.33 MeV.
1The mass unit is anchored to a bound nucleus
Nuclear masses are measured, not calculated: Penning traps compare the cyclotron frequencies of ions, and the best reach a part in 10¹⁰. The tables list neutral atoms, electrons included, on a scale fixed by a bound system: 1 u is one-twelfth of a neutral ¹²C atom, 1.660539 × 10⁻²⁷ kg. Its rest energy follows from the constants: 1.660539 × 10⁻²⁷ kg × (2.997925 × 10⁸ m s⁻¹)² = 1.492418 × 10⁻¹⁰ J, and dividing by 1.602177 × 10⁻¹³ J per MeV gives 931.494 MeV. Because the anchor is bound, free constituents weigh more than 1 u: M(¹H) = 1.007825 u and mₙ = 1.008665 u. That excess is carbon's binding in disguise. Six hydrogen atoms and six neutrons average 1.008245 u per nucleon, against exactly 1 u inside the carbon atom, and 0.008245 u × 931.494 MeV u⁻¹ = 7.680 MeV is B/A for ¹²C. The deficits are near 1% of the mass, so carry masses to six decimal places: 10⁻⁶ u is 0.93 keV.
2A bound system weighs less than its parts
To pull a nucleus apart into free nucleons at rest you must supply an energy B. Energy is conserved and rest energy is energy, so the separated nucleons have a total rest energy larger than the nucleus's by exactly B: Σmᵢc² = Mc² + B. The rest energy of a composite counts the kinetic and potential energy of its parts, and for a bound state that sum is negative. The deuteron lets you test this without a mass table. Its deficit, 1.00782503 + 1.00866492 − 2.01410178 = 0.00238817 u, gives B = 2.2246 MeV. A slow neutron captured by a proton emits a 2.2233 MeV photon, short of B by the 1.3 keV recoil of the deuteron, and no photon below 2.2259 MeV can break a deuteron apart. Scale supplies the correspondence check. Nuclear binding is 1.2 × 10⁻³ of the deuteron's rest energy; the 4.48 eV bond of H₂ is 2.4 × 10⁻⁹ of the molecule's, below what any chemical balance can resolve. Conservation of mass in chemistry is the B/Mc² → 0 limit of the same rule.
3Choose masses so the electrons cancel
A neutral atom's rest energy is M(ᴬX)c² = mnuc c² + Z mₑ c² − Bₑₗ, where mnuc is the bare nucleus and Bₑₗ the total binding energy of the atom's electrons. Write B with bare proton masses, Z mₚ + N mₙ − M(ᴬX), and the atom's Z electrons are left uncancelled: B comes out low by Z mₑ c² = Z × 0.511 MeV, which is 13.3 MeV for ⁵⁶Fe, 2.7% of its 492.3 MeV. Use Z hydrogen atoms instead and each brings its own electron, so Z M(¹H) + N mₙ − M(ᴬX) counts nuclear binding alone. What survives is the mismatch in atomic binding: the formula overstates B by Bₑₗ − Z × 13.6 eV, about 34 keV for iron and 0.76 MeV for uranium, at most a few parts in 10⁴. That is invisible at the precision of this unit, and precision mass work corrects for it. Converting to nuclear masses first, mnuc ≈ M − Z mₑ, and then using mₚ works equally well. Mixing the two, stripping the electrons from atom X while keeping the hydrogen atoms, overstates B by Z × 0.511 MeV.
4The measured curve, and what its shape suggests
Divided by A, nuclei of every size share one scale. B/A climbs steeply through the lightest nuclei, 1.112 MeV for ²H and 2.573 MeV for ³He, with ⁴He leaping to 7.074 MeV, far above ⁶Li at 5.332 MeV. From ¹²C at 7.680 MeV it rises more slowly to a broad maximum of 8.8 MeV at A ≈ 56 to 62, then falls gently: 8.504 MeV at ¹²⁰Sn, 7.867 MeV at ²⁰⁸Pb, 7.570 MeV at ²³⁸U. The flatness is informative. If every nucleon attracted every other, B would grow with the number of pairs, A(A − 1)/2, and B/A would rise in proportion to A; instead it stays between 7.5 and 8.8 MeV from A = 16 to 238, so each nucleon binds only to near neighbours and the attraction saturates. The slow fall at large A is what a long-range repulsion among Z protons would produce. These are readings of data, not a model, and this survey does not derive the curve. B/A is also an average. The marginal cost of one nucleon is the separation energy, a difference of two binding energies: ⁵⁶Fe gives Sₙ = 11.20 MeV against its mean of 8.79 MeV.
5Energy runs toward the peak, counted in total binding
When Z and N are each conserved, every Z M(¹H) + N mₙ term cancels between the two sides, so the energy released is the gain in total binding, Q = ΣB(products) − ΣB(reactants), and the rest mass falls by Q/c². ¹²C + ¹²C → ²⁴Mg releases 198.257 − 2 × 92.162 = 13.93 MeV. Energy comes out whenever the nucleons end up higher on the curve on average, which is why fusing light nuclei and splitting heavy ones both pay: a nucleus near A = 236, at 7.59 MeV per nucleon, split into halves near 8.5 MeV, frees roughly 236 × 0.9 ≈ 210 MeV. Read the curve with care. The total decides, so one product high on the curve guarantees nothing if another sits low. The curve plots the most tightly bound nucleus of each A, while real products keep the reactants' Z and N and may lie below it, so a curve reading estimates direction and size, not Q. And if weak decays can change Z, B is no longer the whole ledger: a neutron outweighs a hydrogen atom by 0.782 MeV/c², so ⁶²Ni has the largest B/A but ⁵⁶Fe the smallest mass per nucleon.
6What the curve cannot tell you
The curve says whether energy is available, not whether it is released. ²³⁸U would release roughly 200 MeV by splitting in two, yet its spontaneous-fission half-life is near 10¹⁶ years; it decays by alpha emission instead, with a half-life of 4.5 × 10⁹ years. Deuterium in seawater could fuse with a Q of 3 to 4 MeV per reaction and never does at room temperature. Rates are set by Coulomb barriers and tunnelling, which later topics in this unit treat. The smooth trend also hides structure. Nuclei with even Z and even N are more tightly bound than their odd-numbered neighbours, and extra binding at particular proton and neutron numbers lifts ⁴He, ¹⁶O and ²⁰⁸Pb above the trend. Nor does the curve say which fragments a nucleus splits into, how often each channel occurs, or how fast a product decays: those are cross sections, branching ratios and lifetimes, and they need rate theory. The curve is a measured fact, and explaining its shape needs a model of the nuclear force that this survey does not build.
Key equations
Read the units, assumptions and limits before substituting.
Energy equivalent of the unified mass unit
1 u × c² = 931.494 MeV
Turns a mass difference read from a table straight into MeV. Since 10⁻⁶ u is 0.93 keV, a mass carried to six decimal places is good to under a keV, and B, summed over up to 240 of them, to within about 10 keV.
1 u = 1.660539 × 10⁻²⁷ kg, one-twelfth of a neutral ¹²C atom; c = 2.997925 × 10⁸ m s⁻¹; 1 MeV = 1.602177 × 10⁻¹³ JBinding energy from neutral-atom masses
B = [Z M(¹H) + N mₙ − M(ᴬX)] c²
Each hydrogen atom brings one electron, so the Z electrons of atom X cancel. The bare proton mass in their place undercounts B by Z × 0.511 MeV.
Z protons, N = A − Z neutrons; M(¹H) = 1.007825 u, mₙ = 1.008665 u, M(ᴬX) the neutral atom, all in u; B in MeVBinding energy per nucleon
b = B/A
Puts nuclei of every size on one scale. From A = 16 to 238 it stays between 7.5 and 8.8 MeV, so B grows nearly in proportion to A.
b in MeV; measured 1.112 for ²H, 7.074 for ⁴He, 7.680 for ¹²C, 8.790 for ⁵⁶Fe, 8.795 for ⁶²Ni, 7.570 for ²³⁸UEnergy released by regrouping nucleons
Q = ΣB(products) − ΣB(reactants)
D + T → ⁴He + n gives 28.296 − 2.225 − 8.482 = 17.59 MeV: the alpha particle sits far higher on the curve than either reactant.
Only when Z and N are each conserved, so every Z M(¹H) + N mₙ term cancels; free neutrons have B = 0; Q in MeVMass per nucleon
M(ᴬX)/A = [Z M(¹H) + N mₙ]/A − b/c²
Two effects, two maxima: ⁶²Ni has the largest B/A, but ⁵⁶Fe, richer in the lighter hydrogen atoms, has the smallest mass per nucleon.
M in u, b = B/A in MeV, c² = 931.494 MeV u⁻¹; mₙ − M(¹H) = 0.000840 u, so more protons per nucleon lowers the first termNeutron separation energy
Sₙ = [M(ᴬ⁻¹X) + mₙ − M(ᴬX)] c² = B(Z, N) − B(Z, N − 1)
The cost of the last nucleon, not the average one: ⁵⁶Fe has Sₙ = 11.20 MeV and Sₚ = 10.18 MeV against B/A = 8.79 MeV.
M(ᴬ⁻¹X) the neutral atom with one neutron fewer, in u; Sₙ in MeV. The proton version uses M(¹H) and the Z − 1 atomInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at 118 + 118: the arrow drops about 0.93 MeV, so fusing costs about 220 MeV and splitting A = 236 releases it. Drop both to 12 and the arrow flips up: Q ≈ +14 MeV, as for ¹²C + ¹²C. On this curve 45 + 45 breaks even, though A = 90 is past the peak.
Live results
- FRAGMENTS' MEAN B/A
- 8.520 MeV
- FUSED NUCLEUS B/A
- 7.594 MeV
- Q TO FUSE = A × ARROW
- -218.6 MeV
- REST MASS LOST, Q/c²
- -0.2347 u
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Fragment mass number A₁: 118; Fragment mass number A₂: 118
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
“In fission or fusion a little of the matter is destroyed and turned into energy.” No particle is destroyed. Count across ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n: 236 nucleons before and after, 92 protons on each side, every one of them still a proton or a neutron. What changes is how tightly they are held. A nucleus's rest energy includes the negative energy of its binding, so when nucleons regroup into more tightly bound nuclei the rest mass of the whole system falls by Q/c², and the same Q appears as kinetic energy of the fragments and as radiation. The mass defect belongs to the bound arrangement, not to any particle in it. A burning match obeys the same accounting; its fractional change in mass is roughly 10⁷ times smaller, which is why chemistry could treat mass as conserved.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
The neutral-atom mass of ²⁰Ne (Z = 10) is 19.992440 u. Using M(¹H) = 1.007825 u, mₙ = 1.008665 u and 1 u × c² = 931.494 MeV, find the mass deficit, the binding energy and the binding energy per nucleon, and express the deficit as a percentage of the separated nucleons' mass.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Count: N = A − Z = 20 − 10 = 10. Use hydrogen atoms for the protons, so their ten electrons match the ten in the neon atom.
- Separated mass: 10 × 1.007825 u + 10 × 1.008665 u = 10.078250 u + 10.086650 u = 20.164900 u.
- Deficit: Δm = 20.164900 u − 19.992440 u = 0.172460 u, which is 0.172460 ÷ 20.164900 = 0.855% of the separated mass.
- B = 0.172460 u × 931.494 MeV u⁻¹ = 160.65 MeV, so B/A = 160.65 MeV ÷ 20 = 8.032 MeV. Check: that lies between ¹⁶O at 7.976 MeV and ²⁴Mg at 8.261 MeV, on the rising side of the curve where A = 20 belongs.
Answer. Δm = 0.172460 u, 0.855% of the separated mass; B = 160.6 MeV and B/A = 8.03 MeV.
MediumExample 2
Measured binding energies per nucleon are 1.1123 MeV for ²H, 7.0739 MeV for ⁴He and 5.3323 MeV for ⁶Li. A student argues that because ⁴He is far more tightly bound per nucleon than ⁶Li, a ⁶Li nucleus ought to fall apart into ⁴He + ²H. Test the claim, then find the energy released when a deuteron fuses with ⁶Li to make two alpha particles, and the rest mass the system loses.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Check the ledgers. ⁶Li has Z = 3 and N = 3, and ⁴He + ²H has 2 + 1 of each; ⁶Li + ²H and two ⁴He both have 4 of each. Z and N are conserved separately in both, so Q = ΣB(products) − ΣB(reactants).
- Totals: B(²H) = 2 × 1.1123 = 2.2246 MeV, B(⁴He) = 4 × 7.0739 = 28.2956 MeV and B(⁶Li) = 6 × 5.3323 = 31.9938 MeV.
- Break-up: Q = 28.2956 + 2.2246 − 31.9938 = −1.4736 MeV, negative, so ⁶Li cannot do it. The four nucleons that end in the alpha particle gain 4 × 1.7416 = 6.9664 MeV, but the two left in the deuteron lose 2 × 4.2200 = 8.4400 MeV. The test is total B, not the best B/A among the products.
- Fusion: Q = 2 × 28.2956 − (31.9938 + 2.2246) = 56.5912 − 34.2184 = 22.3728 MeV, positive, because all eight nucleons end up in alpha particles.
- Rest mass lost: Δm = Q/c² = 22.3728 MeV ÷ 931.494 MeV u⁻¹ = 0.024018 u, about 0.30% of a reactant mass near 8.03 u.
Answer. ⁶Li → ⁴He + ²H has Q = −1.474 MeV and cannot happen on its own; ⁶Li + ²H → 2 ⁴He releases 22.37 MeV, a rest-mass loss of 0.02402 u.
HardExample 3
Silicon burning in a massive star ends, through a chain of alpha captures, in the net change 2 ²⁸Si → ⁵⁶Ni. Atomic masses: ²⁸Si 27.976927 u, ⁵⁶Ni 55.942128 u, ⁵⁶Fe 55.934936 u; M(¹H) = 1.007825 u, mₙ = 1.008665 u. Find Q from binding energies. Then compare the estimate a student gets by reading the B/A curve, whose A = 56 point is ⁵⁶Fe, and account for the difference.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Ledger: two ²⁸Si hold 28 protons and 28 neutrons, exactly as ⁵⁶Ni does, so Z and N are each conserved and Q = B(⁵⁶Ni) − 2B(²⁸Si).
- B(²⁸Si) = [14 × 1.007825 + 14 × 1.008665 − 27.976927] u × 931.494 MeV u⁻¹ = 0.253933 u × 931.494 MeV u⁻¹ = 236.537 MeV, or 8.448 MeV per nucleon.
- B(⁵⁶Ni) = [28 × 1.007825 + 28 × 1.008665 − 55.942128] u × 931.494 MeV u⁻¹ = 0.519592 u × 931.494 MeV u⁻¹ = 483.997 MeV, or 8.643 MeV per nucleon. So Q = 483.997 − 2 × 236.537 = 10.923 MeV. Check: 2 × 27.976927 u − 55.942128 u = 0.011726 u, and 0.011726 u × 931.494 MeV u⁻¹ = 10.92 MeV.
- Curve reading: B(⁵⁶Fe) = [26 × 1.007825 + 30 × 1.008665 − 55.934936] u × 931.494 MeV u⁻¹ = 0.528464 u × 931.494 MeV u⁻¹ = 492.261 MeV, so the curve suggests 492.261 − 473.074 = 19.187 MeV, too high by 8.264 MeV. ⁵⁶Fe has 26 protons, not 28, so it is not a product of this reaction and its ΔB is not this Q.
- The gap is paid out later, when ⁵⁶Ni → ⁵⁶Co → ⁵⁶Fe turns two protons into neutrons. Those decays release [M(⁵⁶Ni) − M(⁵⁶Fe)] c² = 0.007192 u × 931.494 MeV u⁻¹ = 6.699 MeV. That is the 8.264 MeV rise in B less 2 × (mₙ − M(¹H)) c² = 2 × 0.000840 u × 931.494 MeV u⁻¹ = 1.565 MeV, the price of two neutrons heavier than the hydrogen atoms they replace.
- Close the books: 10.923 + 6.699 = 17.622 MeV, which equals [2 M(²⁸Si) − M(⁵⁶Fe)] c² = 0.018918 u × 931.494 MeV u⁻¹. The curve's 19.187 MeV misses that by 1.565 MeV, the neutron–hydrogen mass difference that binding energies alone cannot see.
Answer. Q = 10.92 MeV for 2 ²⁸Si → ⁵⁶Ni. The curve reading of 19.19 MeV is 8.26 MeV too high: 6.70 MeV of that is released later by the decays to ⁵⁶Fe, and the other 1.56 MeV is the neutron–hydrogen mass difference, never released.
Quick check
Select an answer and explain it before checking.
Review & continue
The binding energy B of a nucleus is the energy needed to separate it into free protons and neutrons at rest; it equals the mass deficit Δm, the free nucleons' mass minus the nucleus's, times c², and B/A is its average over the A nucleons.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics III completion tasks.