University Physics V · Nuclear Physics · 14.4
Mass Defect & Separation Energies
Weigh a nucleus, weigh its nucleons separately, and the gap is the energy holding it together. This lesson turns that one subtraction into a working instrument: total binding, binding per nucleon, and the first differences that say what it actually costs to pull one nucleon out — and what the mass table still refuses to tell you.
Build the model
Connect the measurement to the mechanism.
A bound nucleus weighs less than the nucleons inside it, and E = mc² turns that missing mass into the energy needed to take it apart: B(Z, N) = [Z m(¹H) + N mₙ − Mₐₜ(Z, A)]c². Using neutral-atom masses rather than nuclear ones is deliberate — the Z electron rest masses cancel side to side, leaving only their binding, about 1 keV in carbon and 0.5 MeV in lead against 1636 MeV of nuclear binding. That single subtraction is the whole model, and it inherits the absolute precision of mass spectrometry: sub-keV on a 492 MeV number for ⁵⁶Fe.
Everything else comes from differencing it. Divide by A for the intensive curve, which peaks at ⁶²Ni with 8794.6 keV per nucleon and falls to 7867 keV at ²⁰⁸Pb. Take first differences for the separation energies Sₙ = B(Z, N) − B(Z, N−1) and Sₚ = B(Z, N) − B(Z−1, N), which expose what the smooth B/A curve hides: a 3 MeV odd-even zigzag that is the pairing force read straight off measured masses, and multi-MeV cliffs wherever N or Z crosses a magic number.
The cost is a hard boundary. Binding is a property of two ground states, so it says what is energetically allowed and never how fast it happens — ²³²Th (Qα = +4.08 MeV, 1.4 × 10¹⁰ yr) and ²¹²Po (Qα = 8.95 MeV, 0.29 µs) draw the same verdict from the mass table and differ by 24 orders of magnitude in rate.
- Simple definition
- The binding energy of a nucleus is the energy equivalent of its mass defect — how much less the bound nucleus weighs than the free protons and neutrons it contains — and a separation energy is the change in that binding when one nucleon is removed.
- Example
- ⁴He weighs 4.0026033 u while two ¹H atoms and two neutrons weigh 4.0329799 u; the 0.0303766 u defect is 28.30 MeV of binding, 7074 keV per nucleon.
Atomic masses go in unchanged: the Z electron rest masses cancel side to side, so only their binding is left uncorrected.
m(¹H) = 1.0078250 u, mₙ = 1.0086649 u, 1 u = 931.494 MeV/c²; Mₐₜ in u for the neutral atom
The A u terms cancel identically, so B is one integer dot product with the mass-excess column: (Z, N, −1)⋅(ΔH, Δₙ, Δ).
ΔH = 7.2890 MeV, Δₙ = 8.0713 MeV. Tables list Δ in keV, so no ten-digit masses are needed.
The only binding number comparable across different A. ⁶²Ni beats ⁵⁶Fe in total B by 53 MeV purely by having six more nucleons.
⁵⁸Fe 8792.2, ⁵⁶Fe 8790.4, ²⁰⁸Pb 7867, ¹²C 7680, ⁴He 7074, ²H 1112 keV per nucleon
The cost of lifting one nucleon to infinity at rest — a nuclear ionisation energy, set by the topmost occupied level, not by the average.
In mass excess: Sₙ = Δ(Z, A−1) + Δₙ − Δ(Z, A) and Sₚ = Δ(Z−1, A−1) + ΔH − Δ(Z, A)
The symmetric form cancels any linear trend in Sₙ, which biases the one-sided ½[Sₙ(N+1) − Sₙ(N)] by half the slope.
Typically 1–2 MeV, trending down with mass number roughly as A(−1/2), and under 1 MeV in the heaviest nuclei
Plot this against N when you want shell closures without pairing noise: it runs smoothly, then drops sharply past 50, 82 and 126.
Removes a full pair, so the odd-even zigzag cancels between the two terms
One subtraction, and why the electrons cancel
Take a neutral ⁵⁶Fe atom, Mₐₜ = 55.9349363 u. Take 26 hydrogen ATOMS and 30 free neutrons: 26(1.0078250) + 30(1.0086649) = 26.2034508 + 30.2599476 = 56.4633983 u. The parts outweigh the whole by Δm = 0.5284620 u, and at 931.494 MeV per u that defect is B = 492.26 MeV — the work needed to disassemble the nucleus into free nucleons at rest. Atomic rather than nuclear masses are used on purpose. Each m(¹H) carries one electron, so Z of them stand on the parts side against the Z electrons already inside Mₐₜ, and the rest masses cancel exactly. What does not cancel is the difference between the atom's total electron binding Bₑ(Z) and Z times hydrogen's 13.6 eV: roughly 1 keV for carbon, about 34 keV for iron, some 0.5 MeV for lead against 1636 MeV of nuclear binding. Standard tables ignore it. Say so, rather than pretending it is zero.
Mass excess turns the calculation into a table lookup
Nobody types ten-digit masses. Tables list the mass excess Δ ≡ (Mₐₜ − A u)c² in keV, and because Z + N = A the A u terms cancel identically, leaving B = Z ΔH + N Δₙ − Δ(Z, A) with ΔH = 7.2890 MeV and Δₙ = 8.0713 MeV. For ¹⁴C, Δ = 3.0199 MeV, so B = 6(7.2890) + 8(8.0713) − 3.0199 = 43.734 + 64.570 − 3.020 = 105.284 MeV. In NumPy the table is one array and every binding energy is a dot product with the integer vector (Z, N, −1), so a whole isotope chain costs one line. The precision this preserves is worth naming: B is a small difference of large numbers — 0.53 u out of 56 u for ⁵⁶Fe — so it inherits the absolute, not the relative, uncertainty of the masses. A few tenths of a keV on Mₐₜ(⁵⁶Fe) lands as a few tenths of a keV on 492.26 MeV, better than one part in 10⁶. Every fit and every drip-line criterion later in this unit rests on that.
B/A is an average, and its peak is nickel, not iron
Divide by A and you get the only binding number comparable between different nuclides. B/A climbs steeply and unevenly through the light nuclei — ²H 1112, ⁴He 7074, ¹²C 7680 keV per nucleon — flattens across the iron–nickel region and falls slowly to 7867 keV at ²⁰⁸Pb. The maximum is ⁶²Ni at 8794.6 keV per nucleon, with ⁵⁸Fe at 8792.2 and ⁵⁶Fe at 8790.4: the top three lie within 4.2 keV per nucleon of one another, 0.05% of B/A, yet with sub-keV mass uncertainties that margin is hundreds of standard deviations. The ordering is measurement, not model. Never rank total B across different A: ⁶²Ni's 545.26 MeV beats ⁵⁶Fe's 492.26 MeV by 53 MeV solely because it has six more nucleons. And that ⁵⁶Fe still dominates the observed iron peak is a statement about nuclear statistical equilibrium and the neutron-to-proton ratio at freeze-out, not about which nuclide is most bound.
A separation energy is a first difference, not an average
How much does it cost to pull one neutron off? Not B/A. The removal energy is the first difference Sₙ = B(Z, N) − B(Z, N−1), or Sₙ = Δ(Z, A−1) + Δₙ − Δ(Z, A) directly from the table; the proton version is Sₚ = B(Z, N) − B(Z−1, N). These are nuclear ionisation energies: the departing nucleon sits in the highest occupied level, not at the average depth. Doubly magic ¹⁶O has B/A = 7976 keV but Sₙ = 127.619 − 111.955 = 15.66 MeV and Sₚ = 127.619 − 115.492 = 12.13 MeV, both far above that average. Add one neutron and ¹⁷O's Sₙ collapses to 131.762 − 127.619 = 4.14 MeV, a factor of 3.8 down, while B/A barely moves, from 7976 to 7751 keV per nucleon. The average is smooth because it averages; the difference is where the level structure shows.
The odd-even sawtooth is pairing, read before any fit
Plot Sₙ along an isotope chain and it zigzags. Carbon: 18.72 MeV at N = 6, 4.95 MeV at N = 7, 8.18 MeV at N = 8. Even N is always the expensive one, because removing a neutron from an even-N nucleus breaks a J = 0 coupled pair while removing the odd neutron from an odd-N nucleus does not. The amplitude of that zigzag is the pairing energy, and it comes off measured masses before any model is fitted — the semi-empirical mass formula's ±aP A(−1/2) term exists to reproduce this feature, not to explain it. The one-sided estimate ½[Sₙ(N+1) − Sₙ(N)] gives 1.62 MeV for carbon but is biased by whatever smooth slope Sₙ carries; the symmetric Δ⁽³⁾(N) = ¼[Sₙ(N−1) + Sₙ(N+1) − 2 Sₙ(N)] at odd N cancels a linear trend exactly. When you want shell structure without the pairing noise, difference by two instead: S₂n = B(Z, N) − B(Z, N−2) = Sₙ(N) + Sₙ(N−1) removes a whole pair, runs smoothly with N, and drops sharply only where N crosses 8, 20, 28, 50, 82 or 126.
Binding fixes energetics and never a rate
Every number in this topic is a difference of two ground-state masses. That fixes what is energetically permitted and nothing else. Sₙ < 0 means the nucleus cannot hold that neutron and it leaves in about 10⁻²¹ s; Sₙ > 0 means neutron emission is closed, and the mass table stops there. Alpha decay makes the point sharply: ²³²Th has Qα = +4.08 MeV and a half-life of 1.4 × 10¹⁰ years, ²¹²Po has Qα = 8.95 MeV and 0.29 µs. Both are allowed by the same table, and their rates differ by 24 orders of magnitude, because a rate needs the Coulomb barrier and a tunnelling exponent that no mass difference contains. The same discipline applies in reverse: a large Sₙ does not make a nuclide long-lived against beta decay, which is governed by an entirely different mass difference and by a matrix element the masses know nothing about.
Change one variable at a time
Make the relationship visible.
Drag the pairing gap to zero and the zigzag collapses onto the dashed trend, so the sawtooth is the entire pairing signal. Then raise the smooth slope: the one-sided half-step drifts below the true gap by half the slope, while the symmetric three-point readout stays exactly on it.
Sₙ AT EVEN N8.60 MeV
Sₙ AT ODD N4.90 MeV
ONE-SIDED HALF-STEP1.35 MeV
SYMMETRIC 3-POINT1.60 MeV
Live interpretationSₙ AT EVEN N: 8.60 MeV. Sₙ AT ODD N: 4.90 MeV. ONE-SIDED HALF-STEP: 1.35 MeV. SYMMETRIC 3-POINT: 1.60 MeV
Catch the common trap
Explain before calculating.
Binding energies are computed from neutral-atom masses as B = [Z m(¹H) + N mₙ − Mₐₜ]c². A student objects that the tabulated Mₐₜ for ⁵⁶Fe already contains 26 atomic electrons, and so replaces m(¹H) with the bare proton mass mₚ. What happens to the computed binding energy, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyNeutral ⁴He has atomic mass 4.00260325 u. With m(¹H) = 1.00782503 u, mₙ = 1.00866492 u and 1 u = 931.494 MeV/c², find the mass defect, the total binding energy and the binding per nucleon.
- ⁴He has Z = 2 and N = 2, so the parts are two hydrogen ATOMS and two free neutrons: 2(1.00782503) + 2(1.00866492) = 2.01565006 + 2.01732984 = 4.03297990 u.
- Mass defect: Δm = 4.03297990 − 4.00260325 = 0.03037665 u. The whole is lighter than its parts, which is what bound means.
- Convert: B = 0.03037665 × 931.494 = 28.296 MeV. Hydrogen atoms rather than bare protons, so the two electrons on the parts side cancel the two already inside the neutral helium atom.
- Per nucleon: B/A = 28.296/4 = 7.074 MeV, or 7074 keV. That is why ⁴He is so tightly bound for its size — its neighbour ⁶Li manages only 31.994/6 = 5332 keV per nucleon.
AnswerΔm = 0.0303766 u; B = 28.30 MeV; B/A = 7074 keV per nucleon.
MediumMass excesses: Δ(¹²C) = 0 by definition, Δ(¹³C) = 3.1250 MeV, Δ(¹⁴C) = 3.0199 MeV, with Δₙ = 8.0713 MeV. Find the neutron separation energies of ¹³C and ¹⁴C, and use them to estimate the neutron pairing gap in this chain.
- Work in mass excess and the A u terms cancel: Sₙ(Z, A) = B(Z, A) − B(Z, A−1) = Δ(Z, A−1) + Δₙ − Δ(Z, A). No ten-digit masses are needed.
- ¹³C has N = 7, odd: Sₙ = Δ(¹²C) + Δₙ − Δ(¹³C) = 0 + 8.0713 − 3.1250 = 4.9463 MeV.
- ¹⁴C has N = 8, even: Sₙ = Δ(¹³C) + Δₙ − Δ(¹⁴C) = 3.1250 + 8.0713 − 3.0199 = 8.1764 MeV.
- The even-N neutron is 8.1764 − 4.9463 = 3.2301 MeV dearer to remove, because it is one of a coupled J = 0 pair while ¹³C's seventh neutron is unpaired. Half that step estimates the gap: Δ ≈ ½(3.2301) = 1.62 MeV.
- Sanity check against the averages: B/A is 7470 keV for ¹³C and 7520 keV for ¹⁴C, 0.7% apart. The averages hide entirely what the first differences show — a 65% change in what the last neutron costs.
AnswerSₙ(¹³C) = 4.946 MeV and Sₙ(¹⁴C) = 8.176 MeV; the 3.230 MeV odd-even step gives a neutron pairing gap of about 1.6 MeV.
HardAtomic masses: M(⁵⁶Fe) = 55.9349363 u, M(⁵⁸Fe) = 57.9332744 u, M(⁶²Ni) = 61.9283451 u. Using m(¹H) = 1.00782503 u, mₙ = 1.00866492 u and 1 u = 931.494 MeV/c², decide which is the most tightly bound nuclide per nucleon, quantify the margin, and say whether mass uncertainties of a few tenths of a keV could overturn the ordering.
- ⁵⁶Fe (Z = 26, N = 30): parts = 26.2034508 + 30.2599476 = 56.4633983 u, so Δm = 0.5284620 u, B = 492.259 MeV and B/A = 492259/56 = 8790.3 keV.
- ⁵⁸Fe (Z = 26, N = 32): parts = 26.2034508 + 32.2772774 = 58.4807281 u, so Δm = 0.5474537 u, B = 509.950 MeV and B/A = 509950/58 = 8792.2 keV.
- ⁶²Ni (Z = 28, N = 34): parts = 28.2191009 + 34.2946071 = 62.5137080 u, so Δm = 0.5853629 u, B = 545.262 MeV and B/A = 545262/62 = 8794.5 keV.
- Rank on B/A, never on B. ⁶²Ni's total binding beats ⁵⁶Fe's by 545.262 − 492.259 = 53.0 MeV only because it has six more nucleons. Per nucleon the order is ⁶²Ni 8794.5 > ⁵⁸Fe 8792.2 > ⁵⁶Fe 8790.3 keV, margins of 2.3 and 4.2 keV per nucleon.
- Uncertainty: 0.4 keV on a total binding energy becomes 0.4/62 ≈ 0.007 keV per nucleon, so the 4.2 keV per nucleon gap runs to some hundreds of standard deviations. The ordering is a measurement, and no fitted mass formula is entitled to overturn it.
- Do not over-read it. That 4.2 keV per nucleon is 0.05% of B/A, far below the MeV-scale barriers and photodisintegration thresholds that actually set stellar-ash abundances — which is why the observed iron peak is ⁵⁶Fe and not ⁶²Ni.
Answer⁶²Ni, at 8794.5 keV per nucleon, ahead of ⁵⁸Fe (8792.2) and ⁵⁶Fe (8790.3). The 4.2 keV per nucleon margin is hundreds of σ, so the ordering is safe.