
Nuclear, particle & quantum-information sequence · Interactive lesson
Binding energy and the fission barrier
Fission is not a new force — it is the liquid drop's surface tension losing an argument with its own electrostatic repulsion. This topic teaches you to price that argument: read binding from the mass formula, deform the drop, and find exactly where a second-order expansion stops being able to predict a barrier.
UPV-16.01 · Lesson 1 of 9 in this unit. Position is not a completion score.Nuclear, particle & quantum-information sequence. These units continue into nuclear and particle physics and quantum information. Follow your university’s prerequisites, mathematical depth and assessment; this independent pathway is not a universal programme specification.
Build the model
Surface energy resists deformation; Coulomb energy pays for it.
The key idea
The fission barrier is the extra energy a nucleus must acquire to reach the saddle-point shape beyond which surface tension can no longer hold it together against its own Coulomb repulsion.
Example. For ²³⁶U the barrier is about 6 MeV, and capturing a thermal neutron on ²³⁵U delivers 6.545 MeV of excitation — just over it, which is why ²³⁵U fissions with slow neutrons.
The complete physical model and assumptions
The nucleus is modelled as an incompressible charged liquid drop, and the semi-empirical mass formula is that drop's energy budget: a volume term proportional to A because the strong force saturates, a surface term −aS A²⁄³ because nucleons at the skin are under-bonded, a Coulomb term −aC Z²/A¹⁄³ because every proton repels every other across the whole volume, an asymmetry term charging the Pauli cost of unequal neutron and proton Fermi seas, and a pairing term. Five fitted coefficients reproduce B/A to about one percent everywhere but the lightest nuclei and put its maximum near A = 56–62. Fission is then a question about that budget under a change of shape. Deform the drop into a volume-conserving quadrupole, R(θ) = R₀[1 + α P₂(cos θ)], and to second order the surface energy rises by a fraction (2/5)α² while the Coulomb energy falls by (1/5)α², so ΔE = (α²/5)(2ES − EC). The whole of small-amplitude stability sits in the sign of that bracket: the drop resists deformation while the fissility x = EC/2ES ≈ (Z²/A)/50 stays below one. What the expansion cannot do is give a barrier. At order α² the energy is a monotone parabola — no maximum, so no saddle height, no WKB tunnelling exponent, no spontaneous-fission lifetime. The measured ~6 MeV saddle needs the higher-order terms and necked shapes this expansion threw away plus shell corrections the liquid drop never had, and those same shell corrections make its double hump and the asymmetric mass yields.
1Five terms, and three of them are arguments about shape
Write the binding energy as a budget. The volume term aV A says the strong force saturates: a nucleon deep inside binds only to its neighbours, so binding grows with the nucleon count, not with the number of pairs. The surface term −aS A²⁄³ corrects for nucleons at the skin, which are short of neighbours; it scales as area, and a sphere minimises it. The Coulomb term −aC Z²/A¹⁄³ is the electrostatic self-energy of a uniformly charged sphere, 3Z²e²/20πε₀R with R = r₀A¹⁄³; it is the only term that does not saturate. Its coefficient is not free — (3/5)(e²/4πε₀)/r₀ = (0.6)(1.44 MeV fm)/(1.2 fm) = 0.72 MeV, against the fitted 0.711. The asymmetry term −aA(A−2Z)²/A is not electrostatic at all but the Pauli cost of filling neutron and proton Fermi seas to different depths, and δ is the residual pairing of like nucleons in time-reversed orbits.
2B/A peaks near A = 56–62, and both sides run downhill
Divide the budget by A. The volume term is a constant 15.75 MeV; the surface penalty falls off as A(−1/3), so light nuclei suffer most; the Coulomb penalty grows as Z²/A⁴⁄³, so heavy nuclei suffer most. The two crossing penalties give B/A a steep rise to a broad maximum and a slow decline after it. Measured, the peak is ⁶²Ni at B/A = 8.795 MeV, with ⁵⁸Fe at 8.792 and ⁵⁶Fe at 8.790 — a plateau, not a spike, and the ranking flips if you order by mass per nucleon instead, where ⁵⁶Fe wins because it carries fewer of the heavier neutrons. Both downhill sides release energy: run up the light side by fusing, the heavy side by splitting. ²³⁶U sits at B/A = 7.59 MeV and its fragments land near 8.4 MeV, not on the 8.79 plateau: at A ≈ 95 and 140 they sit on the slow decline past the peak, and they inherit the parent's neutron excess besides. That gain of about 0.8 MeV per nucleon over 236 nucleons is the ≈190 MeV scale of fission energy.
3Deform the sphere and expand to second order
Take a volume-conserving quadrupole, R(θ) = R₀[1 + αP₂(cos θ)], with R₀ rescaled at order α² so the enclosed volume never changes. That choice matters: it freezes the volume and asymmetry terms and isolates the two that care about shape. Compute the surface area and the electrostatic self-energy of that shape and expand. A prolate spheroid has more area than the sphere of equal volume, so ES = ES⁰(1 + (2/5)α² + …). Its charge sits further apart, so EC = EC⁰(1 − (1/5)α² + …). Both series start at α², never at α: a sphere is already an extremum, the first derivative vanishes, and the second derivative decides. Adding them gives ΔE = (α²/5)(2ES⁰ − EC⁰). For ²³⁶U, ES⁰ = 17.8 × 236²⁄³ = 680 MeV and EC⁰ = 0.711 × 92²/236¹⁄³ = 974 MeV, so ΔE = (α²/5)(1360 − 974) = 77α² MeV — positive, and worth only 0.77 MeV at α = 0.1.
4One number decides: the fissility x
Divide the bracket by 2ES⁰ and the geometry collapses. ΔE = (2/5)ES⁰(1 − x)α² with x = EC⁰/2ES⁰ = (aC/2aS)(Z²/A). With aC = 0.711 and aS = 17.8 MeV that prefactor is 0.711/35.6 = 0.01997, so x ≈ (Z²/A)/50.1 and nothing else about the nucleus enters: two nuclei with the same Z²/A have the same small-amplitude stability whatever their size. ⁵⁶Fe has Z²/A = 12.1 and x = 0.24, deeply stable. ²³⁵U has 36.0 and x = 0.72; ²³⁹Pu has 37.0 and x = 0.74. The condition x = 1 needs Z²/A = 50.1, and it is approached long before it is reached: ²⁹⁴Og has Z²/A = 47.4 and x = 0.95, leaving the liquid drop a barrier of a fraction of an MeV — yet it survives about a millisecond. Shell corrections, absent from this model entirely, supply what the drop does not.
5A parabola has no maximum, so it has no barrier
This is where the expansion stops, and the stopping point is the lesson. ΔE = (2/5)ES⁰(1 − x)α² is a parabola through the origin: for x < 1 it rises without bound, for x > 1 it falls without bound. Neither has a stationary point away from α = 0, so the model as expanded supplies no saddle height — and therefore no WKB tunnelling exponent, no spontaneous-fission half-life, and no threshold for induced fission. For ²³⁶U it predicts 77α² MeV, which passes 6 MeV at α = 0.28 and climbs to 77 MeV at α = 1, where the polar radius is already four times the equatorial one — plainly absurd beside a measured barrier of about 6 MeV. The barrier appears only once the expansion is abandoned: full liquid-drop energies over a multi-parameter shape family that includes necking, plus Strutinsky shell corrections added back onto the smooth drop. That gives ²³⁶U a saddle near 6 MeV, splits it into two humps with a shallow second minimum between them — the fission isomers — and makes the mass yield asymmetric.
6Shell corrections build the double hump
Carried to all orders and over necked shapes, the smooth drop gives an actinide one broad hump and a symmetric split; the measured barrier has two humps and the measured split is lopsided. What the drop lacks is shell structure. Strutinsky's method keeps it for the bulk energy and adds the missing piece, E(shape) = ELD(shape) + δEₛₕₑₗₗ(shape), where δEₛₕₑₗₗ is the sum of the occupied single-particle energies in the deformed well minus the same sum over a level density smoothed across several MeV. It is negative where levels near the Fermi surface are sparse and positive where they bunch, and as the well stretches it oscillates by a few MeV, comparable to the barrier itself. In the actinides it deepens the deformed ground state and carves a second well near the top of the smooth barrier, at an axis ratio near 2:1 and roughly 2–3 MeV above the ground state. States trapped there are the fission isomers, such as the 14 ms shape isomer of ²⁴²Am, which fission quickly because only the outer hump remains to cross. Shells act again at scission: closures near doubly magic ¹³²Sn stabilise a heavy fragment of A ≈ 135–140, so the thermal-fission yield of ²³⁵U peaks near A ≈ 95 and 140 and the symmetric split near A = 118 is suppressed several-hundred-fold.
Key equations
Read the units, assumptions and limits before substituting.
Semi-empirical mass formula
B = aV A − aS A²⁄³ − aC Z²/A¹⁄³ − aA (A−2Z)²/A ± δ
Five fitted numbers reproduce B/A to about 1% beyond the lightest nuclei, and every fission argument below is a difference of two of these budgets.
aV 15.75, aS 17.8, aC 0.711, aA 23.7 MeV; δ = 11.18/√A MeV, + even-even, 0 odd A, − odd-oddVolume-conserving quadrupole shape
R(θ) = R₀[1 + α P₂(cos θ)], P₂ = (3cos²θ − 1)/2
One parameter stretches the sphere towards a prolate spheroid (α > 0) or flattens it towards an oblate one (α < 0), so small-deformation stability becomes a function of a single number.
α is the dimensionless quadrupole amplitude; R₀ is rescaled at O(α²) so the enclosed volume is fixed.Second-order shape energies
ES(α) = ES⁰(1 + (2/5)α²), EC(α) = EC⁰(1 − (1/5)α²)
Surface area grows while the charge spreads apart, so the two terms pull opposite ways — the whole competition in two coefficients.
ES⁰ = aS A²⁄³ and EC⁰ = aC Z²/A¹⁄³, both in MeV; the linear terms vanish at a sphere.Stability against small deformation
ΔE = (α²/5)(2ES⁰ − EC⁰) = (2/5)ES⁰(1 − x)α²
The sign of the bracket decides, not its size: with 2ES⁰ > EC⁰ the drop pushes back on any small quadrupole squeeze.
ΔE in MeV. Positive means the sphere is a local minimum; negative means it is not.Fissility parameter
x = EC⁰/2ES⁰ = (aC/2aS)(Z²/A) ≈ (Z²/A)/50.1
Collapses two variables into one, and marks where the liquid drop stops offering any barrier at all.
Dimensionless. ²³⁵U: Z²/A = 36.0, x = 0.72. ⁵⁶Fe: 12.1, x = 0.24. x = 1 needs Z²/A = 50.1.Fission Q from the binding budget
Q = 2B(A/2, Z/2) − B(A, Z)
Volume and asymmetry cancel exactly on a symmetric split, so apart from about 1 MeV of pairing the ≈200 MeV is a Coulomb-versus-surface ledger.
For ²³⁶U → two ¹¹⁸Pd the SEMF gives Q = 185 MeV: +360 MeV of Coulomb against −177 MeV of new surface.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at Z = 92, A = 236: x = 0.716, the solid net curve rises, and uranium resists small deformation. Slide Z to 118 and x passes 1, so the net curve turns downward instead. At no setting does it ever turn over — no maximum means no barrier height.
Live results
- FISSILITY x
- 0.716
- SURFACE COST
- 3.92 MeV
- COULOMB GAIN
- 2.80 MeV
- NET ΔE
- 1.11 MeV
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. Proton number Z: 92; Mass number A: 236; Quadrupole deformation α: 0.12
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest error is to read x = EC/2ES < 1 as "this nucleus has a fission barrier", x > 1 as "it has none", and then to quote a barrier height the criterion never contained. The second-order expansion tests one thing: whether the sphere is a local minimum against an infinitesimal quadrupole squeeze. It is a curvature test, and a curvature says nothing about how far the hill extends or how tall it gets. ²³⁵U and ²³⁹Pu have fissilities differing by 0.02, while their spontaneous-fission half-lives differ by more than three orders of magnitude, because the half-life depends exponentially on a saddle height this model does not produce. Quote x for stability against small deformation. Quote a saddle height only from a calculation that went past α².
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Compute the fissility parameter x for ²³²Th (Z = 90) and for ²⁵²Cf (Z = 98) using x = (Z²/A)/50.07, then compare their stiffness (2/5)ES⁰(1 − x) against a small quadrupole deformation.
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.
- Fissility depends only on Z²/A, because x = (aC/2aS)(Z²/A) with aC = 0.711 MeV and aS = 17.8 MeV, giving aC/2aS = 0.711/35.6 = 0.01997, i.e. x = (Z²/A)/50.07.
- ²³²Th: Z² = 8100, so Z²/A = 8100/232 = 34.91 and x = 34.91/50.07 = 0.697.
- ²⁵²Cf: Z² = 9604, so Z²/A = 9604/252 = 38.11 and x = 38.11/50.07 = 0.761.
- Both lie below 1, so both spheres are local minima. Stiffness is (2/5)ES⁰(1 − x): for Th, ES⁰ = 17.8 × 232²⁄³ = 672 MeV gives 0.4 × 672 × 0.303 = 81.4 MeV per unit α²; for Cf, ES⁰ = 17.8 × 252²⁄³ = 710 MeV gives 0.4 × 710 × 0.2388 = 67.8 MeV.
Answer. x(²³²Th) = 0.697 and x(²⁵²Cf) = 0.761. Thorium is the stiffer, 81.4 MeV per unit α² against 67.8 MeV, matching the order of their spontaneous-fission half-lives — 10²¹ y against 86 y — though the size of that gap is set by the saddle height, not by x.
MediumExample 2
Using atomic masses M(²³⁵U) = 235.043930 u, M(¹⁴¹Ba) = 140.914411 u, M(⁹²Kr) = 91.926156 u and mₙ = 1.008665 u, find Q for ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n. Then say why the energy quoted per fission is nearer 200 MeV.
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 electrons before using atomic masses: the left side carries uranium's 92, the right side barium's 56 plus krypton's 36 = 92. They cancel, so atomic masses can be used directly with no electron-mass correction.
- Initial mass: 235.043930 + 1.008665 = 236.052595 u.
- Final mass: 140.914411 + 91.926156 + 3 × 1.008665 = 232.840567 + 3.025995 = 235.866562 u.
- Δm = 236.052595 − 235.866562 = 0.186033 u, and Q = 0.186033 × 931.494 MeV/u = 173.3 MeV.
- That is the prompt Q for one specific split. Averaged over the yield distribution the prompt release is about 181 MeV (169 MeV of fragment kinetic energy, 5 MeV of prompt neutrons, 7 MeV of prompt gammas), and the beta decays of the neutron-rich fragments add roughly 22 MeV more.
Answer. Q = 173.3 MeV for that split. The familiar ≈202 MeV per fission is the yield-averaged total including delayed β and γ emission; about 8.8 MeV of it leaves with antineutrinos, so roughly 193 MeV is recoverable as heat.
HardExample 3
Use the SEMF with aV = 15.75, aS = 17.8, aC = 0.711, aA = 23.7 MeV and δ = 11.18/√A MeV to estimate Q for the symmetric split ²³⁶U → 2 ¹¹⁸Pd, term by term, and identify which term pays for it.
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.
- Volume: 15.75 × 236 = 3717.0 MeV against 2 × 15.75 × 118 = 3717.0 MeV. It contributes exactly nothing, because the strong force saturates and the nucleon count is conserved.
- Asymmetry: the parent has (A − 2Z)²/A = 52²/236 = 11.458; each fragment has 26²/118 = 5.729, and two of them give 11.458 as well. The term is invariant under symmetric division, so it also contributes nothing.
- Surface: 236²⁄³ = 38.19 and 118²⁄³ = 24.06, so the parent costs 17.8 × 38.19 = 679.8 MeV and the pair costs 2 × 17.8 × 24.06 = 856.5 MeV. Splitting creates 176.7 MeV of new surface, which is a cost and reduces Q.
- Coulomb: 236¹⁄³ = 6.180 and 118¹⁄³ = 4.905, so the parent holds 0.711 × 8464/6.180 = 973.8 MeV and the pair holds 2 × 0.711 × 2116/4.905 = 613.5 MeV. Halving Z and separating the charge releases 360.3 MeV.
- Pairing: all three nuclei are even-even, so the parent gains 11.18/√236 = 0.73 MeV and the pair 2 × 11.18/√118 = 2.06 MeV, a further +1.3 MeV.
- Q = 0 − 176.7 + 360.3 + 1.3 = 185 MeV. Fission is a single ledger entry: Coulomb energy released, minus the surface energy of the two new skins.
Answer. Q ≈ 185 MeV, from +360 MeV of Coulomb energy released against −177 MeV of new surface, with volume and asymmetry cancelling exactly and pairing worth +1.3 MeV. The same two terms, in the same competition, are what set the fissility x = EC/2ES.
Quick check
Select an answer and explain it before checking.
Review & continue
The fission barrier is the extra energy a nucleus must acquire to reach the saddle-point shape beyond which surface tension can no longer hold it together against its own Coulomb repulsion.
- 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.