University Physics V · Nuclear Fission and Fusion · 16.1
Binding Energy & 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.
Build the model
Connect the measurement to the mechanism.
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 across the chart to better than one percent 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, its double hump, and the asymmetric mass yields all live in the terms this expansion threw away.
- Simple definition
- 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.
Five fitted numbers reproduce B/A to under 1%, 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-odd
One parameter turns the sphere into a prolate spheroid, so the whole barrier question 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.
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.
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.
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.
Volume and asymmetry cancel exactly on a symmetric split, so the 200 MeV is a Coulomb-versus-surface ledger and nothing else.
For ²³⁶U → two ¹¹⁸Pd the SEMF gives Q = 185 MeV: +360 MeV of Coulomb against −177 MeV of new surface.
Five 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.
B/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, because they inherit the parent's neutron excess. That 0.8 MeV per nucleon over 236 nucleons is the ≈185 MeV of prompt fission energy.
Deform 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.
One 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.
A 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, plainly absurd for a shape that is already two fragments flying apart. 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.
What the barrier is measured against
A barrier only matters beside the energy actually delivered. Capturing a neutron excites the compound nucleus by the neutron separation energy Sₙ = [M(A−1, Z) + mₙ − M(A, Z)]c². From the mass table Sₙ(²³⁶U) = 6.545 MeV and Sₙ(²³⁹U) = 4.806 MeV. Against a saddle near 6 MeV the first clears it and the second does not, which is why a thermal neutron fissions ²³⁵U while ²³⁸U needs about 1 MeV of kinetic energy on top. The 1.74 MeV gap is mostly pairing: ²³⁶U is even-even and collects δ = 11.18/√236 = 0.73 MeV, while ²³⁸U — the target, whose binding is subtracted — is also even-even and gives up 0.72 MeV, so the pairing term alone drives the two separation energies 1.45 MeV apart. Fissile against fertile is therefore not a statement about the barrier at all; it is a statement about which side of a pairing gap the target sits on.
Change one variable at a time
Make the relationship visible.
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.
FISSILITY x0.716
SURFACE COST3.92 MeV
COULOMB GAIN2.80 MeV
NET ΔE1.11 MeV
Live interpretationFISSILITY x: 0.716. SURFACE COST: 3.92 MeV. COULOMB GAIN: 2.80 MeV. NET ΔE: 1.11 MeV
Catch the common trap
Explain before calculating.
A student computes the fissility x = EC/2ES = 0.72 for ²³⁶U from the second-order deformation expansion and concludes that its fission barrier is about 6 MeV. What is wrong with that step?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyCompute 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.
- 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 × 0.239 = 67.8 MeV.
Answerx(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.
MediumUsing 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.
- 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.
AnswerQ = 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.
HardUse 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.
- 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.
AnswerQ ≈ 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.