University Physics V · Nuclear Physics · 14.9
Fission & Fusion from the B/A Curve
Every number in the energy budget of a reactor or a tokamak comes out of one subtraction on one curve. This lesson teaches you to take it correctly — weighting each fragment by its own mass number, charging the free neutrons — and to say where the curve stops predicting and starts merely permitting.
Build the model
Connect the measurement to the mechanism.
Nucleon number is conserved in every strong rearrangement, so the only thing that can change is how tightly those nucleons are held. That reduces the reaction energy to a single subtraction, Q = ΣB(final) − ΣB(initial), with every free nucleon contributing zero. Plot B/A against A and the whole of nuclear energetics becomes a picture: a steep climb, a broad maximum of 8.7945 MeV per nucleon at Ni-62, then a slow fall to 7.570 at U-238.
Anything that carries nucleons toward the peak — light nuclei fusing, heavy nuclei splitting — releases the mass-weighted height they gain. The cost of that convenience is that the curve is a statement about ground-state masses and nothing else. It says n + U-235 can release 173.3 MeV; it says nothing about the 5.8 MeV barrier, nothing about a cross-section, nothing about a rate.
Where it does speak about stability it speaks through the liquid-drop coefficients you fitted by SVD two topics earlier: the deformation barrier vanishes once Z²/A exceeds 2aS/aC ≈ 50, a number your own fit produces rather than one you look up.
- Simple definition
- Reading fission and fusion off the B/A curve means computing a reaction's Q value as the change in total binding energy, ΣB(final) − ΣB(initial), with every nucleus contributing its own B/A multiplied by its own mass number.
- Example
- For n + U-235 → Ba-141 + Kr-92 + 3n: 141 × 8.326 + 92 × 8.513 − 235 × 7.591 = 173.3 MeV, the four free neutrons contributing nothing because an unbound nucleon has B = 0.
Nucleon number is conserved, so only the binding changes — one subtraction, no reaction model required.
B in MeV; a free nucleon has B = 0, so spare neutrons drop out of both sums
Runs the opposite way to the binding route: more binding means less mass, so the two agree.
atomic masses in u; the electron count balances when no beta particle is emitted
Each product's mass number times its height above the parent's level — the arrows in the figure.
sum over all products with Σ Af = A; heights in MeV per nucleon, Af a pure number
3.518 MeV per nucleon, nearly five times fission's 0.734 — but 17.6 MeV per event, not 173.
the emitted neutron is unbound; momentum conservation hands it 14.05 MeV of the 17.589
At x ≥ 1 the surface cannot resist quadrupole deformation at all: no barrier, lifetime ~10⁻²⁰ s.
aS ≈ 17.80 MeV and aC ≈ 0.711 MeV from your own SVD fit; x is dimensionless
Zero at Z²/A ≈ 17.6, so splitting pays above A ≈ 90 — long before it is fast.
coefficients are 1 − 2(−2/3) and 2¹⁄³ − 1; volume and asymmetry terms cancel exactly
Q is a difference of total binding energy
A strong-interaction rearrangement moves nucleons; it never creates or destroys them. Z and N are separately conserved in fission and fusion, so every term depending only on nucleon count cancels and the reaction energy collapses to Q = ΣB(final) − ΣB(initial). Two bookkeeping rules make that reliable. First, a free nucleon is unbound: B = 0. The neutron entering a U-235 fission and the three leaving it contribute nothing to either sum, so they may be dropped from both without changing Q. Second, watch the sign. Binding energy is a positive number that enters the mass negatively, M = Zm(¹H) + Nmₙ − B/c², so the mass ledger runs the other way: Q = [ΣM(initial) − ΣM(final)]c². Products that are more tightly bound are products that are less massive. Reverse one of those two conventions and you have built a refrigerator instead of a reactor.
Weight every product by its own mass number
The vertical axis is B/A, not B, so a height read off the curve must be multiplied by a mass number before it is an energy. Because the products' mass numbers sum to the parent's, the subtraction rearranges into Q = Σf Af[(B/A)f − (B/A)ₚₐᵣₑₙₜ]: each product's mass times its height above the parent's level. Take U-236 at 7.586 MeV per nucleon splitting into Ba-144 at 8.265 and Kr-90 at 8.591, plus two free neutrons. The two bound fragments give 144(0.679) + 90(1.005) = 97.8 + 90.5 = 188.2 MeV. Now charge the neutrons: each leaves the parent's 7.586 MeV per nucleon and lands at B = 0, a height of −7.586, costing 2(7.586) = 15.2 MeV. Q = 188.2 − 15.2 = 173.0 MeV. The identity only closes when every outgoing nucleon is counted, bound or not.
One fission, all the way to the quoted 200 MeV
Run n + U-235 → Ba-141 + Kr-92 + 3n through mass excesses, which is the same subtraction with the A u terms already cancelled: Q = (8.071 + 40.921) − (−79.733 − 68.785 + 3 × 8.071) = 48.992 + 124.305 = 173.30 MeV. That is the prompt release for this channel — fragment kinetic energy plus the excitation that comes off as prompt neutrons and gammas. But Ba-141 and Kr-92 sit at Z/A ≈ 0.40, well left of stability, because fission preserves the parent's neutron excess. Repeat the subtraction with the stable end-points Pr-141 and Zr-92 and you get 199.25 MeV. The extra 26 MeV is the two β-decay chains, of which roughly 9 MeV leaves as antineutrinos and heats nothing. That is how 173 becomes the ~200 MeV per fission you see quoted, and why a reactor collects only about 194.
The other end of the curve
Fusion is the same subtraction taken from the light side. For ²H + ³H → ⁴He + n, the neutron is unbound, so Q = 28.2957 − (2.2246 + 8.4818) = 17.589 MeV. Per nucleon that is 17.589/5 = 3.518 MeV, nearly five times the 0.734 MeV per nucleon of a U-235 fission — the light end of the curve is far steeper than the heavy end is. Per event it is the reverse, and that is what sets the engineering. A 1 GW thermal core needs 3.2 × 10¹⁹ fissions per second at 194 MeV each; the same power from D–T needs 3.5 × 10²⁰ reactions, eleven times as many. Momentum conservation from a target at rest splits Q inversely with mass: Eₙ = Q mα/(mα + mₙ) = 14.05 MeV, leaving 3.54 MeV to the alpha. Only that charged 20% stays in the plasma, which is the whole ignition problem in one number.
Which peak you mean: Ni-62 or Fe-56
The maximum of B/A is Ni-62 at 8.7945 MeV, with Fe-58 at 8.7921 and Fe-56 at 8.7903 — a spread of 4 keV per nucleon, so 'iron peak' is a region, not a point, and which nuclide wins depends on the quantity you rank. Rank instead by atomic mass per nucleon and the order inverts: Fe-56 gives 0.9988382 u, Ni-62 gives 0.9988443 u, Fe-58 gives 0.9988496 u. Fe-56 is the least massive per nucleon even though it is not the most bound per nucleon, because converting a neutron into a proton plus an electron is worth mₙ − m(¹H) = 0.782 MeV, and Fe-56 carries a higher proton fraction (Z/A = 0.464) than Ni-62 (0.452). Silicon burning reaches nuclear statistical equilibrium, which minimises mass per nucleon at the core's electron fraction — so the ash sits in the Fe-56/Ni-56 region, not on the B/A champion.
A positive Q sets no rate
Symmetric splitting turns exothermic once Z²/A passes (2¹⁄³ − 1)/(1 − 2(−2/3)) × aS/aC = 17.6, which on the stability line is A ≈ 90. Zr-90 sits at Z²/A = 17.8 and is perfectly stable, because at that fissility the liquid-drop barrier is tens of MeV. The barrier only disappears when the surface and Coulomb responses to quadrupole deformation cancel, at Z²/A = 2aS/aC ≈ 50.1. U-236 sits between the two at 35.9, fissility x = 0.716, with a barrier near 5.8 MeV. Now the reactor physics falls out: capturing a neutron on U-235 releases Sₙ = 6.55 MeV, which clears that barrier, so thermal neutrons fission it; capturing on U-238 releases only 4.81 MeV against a U-239 barrier near 6.3 MeV, so U-238 needs fast neutrons. Everything past this point — WKB penetration, cross-sections, half-lives — belongs to the next unit.
Change one variable at a time
Make the relationship visible.
Set the split to 0.50: the drop model pays most for a symmetric break, yet U-236 really splits near 141/95, because shell structure at the saddle picks that, not energetics. Then drag A down toward 95 and watch Q cross zero.
Q RELEASED213.6 MeV
Q PER NUCLEON0.905 MeV
PARENT B/A7.592 MeV
FISSILITY x = (Z²/A)/50.10.705
Live interpretationQ RELEASED: 213.6 MeV. Q PER NUCLEON: 0.905 MeV. PARENT B/A: 7.592 MeV. FISSILITY x = (Z²/A)/50.1: 0.705
Catch the common trap
Explain before calculating.
U-236 has B/A = 7.586 MeV. It splits into Ba-144 (B/A = 8.265 MeV), Kr-90 (B/A = 8.591 MeV) and two free neutrons. What is the prompt Q?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe deuteron, the triton and the alpha particle have binding energies 2.2246, 8.4818 and 28.2957 MeV. For ²H + ³H → ⁴He + n, find Q, the release per nucleon, and how that energy divides between the two products.
- Two protons and three neutrons appear on each side, so Q is a pure binding difference, Q = ΣB(final) − ΣB(initial). The emitted neutron is unbound, B = 0, and contributes nothing to the final sum.
- ΣB(initial) = 2.2246 + 8.4818 = 10.7064 MeV; ΣB(final) = 28.2957 MeV.
- Q = 28.2957 − 10.7064 = 17.5893 MeV. Divided over the five nucleons that is 3.518 MeV per nucleon, nearly five times a fission's 0.734.
- The products leave back-to-back from a target at rest, so momentum conservation splits Q inversely with mass: Eₙ = Q mα/(mα + mₙ) = 17.589 × 4.0015/5.0102 = 14.05 MeV, leaving 3.54 MeV to the alpha.
AnswerQ = 17.589 MeV, or 3.518 MeV per nucleon. The neutron carries 14.05 MeV and the α 3.54 MeV — and only that charged 3.54 MeV stays in the plasma to heat it.
MediumFor n + U-235 → Ba-141 + Kr-92 + 3n, the mass excesses in MeV are Δ(n) = 8.071, Δ(U-235) = +40.921, Δ(Ba-141) = −79.733 and Δ(Kr-92) = −68.785. Find Q from the mass excesses, then confirm it through binding energies using Δ(¹H) = 7.289.
- Mass number is 236 on both sides, so the A u terms cancel and Q is a difference of mass excesses: Q = ΣΔ(initial) − ΣΔ(final).
- ΣΔ(initial) = 8.071 + 40.921 = 48.992 MeV. ΣΔ(final) = −79.733 − 68.785 + 3(8.071) = −148.518 + 24.213 = −124.305 MeV.
- Q = 48.992 − (−124.305) = 173.30 MeV, which is 173.30/236 = 0.734 MeV per nucleon.
- Cross-check with B = ZΔ(¹H) + NΔ(n) − Δ: B(U-235) = 92(7.289) + 143(8.071) − 40.921 = 1783.82 MeV, B(Ba-141) = 1173.95 MeV, B(Kr-92) = 783.17 MeV.
- Q = 1173.95 + 783.17 − 1783.82 = 173.30 MeV. The four spare neutrons carry B = 0 and cancel identically between the two routes.
AnswerQ = 173.30 MeV prompt, or 0.734 MeV per nucleon. The quoted ~200 MeV per fission is the same subtraction taken to the stable end-points Pr-141 and Zr-92, which gives 199.25 MeV; about 9 MeV of that escapes as antineutrinos.
HardYour SVD fit of the mass formula returned aV = 15.75, aS = 17.80, aC = 0.711 and aA = 23.70 MeV. For U-236 (Z = 92), predict Q for symmetric fission, find the fissility x, and state the two Z²/A thresholds the model contains.
- Split A into equal halves at fixed Z/A. The volume term goes as A and cancels; so does the asymmetry term, since aA[(A−2Z)/2]²/(A/2) doubled is aA(A−2Z)²/A again. Only surface and Coulomb survive.
- Surface costs binding: 2aS(A/2)²⁄³ − aS A²⁄³ = (2¹⁄³ − 1) aS A²⁄³ = 0.2599 aS A²⁄³. Coulomb returns binding: aC Z²/A¹⁄³ − 2aC(Z/2)²/(A/2)¹⁄³ = (1 − 2(−2/3)) aC Z²/A¹⁄³ = 0.3700 aC Z²/A¹⁄³.
- With A¹⁄³ = 6.180 and A²⁄³ = 38.19: Coulomb gain = 0.3700 × 0.711 × 8464/6.180 = 360.4 MeV, surface cost = 0.2599 × 17.80 × 38.19 = 176.7 MeV, so Qsym = 360.4 − 176.7 = 183.7 MeV.
- Qsym vanishes when Z²/A = (0.2599/0.3700)(aS/aC) = 0.7024 × 25.04 = 17.6, which the stability line reaches near A = 90.
- The barrier vanishes where the second-order surface and Coulomb responses cancel, at Z²/A = 2aS/aC = 50.1. U-236 has Z²/A = 8464/236 = 35.9, so x = 35.9/50.1 = 0.716.
AnswerQsym ≈ 184 MeV and x = 0.716. Splitting is exothermic above Z²/A ≈ 17.6 and barrierless above 50.1; U-236 sits between them, which is why it needs the 6.55 MeV a captured neutron brings to clear its 5.8 MeV barrier.