Worked examples — Fusion — D-T reaction, solar fusion (p-p chain), tokamak - ICF

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The scenario matrix

Every fusion/nuclear-energy problem lives in one of these cells. The examples below are labelled by the cell they cover, so together they leave no gap.

Cell	Case class	What's tricky about it	Example
A	$Q>0$, two products (fusion)	standard mass-defect arithmetic	Ex 1
B	Energy split between two products	momentum conservation, not equal split	Ex 2
C	$Q<0$ (endothermic / "uphill")	negative answer means needs energy	Ex 3
D	$Q=0$ (degenerate, at the B/A peak)	fusing near iron gives ~nothing	Ex 4
E	Multi-step chain, net $Q$	intermediate species cancel; neutrinos leak	Ex 5
F	Coulomb barrier (limiting energy scale)	why we need millions of kelvin	Ex 6
G	Real-world word problem	reactor power ↔ reaction rate	Ex 7
H	Exam twist: $B/A$ given, not masses	back out $Q$ from binding energies	Ex 8

Conversion we will reuse constantly: $1u=931.5\text{MeV}/c^2,1\text{MeV}=1.602\times 10^{-13}\text{J}$

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Cell A — Standard exothermic fusion ($Q>0$)

Forecast: two hydrogens fusing move uphill on the B/A curve → energy out → I expect a small positive number, a few MeV (smaller than D–T's 17.6, because we're not making the super-stable ${}^4$He).

1. Sum the reactants. $2\times 2.014102=4.028204u$. Why this step? By the master formula, $Q$ depends only on total mass before vs after — reactants first.
2. Sum the products. $3.016029+1.008665=4.024694u$. Why this step? Same rule, the "after" side.
3. Mass defect. $\Delta m=4.028204-4.024694=0.003510u$. Why this step? The mass that vanished is what becomes energy — this is $E=mc^2$ in action.
4. Convert. $Q=0.003510\times 931.5\approx 3.27\text{MeV}$. Why this step? Turn the missing mass into MeV using the $u$↔MeV bridge.

Verify: $Q>0$ ✓ (fusion should release energy), and $3.27$ MeV $<$ $17.6$ MeV ✓ (D–D is a weaker step than D–T, matching the forecast). Units: $u\times (\text{MeV}/u)=\text{MeV}$ ✓.

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Cell B — How the energy splits between products

When a reaction starts with the reactants essentially at rest, the two products fly apart back-to-back with equal and opposite momentum. They do not share energy equally — the lighter one is faster and carries more kinetic energy.

The figure below draws exactly this picture: the reaction point at rest, a long cyan arrow for the light, fast neutron flying one way, and a short amber arrow for the heavy, slow ${}^3$He recoiling the other way. Equal-length momentum, unequal kinetic energy — look at how much longer the cyan velocity arrow is.

Figure s01 — Cell B: two products recoil back-to-back with equal momentum ($p_n=p_{\text{He}}$); the lighter neutron (cyan) carries the larger kinetic energy because $E=p^2/2m$.

Forecast: neutron is ~3× lighter, so it should grab ~3× the energy — roughly three-quarters of the total.

1. Set the ratio. $E_n:E_{\text{He}}=m_{\text{He}}:m_n\approx 3:1$. Why this step? Inverse-mass rule from the box above.
2. Total the parts. $3+1=4$ shares of $3.27$ MeV → each share $=3.27/4=0.8175$ MeV. Why this step? Divide the pie into $m_{\text{He}}+m_n$ equal slices, then hand them out.
3. Assign. $E_n=3\times 0.8175\approx 2.45$ MeV; $E_{\text{He}}=1\times 0.8175\approx 0.82$ MeV. Why this step? Give 3 slices to the lighter (faster) neutron, 1 to the heavier ${}^3$He.

Verify: $2.45+0.82=3.27$ MeV = total $Q$ ✓. Neutron got the larger share ✓. Same logic gives D–T's famous $14.1:3.5$ split (a $4:1$ ratio, since the $\alpha$ — the ${}^4$He nucleus — is 4× the neutron mass) — reassuring consistency.

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Cell C — Endothermic reaction ($Q<0$)

Forecast: ${}^8$Be is famously unstable — it falls apart almost instantly. That hints the product is not more bound, so I expect $Q\le 0$ (nature won't hand out free energy here).

1. Sum reactants. $2\times 4.002602=8.005204u$. Why this step? Two alpha particles (${}^4$He nuclei) combining — total their mass.
2. Product mass. $m{(}^8\text{Be})=8.005305u$. Why this step? The single product's mass is the "after" side.
3. Mass defect. $\Delta m=8.005204-8.005305=-0.000101u$. Why this step? Negative! Mass went up, so energy was absorbed, not released.
4. Convert. $Q=-0.000101\times 931.5\approx -0.094\text{MeV}=-94\text{keV}$. Why this step? Same bridge; the sign carries through.

Verify: $Q<0$ ✓, matching that ${}^8$Be is unbound relative to two alphas (it lives ~$10^{-16}$ s). This is exactly why stars need the triple-alpha trick — a third helium must arrive during that fleeting instant. See Stellar Nucleosynthesis. Units ✓.

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Cell D — Degenerate case: fusion at the peak ($Q\approx 0$)

Forecast: both sit near the peak (~8.5–8.8 MeV/nucleon), so the gain per nucleon is tiny; total should be a mere few MeV — vanishing compared to D–T, and essentially "worthless" as fuel.

1. Binding of reactants. $B_{\text{react}}=2\times (28\times 8.45)=2\times 236.6=473.2$ MeV. Why this step? Total binding $=A\times (B/A)$ for each nucleus. More binding = more stable.
2. Binding of product. $B_{\text{prod}}=56\times 8.64=483.84$ MeV. Why this step? Same, for the single heavy product.
3. $Q$ = gain in binding. $Q=B_{\text{prod}}-B_{\text{react}}=483.84-473.2=10.6$ MeV. Why this step? Energy released = how much more tightly bound the product is (binding released as KE).

Verify: $Q$ is positive but tiny per nucleon: $10.6/56\approx 0.19$ MeV/nucleon vs D–T's $17.6/5\approx 3.5$ MeV/nucleon — about 18× less useful. And just past ${}^{56}$Fe/Ni the peak flattens and turns over, so the next step would give $Q\to 0$ then negative. This is the "degenerate" cell: at the very peak, fusion yields essentially nothing. See Nuclear Fission — beyond the peak, it's splitting that pays.

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Cell E — Multi-step chain, net $Q$ (with leaking neutrinos)

Forecast: the parent note quotes ~26.7 MeV; I expect the mass-defect route to land there, with a couple percent then vanishing as neutrinos escape.

1. Reactant mass. $4\times 1.007825=4.031300u$. Why this step? Four hydrogen atoms in, total their atomic mass.
2. Product mass. $m{(}^4\text{He})=4.002602u$. Why this step? Using atomic helium mass already includes 2 electrons, cancelling the 2 positrons' annihilation — a neat bookkeeping shortcut.
3. Mass defect. $\Delta m=4.031300-4.002602=0.028698u$. Why this step? This total missing mass is the entire chain's energy budget.
4. Convert. $Q=0.028698\times 931.5\approx 26.7\text{MeV}$. Why this step? Bridge to MeV.
5. Subtract the leak. Neutrinos carry off ~2% (~0.5 MeV); usable sunlight ≈ $26.2$ MeV. Why this step? Neutrinos barely interact — that energy leaves the Sun instantly, never becoming heat.

Verify: $26.7$ MeV matches Worked Example 2 of the parent ✓. The slow first step (a proton must ${\beta }^{+}$-decay mid-collision, a rare weak-interaction event) is why the Sun burns for ~10 billion years. Units ✓.

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Cell F — The Coulomb barrier (limiting energy scale)

The figure below plots potential energy versus separation $r$: a cyan $1/r$ wall rising as the nuclei approach (Coulomb repulsion), then a sudden drop into a deep amber well once they touch at nuclear range (the strong force). The white dot marks the barrier top — the ~288 keV summit a nucleus must climb (or tunnel through) before it can fall into the well and fuse.

Figure s02 — Cell F: the Coulomb barrier (cyan, $\sim 1/r$) must be surmounted before the short-range strong-force well (amber) captures the two nuclei.

Forecast: the parent said "hundreds of keV." I expect a few hundred keV.

1. Write the barrier. $U_C=\frac{1}{4\pi {\epsilon }_0}\frac{Z_1{Z}_2{e}^2}{r_0}$. Why this step? Electrostatic potential energy of two point charges at separation $r_0$ — the wall's peak.
2. Plug numbers (SI, joules). $U_C=8.99\times 10^9\times \frac{(1)(1)(1.602\times 10^{-19}{)}^2}{5\times 10^{-15}}$ $=8.99\times 10^9\times \frac{2.566\times 10^{-38}}{5\times 10^{-15}}\approx 4.61\times 10^{-14}\text{J}$. Why this step? Do it in SI to avoid unit traps, convert at the end.
3. Convert to keV. $\frac{4.61\times 10^{-14}}{1.602\times 10^{-16}}\approx 288\text{keV}$. Why this step? $1\text{keV}=1.602\times 10^{-16}$ J; report in nuclear units.

Verify: $\approx 288$ keV — squarely in the "hundreds of keV" range the parent claimed ✓. A gas at temperature $T$ has typical thermal energy $\sim k_{B}T$ (with the Boltzmann constant $k_B=1.381\times 10^{-23}$ J/K); to reach $\sim$keV energies you'd naively need $T\sim 10^9$ K, yet reactors run at "only" $\sim 10^8$ K. The gap is bridged by the high-energy tail plus quantum tunnelling through the wall. Units ✓.

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Cell G — Real-world word problem (power ↔ reaction rate)

Forecast: MeV is a minuscule joule amount, so we'll need a colossal rate — but the fuel mass will be tiny because nuclei are so light. That's fusion's headline: little fuel, huge power.

1. Energy per reaction in joules. $17.6\text{MeV}=17.6\times 10^6\times 1.602\times 10^{-19}=2.82\times 10^{-12}$ J. Why this step? Power is in J/s, so convert the per-reaction energy to J.
2. Reactions per second. $\text{rate}=\frac{10^9\text{J/s}}{2.82\times 10^{-12}\text{J}}\approx 3.55\times 10^{20}{\text{s}}^{-1}$. Why this step? Power = (energy per event) × (events per second) ⇒ divide.
3. Deuterons per day. One D per reaction: $3.55\times 10^{20}\times 86400\text{s}\approx 3.07\times 10^{25}$ deuterons/day. Why this step? Multiply the per-second rate by seconds in a day ($86400$).
4. Mass of deuterium. $3.07\times 10^{25}\times (2.014\times 1.66\times 10^{-27}\text{kg})\approx 1.03\times 10^{-1}$ kg. Why this step? Number of atoms × mass per atom = total mass.

Verify: $\approx 0.10$ kg of deuterium per day for a gigawatt — about 100 grams, roughly the mass of an apple, powering a city ✓. This is the "little fuel, huge power" punchline, and it's why confinement research is worth billions. Units: J ÷ (J/reaction) = reactions ✓.

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Cell H — Exam twist: $Q$ from $B/A$ values, not masses

Forecast: should reproduce ~17.6 MeV — that's the consistency check the examiner is fishing for.

1. Reactant binding. $B_D=2\times 1.112=2.224$ MeV; $B_T=3\times 2.827=8.481$ MeV; total $=10.705$ MeV. Why this step? $B=A\times (B/A)$; sum both reactant nuclei.
2. Product binding. $B_{\text{He}}=4\times 7.074=28.296$ MeV; the free neutron has $B=0$; total $=28.296$ MeV. Why this step? Same rule; the loose neutron adds nothing (nothing binds it).
3. $Q$ = gain in binding. $Q=B_{\text{prod}}-B_{\text{react}}=28.296-10.705=17.591\approx 17.6\text{MeV}$. Why this step? Energy released equals how much more bound the products are — that binding surplus emerges as kinetic energy of the $\alpha$ and neutron.

Verify: $17.59$ MeV matches the mass-defect $17.6$ MeV ✓ — two independent routes to the same number, the hallmark that you understand why $Q$ works, not just the formula. Units: MeV throughout ✓.

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Recall Self-test the whole matrix

Which cell has $Q<0$, and what does that physically mean? ::: Cell C — the products are less bound (e.g. ${}^8$Be from two alphas); the reaction absorbs energy instead of releasing it. In an at-rest reaction, which product carries more kinetic energy? ::: The lighter one — energy splits inversely to mass, since $p_1=p_2$ and $E=p^2/2m$. Two ways to compute $Q$? ::: (1) mass defect $\times c^2$; (2) (binding of products) − (binding of reactants). They must agree. Why must we heat plasma to $\sim 10^8$ K even though $Q>0$? ::: The ~288 keV Coulomb barrier repels the nuclei; only the tail particles + tunnelling make it through.