Fusion — solar fusion, tokamak (concept): Worked examples

This page is a drill deck for the parent Fusion note. We will not re-derive theory here — instead we hunt down every kind of number a fusion problem can hand you, and solve one of each. If a symbol feels unfamiliar, jump back to the parent, or to Mass-Energy Equivalence (E=mc^2) and the Binding Energy per Nucleon Curve.

The scenario matrix

#	Cell (case class)	What makes it distinct	Worked in
A	Standard exothermic fusion	$Δ m > 0$ , energy released	Ex. 1
B	Sign flip — endothermic	$Δ m < 0$ , energy must be supplied	Ex. 2
C	Degenerate / zero case	$Δ m = 0$ , no net energy	Ex. 3
D	Using binding energies instead of masses	$Q$ from $B / A$ curve, not mass tables	Ex. 4
E	Per-nucleon / scaling	energy per kilogram of fuel, "how much coal?"	Ex. 5
F	Limiting behaviour	Coulomb barrier vs. temperature — order-of-magnitude	Ex. 6
G	Confinement (Lawson)	triple product $n T τ_{E}$ , does the plasma ignite?	Ex. 7
H	Real-world word problem	Sun's lifetime from mass budget	Ex. 8
I	Exam-style twist	fraction of mass converted; unit trap	Ex. 9

The three tools we lean on:

The sign of $Q$ is the whole story: $Q > 0$ ⇒ energy released (fusion works); $Q < 0$ ⇒ energy absorbed (won't happen spontaneously).

Alt-text: A horizontal number line for the Q-value in MeV. The origin is marked "Q = 0 (degenerate, Cell C)". The right half is shaded green and labelled "energy RELEASED (fusion works)" with a green dot at +17.6 MeV tagged Cells A/D/E. The left half is shaded red and labelled "energy ABSORBED" with a red dot near −0.09 MeV tagged Cell B.

Example 1 — Standard exothermic fusion (Cell A)

Worked example D–T reaction

$Q$ -value $_{1}^{2} D +_{1}^{3} T \to_{2}^{4} He + n$ . Masses (u): $m_{D} = 2.014102$ , $m_{T} = 3.016049$ , $m_{H e} = 4.002603$ , $m_{n} = 1.008665$ . Find $Q$ .

Forecast: Products are one big tightly-bound helium plus a neutron — reactants are two loose light nuclei. Guess: mass drops, so $Q$ should be a healthy positive number, tens of MeV.

Step 1 — Add the reactant masses. $2.014102 + 3.016049 = 5.030151$ u. Why this step? Rest mass stores rest energy; the total "energy account" of the inputs lives in this sum.

Step 2 — Add the product masses. $4.002603 + 1.008665 = 5.011268$ u. Why? Same accounting for what comes out.

Step 3 — Mass defect. $Δ m = 5.030151 - 5.011268 = 0.018883$ u. Why subtract this way? The missing mass is what turned into kinetic energy of the products.

Step 4 — Convert to energy. $Q = 0.018883 \times 931.5 = 17.59 MeV$ . Why $\times 931.5$ ? That is $1 u \cdot c^{2}$ expressed in MeV — it does the $\times c^{2}$ for us.

Verify: $Q > 0$ ✓ (matches forecast). Units: (u) $\times$ (MeV/u) = MeV ✓. This is the famous $\approx 17.6$ MeV of the ITER fuel.

Example 2 — Sign flip: an endothermic "fusion" (Cell B)

Worked example When joining

costs energy Consider $_{2}^{4} He +_{2}^{4} He \to_{4}^{8} Be$ . Masses (u): $m_{H e} = 4.002603$ , $m_{B e - 8} = 8.005305$ . Find $Q$ and interpret its sign.

Forecast: Beryllium-8 is famously unstable — sitting just above two heliums in energy. Guess: mass might actually rise, giving $Q < 0$ . If so, this reaction does not release energy.

Step 1 — Reactant mass. $4.002603 \times 2 = 8.005206$ u. Why? Two identical helium nuclei.

Step 2 — Product mass. $8.005305$ u (given). Why? Single beryllium-8 nucleus.

Step 3 — Mass defect. $Δ m = 8.005206 - 8.005305 = - 0.000099$ u. Why keep the negative sign? A negative defect means the product is heavier — mass had to be added, so energy was absorbed.

Step 4 — Convert. $Q = - 0.000099 \times 931.5 = - 0.092 MeV \approx - 92 keV$ .

Verify: $Q < 0$ ✓. Interpretation: $^{8} Be$ flies apart back into two heliums almost instantly. This is exactly why the Sun cannot fuse helium two-at-a-time — it needs the triple-alpha trick. Units: keV ✓, and $∣ Q ∣$ small ✓ (barely bound).

Example 3 — The degenerate zero case (Cell C)

Worked example What if

$Δ m = 0$ ? Imagine a hypothetical rearrangement where the total product mass exactly equals the total reactant mass, e.g. $m_{in} = 6.000000$ u, $m_{out} = 6.000000$ u. Find $Q$ .

Forecast: No mass changes hands ⇒ nothing converts ⇒ $Q$ should be exactly zero. This is the boundary between fusion and non-reaction.

Step 1 — Mass defect. $Δ m = 6.000000 - 6.000000 = 0$ u. Why note this? It sits us exactly at the origin of the number line in the figure above.

Step 2 — Convert. $Q = 0 \times 931.5 = 0 MeV$ . Why it matters: No energy is released or absorbed. On the Binding Energy per Nucleon Curve this is the flat top near iron — combining nuclei there gains you nothing. That is precisely why fusion stops at iron: past it, $Δ m$ goes negative (fission territory).

Verify: $Q = 0$ ✓. This degenerate point is the physical "peak" that separates cells A ( $Q > 0$ , light nuclei) from cell B ( $Q < 0$ , above iron).

Example 4 — $Q$ from binding energies, not masses (Cell D)

Worked example Same answer, different toolbox

For $^{2} D +^{3} T \to^{4} He + n$ , use binding energies instead of masses: $B (D) = 2.224$ MeV, $B (T) = 8.482$ MeV, $B (^{4} He) = 28.296$ MeV, $B (n) = 0$ . Find $Q$ .

Forecast: Binding energy is "energy already released when the nucleus formed." The product is much more bound, so $Q =$ (product binding) $-$ (reactant binding) should land near the same $\approx 17.6$ MeV as Example 1.

Step 1 — Total binding of products. $28.296 + 0 = 28.296$ MeV. Why? A free neutron has no partners, so $B (n) = 0$ .

Step 2 — Total binding of reactants. $2.224 + 8.482 = 10.706$ MeV. Why? Sum the binding already stored in D and T.

Step 3 — Subtract. $Q = 28.296 - 10.706 = 17.59 MeV$ . Why products minus reactants here (opposite order to the mass formula)? More binding = less mass. Extra binding in the product is exactly the energy set free.

Verify: $17.59$ MeV matches Example 1 to two decimals ✓. Two different tables, one physics.

Example 5 — Energy per kilogram of fuel (Cell E)

Worked example How much energy in 1 gram of D–T?

Each D–T fusion gives $17.6$ MeV. One fusion consumes $2.014 + 3.016 = 5.030$ u of fuel. How much energy from 1 gram of D–T mixture? ( $1 u = 1.6605 \times 1 0^{- 27}$ kg, $1 MeV = 1.602 \times 1 0^{- 13}$ J.)

Forecast: A gram is a huge number of atoms, and each gives a whole MeV-scale burst. Guess: hundreds of gigajoules — enough to run a house for years.

Step 1 — Mass per reaction in kg. $5.030 \times 1.6605 \times 1 0^{- 27} = 8.352 \times 1 0^{- 27}$ kg. Why? We need "how many reactions fit in 1 g", so first get the mass one reaction uses.

Step 2 — Number of reactions in 1 g. $N = \frac{1 0 ^{- 3}}{8.352 \times 1 0 ^{- 27}} = 1.197 \times 1 0^{23}$ . Why divide? Total mass ÷ mass-per-reaction = number of reactions.

Step 3 — Energy per reaction in joules. $17.6 \times 1.602 \times 1 0^{- 13} = 2.820 \times 1 0^{- 12}$ J. Why convert? We want a real-world energy in joules, not MeV.

Step 4 — Total energy. $E = N \times (2.820 \times 1 0^{- 12}) = 1.197 \times 1 0^{23} \times 2.820 \times 1 0^{- 12} \approx 3.4 \times 1 0^{11}$ J.

Verify: $\approx 3.4 \times 1 0^{11}$ J = 340 GJ from one gram ✓ (forecast: hundreds of GJ). For scale, burning 1 g of coal gives only about $3 \times 1 0^{4}$ J (≈30 kJ) — so fusion is roughly $1 0^{7}$ × denser in energy per gram. Units: (count) $\times$ (J) = J ✓.

Example 6 — Limiting behaviour: barrier vs. temperature (Cell F)

Worked example Why

$1 0^{7}$ K is "cold" — an order-of-magnitude check Two protons must approach to $r \approx 3 \times 1 0^{- 15}$ m for the strong force to grab them. The Coulomb (electric) potential energy at that separation is $U = \frac{k e ^{2}}{r}$ with $k = 8.99 \times 1 0^{9}$ , $e = 1.602 \times 1 0^{- 19}$ C. Compare $U$ to the thermal energy $\sim k_{B} T$ at $T = 1.5 \times 1 0^{7}$ K ( $k_{B} = 1.381 \times 1 0^{- 23}$ J/K).

Forecast: If $U ≫ k_{B} T$ , protons classically can't reach — the Sun burns only via Quantum Tunnelling. Guess: $U$ beats thermal energy by a factor of hundreds.

Step 1 — Coulomb barrier height. $U = \frac{( 8.99 \times 1 0 ^{9} ) ( 1.602 \times 1 0 ^{- 19} ) ^{2}}{3 \times 1 0 ^{- 15}}$ . Numerator: $8.99 \times 1 0^{9} \times 2.566 \times 1 0^{- 38} = 2.307 \times 1 0^{- 28}$ . Divide by $3 \times 1 0^{- 15}$ : $U = 7.69 \times 1 0^{- 14}$ J. Why this formula? Two like charges repel; their stored electric energy is the "wall" they must climb.

Step 2 — Thermal energy. $k_{B} T = 1.381 \times 1 0^{- 23} \times 1.5 \times 1 0^{7} = 2.07 \times 1 0^{- 16}$ J. Why $k_{B} T$ ? It sets the typical kinetic energy of a particle at temperature $T$ .

Step 3 — Take the ratio. $\frac{U}{k _{B} T} = \frac{7.69 \times 1 0 ^{- 14}}{2.07 \times 1 0 ^{- 16}} \approx 371$ . Why a ratio? It tells us how many times short a typical proton falls of the barrier.

Verify: ratio $\approx 370 ≫ 1$ ✓ (forecast: hundreds). So a typical solar proton has ~1/370 of the energy needed — classically forbidden, permitted only by tunnelling. This is the limiting case that explains the Sun's slowness.

Example 7 — Confinement: does the plasma ignite? (Cell G)

Worked example Lawson triple product

A tokamak run reaches density $n = 1.0 \times 1 0^{20} m^{- 3}$ , temperature $T = 15$ keV, confinement time $τ_{E} = 2.0$ s. The D–T ignition threshold is $n T τ_{E} ≳ 3 \times 1 0^{21} keV\cdots\cdotm^{- 3}$ . Does it ignite?

Forecast: Multiply the three; if the product clears $3 \times 1 0^{21}$ , it ignites. These are near-ITER numbers, so guess: just about reaching threshold.

Step 1 — Multiply the triple product. $n T τ_{E} = (1.0 \times 1 0^{20}) (15) (2.0)$ . Why multiply, not add? Losses fall if you are denser or hotter or leakier-for-longer — the requirements compound, so the criterion is a product (see parent note).

Step 2 — Compute the value. $1.0 \times 1 0^{20} \times 15 \times 2.0 = 1.0 \times 1 0^{20} \times 30 = 3.0 \times 1 0^{21} keV\cdots\cdotm^{- 3}$ . Why this arithmetic? We collapse the three factors into one number so we can compare it directly against the single threshold value — multiply the two small factors ( $15 \times 2.0 = 30$ ) first, then attach the power of ten.

Step 3 — Compare. $3.0 \times 1 0^{21} \geq 3 \times 1 0^{21}$ ⇒ just ignites (marginally). Why "marginal" matters: any drop in $τ_{E}$ (worse confinement) and it fizzles — which is why building bigger machines (longer $τ_{E}$ ) is the whole game.

Verify: product $= 3.0 \times 1 0^{21}$ , exactly at threshold ✓. Units: $m^{- 3} \cdot keV \cdot s = keV\cdots\cdotm^{- 3}$ ✓ — matches the criterion's units.

Example 8 — Real-world word problem: the Sun's lifetime (Cell H)

Worked example How long can the Sun shine?

The Sun ( $M = 2.0 \times 1 0^{30}$ kg) fuses hydrogen in its core. Only about 10% of its mass is core hydrogen available for fusion, and fusion converts about $0.7%$ of that hydrogen's mass into energy. It radiates at $L = 3.8 \times 1 0^{26}$ W. Estimate its main-sequence lifetime in years.

Forecast: Only a fraction of a fraction of the mass becomes energy, but $c^{2}$ is enormous. Guess: about ten billion years (the textbook figure).

Step 1 — Fusible hydrogen mass. $M_{H} = 0.10 \times 2.0 \times 1 0^{30} = 2.0 \times 1 0^{29}$ kg. Why 10%? Only the hot dense core fuses; the outer layers are too cool.

Step 2 — Mass actually converted to energy. $Δ m = 0.007 \times 2.0 \times 1 0^{29} = 1.4 \times 1 0^{27}$ kg. Why $0.7%$ ? That is the mass-defect fraction of the p–p chain (4 protons → helium loses ~0.7% of their mass).

Step 3 — Total energy available. $E = Δ m c^{2} = 1.4 \times 1 0^{27} \times (3.0 \times 1 0^{8})^{2} = 1.26 \times 1 0^{44}$ J. Why $E = m c^{2}$ ? Converts the vanished mass into radiated energy — see Mass-Energy Equivalence (E=mc^2).

Step 4 — Lifetime = energy ÷ power. $t = \frac{1.26 \times 1 0 ^{44}}{3.8 \times 1 0 ^{26}} = 3.3 \times 1 0^{17}$ s. Why divide? Power is energy per second; total energy ÷ rate = time.

Step 5 — Convert to years. $t = \frac{3.3 \times 1 0 ^{17}}{3.156 \times 1 0 ^{7}} \approx 1.05 \times 1 0^{10}$ yr. Why divide by $3.156 \times 1 0^{7}$ ? There are about $3.156 \times 1 0^{7}$ seconds in one year ( $365.25 \times 24 \times 3600$ ), so dividing the total seconds by "seconds per year" leaves a count of years.

Verify: $\approx 1.0 \times 1 0^{10}$ yr = 10 billion years ✓ (forecast met — the standard solar lifetime). Units: J ÷ W = s ✓, s ÷ (s/yr) = yr ✓.

Example 9 — Exam-style twist: fraction of mass converted (Cell I)

Worked example The unit-trap question

In the D–T reaction, what fraction of the initial rest mass is converted to energy? (Use $Δ m = 0.018883$ u from Example 1 and reactant mass $5.030151$ u.) Then confirm this fraction times $c^{2}$ reproduces the energy per unit mass.

Forecast: Fusion is famously more efficient than fission (~0.1%) but still tiny in absolute terms. Guess: a few tenths of a percent.

Step 1 — Fraction converted. $f = \frac{Δ m}{m _{reactants}} = \frac{0.018883}{5.030151} = 3.754 \times 1 0^{- 3}$ . Why divide by reactant mass? "Fraction converted" = mass lost ÷ mass you started with.

Step 2 — As a percentage. $f = 3.754 \times 1 0^{- 3} \times 100% = 0.375%$ . Why note this? It is ~4× the fraction in nuclear fission (~0.09%) — a key reason fusion is prized (compare Nuclear Fission).

Step 3 — Cross-check via energy per kg. Energy released per kg of D–T fuel should be $f c^{2} = 3.754 \times 1 0^{- 3} \times (3.0 \times 1 0^{8})^{2} = 3.38 \times 1 0^{14}$ J/kg. Why? Since $E = (f m) c^{2}$ , dividing by $m$ gives $E / m = f c^{2}$ — the fraction only becomes an energy density after multiplying by $c^{2}$ .

Step 4 — Compare with Example 5. Example 5 gave $3.4 \times 1 0^{11}$ J per gram, which is $3.4 \times 1 0^{14}$ J per kg. This matches $f c^{2} = 3.38 \times 1 0^{14}$ J/kg. ✓

Verify: $f = 0.375%$ ✓; and $f c^{2} = 3.38 \times 1 0^{14}$ J/kg agrees with Example 5's per-kg value ✓ — the twist and the direct method meet. Trap dodged: the "fraction" $f$ is dimensionless; only after $\times c^{2}$ does it carry the units of an energy density (J/kg).

Recall Self-test the matrix

Which cell has $Q < 0$ , and what does it physically mean? ::: Cell B — energy is absorbed; the reaction won't run spontaneously (e.g. $^{4} He +^{4} He \to^{8} Be$ ). Two ways to compute $Q$ ? ::: From masses (reactants − products) $c^{2}$ , or from binding energies (products − reactants). Why is the Lawson condition a product of three quantities? ::: Because losses shrink if you raise density, temperature, or confinement time — the requirements compound, not add. Roughly what fraction of D–T mass becomes energy? ::: About $0.375%$ — a few times more than fission.