Fusion — D-T reaction, solar fusion (p-p chain), tokamak - ICF: Visual walkthrough

This page rebuilds the central result of the parent topic — the 17.6 MeV released when deuterium and tritium fuse — from absolute zero. No symbol is used before it is drawn. Follow the pictures; the equations only label what the pictures already show.

Step 1 — Name the players (what a nucleus is)

WHAT. A nucleus is a clump of two kinds of particles:

a proton — a positively charged ball, symbol $p$ ,
a neutron — an uncharged ball of almost the same weight, symbol $n$ .

We label any nucleus as $_{Z}^{A} X$ , where:

$_{Z}^{A} X$

$Z$ (bottom) = how many protons — this decides which element it is,
$A$ (top) = protons + neutrons together (the "mass number", i.e. how many balls in total),
so the number of neutrons is just $A - Z$ .

WHY these two numbers. We only ever need "how many balls, and how many are charged" — because charge causes the pushing-apart, and total ball-count sets the weight.

PICTURE. Deuterium $_{1}^{2} H$ = 1 proton + 1 neutron. Tritium $_{1}^{3} H$ = 1 proton + 2 neutrons. Both are heavy cousins of hydrogen ( $Z = 1$ ).

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

Step 2 — The reaction: what sticks to what

WHAT. We slam deuterium and tritium together. They merge and then spit out two products:

$deuterium_{1}^{2} H + tritium_{1}^{3} H \to helium-4_{2}^{4} He + a free neutron_{0}^{1} n$

Read every symbol:

Left side has $2 + 3 = 5$ balls total. Right side has $4 + 1 = 5$ balls. Balls are conserved.
Left side charge: $1 + 1 = 2$ protons. Right side: helium has $2$ , the neutron has $0$ , total $2$ . Charge is conserved.

WHY this reaction and not another. Of all the light-nucleus pairs, D–T is the easiest to ignite at temperatures we can reach — but that is a separate story (Quantum Tunnelling, Coulomb barrier). Here we only ask: if it happens, how much energy comes out?

PICTURE. Two small clumps combine; a tight helium clump and a lone neutron fly apart.

Step 3 — Weigh the reactants

WHAT. We add up the masses of the two things going in. Mass here is measured in atomic mass units ( $u$ ):

$m_{before} = m (^{2} H) + m (^{3} H) = 2.014102 u + 3.016049 u = 5.030151 u$

$m (^{2} H) = 2.014102 u$ — the weight of one deuterium,
$m (^{3} H) = 3.016049 u$ — the weight of one tritium,
their sum is the total mass on the shelf before anything happens.

WHY this step. The whole trick of fusion energy is comparing "weight before" with "weight after". So first we need the before-weight.

PICTURE. Two mass-blocks stacked on the left pan of a balance.

Step 4 — Weigh the products

WHAT. Now the masses of the two things coming out:

$m_{after} = m (^{4} He) + m (n) = 4.002602 u + 1.008665 u = 5.011267 u$

$m (^{4} He) = 4.002602 u$ — helium-4,
$m (n) = 1.008665 u$ — the free neutron,
their sum is the total mass after the reaction.

WHY this step. This is the other pan of the balance. Notice already: $5.011267 < 5.030151$ . The products weigh less than the reactants.

PICTURE. The right pan sits higher — it is lighter. Something vanished.

Step 5 — The mass defect: what went missing

WHAT. Subtract after from before:

$Δ m = m_{before} - m_{after} = 5.030151 u - 5.011267 u = 0.018884 u$

$Δ m$ (read "delta-m") means change in mass — here, the mass that disappeared,
it is small: about $0.4%$ of one nucleon.

WHY this happens. The helium nucleus is bound more tightly than the loose D and T were. Being more tightly bound means the parts have "fallen into a deeper well" and lost some binding energy — which came out of their rest mass. This is exactly the rise of the Binding Energy per Nucleon Curve: moving toward the peak, the product is more bound, so it weighs less.

PICTURE. A sliver carved off the mass, labelled " $Δ m$ — becomes energy".

Step 6 — Turn missing mass into energy

WHAT. The lost mass reappears as energy, using Einstein's rule $E = m c^{2}$ . The conversion factor is packaged neatly: $1 u$ of mass is worth $931.5$ MeV of energy.

$Q = Δ m \times 931.5 \frac{MeV}{u} = 0.018884 u \times 931.5 \frac{MeV}{u} \approx 17.6 MeV$

$Q$ = the energy released by one reaction (the "Q-value"),
$Δ m$ = the missing mass from Step 5,
$931.5 MeV / u$ = how much energy one $u$ is worth — the $c^{2}$ already baked in,
$Q > 0$ tells us energy comes out (exothermic). If $Q$ were negative we would have to pay energy in.

WHY $E = m c^{2}$ and not some other rule. We need a bridge from "kilograms of missing mass" to "joules of released energy". $E = m c^{2}$ is exactly that bridge — mass and energy are the same currency, and $c^{2}$ is the exchange rate.

PICTURE. The sliver of mass flows through an " $\times 931.5$ " gate and comes out as a burst labelled $17.6$ MeV.

WHAT. The $17.6$ MeV does not split evenly. The two products fly apart back-to-back with equal and opposite momentum. Since the D and T started nearly at rest, total momentum was zero, so it must stay zero:

$p_{α} = p_{n} (equal in size, opposite in direction)$

$p_{α}$ = momentum of the helium nucleus (often called an $α$ -particle),
$p_{n}$ = momentum of the neutron.

Kinetic energy relates to momentum by $E = \frac{p ^{2}}{2 m}$ , so for the same $p$ :

$E \propto \frac{1}{m} \Rightarrow \frac{E _{n}}{E _{α}} = \frac{m _{α}}{m _{n}} \approx \frac{4}{1}$

the lighter particle (neutron, mass ≈ 1) gets the bigger share of energy,
the heavier particle (helium, mass ≈ 4) gets the smaller share.

WHY use momentum here. Energy alone can't tell us who gets what — two unknowns need two equations. Momentum conservation is the second equation, and it fixes the split.

Splitting $17.6$ MeV in the ratio $4 : 1$ :

$E_{n} \approx \frac{4}{5} \times 17.6 \approx 14.1 MeV, E_{α} \approx \frac{1}{5} \times 17.6 \approx 3.5 MeV$

PICTURE. Light neutron zooms far right with a long arrow ( $14.1$ MeV); heavy helium recoils left with a short arrow ( $3.5$ MeV); the two momentum arrows are the same length.

Step 8 — Edge & degenerate cases (never get surprised)

WHAT. Check the boundaries so no scenario ambushes you.

What if $Δ m < 0$ ? Then $Q < 0$ : the products would be heavier, energy must be supplied, no fusion power. This is what happens past iron on the Binding Energy per Nucleon Curve — fusing iron costs energy.
What if the nuclei start with kinetic energy? They must, to beat the Coulomb repulsion. But that input KE just adds to both sides equally; the net released $Q$ from mass defect is unchanged. $Q$ is a property of the masses, not of how hard you threw them.
What if we mis-share as $1 : 1$ ? Momentum would no longer be zero — forbidden. The $4 : 1$ split is the only one consistent with starting at rest.
Degenerate limit — equal masses. If both products had the same mass, the $4 : 1$ becomes $1 : 1$ and energy would split evenly. D–T is not this case, but it shows the rule is general.

WHY include this. The formula $Q = Δ m c^{2}$ and the $E \propto 1/ m$ split are universal — knowing where they'd give zero or flip tells you exactly when fusion stops paying.

PICTURE. A number line of $Δ m$ : right of zero → energy out (fusion works, green); at zero → break-even; left of zero → energy in (fusion fails, past the peak).

The one-picture summary

Everything on one board: weigh in, weigh out, the missing sliver becomes $17.6$ MeV, which splits $4 : 1$ into a fast neutron and a slow helium.

Recall Feynman: the whole walkthrough in plain words

We put two little hydrogen-cousins on a scale — deuterium (1 proton, 1 neutron) and tritium (1 proton, 2 neutrons). Together they weigh $5.030151$ tiny units. We let them snap together into helium plus one leftover neutron, and weigh that — only $5.011267$ units. A sliver of mass, $0.018884$ units, simply vanished. Einstein tells us mass is energy, at an exchange rate of $931.5$ MeV per unit, so the sliver reappears as $17.6$ MeV of motion energy — that's the payoff of fusion. Now, who gets the energy? The two pieces fly apart like a fired gun and its recoil: equal push both ways. The light neutron, being four times lighter, races off with four-fifths of the energy — about $14.1$ MeV — while the heavier helium recoils gently with the last $3.5$ MeV. That single missing sliver, and that unfair-looking split, are the entire physics of a D–T reactor.

Recall Quick self-test

Why do the products weigh less than the reactants? ::: Helium is more tightly bound; the lost binding energy is subtracted from the rest mass — the mass defect $Δ m$ . Convert $Δ m = 0.018884 u$ to energy. ::: $0.018884 \times 931.5 \approx 17.6$ MeV. Why does the neutron carry more energy than the helium? ::: Equal momentum (started at rest) with smaller mass means larger kinetic energy, since $E = p^{2} /2 m \Rightarrow E \propto 1/ m$ ; ratio $\approx 4 : 1$ . When would fusion cost energy instead? ::: When $Δ m < 0$ (products heavier), i.e. fusing nuclei beyond the iron peak of the $B / A$ curve.