Nuclear Fusion & Stellar Energy

Fusion is the power source of the universe. Every star, including the Sun, runs on it; the carbon in your bones, the oxygen you breathe, and the iron in your blood were all forged by fusion in stellar cores and scattered by exploding stars — this is stellar nucleosynthesis, and it means you are quite literally made of star-stuff. On Earth, controlled fusion is one of the great engineering quests: reactors such as the ITER project and the world's tokamaks aim to hold a deuterium–tritium plasma at over a hundred million kelvin, hotter than the Sun's core, long enough to release more energy than it takes to heat it. The prize is enormous — fusion fuel is effectively limitless (deuterium comes from seawater), it makes no long-lived radioactive waste and cannot melt down or run away, and it emits no carbon. In 2022 the National Ignition Facility, using lasers rather than magnets, became the first experiment to get more fusion energy out than the laser delivered to the fuel, crossing the long-sought ignition threshold. Fusion has a darker sibling too — the thermonuclear (hydrogen) bomb uses uncontrolled fusion for destruction. Whether or not fusion powers our grid this century, it already powers everything else.

Picture the previous article's binding-energy curve: binding energy per nucleon, $BE/A$, plotted against mass number $A$. It rises steeply from hydrogen, climbs through helium and carbon and oxygen, reaches a broad summit near iron-56 at about $8.8\ \text{MeV}$ per nucleon, and then falls gently away to uranium at about $7.6\ \text{MeV}$ per nucleon. The higher a nucleus sits on this curve, the more tightly each of its nucleons is bound, and the lower its energy — a nucleus near the peak is the nuclear equivalent of a ball at the bottom of a valley. Energy is released by any process that moves nucleons up this curve, toward the peak. There are two ways to do it. Split a heavy nucleus on the right-hand slope into two middleweight fragments closer to the peak — that is nuclear fission. Or join two light nuclei on the steep left-hand slope into a single heavier one further up — that is nuclear fusion. The left slope is far steeper than the right, which is the deep reason fusion, nucleon for nucleon, releases several times more energy than fission.

Two facts about this picture matter for everything below. First, the left slope is steep: going from two hydrogen isotopes at barely $1$–$3\ \text{MeV}$ per nucleon up to helium-4 at $7.07\ \text{MeV}$ per nucleon is an enormous climb, so light-nucleus fusion is exceptionally energetic per kilogram of fuel. Second, the curve turns over at iron. Nuclei up to iron-56 release energy when they fuse; past iron, fusing would move nucleons down the far slope and would cost energy. Iron is the ash at the end of fusion's road — a star cannot power itself by fusing iron, and that single fact ultimately dictates how massive stars die.

The bookkeeping is the same conservation-of-energy statement used for every nuclear reaction. Weigh the reactants, weigh the products, and any mass that has gone missing has been converted to energy through $E = mc^2$ (quoted here from relativity, not derived — the mass–energy article owns that). The energy released is called the Q-value:

where: the sums are over the rest masses of the nuclei going in and coming out, and $c = 3.00\times10^{8}\ \text{m s}^{-1}$. A positive $Q$ means the products are lighter than the reactants, mass has been liberated as kinetic energy of the products (and any photons or neutrinos), and the reaction is exothermic — it releases energy. Reading it physically: the fused nucleus is more tightly bound, so its nucleons have sunk to a lower energy, and the energy they shed shows up as motion. In practice we use the convenient unit $1\ \text{u} = 931.5\ \text{MeV}/c^2$, so a mass defect measured in atomic mass units converts to megaelectronvolts just by multiplying by $931.5$. A crucial bookkeeping convenience: use neutral-atom masses, not bare-nucleus masses, whenever the number of electrons balances across the reaction. The tabulated atomic mass of an atom already includes its $Z$ electrons; if the same number of electrons appears on both sides of the equation, those electron masses cancel exactly and you may use atomic masses directly. This is why physics tables list atomic masses — they are built for exactly this cancellation.

Fusion's energy is free for the taking — but only after you pay an entrance fee. Both nuclei carry positive charge, so as they approach they repel through the electrostatic (Coulomb) force, which grows as they get closer. The height of this repulsive wall, at the separation where the nuclei just touch, is

where: $Z_1, Z_2$ are the proton numbers of the two nuclei, $e = 1.60\times10^{-19}\ \text{C}$ is the elementary charge, $r$ is their centre-to-centre separation at contact, and the combination $ke^2 = 1.44\ \text{MeV}\,\cdot\,\text{fm}$ is a handy nuclear-scale constant. For deuterium and tritium ($Z_1 = Z_2 = 1$) touching at about $r \approx 3.24\ \text{fm}$, this barrier is roughly $0.44\ \text{MeV}$ — a few hundred kiloelectronvolts. The strong nuclear force, which does the actual binding, only switches on once the nuclei are inside this wall, at femtometre range; the whole game is getting them there. Now compare with heat. In a gas at temperature $T$, a typical particle carries thermal energy of order $k_{\text{B}}T$, where $k_{\text{B}} = 1.38\times10^{-23}\ \text{J K}^{-1}$ is Boltzmann's constant. At ten million kelvin — the temperature of a stellar core — this is only about $0.86\ \text{keV}$, roughly five hundred times smaller than the Coulomb barrier. Classically, then, fusion in the Sun should be impossible: no nucleus is moving fast enough to climb the wall. Two effects rescue it: a lucky few nuclei in the high-speed tail move far faster than average, and — decisively — quantum mechanics lets a nucleus tunnel through the barrier rather than over it, a purely quantum effect (worked out by George Gamow, and beyond this article's algebra). Between them, a tiny but steady trickle of nuclei fuse — enough to light a star. The lesson is fixed: fusion needs extreme temperature, of order $10^7\ \text{K}$ and up, which is why the fuel is always a fully ionised plasma and why confining it is so hard.

Everything now reduces to the mass-difference calculation of equation (1), applied to a specific reaction. The two reactions worth knowing cold are the laboratory reaction that fusion reactors chase, and the stellar reaction that lights the Sun. The laboratory favourite is deuterium–tritium (D–T) fusion, joining the two heavy hydrogen isotopes:

where: deuterium $^2\mathrm{H}$ has one proton and one neutron, tritium $^3\mathrm{H}$ one proton and two neutrons, and they fuse into a helium-4 nucleus (an alpha particle) plus a spare neutron. Check the conservation: nucleons $2 + 3 = 4 + 1$, and charge $1 + 1 = 2 + 0$ — both balance. D–T is the reaction of choice because it has the lowest Coulomb barrier of any fusion fuel and the largest cross-section at accessible temperatures; it is what ITER and every experimental reactor burns. The stellar reaction is the proton–proton chain, whose net effect turns four ordinary protons into one helium-4 nucleus:

where: $e^+$ is a positron (an anti-electron) and $\nu$ is a neutrino, both required so that charge and lepton number balance as two of the protons effectively convert to neutrons. This does not happen in one collision — it proceeds through several steps, the first and slowest being two protons fusing to deuterium — but the bookkeeping of the whole chain is captured by equation (4). It is the reason the Sun has shone for four and a half billion years and will shine for five billion more.

Board level — the D–T reaction energy (CBSE, ISC and NEET pattern)

Calculate the energy released in the fusion reaction $^2\mathrm{H} + {}^3\mathrm{H} \to {}^4\mathrm{He} + n$. Take atomic masses $M(^2\mathrm{H}) = 2.014102\ \text{u}$, $M(^3\mathrm{H}) = 3.016049\ \text{u}$, $M(^4\mathrm{He}) = 4.002603\ \text{u}$, neutron mass $1.008665\ \text{u}$, and $1\ \text{u} = 931.5\ \text{MeV}$.

Step 1. State why atomic masses are safe here: deuterium and tritium contribute one electron each (two total) and helium-4 contributes two, so electrons balance and cancel. Use the atomic masses as given.

Step 2. Mass of reactants: $2.014102 + 3.016049 = 5.030151\ \text{u}$. Mass of products: $4.002603 + 1.008665 = 5.011268\ \text{u}$.

Step 3. Mass defect: $\Delta m = 5.030151 - 5.011268 = 0.018883\ \text{u}$.

Step 4. Energy released: $Q = \Delta m \times 931.5 = 0.018883 \times 931.5 = \boxed{17.6\ \text{MeV}}$. As a mass fraction, $0.018883/5.030151 = 0.38\%$ of the reactant mass has become energy — small, but four times the per-nucleon yield of uranium fission.

Board level — why fusion needs high temperature (qualitative, all boards including IGCSE)

Explain, in terms of forces, why nuclear fusion requires temperatures of tens of millions of kelvin, while nuclear fission of uranium-235 can be triggered by a slow neutron at room temperature.

Step 1. A neutron is uncharged. Approaching a uranium nucleus it feels no electrical repulsion, so even a slow, room-temperature neutron can walk straight in and induce fission. No barrier, no heat needed.

Step 2. Two fusing nuclei are both positive. As they approach, the Coulomb force pushes them apart, and this repulsion peaks at a barrier of order a few hundred keV before the short-range strong force can take over.

Step 3. To overcome that barrier the nuclei must collide at very high speed, and the only way to give a whole fuel such speeds is to heat it to tens of millions of kelvin — turning it into a plasma. This is why fusion demands the conditions of a stellar core. (Even then, classical speeds fall short; quantum tunnelling supplies the rest.)

Competitive level — sharing the energy between neutron and alpha (JEE Advanced pattern)

The D–T reaction releases $Q = 17.6\ \text{MeV}$. Assuming the deuteron and triton fuse essentially at rest, find the kinetic energy carried away by (a) the neutron and (b) the helium-4 nucleus. Use mass numbers $1$ and $4$.

Step 1. With negligible initial momentum, conservation of momentum makes the neutron and helium nucleus recoil back-to-back with equal-magnitude momenta $p$. So $p_n = p_{\alpha}$.

Step 2. Kinetic energy for a given momentum is $KE = p^2/2m$, which is larger for the smaller mass. Hence the energy splits in inverse proportion to mass: the neutron (mass $1$) gets the fraction $4/(4+1)$ and the helium (mass $4$) gets $1/5$.

Step 3. Compute: $KE_n = 17.6 \times \tfrac{4}{5} = \boxed{14.1\ \text{MeV}}$ and $KE_{\alpha} = 17.6 \times \tfrac{1}{5} = \boxed{3.5\ \text{MeV}}$. Check: $14.1 + 3.5 = 17.6\ \text{MeV}$, the full Q-value, as it must. Timing note: about 90 seconds. The trap is to split the energy equally or to give the heavy nucleus the larger share. Fix the rule once — equal momentum means the light particle runs off with most of the energy — and the item is two multiplications. The $14.1\ \text{MeV}$ fusion neutron is worth memorising; it is the neutron a real reactor must survive.

Competitive level — the Sun's fuel bill (JEE Advanced pattern)

The Sun radiates power $L = 3.8\times10^{26}\ \text{W}$. Each helium-4 nucleus built by the proton–proton chain releases $26.7\ \text{MeV} = 4.28\times10^{-12}\ \text{J}$. Find (a) the mass the Sun converts to energy each second, and (b) the mass of hydrogen it fuses each second. Take $1\ \text{MeV} = 1.60\times10^{-13}\ \text{J}$ and proton mass $1.67\times10^{-27}\ \text{kg}$.

Step 1. Mass-to-energy rate straight from $E = mc^2$: the mass lost per second is $\dfrac{L}{c^{2}} = \dfrac{3.8\times10^{26}}{(3.0\times10^{8})^{2}} = \dfrac{3.8\times10^{26}}{9.0\times10^{16}} = \boxed{4.2\times10^{9}\ \text{kg s}^{-1}}$. The Sun loses about four million tonnes of mass every second as pure radiation.

Step 2. Number of helium nuclei made per second: $N = L / (4.28\times10^{-12}) = 3.8\times10^{26}/4.28\times10^{-12} = 8.9\times10^{37}$ per second.

Step 3. Each helium takes four protons, so the hydrogen mass consumed is $N\times 4\times m_p = 8.9\times10^{37}\times4\times1.67\times10^{-27} = \boxed{6\times10^{11}\ \text{kg s}^{-1}}$ — about six hundred million tonnes of hydrogen per second. Only $4.2\times10^{9}/6\times10^{11} = 0.7\%$ of that fused mass actually turns into energy; the rest stays as helium. Timing note: about two minutes. The two headline numbers — 4 million tonnes of mass converted, 600 million tonnes of hydrogen fused, every second — are worth carrying in memory; examiners love the contrast between them, and their ratio is exactly helium-4's $0.7\%$ mass-to-binding fraction. Don't confuse "mass converted to energy" ($L/c^2$) with "mass of fuel fused" — they differ by that factor of a hundred-and-forty.

Getting the direction of energy release wrong. Fusion releases energy only below iron-56; fission releases energy only above it. Do not claim fusing heavy nuclei releases energy, and do not claim splitting light nuclei does. Anchor to the curve: any move toward the iron peak gives out energy. Tested on CBSE, JEE, NEET, IB HL and A-Level.

Using bare-nucleus masses when the table gives atomic masses. For D–T, deuterium and tritium electrons balance helium's two, so atomic masses cancel cleanly — use them directly. Subtracting electron masses by hand here double-counts and gives a wrong $Q$. State the electron-cancellation each time (it is a favourite one-mark justification).

Splitting the reaction energy equally. In $^2\mathrm{H}+{}^3\mathrm{H}\to{}^4\mathrm{He}+n$ the neutron and alpha do not share $Q$ equally. Equal momentum sends most of the energy ($14.1$ of $17.6\ \text{MeV}$) to the lighter neutron. The energy ratio is inverse to the mass ratio.

Confusing "mass converted" with "fuel fused." The Sun converts $4\times10^{9}\ \text{kg s}^{-1}$ to energy but fuses $6\times10^{11}\ \text{kg s}^{-1}$ of hydrogen — a factor of about $140$ apart, because only $0.7\%$ of the fuel mass becomes energy. Reporting one for the other is a standard slip.

Saying the Sun's fusion is "fast" or bomb-like. Solar fusion is astonishingly slow per unit volume — the first proton–proton step is so improbable that the Sun's core generates less power per cubic metre than a compost heap. The Sun is powerful only because it is enormous. Fusion is not a runaway chain reaction the way fission can be.

Board scope. The qualitative story of fusion — light nuclei joining, releasing energy, powering the Sun and stars, needing very high temperature, and the promise of fusion power — is examinable everywhere, IGCSE 0625 included. The quantitative Q-value from mass defect and $931.5\ \text{MeV/u}$, the D–T $17.6\ \text{MeV}$ and pp-chain $26.7\ \text{MeV}$, belong to CBSE, ICSE (ISC), JEE Main and Advanced, NEET, AP Physics 2, IB (SL and HL) and A-Level 9702 — do not ask an IGCSE candidate to compute a mass defect. The Coulomb-barrier estimate and quantum tunnelling are higher-tier enrichment (JEE Advanced, A-Level, IB HL).

Texts. NCERT Physics Class 12, Chapter 13 ("Nuclei"), the nuclear-energy and fusion sections · H. C. Verma, Concepts of Physics Vol. 2, "The Nucleus" · Halliday, Resnick and Walker, "Energy from the Nucleus" · Krane, Introductory Nuclear Physics, for the stellar cycles.

Provenance. LANE_SPEC / anchor_check.py locked values. Recompute: D–T mass defect $\Delta m = 0.018883\ \text{u}$, $Q = 17.6\ \text{MeV}$; neutron/alpha split $14.1\ \text{MeV}$ and $3.5\ \text{MeV}$; pp-chain $\Delta m = 0.028697\ \text{u}$, $Q = 26.7\ \text{MeV}$, helium-4 binding energy $28.3\ \text{MeV}$, neutrino loss about $2\%$; per-nucleon D–T $3.52\ \text{MeV}$ versus fission $0.85\ \text{MeV}$ (ratio $\approx 4$); Coulomb constant $ke^2 = 1.44\ \text{MeV}\,\cdot\,\text{fm}$, D–T barrier $\approx 0.44\ \text{MeV}$ at $r \approx 3.24\ \text{fm}$; thermal energy $k_{\text{B}}T \approx 0.86\ \text{keV}$ at $10^{7}\ \text{K}$ (barrier-to-thermal ratio $\approx 500$); Sun mass-conversion rate $L/c^2 = 4.2\times10^{9}\ \text{kg s}^{-1}$, hydrogen fused $\approx 6\times10^{11}\ \text{kg s}^{-1}$, fuel-mass converted $0.7\%$; energy per kg of D–T fuel $\approx 3.4\times10^{14}\ \text{J}$.

Prerequisites. The Binding-Energy Curve — the single graph that explains why fusing light nuclei releases energy, and where the iron-56 peak comes from. Leads to. Neutron Stars & Degeneracy Pressure (university seam, go-deeper only), where the story picks up after fusion ends: what happens to a stellar core once it can no longer fuse, and the quantum pressure that holds up the densest matter in the universe. Related in this chapter: Nuclear Fission & Chain Reactions, fusion's energy-releasing mirror image; Mass–Energy Equivalence in the Nucleus, for the $E = mc^2$ and $931.5\ \text{MeV/u}$ this article rests on.