Educerie · IB Diploma · Physics
Theme E Nuclear and quantum physics · E.5 Fusion and stars
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Explain the stability of a star as a balance of radiation pressure and gravity | SL, HL | "Explain why a main-sequence star is stable" (2 or 3 marks) |
| Explain that fusion is the energy source of stars, and calculate the energy released | SL, HL | "Calculate the energy released in this fusion reaction" (2 marks) |
| Explain the conditions for fusion in terms of density and temperature | SL, HL | "Explain why fusion requires a high temperature" (2 marks) |
| Describe the effect of stellar mass on a star's evolution | SL, HL | "Describe the evolution of a star of mass 20 M⊙ after it leaves the main sequence" (3 marks) |
| Sketch and interpret the HR diagram, and describe stars in each region | SL, HL | "Sketch the evolutionary path of the Sun on the HR diagram" (2 marks) |
| Use d(parsec) = 1/p(arc-second), and convert between AU, ly and pc | SL, HL | "Calculate the distance to the star in light years" (2 marks) |
| Determine a stellar radius from luminosity and surface temperature | SL, HL | "Determine the radius of the star" (3 marks) |
| Explain how a spectrum gives surface temperature and composition | SL, HL | "Outline how the chemical composition of a star is determined" (2 marks) |
Before you start
From B.1: luminosity L = σAT⁴ with A = 4πR² for a star, apparent brightness b = L/4πd², and Wien's law λₘₐₓT = 2.9 × 10⁻³ m K. From E.1: each element absorbs and emits light at its own set of wavelengths. From E.3 and E.4: binding energy per nucleon and the energy released = Δm × 931.5 MeV per u. From the data booklet: L⊙ = 3.83 × 10²⁶ W, R⊙ = 6.96 × 10⁸ m, 1 AU = 1.50 × 10¹¹ m, 1 ly = 9.46 × 10¹⁵ m, 1 pc = 3.26 ly.
1The idea in one paragraph
A star is a ball of gas trying to collapse under its own gravity and held up by the pressure of the hot gas and radiation inside it. The energy that keeps it hot comes from nuclear fusion in the core: light nuclei join to form heavier ones, the binding energy per nucleon rises, and mass is converted to energy. Fusion needs a very high temperature, so that nuclei collide fast enough to overcome their electric repulsion, and a high density, so that collisions are frequent. A star spends most of its life fusing hydrogen into helium on the main sequence; what happens next depends on its mass. Light stars become red giants and end as white dwarfs; heavy stars fuse elements all the way to iron and explode as supernovae, leaving neutron stars or black holes. We learn all this from light: parallax gives distance, the spectrum gives temperature and composition, and the Hertzsprung–Russell (HR) diagram puts it all on one page.
2Why a star is stable
Every layer of a star feels two things. Gravity pulls it inwards. Pressure from below pushes it outwards: the pressure of the hot gas and the radiation pressure of the enormous flux of photons flowing out from the core. When the two balance at every depth, the star neither shrinks nor grows. This is equilibrium, and Figure 1(a) shows it.
A stable star: outward radiation (and gas) pressure = inward gravitational force, at every depth.
The balance is self-correcting, as Figure 1(b) shows. If the core is squeezed a little, it becomes denser and hotter, the fusion rate rises steeply, the pressure rises, and the core pushes back out. If it expands, it cools, fusion slows, the pressure falls and gravity pulls it back. It works like a thermostat. The balance fails only when the fuel in the core runs out, which is where section 8 begins.
3Fusion: the energy source of stars
Nuclear fusion is the joining of two light nuclei to make a heavier one. Figure 2 shows why it releases energy: on the left of the binding energy curve, the heavier nucleus is more tightly bound, so the total binding energy rises and the mass falls.
The gain is largest for making helium-4, at 7.07 MeV per nucleon, so stars spend most of their lives turning hydrogen into helium. In the Sun this happens through a chain of steps called the proton–proton chain, whose overall result is:
Worked example 1: energy from hydrogen fusion. Atomic masses: hydrogen-1 = 1.007825 u, helium-4 = 4.002603 u. Using atomic masses, the energy of the two positrons annihilating with electrons is included automatically.
Worked example 2: how fast the Sun loses mass. The Sun's luminosity is 3.83 × 10²⁶ W. Every second that energy leaves as radiation, so an equal mass-energy is lost from the core.
Four million tonnes a second, for about ten billion years.
Fusion compared with fission. Both release energy by increasing binding energy per nucleon, and both convert mass to energy by E = mc². Fission splits heavy nuclei, fusion joins light ones. Per nucleon, fusion releases more: about 200 ÷ 236 ≈ 0.85 MeV per nucleon for uranium fission, but 26.7 ÷ 4 ≈ 6.7 MeV per nucleon for turning hydrogen into helium. Fission needs only a slow neutron; fusion needs extreme temperature and density, which is why it is so hard to control on Earth.
4The conditions for fusion: temperature and density
Two nuclei are both positive, so they repel. The strong force can pull them together only when they are about 10⁻¹⁵ m apart, and to get that close they must climb the electric barrier in Figure 3.
Worked example 3: how hot would a gas need to be? Estimate the electric potential energy of two protons 3.0 × 10⁻¹⁵ m apart, and the temperature at which the mean kinetic energy of the particles, (3/2)kBT, equals it.
The Sun's core is about 1.5 × 10⁷ K, some 250 times cooler. Fusion still happens because the particles have a spread of speeds, and a tiny fraction are far faster than average (a quantum effect beyond this course also helps them through the barrier). That tiny fraction is enough only because there are so many nuclei, packed very closely. So the guide's two conditions are:
- High temperature, so that nuclei move fast enough for some collisions to bring them within range of the strong force.
- High density, so that collisions happen often enough for the rate of fusion to be large. The Sun's core is over a hundred times denser than water.
Gravity supplies both. A cloud of gas collapsing under its own weight converts gravitational potential energy into kinetic energy of its particles, so it heats up, and it becomes denser as it shrinks. When the core reaches about 10⁷ K, hydrogen fusion begins, the pressure rises, the collapse stops, and a star is born on the main sequence.
5How elements are created
Hydrogen and most of the helium in the universe were made in the first few minutes after the Big Bang. Almost everything else was made in stars. When hydrogen in the core runs out, the core contracts and heats until helium can fuse into carbon and oxygen, at around 10⁸ K. Each heavier fuel needs a higher temperature, because heavier nuclei carry more charge and repel more strongly. A massive star can climb the whole ladder, and near the end of its life it is layered like an onion, as Figure 4 shows.
The ladder stops at iron. Iron-56 is at the peak of the binding energy curve, so fusing it releases no energy. Elements heavier than iron are made when nuclei capture neutrons during a supernova (and in collisions of neutron stars), and the explosion scatters all these elements into space, where they become part of new stars, planets and people.
6Measuring the stars
Temperature and composition, from the spectrum. A star's continuous spectrum is close to that of a black body, so Wien's law gives its surface temperature: find λₘₐₓ and divide 2.9 × 10⁻³ m K by it. Crossing the continuous spectrum are dark absorption lines: cooler gas in the star's outer layers absorbs photons whose energies match the energy level differences of its atoms (E.1). Each element has its own pattern of lines, so the pattern identifies the elements present. Helium was discovered this way, in the Sun's spectrum, before it was found on Earth. Figure 5 shows a Sun-like spectrum.
Distance, from parallax. Figure 6 shows a nearby star viewed six months apart, from opposite sides of the Earth's orbit. Against the very distant stars it appears to shift, as a finger does when you look with one eye and then the other. The parallax angle p is half that total shift: the angle at the star between the directions to the Sun and to the Earth, with the 1 AU Sun–Earth distance as the baseline.
The parsec is defined from this: 1 pc is the distance at which 1 AU subtends an angle of 1 arc-second (1/3600 of a degree). The smaller the angle, the further the star, which gives the data booklet's equation:
d (parsec) = 1 ÷ p (arc-second)
The three distance units, and how they relate:
| Unit | Definition | In metres |
|---|---|---|
| astronomical unit, AU | mean Earth–Sun distance | 1.50 × 10¹¹ m |
| light year, ly | distance light travels in one year | 9.46 × 10¹⁵ m |
| parsec, pc | distance at which 1 AU subtends 1″ | 3.26 ly = 3.08 × 10¹⁶ m |
Worked example 4. A star's parallax angle is 0.25″. Find its distance.
Parallax works only for nearby stars: at 1000 pc, p = 0.001″, too small to measure well through the atmosphere. Telescopes in space measure far smaller angles.
Radius, from luminosity and temperature. Once the distance d and the brightness b are measured, L = 4πd²b. With T from the spectrum, L = σ4πR²T⁴ gives the radius:
Worked example 5: size of a star. A star's spectrum peaks at 380 nm, its parallax angle is 0.0500″, and its apparent brightness is 4.0 × 10⁻¹⁰ W m⁻². Determine its radius.
A quicker route compares with the Sun, since L ∝ R²T⁴: R/R⊙ = √(L/L⊙) ÷ (T/T⊙)². Use it whenever the data are given as multiples of solar values.
7The Hertzsprung–Russell diagram
Plot luminosity against surface temperature for many stars and they fall into a few groups, each telling you what the stars are doing inside. Figure 7 uses the guide's conventions: luminosity up the vertical axis, surface temperature along the horizontal axis, hotter on the left, both scales logarithmic.
The dashed lines are lines of constant radius. Since L ∝ R²T⁴, stars of the same radius lie on a straight line on logarithmic axes, sloping down to the right. Stars at the top right are huge, stars at the bottom left tiny.
| Region | Temperature | Luminosity | Radius | What is happening inside |
|---|---|---|---|---|
| Main sequence | 3 000 to 40 000 K | 10⁻³ to 10⁵ L⊙ | about 0.1 to 10 R⊙ | hydrogen fusing to helium in the core; about 90% of stars; more massive = hotter and brighter |
| Red giants | 3 500 to 5 000 K, cool | 10² to 10³ L⊙ | 10 to 100 R⊙ | core has run out of hydrogen; shell fusion; outer layers swollen and cool |
| Supergiants | any temperature | 10⁴ to 10⁶ L⊙ | up to 1000 R⊙ | massive stars fusing heavier elements |
| White dwarfs | hot, 10 000 K or more | 10⁻⁴ to 10⁻² L⊙ | about 0.01 R⊙, Earth-sized | no fusion; the hot core left behind, slowly cooling |
| Instability strip | 5 000 to 8 000 K | across the giants | varies | stars that pulsate, so their luminosity rises and falls regularly |
Worked example 6: reading a position. A star has L = 1.0 × 10⁴ L⊙ and T = 4000 K; another has L = 0.010 L⊙ and T = 10 000 K. Take T⊙ = 5800 K.
8How mass decides a star's evolution
The one property that decides a star's life is its mass. A more massive star needs a higher core temperature and pressure to support its weight, so it fuses much faster: a star of 10 M⊙ is thousands of times more luminous than the Sun but lives only tens of millions of years, against the Sun's ten billion. Figure 8 draws two lives on the HR diagram, and Figure 9 sets out the whole story.
A star like the Sun (below about 8 M⊙). When core hydrogen runs out, fusion stops there, the pressure falls and the core contracts and heats. Hydrogen now fuses in a shell around it, and the extra energy swells the outer layers enormously; they cool and redden. The star moves up and to the right: a red giant. The core reaches about 10⁸ K and fuses helium to carbon and oxygen. It never gets hot enough for carbon fusion. The outer layers drift away as a glowing planetary nebula, and the exposed core is a white dwarf: hot, small, dim, with no fusion, slowly cooling for billions of years. A white dwarf cannot have a mass above about 1.4 M⊙.
A massive star (above about 8 M⊙). It becomes a red supergiant and fuses heavier and heavier elements in shells, up to iron (Figure 4). Iron fusion releases no energy, so the iron core has no support. It collapses in less than a second, and the rebound blows the outer layers off in a supernova, briefly as bright as a whole galaxy. What is left depends on the mass of the collapsed core: below about 3 M⊙, a neutron star, a ball of neutrons perhaps 20 km across; above it, a black hole, from which not even light escapes.
That answers the guide's first question: stars turn hydrogen into the rest of the periodic table, and supernovae spread it out to make new stars and planets.
9Where marks are lost
Saying gravity is balanced by "fusion". Fusion is the energy source; the force that balances gravity is the outward pressure of radiation and hot gas.
Explaining the need for high temperature without the reason. The mark is for "nuclei must move fast enough to overcome the electrostatic repulsion and come within range of the strong force". "Fusion needs heat" scores nothing.
Forgetting density. The guide asks for both conditions. High density makes collisions frequent enough.
Drawing the HR diagram with temperature increasing to the right. Hotter is on the left. Luminosity goes up, and both scales are logarithmic.
Using the whole shift as p. The parallax angle is half the total angular shift over six months, and it must be in arc-seconds for d = 1/p to give parsecs.
Mixing up L and b, or R and d. Luminosity (W) is the star's total output; brightness (W m⁻²) is what arrives here. R is the star's radius in L = σ4πR²T⁴; d is its distance in b = L/4πd².
Saying stars "burn" hydrogen, or that a white dwarf is fusing. Fusion is a nuclear process, not burning, and a white dwarf has no fusion at all; it shines because it is hot.
10Draw it right
- HR diagram: luminosity (or L/L⊙) on the vertical axis, surface temperature on the horizontal axis increasing to the left, both logarithmic, with the Sun at about 5800 K and 1 L⊙.
- The main sequence runs from top left (hot, bright) to bottom right (cool, dim); red giants above it on the right; supergiants across the top; white dwarfs below it on the left; the instability strip a near-vertical band between the main sequence and the giants.
- Lines of constant radius slope down to the right, with larger radii further up and right.
- An evolutionary path starts on the main sequence: for the Sun, up and right to the red giant region, then across to the left and down to the white dwarfs.
- A parallax diagram shows two Earth positions six months apart, the star, the distant background, and p marked at the star, opposite a 1 AU baseline.
- A balance diagram for a star shows gravity inwards and radiation pressure outwards, labelled, on the same layer.
11Try it
Marks in brackets. Take L⊙ = 3.83 × 10²⁶ W, T⊙ = 5800 K, 1 pc = 3.26 ly, 1 ly = 9.46 × 10¹⁵ m, 1 u = 931.5 MeV c⁻². Answers are at the end.
Q1. (Paper 1A style) A star lies at the bottom left of the HR diagram. What type of star is it? 1 mark
A. red giant B. white dwarf C. main-sequence star like the Sun D. red supergiant
Q2. A star has a parallax angle of 0.12 arc-seconds.
(a) Calculate its distance in parsecs. 1 mark
(b) Calculate its distance in metres. 2 marks
Q3. The spectrum of a star peaks at 830 nm. Its luminosity is 350 L⊙.
(a) Calculate its surface temperature. 1 mark
(b) Determine its radius in terms of the Sun's radius R⊙. 2 marks
(c) Identify the region of the HR diagram in which the star lies. 1 mark
Q4. (a) Explain why a main-sequence star remains stable for a long time. 2 marks
(b) Describe the evolution of the Sun after the hydrogen in its core has been used up. 3 marks
Q5. (Paper 1B style) An astronomer measures parallax angles for two stars. The values are invented for this question.
| Star | Parallax angle / arc-second |
|---|---|
| X | 0.0200 ± 0.0008 |
| Y | 0.0010 ± 0.0008 |
(a) Determine the distance to star X, with its absolute uncertainty. 3 marks
(b) Comment on the usefulness of the measurement for star Y. 2 marks
Q6. In some stars, two helium-3 nuclei fuse: ³₂He + ³₂He → ⁴₂He + 2 ¹₁H. Atomic masses: helium-3 = 3.016029 u, helium-4 = 4.002603 u, hydrogen-1 = 1.007825 u.
(a) Calculate the energy released in MeV. 2 marks
(b) Explain why this reaction needs a higher temperature than the fusion of two protons. 2 marks
12In one breath
A main-sequence star is stable because outward radiation and gas pressure balance inward gravity at every depth, and the balance corrects itself through the fusion rate. Fusion of light nuclei raises the binding energy per nucleon and releases energy: four hydrogen nuclei make one helium-4 and 26.7 MeV, 0.7% of the mass, and the Sun loses 4 × 10⁹ kg a second this way. Fusion needs a high temperature, so nuclei are fast enough to overcome electric repulsion, and a high density, so collisions are frequent; gravity provides both. Stars build elements up to iron, and heavier ones are made in supernovae. Parallax gives distance, d(pc) = 1/p(″), with 1 pc = 3.26 ly; Wien's law gives temperature; absorption lines give composition; L = 4πd²b and L = σ4πR²T⁴ then give the radius. On the HR diagram, luminosity is up and temperature increases to the left: main sequence diagonally, red giants and supergiants top right, white dwarfs bottom left, instability strip between, radius constant along lines sloping down to the right. Below about 8 M⊙ a star becomes a red giant, then a planetary nebula and a white dwarf; above it, a red supergiant, a supernova, and a neutron star or black hole.
Answers
Q1. B. Bottom left means hot but dim, which is only possible for a very small star. A and D are at the top right; C lies on the diagonal through the middle.
Q2. (a) d = 1 ÷ 0.12 = 8.3 pc. (b) 8.33 × 3.26 = 27.2 ly; 27.2 × 9.46 × 10¹⁵ = 2.6 × 10¹⁷ m. (a) A1. (b) M1 for the conversion via light years (or 3.08 × 10¹⁶ m per pc), A1. Dividing by 3.26 instead of multiplying gives 2.4 × 10¹⁶ m and scores M0.
Q3. (a) T = 2.9 × 10⁻³ ÷ (830 × 10⁻⁹) = 3.5 × 10³ K. (b) R/R⊙ = √(L/L⊙) ÷ (T/T⊙)² = √350 ÷ (3494/5800)² = 18.7 ÷ 0.363 = 52, so R ≈ 52 R⊙. (c) Red giants: cool, luminous and large. (a) A1. (b) M1 for L ∝ R²T⁴ with the squared temperature ratio, A1 for 51 to 52 R⊙. (c) A1; "giant" alone accepted. Forgetting to square the temperature ratio gives 31 R⊙ and loses the A1.
Q4. (a) Gravity pulls the star's layers inwards; the pressure of radiation and hot gas from fusion in the core pushes outwards, and the two are in equilibrium. If the core contracts it heats, fusion speeds up and the pressure restores the balance, so the star stays stable as long as its hydrogen lasts. (b) The core, no longer fusing, contracts and heats; hydrogen fuses in a shell around it and the outer layers expand and cool: the Sun becomes a red giant. Helium fuses to carbon and oxygen in the core. The outer layers are ejected as a planetary nebula, leaving the core as a white dwarf with no fusion, which slowly cools. (a) 1 for gravity balanced by radiation/gas pressure, 1 for the self-correcting mechanism or the fuel supply. (b) 1 for red giant with a reason (expansion and cooling), 1 for planetary nebula, 1 for white dwarf. "The Sun explodes as a supernova" scores 0 in (b).
Q5. (a) d = 1 ÷ 0.0200 = 50 pc. Fractional uncertainty in p = 0.0008 ÷ 0.0200 = 4%, and d = 1/p has the same fractional uncertainty, so Δd = 0.04 × 50 = 2 pc: d = 50 ± 2 pc. (b) The fractional uncertainty is 0.0008 ÷ 0.0010 = 80%, so the distance could lie anywhere from about 560 pc to 5000 pc. The measurement gives only an order of magnitude and is of little use; the angle is too small compared with the precision of the instrument, so a more precise method (a space telescope) or another distance method is needed. (a) M1 for the fractional uncertainty in p, M1 for applying it to d, A1 for 50 ± 2 pc. (b) 1 for the 80% (or the range of possible distances), 1 for the conclusion that the result is unreliable because p is comparable to its uncertainty.
Q6. (a) Δm = 2 × 3.016029 − (4.002603 + 2 × 1.007825) = 6.032058 − 6.018253 = 0.013805 u; E = 0.013805 × 931.5 = 12.9 MeV. (b) Each helium-3 nucleus has charge +2e, so the electrostatic repulsion between two of them is four times that between two protons at the same separation. The nuclei need more kinetic energy to get close enough for the strong force to act, and a higher temperature means a higher mean kinetic energy. (a) M1 for the mass difference with two hydrogen masses, A1. (b) 1 for the larger charge giving a larger repulsion/barrier, 1 for linking temperature to kinetic energy.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section E.5 Fusion and stars. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.