Why is "Confusing fusion with fission" a common mistake in Fusion and stars?

Why it happens: Both are nuclear processes that release energy, so students mix up which one joins nuclei and which one splits them. How to avoid it: Remember: **fuse = join** (light nuclei combine into a heavier one); **fission = split** (a heavy nucleus breaks apart). Stars run on fusion of light nuclei. Source: IB Physics Theme E subject reports

Why is "Drawing the HR diagram with temperature increasing to the right" a common mistake in Fusion and stars?

Why it happens: Every other graph in physics has its horizontal axis increasing to the right, so students apply the same habit. How to avoid it: On an HR diagram the surface temperature **decreases** to the right by convention: hot blue stars on the **left**, cool red stars on the **right**. Draw a small arrow labelled 'decreasing' to remind yourself. Source: IB Physics Theme E subject reports

Why is "Confusing luminosity (total power emitted) with apparent brightness (power received per m²)" a common mistake in Fusion and stars?

Why it happens: In everyday language 'bright' and 'luminous' mean the same thing, and both symbols look similar. How to avoid it: Luminosity **L** is intrinsic (unit W) and fixed; apparent brightness **b** (unit W m⁻²) depends on distance through b = L/(4πd²). A luminous star can look faint if it is far away. Source: IB Physics Theme E subject reports

Why is "Using diameter instead of radius in L = 4πr²σT⁴" a common mistake in Fusion and stars?

Why it happens: Data may quote a star's diameter, and students substitute it directly for r. How to avoid it: The formula uses the **radius**. If given a diameter, halve it first. Substituting the diameter gives a luminosity four times too large. Source: IB Physics Theme E subject reports

Why is "Forgetting to use the Kelvin temperature in the Stefan–Boltzmann or Wien equations" a common mistake in Fusion and stars?

Why it happens: Temperatures are sometimes discussed loosely, and students carry a Celsius value into the formula. How to avoid it: Both L = 4πr²σT⁴ and Wien's law require the **absolute (kelvin)** temperature. Because T is raised to the fourth power in Stefan–Boltzmann, even a small unit error is disastrous. Source: IB Physics Theme E subject reports

Why is "Muddling the parallax relation d = 1/p (wrong angle units)" a common mistake in Fusion and stars?

Why it happens: Students put the parallax angle in degrees, or expect the distance directly in metres or light-years. How to avoid it: In d = 1/p the angle **p must be in arcseconds** and the distance **d comes out in parsecs**. Convert to light-years (×3.26) or metres (×3.09×10¹⁶) only as a final step. Source: IB Physics Theme E subject reports

IB DP Physics (100452)

Fusion and stars

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Detailed Study Notes

Detailed notes on Nuclear and quantum physics for IB DP Physics, covering key concepts, explanations, examples, and exam-focused revision points.

E.5 Fusion and Stars — IB Physics SL Study Notes (Theme E: Nuclear and Quantum Physics)

How stars shine: nuclear fusion of light nuclei releases energy because the product has a higher binding energy per nucleon; the proton–proton chain powers the Sun; a star balances gravity against radiation pressure; and we read a star's luminosity, brightness, temperature and distance using the Stefan–Boltzmann law, Wien's law, the Hertzsprung–Russell diagram and stellar parallax.

At a glance

Fusion joins light nuclei (e.g. hydrogen) into a heavier nucleus; fission splits heavy nuclei. Do not confuse them.
Fusion releases energy because the product nucleus has a higher binding energy per nucleon — mass is lost and converted to energy via $E = \Delta m c^2$ .
Fusion needs extremely high temperature and pressure so that positive nuclei have enough kinetic energy to overcome the Coulomb (electrostatic) repulsion and get close enough for the strong force to act.
The Sun runs on the proton–proton chain: overall $4\,^{1}_{1}\text{H} \rightarrow\,^{4}_{2}\text{He} + 2e^{+} + 2\nu + \text{energy}$ (about 26.7 MeV per helium nucleus).
A star is in equilibrium: inward gravitation is balanced by the outward radiation and gas pressure from fusion in the core.
Luminosity $L$ (total power radiated, W) follows the Stefan–Boltzmann law $L = 4\pi r^2 \sigma T^4$ ; apparent brightness $b = \dfrac{L}{4\pi d^2}$ is the power received per m² at Earth.
The Hertzsprung–Russell (HR) diagram plots luminosity (up) against surface temperature — temperature DECREASES to the right — showing the main sequence, red giants, supergiants and white dwarfs.
Stellar parallax gives distance: $d$ in parsec $= \dfrac{1}{p}$ , with $p$ the parallax angle in arcseconds.

What you’ll learn

Mapped to the 100452 subject guide (2025-onwards).

Explain nuclear fusion using the binding-energy-per-nucleon curve, and state why fusion requires extremely high temperatures and pressures to overcome Coulomb repulsion.
Describe the proton–proton chain as the fusion process powering the Sun and write its overall reaction.
Describe a star as a body in equilibrium between the inward gravitational force and the outward radiation/gas pressure produced by fusion.
Use the Stefan–Boltzmann law $L = 4\pi r^2 \sigma T^4$ , apparent brightness $b = L/(4\pi d^2)$ and Wien's displacement law to relate a star's luminosity, radius, surface temperature, brightness and distance.
Interpret the Hertzsprung–Russell diagram (main sequence, red giants, white dwarfs, supergiants) and use stellar parallax to determine distances in parsecs.

Why fusion releases energy: the binding-energy curve (first principles)

Fusing light nuclei climbs the binding-energy-per-nucleon curve, so mass is lost and energy is released.

Nuclear fusion is the joining of two light nuclei to form a single heavier nucleus. This is the opposite of fission, in which a single heavy nucleus splits into lighter fragments. Getting this distinction right is the first mark in almost every E.5 question: fusion → light nuclei combine; fission → heavy nucleus splits.

The key idea — binding energy per nucleon. The binding energy of a nucleus is the energy needed to pull it completely apart into separate protons and neutrons; equivalently, it is the energy released when those nucleons come together. Dividing by the number of nucleons gives the binding energy per nucleon, which is a measure of how tightly bound — and therefore how stable — a nucleus is.

If you plot binding energy per nucleon against nucleon number (mass number $A$ ), you get the famous curve below. It rises steeply from hydrogen, peaks around iron-56 (the most stable nucleus, about 8.8 MeV per nucleon), then falls slowly for heavier nuclei.

Binding energy per nucleon rises steeply for light nuclei. Fusing hydrogen into helium moves *up* the curve, so the product is more tightly bound — the lost mass is released as energy.

The first-principles argument for why fusion releases energy:

When light nuclei fuse, the product sits higher up the binding-energy-per-nucleon curve — its nucleons are more tightly bound.
A more tightly bound nucleus has less mass than the separate particles that formed it (this "missing" mass is the mass defect $\Delta m$ ).
That lost mass appears as energy, following Einstein's mass–energy equivalence: $E = \Delta m\,c^2$
Because the left side of the curve is very steep, fusing light nuclei releases a large amount of energy per nucleon — which is why fusion powers the stars.

Why fusion needs extreme conditions. Nuclei are all positively charged, so as two nuclei approach they repel through the electrostatic (Coulomb) force. To fuse, they must get close enough (about $10^{-15}$ m) for the very short-range strong nuclear force to take over and bind them. Overcoming this Coulomb barrier requires the nuclei to be moving extremely fast — that is, they must be at very high temperature (high average kinetic energy). A very high pressure/density is also needed so that collisions happen often enough. In the Sun's core the temperature is about $1.5\times10^{7}\,\text{K}$ .

Fusion = light nuclei join; fission = heavy nucleus splits.
Fusion releases energy because the product has a higher binding energy per nucleon (mass defect → E = Δmc²).
High temperature (fast nuclei) and high pressure are needed to overcome the Coulomb repulsion between positive nuclei.

See the full worked example for fusion and stars →

The proton–proton chain: how the Sun makes helium

Overall four hydrogen nuclei fuse into one helium-4 nucleus, releasing positrons, neutrinos and energy.

In stars about the size of the Sun, hydrogen is fused into helium through the proton–proton (p–p) chain. You do not need every intermediate step at SL, but you must know the overall reaction and be able to interpret it.

Overall reaction: $4\,^{1}_{1}\text{H} \;\longrightarrow\; ^{4}_{2}\text{He} + 2e^{+} + 2\nu + \text{energy}$

Reading it out: four hydrogen nuclei (protons) fuse to make one helium-4 nucleus, plus two positrons ( $e^{+}$ , the antiparticles of electrons), two neutrinos ( $\nu$ ), and energy (released as gamma-ray photons and the kinetic energy of the products). The energy released per helium nucleus formed is about 26.7 MeV ( $\approx 4.3\times10^{-12}\,\text{J}$ ).

The net effect of the proton–proton chain: four protons become a helium-4 nucleus (2 protons + 2 neutrons), releasing two positrons, two neutrinos and about 26.7 MeV of energy.

Checking the bookkeeping (a good habit examiners reward):

Charge: left side $= 4(+1) = +4$ ; right side $= 2(+1)$ from the helium protons $+\,2(+1)$ from the positrons $= +4$ . ✓
Nucleons (mass number): left $= 4$ ; right $= 4$ (helium). ✓
The energy comes from the mass defect: the helium-4 nucleus has slightly less mass than the four protons, and that lost mass becomes energy.

This slow, steady fusion is what keeps the Sun shining for billions of years. Because the Sun contains an enormous amount of hydrogen, it has been fusing on the main sequence for about 4.6 billion years and will continue for roughly another 5 billion.

Overall p–p chain: 4 ¹₁H → ⁴₂He + 2e⁺ + 2ν + energy (~26.7 MeV).
Four protons make one helium-4 nucleus (2 protons + 2 neutrons).
Charge and nucleon number are conserved; the energy comes from the mass defect.

See the full worked example for fusion and stars →

A star is a balancing act: gravity vs radiation pressure

A stable star holds its size because inward gravity is balanced by outward radiation and gas pressure.

A star is a huge ball of very hot gas (mostly hydrogen and helium plasma). Two competing effects act on it:

Gravitation pulls inward. Every part of the star is attracted to every other part; gravity tries to make the star collapse.
Radiation and gas pressure push outward. Fusion in the core releases energy, keeping the core extremely hot. The resulting thermal gas pressure and radiation pressure push outward, resisting the collapse.

A star that is stable (like the Sun during its main-sequence life) is in equilibrium: these two effects are balanced, so the star neither collapses nor expands. This balance is often called hydrostatic equilibrium.

A stable star sits in equilibrium: the inward pull of gravity is exactly balanced by the outward radiation and gas pressure generated by fusion in the core.

What happens when the balance breaks (qualitative, for context): when a Sun-like star runs low on hydrogen in its core, fusion in the core slows, the outward pressure drops, and gravity wins — the core contracts and heats up while the outer layers swell and cool, forming a red giant. Eventually the outer layers drift away and the hot, dense core is left behind as a white dwarf, held up not by fusion but by other pressure. You will meet these stages again on the HR diagram below.

A star balances inward gravitation against outward radiation and gas pressure.
A stable (main-sequence) star is in equilibrium — constant size.
When core fusion slows, gravity dominates → the star evolves (red giant, then white dwarf for a Sun-like star).

Luminosity and apparent brightness: deriving the relations

L = 4πr²σT⁴ gives the total power a star emits; spreading that power over a sphere gives b = L/(4πd²) at Earth.

Two quantities are often confused but mean very different things:

Luminosity $L$ is the total power radiated by the star in all directions (unit: watt, W). It is an intrinsic property — it does not depend on how far away we are.
Apparent brightness $b$ is the power received per unit area at the observer (unit: W m⁻²). It does depend on distance: a bright star can look faint if it is far away.

Deriving the Stefan–Boltzmann law for a star. A star radiates almost like a black body. The Stefan–Boltzmann law says a black body emits power per unit surface area of $\sigma T^4$ , where $\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ and $T$ is the surface temperature in kelvin. A star of radius $r$ has surface area $4\pi r^2$ , so its total power (luminosity) is: $L = (\text{surface area}) \times (\text{power per area}) = 4\pi r^2 \times \sigma T^4$ $\boxed{L = 4\pi r^2 \sigma T^4}$ Notice $L$ depends on radius squared and temperature to the fourth power — so a small rise in surface temperature has a huge effect on luminosity.

Deriving apparent brightness. By the time the star's light reaches us at distance $d$ , that power $L$ has spread out over the surface of a giant sphere of radius $d$ , whose area is $4\pi d^2$ . The power arriving per square metre is therefore: $\boxed{b = \frac{L}{4\pi d^2}}$ This is an inverse-square law: double the distance and the brightness falls to a quarter. Combining the two boxed equations lets you find a star's distance from measured quantities.

The star's total power L spreads over a sphere of area 4πd². At distance d the brightness is b; at 2d the same power covers four times the area, so the brightness is only b/4.

Finding surface temperature with Wien's displacement law. A hotter black body glows a bluer colour — its emission peaks at a shorter wavelength. Wien's displacement law links the peak-emission wavelength to the surface temperature: $\lambda_{\text{max}} = \frac{2.90\times10^{-3}}{T} \quad (\lambda_{\text{max}}\text{ in m},\ T\text{ in K})$ So if we measure the colour (peak wavelength) of a star's light, we get its surface temperature — which we then feed into the Stefan–Boltzmann law. This is how astronomers know the temperatures of stars they can never visit. (For the Sun, $T \approx 5800\,\text{K}$ gives $\lambda_{\text{max}} \approx 500\,\text{nm}$ , in the green–yellow part of the visible spectrum.)

Luminosity L = 4πr²σT⁴ — total power emitted (intrinsic); σ = 5.67×10⁻⁸ W m⁻² K⁻⁴, T in kelvin.
Apparent brightness b = L/(4πd²) — power received per m² (inverse-square with distance d).
Wien's law λ_max = 2.90×10⁻³/T gives the surface temperature from the peak wavelength.

See the full worked example for fusion and stars →

The Hertzsprung–Russell diagram (temperature decreases to the right)

A map of stars: luminosity up the axis, surface temperature along it but DECREASING to the right.

The Hertzsprung–Russell (HR) diagram is the single most important chart in stellar astronomy. It plots:

luminosity (usually in solar luminosities $L/L_\odot$ , on a logarithmic scale) up the vertical axis, and
surface temperature along the horizontal axis — but by historical convention the temperature DECREASES to the right (hot blue stars on the left, cool red stars on the right).

This "backwards" temperature axis is examined constantly. Always draw the hottest stars on the LEFT.

The HR diagram. The main sequence is a diagonal band from hot, luminous stars (top left) to cool, dim stars (bottom right). Red giants and supergiants (large, cool but very luminous) sit above it; white dwarfs (small, hot but dim) sit below.

The main regions:

Main sequence — the diagonal band where stars spend most of their lives fusing hydrogen into helium. The Sun is a main-sequence star. Hot, luminous stars sit top-left; cool, dim ones bottom-right.
Red giants — large, cool (so red) but very luminous because they are huge; they sit in the upper right. (Their large radius, via $L = 4\pi r^2 \sigma T^4$ , wins over their low temperature.)
Supergiants — even larger and more luminous, stretched across the top.
White dwarfs — very hot (so white/blue) but very dim because they are tiny; they sit in the lower left.

Stellar evolution of a Sun-like star (qualitative): a star forms and settles onto the main sequence, where it fuses hydrogen for billions of years. When the core hydrogen runs out, it swells and cools into a red giant. Finally it sheds its outer layers and its hot core is left as a white dwarf, which slowly cools. On the HR diagram this traces a path from the main sequence, up to the red-giant region, then down to the white-dwarf region.

HR diagram: luminosity (log, up) vs surface temperature (DECREASING to the right).
Main sequence = hydrogen-fusing stars (diagonal band); the Sun is on it.
Red giants/supergiants = large, cool, luminous (upper right/top); white dwarfs = small, hot, dim (lower left).

Measuring distance: stellar parallax and the units of space

A nearby star appears to shift against distant stars as Earth orbits; the shift angle gives its distance.

To use $b = L/(4\pi d^2)$ we need distances — but stars are far too remote to reach with a tape measure. For relatively nearby stars we use stellar parallax: the apparent shift in a star's position against the very distant background stars as the Earth moves from one side of its orbit to the other (six months apart).

The method. Observe a nearby star in January and again in July, when the Earth is on opposite sides of the Sun (a baseline of 2 astronomical units). The star appears to shift slightly. Half of the total angular shift is the parallax angle $p$ . The nearer the star, the bigger the parallax angle.

As Earth orbits, a nearby star appears to shift against the distant background. The parallax angle p (measured in arcseconds) gives the distance directly: d in parsec = 1/p.

The parallax relation (learn this exactly): $d\,(\text{parsec}) = \frac{1}{p\,(\text{arcsecond})}$

$d$ is the distance in parsecs (pc).
$p$ is the parallax angle in arcseconds (1 arcsecond = $1/3600$ of a degree).
A star with parallax $p = 1\,\text{arcsecond}$ is, by definition, 1 parsec away.

The units of astronomical distance (know what each means):

Unit	Meaning	Rough size
Astronomical unit (AU)	Average Earth–Sun distance	$1.5\times10^{11}\,\text{m}$
Light-year (ly)	Distance light travels in one year	$9.46\times10^{15}\,\text{m}$
Parsec (pc)	Distance giving a parallax of 1 arcsecond	$3.09\times10^{16}\,\text{m} \approx 3.26\,\text{ly}$

A worked mini-example: a star has a measured parallax of $p = 0.20\,\text{arcsec}$ . Its distance is $d = 1/0.20 = 5.0\,\text{pc}$ , which is $5.0 \times 3.26 = 16.3\,\text{light-years}$ . The smaller the parallax, the farther the star — very distant stars have parallax angles too tiny to measure, which is why parallax only works for relatively nearby stars.

Parallax = apparent shift of a nearby star against distant stars as Earth orbits the Sun.
d (parsec) = 1/p (arcsecond): smaller parallax → greater distance.
1 AU = 1.5×10¹¹ m; 1 ly = 9.46×10¹⁵ m; 1 pc ≈ 3.26 ly ≈ 3.09×10¹⁶ m.

See the full worked example for fusion and stars →

Quick recap

Fusion joins light nuclei into a heavier one and releases energy because the product has a higher binding energy per nucleon (mass defect → E = Δmc²); fission splits heavy nuclei.
Fusion needs very high temperature and pressure so nuclei overcome the Coulomb repulsion and the strong force can bind them; the Sun uses the proton–proton chain: 4 ¹₁H → ⁴₂He + 2e⁺ + 2ν + energy.
A stable star is in equilibrium — inward gravity balances the outward radiation and gas pressure from fusion.
Luminosity L = 4πr²σT⁴ (total power emitted); apparent brightness b = L/(4πd²) (power received per m²); Wien's law λ_max = 2.90×10⁻³/T gives surface temperature.
The HR diagram plots luminosity vs surface temperature (decreasing to the right): main sequence, red giants, supergiants, white dwarfs; a Sun-like star evolves main sequence → red giant → white dwarf.
Stellar parallax gives distance: d (parsec) = 1/p (arcsecond); useful units are the AU, light-year and parsec.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

Fusion = light nuclei join (releases energy: product has higher binding energy per nucleon). Fission = heavy nucleus splits.
Fusion needs high temperature + pressure to overcome the Coulomb repulsion between positive nuclei.
Proton–proton chain (the Sun): 4 ¹₁H → ⁴₂He + 2e⁺ + 2ν + energy (~26.7 MeV).
Star equilibrium: inward gravitation = outward radiation + gas pressure.
Stefan–Boltzmann: L = 4πr²σT⁴, σ = 5.67×10⁻⁸ W m⁻² K⁻⁴, T = surface temp in K.
Apparent brightness: b = L/(4πd²) (inverse-square law). Wien: λ_max = 2.90×10⁻³/T.
HR diagram: luminosity (up) vs surface temperature DECREASING to the right; main sequence / red giants / white dwarfs / supergiants.
Parallax: d (parsec) = 1/p (arcsecond). 1 pc ≈ 3.26 ly ≈ 3.09×10¹⁶ m.

How it’s examined

E.5 appears in both papers and links nuclear physics (Theme E) to astrophysics. Paper 1A (MCQ): distinguishing fusion from fission, identifying the products of the p–p chain, recognising the correct HR-diagram layout (temperature decreasing to the right), and inverse-square-law reasoning for brightness — no calculator, so numbers are chosen to be clean. Paper 1B (data-based): using Wien's law to get a surface temperature from a spectrum, and reading positions on an HR diagram. Paper 2: multi-step calculations (3–7 marks) that chain Wien's law → Stefan–Boltzmann → apparent brightness → parallax to find temperature, luminosity, radius or distance, plus 'describe/explain' questions on why fusion needs high temperatures, on stellar equilibrium, and on a Sun-like star's evolution. Command terms: state, describe, explain, determine, calculate, show that. Examiner reports for Theme E repeatedly flag: confusing fusion with fission, drawing the HR temperature axis the wrong way round, muddling luminosity with apparent brightness, using diameter instead of radius in L = 4πr²σT⁴, forgetting to work in kelvin, and mishandling the parallax relation d = 1/p. Always show working and quote units and sensible significant figures — method marks survive an arithmetic slip.

Sources: IB Diploma Programme Physics Guide (first assessment 2025) — Theme E: Nuclear and quantum physics (E.5 Fusion and stars); IB Physics Data Booklet (2025); IB Physics specimen papers and subject reports (2023–2025). Last reviewed 2026-07-21.

Take this whole topic with you

Step-by-step worked examples — Fusion and stars

Step-by-step solutions to past-paper-style questions on fusion and stars, written exactly the way a tutor would explain them at the board.

Question type:

1Fusion or fission? Classifying a nuclear reaction
Getting startedIdentify & classify• fusion, AO1
Question
State whether each process is nuclear fusion or nuclear fission, and in one sentence explain why energy is released in the proton–proton chain. (a) $^{235}_{92}\text{U} + \,^{1}_{0}\text{n} \rightarrow \,^{141}_{56}\text{Ba} + \,^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n}$ (b) $^{2}_{1}\text{H} + \,^{3}_{1}\text{H} \rightarrow \,^{4}_{2}\text{He} + \,^{1}_{0}\text{n}$ (3 marks)
Step-by-step solution
1. Step 1
  In (a) a heavy nucleus (uranium-235) splits into lighter fragments → this is fission.
2. Step 2
  In (b) two light nuclei (deuterium and tritium) combine into a heavier nucleus (helium) → this is fusion.
3. Step 3
  Energy is released in the p–p chain because the helium-4 product has a higher binding energy per nucleon than the hydrogen nuclei, so mass is lost and converted to energy via $E = \Delta m c^2$ .
Answer
(a) fission (heavy nucleus splits); (b) fusion (light nuclei combine); energy is released because the product is more tightly bound (higher binding energy per nucleon), so the lost mass becomes energy.
Examiner tip
Mark scheme: (1) (a) = fission; (2) (b) = fusion; (3) energy from increased binding energy per nucleon / mass defect. The classifier is simple: heavy-splits = fission, light-combine = fusion.
2Surface temperature from Wien's displacement law
Getting startedDirect calculation• Wien's law, AO2
Question
The light from a star peaks in intensity at a wavelength of 580 nm. Calculate the surface temperature of the star. (Wien constant $= 2.90\times10^{-3}\,\text{m K}$ ) (2 marks)
Step-by-step solution
1. Step 1
  Rearrange Wien's law $\lambda_{\text{max}} = \dfrac{2.90\times10^{-3}}{T}$ for $T$ , and convert 580 nm to metres: $580\,\text{nm} = 5.80\times10^{-7}\,\text{m}$ .
  $T = \frac{2.90\times10^{-3}}{\lambda_{\text{max}}}$
2. Step 2
  Substitute the peak wavelength in metres.
  $T = \frac{2.90\times10^{-3}}{5.80\times10^{-7}} = 5.00\times10^{3}\,\text{K}$
Answer
Surface temperature T = 5.0×10³ K (5000 K).
Examiner tip
Mark scheme: (1) correct rearrangement and wavelength in metres; (2) T = 5000 K. Forgetting to convert nm → m (a factor of 10⁹) is the classic slip and gives a wildly wrong temperature.
3Distance from stellar parallax
Getting startedDirect calculation• parallax, AO2
Question
A nearby star has a measured parallax angle of 0.125 arcseconds. Calculate its distance (a) in parsecs and (b) in light-years. (1 pc = 3.26 ly) (2 marks)
Step-by-step solution
1. Step 1
  Use the parallax relation $d\,(\text{parsec}) = \dfrac{1}{p\,(\text{arcsecond})}$ .
  $d = \frac{1}{0.125} = 8.00\,\text{pc}$
2. Step 2
  Convert parsecs to light-years using 1 pc = 3.26 ly.
  $d = 8.00 \times 3.26 = 26.1\,\text{ly}$
Answer
(a) d = 8.00 pc; (b) d = 26.1 ly.
Examiner tip
Mark scheme: (1) d = 1/p = 8.00 pc; (2) 26.1 ly. The relation gives parsecs only when p is in arcseconds — the conversion to light-years is a separate final step.
4Luminosity from the Stefan–Boltzmann law
Building confidenceDirect calculation• Stefan–Boltzmann, luminosity, AO2
Question
A star has a radius of $7.0\times10^{8}\,\text{m}$ and a surface temperature of 5800 K. Calculate its luminosity. ( $\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ ) (3 marks)
Step-by-step solution
1. Step 1
  Use the Stefan–Boltzmann law $L = 4\pi r^2 \sigma T^4$ . First evaluate the pieces: $r^2 = (7.0\times10^{8})^2 = 4.9\times10^{17}\,\text{m}^2$ and $T^4 = 5800^4 = 1.13\times10^{15}\,\text{K}^4$ .
  $L = 4\pi r^2 \sigma T^4$
2. Step 2
  Substitute all values (radius, not diameter!).
  $L = 4\pi (4.9\times10^{17})(5.67\times10^{-8})(1.13\times10^{15})$
3. Step 3
  Evaluate to get the luminosity in watts.
  $L = 3.9\times10^{26}\,\text{W}$
Answer
L ≈ 3.9×10²⁶ W (about one solar luminosity).
Examiner tip
Mark scheme: (1) correct equation with r squared and T to the fourth; (2) correct substitution using the radius; (3) L ≈ 3.9×10²⁶ W. Using diameter instead of radius (a factor of 4 error) and forgetting T is to the fourth power are the two most common failures.
5Apparent brightness at Earth
Building confidenceDirect calculation• apparent brightness, AO2
Question
A star has a luminosity of $4.0\times10^{26}\,\text{W}$ and is at a distance of $8.0\times10^{17}\,\text{m}$ from Earth. Calculate its apparent brightness at Earth. (3 marks)
Step-by-step solution
1. Step 1
  Apparent brightness is the power spread over a sphere of radius $d$ : $b = \dfrac{L}{4\pi d^2}$ .
  $b = \frac{L}{4\pi d^2}$
2. Step 2
  Compute the sphere's area: $4\pi d^2 = 4\pi (8.0\times10^{17})^2 = 8.04\times10^{36}\,\text{m}^2$ .
  $4\pi d^2 = 4\pi (6.4\times10^{35}) = 8.04\times10^{36}\,\text{m}^2$
3. Step 3
  Divide the luminosity by that area.
  $b = \frac{4.0\times10^{26}}{8.04\times10^{36}} = 5.0\times10^{-11}\,\text{W m}^{-2}$
Answer
b ≈ 5.0×10⁻¹¹ W m⁻².
Examiner tip
Mark scheme: (1) select b = L/(4πd²); (2) correct 4πd²; (3) b ≈ 5.0×10⁻¹¹ W m⁻² with unit. Remember to square the distance — it is an inverse-square law — and keep b (W m⁻²) distinct from L (W).
6Energy released in the proton–proton chain
Building confidenceMulti-step problem• fusion, mass–energy, AO2
Question
In the overall proton–proton chain, four hydrogen atoms (mass 1.007825 u each) fuse to form one helium-4 atom (mass 4.002602 u). Calculate (a) the mass defect in kilograms and (b) the energy released, in joules. (1 u $= 1.66\times10^{-27}\,\text{kg}$ , $c = 3.0\times10^{8}\,\text{m s}^{-1}$ ) (4 marks)
Step-by-step solution
1. Step 1
  Mass defect $\Delta m$ = mass of 4 hydrogen atoms − mass of helium-4 atom (in u).
  $\Delta m = 4(1.007825) - 4.002602 = 4.031300 - 4.002602 = 0.028698\,\text{u}$
2. Step 2
  Convert the mass defect to kilograms.
  $\Delta m = 0.028698 \times 1.66\times10^{-27} = 4.76\times10^{-29}\,\text{kg}$
3. Step 3
  Use $E = \Delta m c^2$ to find the energy released.
  $E = (4.76\times10^{-29})(3.0\times10^{8})^2 = (4.76\times10^{-29})(9.0\times10^{16})$
4. Step 4
  Evaluate the energy in joules (this equals about 26.7 MeV).
  $E = 4.3\times10^{-12}\,\text{J}$
Answer
(a) Δm = 0.028698 u = 4.76×10⁻²⁹ kg; (b) E ≈ 4.3×10⁻¹² J (≈ 26.7 MeV).
Examiner tip
Mark scheme: (1) mass defect = 4(1.007825) − 4.002602; (2) Δm in kg; (3) E = Δmc² with c squared; (4) E ≈ 4.3×10⁻¹² J. Using atomic masses automatically accounts for the electrons/positrons. Squaring c (not just multiplying by it) is essential.
7Finding a star's distance by combining laws
StretchMulti-step problem• Stefan–Boltzmann, apparent brightness, AO2, AO3
Question
A star has a radius of $1.0\times10^{9}\,\text{m}$ and a surface temperature of 6000 K. Its apparent brightness measured at Earth is $2.0\times10^{-9}\,\text{W m}^{-2}$ . Determine the distance of the star from Earth. ( $\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ ) (5 marks)
Step-by-step solution
1. Step 1
  First find the luminosity from the Stefan–Boltzmann law: $r^2 = 1.0\times10^{18}\,\text{m}^2$ and $T^4 = 6000^4 = 1.296\times10^{15}\,\text{K}^4$ .
  $L = 4\pi (1.0\times10^{18})(5.67\times10^{-8})(1.296\times10^{15}) = 9.2\times10^{26}\,\text{W}$
2. Step 2
  Now rearrange the apparent-brightness relation $b = \dfrac{L}{4\pi d^2}$ for the distance $d$ .
  $d = \sqrt{\frac{L}{4\pi b}}$
3. Step 3
  Substitute L and the measured brightness b.
  $d = \sqrt{\frac{9.2\times10^{26}}{4\pi (2.0\times10^{-9})}} = \sqrt{3.66\times10^{34}}$
4. Step 4
  Take the square root to get the distance in metres.
  $d = 1.9\times10^{17}\,\text{m}$
Answer
d ≈ 1.9×10¹⁷ m (about 20 light-years).
Examiner tip
Mark scheme: (1) L from Stefan–Boltzmann ≈ 9.2×10²⁶ W; (2) rearrange b = L/(4πd²) for d; (3) correct substitution; (4) d ≈ 1.9×10¹⁷ m. The skill being tested is chaining two relations: the star's own properties give L, then L with the measured b gives d. Remember to take the square root at the end.
8How big is a red giant? Comparing radii on the HR diagram
StretchMulti-step problem• HR diagram, Stefan–Boltzmann, AO2, AO3
Question
A red giant has a luminosity of $1.0\times10^{4}$ solar luminosities and a surface temperature of 3500 K. The Sun has a surface temperature of 5800 K. Using $L = 4\pi r^2 \sigma T^4$ , determine the radius of the red giant as a multiple of the Sun's radius, and comment on your answer. (4 marks)
Step-by-step solution
1. Step 1
  Take the ratio of the two Stefan–Boltzmann equations; $\sigma$ and $4\pi$ cancel, leaving $\dfrac{L_{\text{g}}}{L_{\odot}} = \left(\dfrac{r_{\text{g}}}{r_{\odot}}\right)^2 \left(\dfrac{T_{\text{g}}}{T_{\odot}}\right)^4$ .
  $\frac{r_{\text{g}}}{r_{\odot}} = \sqrt{\frac{L_{\text{g}}}{L_{\odot}}} \left(\frac{T_{\odot}}{T_{\text{g}}}\right)^2$
2. Step 2
  Substitute $L_{\text{g}}/L_{\odot} = 10^4$ and the two temperatures.
  $\frac{r_{\text{g}}}{r_{\odot}} = \sqrt{10^4}\left(\frac{5800}{3500}\right)^2 = 100 \times (1.657)^2$
3. Step 3
  Evaluate the numerical factor.
  $\frac{r_{\text{g}}}{r_{\odot}} = 100 \times 2.75 = 275$
4. Step 4
  The red giant's radius is about 275 times the Sun's. Despite being cooler than the Sun, it is enormously more luminous because its radius is so large — exactly why red giants sit in the upper right of the HR diagram.
Answer
r_giant ≈ 275 × r_Sun. The star is cool yet very luminous because its huge radius dominates L = 4πr²σT⁴.
Examiner tip
Mark scheme: (1) form the ratio, cancelling σ and 4π; (2) correct substitution with T ratio to the fourth (radius ratio squared); (3) r_g/r_⊙ ≈ 275; (4) comment linking large radius to high luminosity despite low temperature. Using ratios avoids ever needing σ or the Sun's absolute radius.

Model Answers — Fusion and stars

High-scoring sample answers for fusion and stars on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define nuclear fusion and state one condition required for it to occur in the core of a star.
Model answer
Nuclear fusion is the process in which two light nuclei combine (join) to form a single heavier nucleus, releasing energy.

Condition: it requires a very high temperature (so the nuclei move fast enough to overcome the electrostatic/Coulomb repulsion between them) — a very high pressure/density is also required.
Why this scores
Why this scores 2/2. (1) fusion = light nuclei combine into a heavier nucleus with energy released; (2) one valid condition (high temperature, or high pressure/density) with the reason being to overcome Coulomb repulsion. Saying 'nuclei join' without 'light' or 'heavier' risks the first mark.
Question 2
3 marks
Q (3 marks). A star has a radius of $2.0\times10^{9}\,\text{m}$ and a surface temperature of 4000 K. Calculate its luminosity. ( $\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ )
Model answer
Use the Stefan–Boltzmann law $L = 4\pi r^2 \sigma T^4$ .

$r^2 = (2.0\times10^{9})^2 = 4.0\times10^{18}\,\text{m}^2, \qquad T^4 = 4000^4 = 2.56\times10^{14}\,\text{K}^4$

$L = 4\pi (4.0\times10^{18})(5.67\times10^{-8})(2.56\times10^{14})$

$L = 7.3\times10^{26}\,\text{W}$
Why this scores
Why this scores 3/3. (1) correct equation with r² and T⁴; (2) correct substitution using the radius; (3) L ≈ 7.3×10²⁶ W with unit. The two guardrails are: use radius (not diameter) and raise T to the fourth power.
Question 3
3 marks
Q (3 marks). (a) Explain what is meant by stellar parallax. (b) A star has a parallax angle of 0.040 arcseconds. Calculate its distance in parsecs.
Model answer
(a) Stellar parallax is the apparent shift in the position of a nearby star relative to much more distant background stars, observed as the Earth moves from one side of its orbit to the other (six months apart). The parallax angle is half of the total apparent angular shift.

(b) Using $d\,(\text{parsec}) = \dfrac{1}{p\,(\text{arcsecond})}$ :

$d = \frac{1}{0.040} = 25\,\text{pc}$
Why this scores
Why this scores 3/3. (a) 2 marks: apparent shift of a nearby star against distant background (1) as Earth orbits the Sun (1). (b) 1 mark: d = 1/p = 25 pc. The relation gives parsecs directly because p is in arcseconds.
Question 4
4 marks
Q (4 marks). Write the overall reaction equation for the proton–proton chain that powers the Sun, and explain why energy is released.
Model answer
Overall reaction:

$4\,^{1}_{1}\text{H} \;\longrightarrow\; ^{4}_{2}\text{He} + 2e^{+} + 2\nu + \text{energy}$

That is, four hydrogen nuclei (protons) fuse to form one helium-4 nucleus, releasing two positrons, two neutrinos and energy.

Why energy is released:

The helium-4 nucleus has a higher binding energy per nucleon than the four separate protons — it is more tightly bound (more stable).

This means the helium nucleus has less mass than the four protons; the difference is the mass defect $\Delta m$ .

The lost mass is converted to energy according to $E = \Delta m c^2$ (released as gamma photons and kinetic energy of the products).
Why this scores
Why this scores 4/4. (1) correct overall equation 4¹H → ⁴He …; (2) products include positrons and neutrinos; (3) product has higher binding energy per nucleon / is more tightly bound; (4) mass defect converted to energy via E = Δmc². Writing electrons instead of positrons breaks charge conservation and loses a mark.
Question 5
4 marks
Q (4 marks). Explain how a main-sequence star such as the Sun remains stable in size for billions of years, and describe what eventually happens to a Sun-like star when the hydrogen in its core runs out.
Model answer
Stability (equilibrium):

The star's own gravitation pulls its material inward, tending to make it collapse.

Fusion in the core keeps it extremely hot, producing an outward radiation and gas (thermal) pressure.

In a stable main-sequence star these are balanced — the inward gravitational force equals the outward pressure force — so the star stays the same size (it is in equilibrium).

When core hydrogen runs out:

Core fusion slows, so the outward pressure falls and gravity dominates; the core contracts and heats while the outer layers expand and cool, forming a red giant.

Eventually the outer layers are shed and the hot, dense core is left as a white dwarf.
Why this scores
Why this scores 4/4. (1) gravity acts inward; (2) radiation/gas pressure from fusion acts outward; (3) these are balanced → equilibrium → constant size; (4) correct evolution: red giant then white dwarf. The assessed idea is the balance of forces, not just 'gravity holds it together'.
Question 6
5 marks
Q (5 marks). Star A and Star B have the same luminosity, but Star B is three times as far from Earth as Star A. (a) State what is meant by apparent brightness. (b) Determine the ratio of the apparent brightness of Star A to that of Star B. (c) One student says 'Star A must be more luminous because it looks brighter.' Comment on this statement.
Model answer
(a) Apparent brightness is the power received per unit area at the observer on Earth (unit W m⁻²).

(b) Apparent brightness follows $b = \dfrac{L}{4\pi d^2}$ . With equal luminosity $L$ , brightness depends only on $1/d^2$ :

$\frac{b_{\text{A}}}{b_{\text{B}}} = \frac{L/(4\pi d_{\text{A}}^2)}{L/(4\pi d_{\text{B}}^2)} = \frac{d_{\text{B}}^2}{d_{\text{A}}^2} = \left(\frac{3d_{\text{A}}}{d_{\text{A}}}\right)^2 = 9$

So Star A appears 9 times brighter than Star B.

(c) The statement is wrong. The two stars have the same luminosity (same total power emitted). Star A only looks brighter because it is closer; apparent brightness depends on distance as well as luminosity, so a brighter-looking star is not necessarily more luminous.
Why this scores
Why this scores 5/5. (a) apparent brightness = power received per m² (1); (b) uses inverse-square, ratio = (d_B/d_A)² = 9 (2); (c) identifies that equal luminosity means the difference is purely distance (2). The whole question turns on separating intrinsic luminosity from distance-dependent brightness.
Question 7
6 marks
Q (6 marks). (a) Sketch a Hertzsprung–Russell diagram, labelling the axes and marking the main sequence, red giants and white dwarfs. (b) Explain why a red giant can be much more luminous than the Sun even though its surface is cooler.
Model answer
(a) The HR diagram should show:

Vertical axis: luminosity (in solar luminosities, on a logarithmic scale), increasing upward.

Horizontal axis: surface temperature in kelvin, decreasing to the right (hot stars on the left, cool stars on the right).

Main sequence: a diagonal band running from hot, luminous stars (top left) to cool, dim stars (bottom right), with the Sun on it.

Red giants: upper right (cool but very luminous).

White dwarfs: lower left (hot but dim).

(b) From the Stefan–Boltzmann law $L = 4\pi r^2 \sigma T^4$ , luminosity depends on both radius and temperature. A red giant has a much lower surface temperature than the Sun, which alone would make it dimmer. However, its radius is enormous — tens or hundreds of times the Sun's. Because $L$ depends on $r^2$ , this huge radius more than compensates for the lower $T^4$ , so the red giant's overall luminosity is far greater than the Sun's.
Why this scores
Why this scores 6/6. (a) 3 marks: correct axes with temperature decreasing to the right (1), main sequence as a diagonal band (1), red giants upper-right and white dwarfs lower-left (1). (b) 3 marks: quote L = 4πr²σT⁴ (1), note the red giant's lower T (1), explain the very large radius (r² term) dominates (1). Drawing the temperature axis the wrong way round is the single most common way to lose marks here.
Question 8
6 marks
Q (6 marks). The light from a star peaks at a wavelength of 480 nm, and the star has a radius of $8.4\times10^{8}\,\text{m}$ . (a) Calculate the surface temperature of the star. (b) Calculate its luminosity. (Wien constant $= 2.90\times10^{-3}\,\text{m K}$ ; $\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ )
Model answer
(a) Surface temperature — Wien's displacement law, with $\lambda_{\text{max}} = 480\,\text{nm} = 4.80\times10^{-7}\,\text{m}$ :

$T = \frac{2.90\times10^{-3}}{\lambda_{\text{max}}} = \frac{2.90\times10^{-3}}{4.80\times10^{-7}} = 6.04\times10^{3}\,\text{K}$

(b) Luminosity — Stefan–Boltzmann law with $r = 8.4\times10^{8}\,\text{m}$ and $T = 6040\,\text{K}$ :

$T^4 = (6040)^4 = 1.33\times10^{15}\,\text{K}^4, \qquad r^2 = (8.4\times10^{8})^2 = 7.06\times10^{17}\,\text{m}^2$

$L = 4\pi r^2 \sigma T^4 = 4\pi (7.06\times10^{17})(5.67\times10^{-8})(1.33\times10^{15})$

$L = 6.7\times10^{26}\,\text{W}$
Why this scores
Why this scores 6/6. (a) 2 marks: Wien's law rearranged, wavelength in metres, T ≈ 6040 K. (b) 4 marks: correct equation (1), T⁴ evaluated (1), correct substitution with radius (1), L ≈ 6.7×10²⁶ W with unit (1). The chain is Wien → T, then Stefan–Boltzmann → L; carrying the temperature forward accurately is essential.
Question 9
10 marks
Q (10 marks — extended response). Astronomers study a star and measure the following: its light peaks in intensity at a wavelength of 500 nm; its parallax angle is 0.050 arcseconds; and its apparent brightness at Earth is $1.3\times10^{-8}\,\text{W m}^{-2}$ . Using this data: (a) determine the star's surface temperature; (b) determine its distance from Earth in metres (1 pc $= 3.09\times10^{16}\,\text{m}$ ); (c) determine the star's luminosity; (d) hence estimate the star's radius; (e) state, with a reason, where this star lies on the Hertzsprung–Russell diagram. (Wien constant $= 2.90\times10^{-3}\,\text{m K}$ ; $\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ )
Model answer
(a) Surface temperature (Wien's law, $\lambda_{\text{max}} = 5.00\times10^{-7}\,\text{m}$ ): $T = \frac{2.90\times10^{-3}}{5.00\times10^{-7}} = 5.80\times10^{3}\,\text{K}$

(b) Distance (parallax first gives parsecs, then convert to metres): $d = \frac{1}{p} = \frac{1}{0.050} = 20\,\text{pc}$ $d = 20 \times 3.09\times10^{16} = 6.18\times10^{17}\,\text{m}$

(c) Luminosity (rearrange apparent brightness $b = \dfrac{L}{4\pi d^2}$ ): $L = 4\pi d^2 b = 4\pi (6.18\times10^{17})^2 (1.3\times10^{-8})$ $L = 4\pi (3.82\times10^{35})(1.3\times10^{-8}) = 6.2\times10^{28}\,\text{W}$

(d) Radius (rearrange the Stefan–Boltzmann law $L = 4\pi r^2 \sigma T^4$ ): $r = \sqrt{\frac{L}{4\pi \sigma T^4}} = \sqrt{\frac{6.2\times10^{28}}{4\pi (5.67\times10^{-8})(5800^4)}}$ $T^4 = 5800^4 = 1.13\times10^{15}, \qquad r = \sqrt{\frac{6.2\times10^{28}}{8.05\times10^{8}}} = \sqrt{7.70\times10^{19}}$ $r = 8.8\times10^{9}\,\text{m}$

(e) Position on the HR diagram. The surface temperature (5800 K) is Sun-like, but the luminosity ( $6.2\times10^{28}\,\text{W}$ ) is about 160 times the Sun's ( $\approx 3.8\times10^{26}\,\text{W}$ ) and the radius ( $8.8\times10^{9}\,\text{m}$ ) is roughly 13 times the Sun's. A star that is far more luminous and much larger than the Sun at a similar temperature lies above the main sequence, in the giant region (upper right).
Why this scores
Why this scores 10/10. (a) Wien → T = 5800 K (2); (b) parallax → 20 pc → 6.18×10¹⁷ m (2); (c) rearrange b = L/(4πd²) → L ≈ 6.2×10²⁸ W (2); (d) rearrange Stefan–Boltzmann → r ≈ 8.8×10⁹ m (2); (e) compare with the Sun and place above the main sequence as a giant, with reasoning (2). This is a model IB Paper 2 extended-response: the grade-9 discriminators are (i) chaining four separate relations without dropping a power of ten, and (ii) justifying the HR-diagram position by comparing L and r with the Sun rather than merely guessing.

Key Formulae — Fusion and stars

The formulae you need to memorise for fusion and stars on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Stefan–Boltzmann law (luminosity of a star)
$L = 4\pi r^2 \sigma T^4$
$L$
luminosity — total power radiated (W)
$r$
radius of the star (m)
$\sigma$
Stefan–Boltzmann constant = 5.67×10⁻⁸ W m⁻² K⁻⁴
$T$
surface temperature (K)
When to use
To find a star's total emitted power from its radius and surface temperature (or to compare two stars via ratios). Use radius, not diameter, and T in kelvin.
Example
r = 7.0×10⁸ m, T = 5800 K: L = 4π(7.0×10⁸)²(5.67×10⁻⁸)(5800⁴) ≈ 3.9×10²⁶ W.
Apparent brightness
$b = \frac{L}{4\pi d^2}$
$b$
apparent brightness — power received per m² (W m⁻²)
$L$
luminosity of the star (W)
$d$
distance from the star to the observer (m)
When to use
To relate the power received at Earth to the star's luminosity and distance. It is an inverse-square law — rearrange for d when b and L are known.
Example
L = 4.0×10²⁶ W, d = 8.0×10¹⁷ m: b = 4.0×10²⁶ / [4π(8.0×10¹⁷)²] ≈ 5.0×10⁻¹¹ W m⁻².
Wien's displacement law
$\lambda_{\text{max}} = \frac{2.90\times10^{-3}}{T}$
$\lambda_{\text{max}}$
wavelength of peak emission (m)
$T$
surface temperature (K)
$2.90\times10^{-3}$
Wien constant (m K)
When to use
To find a star's surface temperature from the peak wavelength of its spectrum (its colour). Convert the wavelength to metres and give T in kelvin.
Example
λ_max = 580 nm = 5.80×10⁻⁷ m: T = 2.90×10⁻³ / 5.80×10⁻⁷ = 5000 K.
Stellar parallax–distance relation
$d\,(\text{parsec}) = \frac{1}{p\,(\text{arcsecond})}$
$d$
distance to the star (parsec, pc)
$p$
parallax angle (arcsecond)
When to use
To find the distance to a relatively nearby star from its measured parallax angle. The answer is in parsecs only when p is in arcseconds; convert to ly or m afterwards.
Example
p = 0.125 arcsec: d = 1/0.125 = 8.0 pc (= 8.0 × 3.26 = 26 ly).
Proton–proton chain (overall reaction)
$4\,^{1}_{1}\text{H} \rightarrow \,^{4}_{2}\text{He} + 2e^{+} + 2\nu + \text{energy}$
$^{1}_{1}\text{H}$
hydrogen nucleus (proton)
$^{4}_{2}\text{He}$
helium-4 nucleus (2 protons + 2 neutrons)
$e^{+}$
positron (antiparticle of the electron)
$\nu$
neutrino
When to use
To describe how the Sun fuses hydrogen into helium. Charge (+4 both sides) and nucleon number (4 both sides) are conserved; ~26.7 MeV is released per helium nucleus.
Example
Mass defect Δm = 4(1.007825) − 4.002602 = 0.0287 u → E = Δmc² ≈ 4.3×10⁻¹² J (≈ 26.7 MeV).
Mass–energy equivalence (fusion energy)
$E = \Delta m\,c^2$
$E$
energy released (J)
$\Delta m$
mass defect — mass lost in the reaction (kg)
$c$
speed of light in vacuum = 3.0×10⁸ m s⁻¹
When to use
To calculate the energy released in fusion from the mass defect. Convert the mass defect to kilograms first and remember to square c.
Example
Δm = 4.76×10⁻²⁹ kg: E = (4.76×10⁻²⁹)(3.0×10⁸)² = 4.3×10⁻¹² J.

Key Definitions and Keywords — Fusion and stars

Definitions to memorise and the exact keywords mark schemes credit for fusion and stars answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Nuclear fusion
Examiner keyword
The joining of two light nuclei to form a single heavier nucleus, releasing energy because the product has a higher binding energy per nucleon.
Nuclear fission
Examiner keyword
The splitting of a heavy nucleus into two lighter nuclei (plus neutrons), releasing energy. The opposite of fusion — do not confuse the two.
Binding energy per nucleon
Examiner keyword
The binding energy of a nucleus divided by its number of nucleons — a measure of how tightly bound (stable) the nucleus is. It peaks around iron-56.
Mass defect
Examiner keyword
The difference between the total mass of the separate nucleons and the mass of the assembled nucleus; the 'missing' mass released as energy via E = Δmc².
Coulomb barrier
Examiner keyword
The electrostatic repulsion between two positively charged nuclei that must be overcome (by high kinetic energy at high temperature) before they can fuse.
Proton–proton chain
Examiner keyword
The sequence of fusion reactions powering Sun-like stars, with overall effect 4 ¹₁H → ⁴₂He + 2e⁺ + 2ν + energy.
Stellar (hydrostatic) equilibrium
Examiner keyword
The stable state of a star in which the inward gravitational force is balanced by the outward radiation and gas pressure from core fusion, keeping the star a constant size.
Luminosity (L)
Examiner keyword
The total power radiated by a star in all directions, measured in watts (W). An intrinsic property, independent of distance.
Apparent brightness (b)
Examiner keyword
The power received from a star per unit area at the observer, measured in W m⁻². It depends on both luminosity and distance (b = L/4πd²).
Stefan–Boltzmann law
Examiner keyword
The relation L = 4πr²σT⁴ giving a star's luminosity from its radius and surface temperature (σ = 5.67×10⁻⁸ W m⁻² K⁻⁴).
Wien's displacement law
Examiner keyword
λ_max = 2.90×10⁻³/T — the peak-emission wavelength of a black body is inversely proportional to its surface temperature, used to find stellar temperatures.
Black body
An idealised object that absorbs all incident radiation and emits a characteristic thermal spectrum depending only on its temperature. Stars behave approximately as black bodies.
Hertzsprung–Russell (HR) diagram
Examiner keyword
A plot of stellar luminosity (vertical, log scale) against surface temperature (horizontal, decreasing to the right) showing the main sequence, red giants, supergiants and white dwarfs.
Main sequence
Examiner keyword
The diagonal band on the HR diagram where stars, including the Sun, spend most of their lives fusing hydrogen into helium.
Red giant / white dwarf
Examiner keyword
A red giant is a large, cool but very luminous late stage of a star (upper right of the HR diagram); a white dwarf is the small, hot but dim remnant core left behind (lower left).
Stellar parallax / parsec
Examiner keyword
Parallax is the apparent shift of a nearby star against distant stars as Earth orbits the Sun; a parsec (pc) is the distance giving a parallax of 1 arcsecond, so d(pc) = 1/p(arcsec). 1 pc ≈ 3.26 ly ≈ 3.09×10¹⁶ m.

Common Mistakes and Misconceptions — Fusion and stars

The traps other students keep falling into on fusion and stars questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Confusing fusion with fission
IB Physics Theme E subject reports
Why it happens
Both are nuclear processes that release energy, so students mix up which one joins nuclei and which one splits them.
How to avoid it
Remember: fuse = join (light nuclei combine into a heavier one); fission = split (a heavy nucleus breaks apart). Stars run on fusion of light nuclei.
✕Drawing the HR diagram with temperature increasing to the right
IB Physics Theme E subject reports
Why it happens
Every other graph in physics has its horizontal axis increasing to the right, so students apply the same habit.
How to avoid it
On an HR diagram the surface temperature decreases to the right by convention: hot blue stars on the left, cool red stars on the right. Draw a small arrow labelled 'decreasing' to remind yourself.
✕Confusing luminosity (total power emitted) with apparent brightness (power received per m²)
IB Physics Theme E subject reports
Why it happens
In everyday language 'bright' and 'luminous' mean the same thing, and both symbols look similar.
How to avoid it
Luminosity L is intrinsic (unit W) and fixed; apparent brightness b (unit W m⁻²) depends on distance through b = L/(4πd²). A luminous star can look faint if it is far away.
✕Using diameter instead of radius in L = 4πr²σT⁴
IB Physics Theme E subject reports
Why it happens
Data may quote a star's diameter, and students substitute it directly for r.
How to avoid it
The formula uses the radius. If given a diameter, halve it first. Substituting the diameter gives a luminosity four times too large.
✕Forgetting to use the Kelvin temperature in the Stefan–Boltzmann or Wien equations
IB Physics Theme E subject reports
Why it happens
Temperatures are sometimes discussed loosely, and students carry a Celsius value into the formula.
How to avoid it
Both L = 4πr²σT⁴ and Wien's law require the absolute (kelvin) temperature. Because T is raised to the fourth power in Stefan–Boltzmann, even a small unit error is disastrous.
✕Muddling the parallax relation d = 1/p (wrong angle units)
IB Physics Theme E subject reports
Why it happens
Students put the parallax angle in degrees, or expect the distance directly in metres or light-years.
How to avoid it
In d = 1/p the angle p must be in arcseconds and the distance d comes out in parsecs. Convert to light-years (×3.26) or metres (×3.09×10¹⁶) only as a final step.

Fusion and stars

Detailed Study Notes

At a glance

What you’ll learn

Why fusion releases energy: the binding-energy curve (first principles)

The proton–proton chain: how the Sun makes helium

A star is a balancing act: gravity vs radiation pressure

Luminosity and apparent brightness: deriving the relations

The Hertzsprung–Russell diagram (temperature decreases to the right)

Measuring distance: stellar parallax and the units of space

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Fusion or fission? Classifying a nuclear reaction

2Surface temperature from Wien's displacement law

3Distance from stellar parallax

4Luminosity from the Stefan–Boltzmann law

5Apparent brightness at Earth

6Energy released in the proton–proton chain

7Finding a star's distance by combining laws

8How big is a red giant? Comparing radii on the HR diagram

Stefan–Boltzmann law (luminosity of a star)

Apparent brightness

Wien's displacement law

Stellar parallax–distance relation

Proton–proton chain (overall reaction)

Mass–energy equivalence (fusion energy)

Nuclear fusion

Nuclear fission

Binding energy per nucleon

Mass defect

Coulomb barrier

Proton–proton chain

Stellar (hydrostatic) equilibrium

Luminosity (L)

Apparent brightness (b)

Stefan–Boltzmann law

Wien's displacement law

Black body

Hertzsprung–Russell (HR) diagram

Main sequence

Red giant / white dwarf

Stellar parallax / parsec

✕Confusing fusion with fission

✕Drawing the HR diagram with temperature increasing to the right

✕Confusing luminosity (total power emitted) with apparent brightness (power received per m²)

✕Using diameter instead of radius in L = 4πr²σT⁴

✕Forgetting to use the Kelvin temperature in the Stefan–Boltzmann or Wien equations

✕Muddling the parallax relation d = 1/p (wrong angle units)