E.5 Fusion and stars

Syllabus
First assessment 2025
Topic
Level
SL

Model Stellar Equilibrium

Balance inward and outward effects

A stable main-sequence star is in hydrostatic equilibrium: inward gravitational force is balanced by outward thermal or radiation pressure. The star’s radius remains approximately stable while the balance holds.

Link the balance to fusion

Fusion in the core releases energy. The energy transported outward produces thermal and radiation pressure that resists gravitational collapse.

Predict imbalance

If the outward pressure falls, gravity compresses the star and raises core temperature; if pressure grows, the star expands until a new balance is reached.

Common trap

Do not write only “gravity balances pressure” without directions. State that gravity acts inward and thermal/radiation pressure acts outward, with fusion supplying the energy.

E.5.1 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

Questions state how main-sequence stability is maintained or explain fusion’s role in the Sun’s stable radius.

Command terms

State / Outline

What earns marks

Name both forces or pressures and their directions, then link outward pressure to energy released by fusion.

Watch for

Mentioning fusion without the force balance or saying gravity acts outward.

Trace Fusion in Stars

Use fusion as a stellar source

In stellar fusion, light nuclei combine to form more tightly bound nuclei. The mass difference is released as energy, which powers the star and supports its pressure balance.

E_{\text{released}}=B_{\text{products}}-B_{\text{reactants}}=\Delta mc^2

Worked example — deuterium–tritium fusion

For 2H+3H4He+n^2\mathrm{H}+{}^3\mathrm{H}\rightarrow{}^4\mathrm{He}+n, the binding energies are 2(1.11)=2.22MeV2(1.11)=2.22\,\mathrm{MeV}, 3(2.83)=8.49MeV3(2.83)=8.49\,\mathrm{MeV} and 4(7.07)=28.28MeV4(7.07)=28.28\,\mathrm{MeV}. Therefore E=28.282.228.49=17.57MeVE=28.28-2.22-8.49=17.57\,\mathrm{MeV}. The product is more tightly bound, so energy is released.

Follow nucleosynthesis

Fusion in stars can build elements up to iron through successive reactions. Elements heavier than iron are mainly formed in explosive environments and neutron-capture processes rather than ordinary core fusion.

Compare fusion and fission

Fusion can offer high energy per mass and potentially fewer long-lived waste products, but it requires extreme temperature and confinement conditions.

Common trap

Do not say all heavy elements are made by fusion in ordinary stars. The pathway changes around iron.

E.5.2 Exam Analysis

Assessment in practice

1–3 marks
How it is assessed

Questions compare fusion with fission or outline how elements heavier than hydrogen and helium formed.

Command terms

Outline / State

What earns marks

Mention stellar nucleosynthesis/fusion for elements up to iron, then supernova or neutron capture for heavier elements.

Watch for

Claiming fusion alone forms every element or ignoring the comparison condition in an advantage question.

Explain Fusion Conditions

Use high temperature

Nuclei are positively charged and repel electrically. High core temperature gives them enough kinetic energy and speed to approach despite this repulsion.

Use high density

High density puts more nuclei into a given volume, increasing the collision frequency and the probability of close encounters.

Reach the strong-force range

Fusion requires nuclei to approach closely enough for the attractive strong interaction to act and for a bound product to form.

Common trap

Do not use surface temperature or star size as the direct fusion condition. The relevant evidence is high core temperature and density.

E.5.3 Exam Analysis

Assessment in practice

2–3 marks
How it is assessed

Questions explain why fusion occurs in stellar cores or identify which solar features make fusion possible.

Command terms

Explain / State

What earns marks

Mention both temperature and density, then link each to a distinct physical role.

Watch for

Giving only high temperature or citing surface temperature instead of core conditions.

Relate Mass to Evolution

Start when core hydrogen is depleted

Reduced core fusion lowers outward pressure, so gravity contracts and heats the core. Hydrogen shell fusion then expands the outer layers: lower-mass stars become red giants, while high-mass stars become red supergiants.

Initial mass Later pathway Final remnant
Lower or Sun-like Main sequence → red giant → planetary nebula White dwarf
High mass Main sequence → red supergiant → supernova Neutron star or, for a sufficiently massive remnant, black hole

Connect mass to lifetime

A more massive main-sequence star has more fuel, but its fusion rate and luminosity rise much more strongly. It therefore uses core hydrogen faster and has a shorter main-sequence lifetime.

Keep the path conditional

Mass controls the pathway; not every star becomes a supernova, and not every supernova leaves a black hole. A planetary nebula is expelled gas from a red giant, not a planet-forming stage.

E.5.4 Exam Analysis

Assessment in practice

2–3 marks
How it is assessed

Questions compare main-sequence lifetimes or describe stages after a massive star leaves the main sequence.

Command terms

Describe / Compare

What earns marks

Link mass to luminosity and lifetime, then give the ordered evolution and conditional remnant endpoint.

Watch for

Giving the evolution sequence without the mass/luminosity reasoning or naming only one remnant without its condition.

Read the HR Diagram

Read the axes

An HR diagram compares luminosity with surface temperature. Temperature usually decreases from left to right, so the hottest stars are on the left.

HR region Surface temperature Luminosity and size
Main sequence Hot at upper left to cool at lower right Luminosity and typical radius decrease along the sequence
Red giants / supergiants Cool, right side Luminous because their radii are very large
White dwarfs Hot, lower left Dim because their radii are very small
Instability strip Narrow diagonal region crossing the diagram Pulsating stars whose luminosity varies periodically

Use constant-radius lines

From L=4πR2σT4L=4\pi R^2\sigma T^4, a constant-radius line links luminosity and temperature. At the same temperature, greater luminosity means greater radius; at the same luminosity, the hotter star has the smaller radius.

Common trap

Do not read the temperature axis as increasing to the right, and do not identify a white dwarf from temperature alone; its low luminosity is also essential.

E.5.5 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

Questions identify a white dwarf or compare temperatures, radii and luminosities of plotted stars.

Command terms

Identify / Compare

What earns marks

Read both axes with their directions and use the appropriate region or constant-radius relation.

Watch for

Reading the horizontal temperature direction incorrectly or using only one of luminosity and temperature.

Measure Stellar Parallax

Use the Earth’s orbit

Observe a nearby star from opposite sides of Earth’s orbit at different times of year. Its apparent position shifts against distant background stars; half of the total angular shift is the parallax angle pp.

d(\mathrm{pc})=\frac{1}{p(\mathrm{arcsec})}

Calculate distance

When pp is measured in arcseconds, distance in parsecs is d=1/pd=1/p. For p=0.25p=0.25 arcsec, d=4d=4 pc, which can then be converted to light-years.

Distance unit Conversion
1 astronomical unit (AU) 1.50×1011m1.50\times10^{11}\,\mathrm{m}
1 light-year (ly) 9.46×1015m9.46\times10^{15}\,\mathrm{m}
1 parsec (pc) 3.09×1016m=3.26ly=2.06×105AU3.09\times10^{16}\,\mathrm{m}=3.26\,\mathrm{ly}=2.06\times10^5\,\mathrm{AU}

Know the range

Parallax is a geometric distance method and is most useful for relatively nearby stars. It does not use a star’s spectrum or brightness directly.

Common trap

Do not use the full annual position shift as p if the diagram shows the total displacement from one side of Earth’s orbit to the other.

E.5.6 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

Questions calculate distance from a parallax angle or identify what is measured in the method.

Command terms

Calculate / Identify

What earns marks

Identify the positional shift, use p in arcseconds, calculate parsecs, then convert units if requested.

Watch for

Choosing spectral wavelength or intensity as the measured quantity, or forgetting the inverse relation.

Calculate Stellar Radius

Model a star as a spherical black body

Its luminosity LL is the total power radiated from surface area 4πR24\pi R^2 at absolute surface temperature TT. This approximation connects observable luminosity and spectrum-derived temperature to radius.

L=4\pi R^2\sigma T^4\qquad\Rightarrow\qquad R=\sqrt{\frac{L}{4\pi\sigma T^4}}

Form the ratio

R1R2=L1L2(T2T1)4\frac{R_1}{R_2}=\sqrt{\frac{L_1}{L_2}\left(\frac{T_2}{T_1}\right)^4}. A hotter star can have a smaller radius at the same luminosity, while a very luminous cool star must be large.

Worked example — Canopus

For L=10700L=4.12×1030WL=10700L_\odot=4.12\times10^{30}\,\mathrm{W}, T=7400KT=7400\,\mathrm{K} and σ=5.67×108Wm2K4\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}, R=L/(4πσT4)=4.4×1010mR=\sqrt{L/(4\pi\sigma T^4)}=4.4\times10^{10}\,\mathrm{m}. The large radius explains high luminosity despite a moderate surface temperature.

Check powers

Temperature enters to the fourth power and radius enters squared. Keep the temperature ratio in the inverse order shown before taking the square root.

Common trap

Do not use RL/TR\propto L/T; the correct scaling is RL/T2R\propto\sqrt L/T^2.

E.5.7 Exam Analysis

Assessment in practice

2–3 marks
How it is assessed

Questions calculate radius ratios for stars with known luminosities and temperatures.

Command terms

Calculate

What earns marks

Write the law or form a ratio, use the fourth-power temperature term, then take the square root for the radius ratio.

Watch for

Using temperature to the second power or reversing the temperature ratio.

Retrieve the Stellar Model

Retrieve stellar balance

Fusion releases energy, outward thermal/radiation pressure balances inward gravity, and high temperature and density allow fusion in the core.

Retrieve stellar inference

Mass controls evolution; HR regions classify stars; parallax gives distance; and L=4πR2σT4L=4\pi R^2\sigma T^4 gives stellar radius from luminosity and temperature.

Objective notes

7 learning objectives