IB DP Physics: Fusion, stellar evolution, HR diagrams and stellar distances (DP Physics)

This guide covers sub-topic E.5, Fusion and stars, from Topic E, Nuclear and quantum physics, in the IB Physics guide, first assessment 2025. The content is the same at Standard Level and Higher Level, and E.5 has no additional higher level material. It is aligned to the guide for first assessment 2025 onwards.

Work through the sections in order, then use the revision notes for recall and the practice questions for timed work. Course page: IB DP Physics. Checklist: IB DP Physics checklist. Test yourself with the free diagnostics.

Two earlier guides feed into this one. The mass defect and binding energy work is in nuclear structure and binding energy. The luminosity, Stefan-Boltzmann and Wien relationships are in thermal radiation.

What E.5 asks you to do

Guide point	What you must be able to do	Level
Stability of stars	Explain stability as a balance of outward radiation pressure and inward gravitational force	SL and HL
Fusion as an energy source	Explain that stars shine because of fusion, and calculate energy released	SL and HL
Conditions for fusion	Link fusion to density and temperature	SL and HL
Effect of stellar mass	Compare how stars of different mass evolve	SL and HL
HR diagram	Sketch and interpret it: main sequence, red giants, super giants, white dwarfs, instability strip, lines of constant radius	SL and HL
Stellar parallax	Use d (parsec) = 1/p (arc-second); convert between AU, ly and pc	SL and HL
Stellar radii	Find a radius from luminosity and surface temperature	SL and HL
Spectra	Say what a stellar spectrum shows about surface temperature and composition	SL and HL

Cepheid variables are not required. Every understanding above applies to both levels.

Stability: gravity against pressure

A star is a huge ball of gas. Gravity pulls every layer inward. The hot gas and the radiation flowing out from the core push every layer outward. A stable star sits in equilibrium: at each depth, the outward pressure from the gas and the radiation balances the inward gravitational force on the material above.

This balance corrects itself. If the core cooled slightly, the outward push would fall, the star would shrink, the core would be squeezed and heated, and fusion would speed up until the push returned. If the core overheated, the star would expand, the core would cool and fusion would slow. This feedback is why a main sequence star can shine steadily for a very long time.

When the core runs short of fuel, the outward push can no longer match gravity. The balance shifts, and the star changes its structure. That is the start of stellar evolution.

Fusion as the energy source

In a star’s core, light nuclei join to form heavier nuclei. The product has a smaller total mass than the parts. The difference is the mass defect Δm, and the energy released is

E = Δm c²

You can work in joules with Δm in kilograms, or in MeV using the fact that 1 u is equivalent to 931.5 MeV c⁻². The nuclear masses will be given in the question in kg, MeV c⁻² or u.

Hydrogen fusion is the main source of energy for a main sequence star. The overall result is that four hydrogen nuclei become one helium-4 nucleus, with positrons and neutrinos emitted.

Worked example 1: energy from hydrogen fusion. Use these atomic masses: hydrogen-1 is 1.007825 u and helium-4 is 4.002603 u. Find the energy released when four hydrogen atoms become one helium-4 atom, and the fraction of the mass converted.

Δm = 4 × 1.007825 − 4.002603 = 0.028697 u
E  = 0.028697 × 931.5 = 26.7 MeV
fraction = 0.028697 / (4 × 1.007825) = 0.00712, about 0.71 %

Using atomic masses includes the electrons, so the figure counts the energy from positron annihilation. A small part leaves with the neutrinos.

Worked example 2: how much hydrogen does a star burn? The Sun has luminosity 3.83 × 10²⁶ W. Take 26.7 MeV per helium nucleus, with 1 MeV = 1.602 × 10⁻¹³ J.

E per fusion = 26.7 × 1.602 × 10⁻¹³ = 4.28 × 10⁻¹² J
fusions per second = 3.83 × 10²⁶ / 4.28 × 10⁻¹² = 8.94 × 10³⁷ s⁻¹
hydrogen used per second = 8.94 × 10³⁷ × 4 × 1.007825 × 1.661 × 10⁻²⁷ kg
                         = 6.0 × 10¹¹ kg per second
mass converted to energy = L / c² = 3.83 × 10²⁶ / (3.00 × 10⁸)² = 4.26 × 10⁹ kg per second

The second figure is much smaller than the first. Almost all the hydrogen ends up as helium. Only the 0.7 % mass difference becomes energy.

Conditions for fusion: density and temperature

Nuclei are positively charged, so they repel. To fuse they must get close enough for the strong nuclear force to win. That needs two things.

• High temperature, so nuclei move fast enough to approach one another against the electric repulsion. Hydrogen fusion in a star’s core needs a temperature of the order of 10⁷ K.
• High density, so collisions between nuclei are frequent enough that the fusion rate is large.

Worked example 3: the barrier. Two protons at a separation of 1.0 × 10⁻¹⁵ m have electric potential energy

U = k e² / r = 8.99 × 10⁹ × (1.602 × 10⁻¹⁹)² / 1.0 × 10⁻¹⁵
  = 2.31 × 10⁻¹³ J = 1.44 MeV

The mean thermal energy at 1.5 × 10⁷ K is 3/2 kB T = 1.5 × 1.38 × 10⁻²³ × 1.5 × 10⁷ = 3.1 × 10⁻¹⁶ J, about 1.9 keV. The barrier is roughly 740 times larger. Fusion still happens because some nuclei are far faster than average, and because quantum effects let nuclei fuse without climbing the whole barrier. The huge number of nuclei does the rest.

Heavier fuels need higher temperatures, because the charges are larger and the repulsion is stronger.

Stellar mass and evolution

A star forms from a collapsing cloud of gas. When the core is hot and dense enough, hydrogen fusion starts and the star joins the main sequence. Mass decides almost everything that follows.

• A more massive star has a stronger gravitational squeeze, a hotter core and a higher luminosity. It burns its fuel much faster, so it spends less time on the main sequence.
• A star of low or moderate mass (up to roughly eight times the mass of the Sun) swells into a red giant when core hydrogen runs out. Later its outer layers drift away and the core remains as a white dwarf.
• A star of high mass becomes a super giant. Its core can fuse heavier elements in stages. Fusion up to iron releases energy, but fusing iron does not, so the core cannot hold itself up. It collapses and the star explodes as a supernova, leaving a neutron star or a black hole.

The mass limits are approximate. The guide asks for the effect of mass, so concentrate on the pattern: low mass gives a long life ending as a white dwarf, and high mass gives a short life ending in a supernova and a compact remnant.

The HR diagram

The Hertzsprung-Russell diagram plots luminosity on the vertical axis against surface temperature on the horizontal axis. Both axes are logarithmic, and temperature increases to the left, so hot stars are on the left.

Region	Position	Typical properties
Main sequence	Diagonal band, top left to bottom right	Fusing hydrogen in the core; most stars; size and luminosity rise with mass
Red giants	Upper right	Cool surface, luminosity of order 10² to 10³ times the Sun, radius many times solar
Super giants	Top	Very luminous, very large; hot blue ones and cool red ones
White dwarfs	Bottom left	Hot but faint, so small, about the size of a planet
Instability strip	Near-vertical band crossing the giant region	Stars here are unstable and vary in brightness

Lines of constant radius. From L = 4πσR²T⁴, a star of fixed R has L ∝ T⁴. On logarithmic axes this is a straight line. Stars above and to the right of a line are larger than stars on it. White dwarfs sit on lines of small radius, and giants and super giants on lines of large radius.

To sketch the diagram: label the axes, draw the main sequence band, put giants and super giants above it on the right, put white dwarfs below it on the left, mark the instability strip, and add a few constant-radius lines running down to the right with radius increasing towards the top right.

Measuring distance: parallax

As Earth orbits the Sun, a nearby star appears to shift against the background of distant stars. The parallax angle p is half the total shift, measured against a baseline of 1 AU. One parsec is the distance at which p is one arc-second.

d (parsec) = 1 / p (arc-second)

Units to know, given as data: 1 pc = 3.09 × 10¹⁶ m = 3.26 ly = 2.06 × 10⁵ AU, 1 ly = 9.46 × 10¹⁵ m, and 1 AU = 1.50 × 10¹¹ m. Also 1 degree = 3600 arc-seconds.

Worked example 4. A star has p = 0.0832 arc-second.

d = 1 / 0.0832 = 12.0 pc
  = 12.0 × 3.26 = 39.1 ly
  = 12.0 × 3.09 × 10¹⁶ = 3.71 × 10¹⁷ m

Worked example 5. A star is 7.5 pc away. Then p = 1/7.5 = 0.13 arc-second. A smaller angle means a greater distance, so parallax works only for relatively near stars.

Stellar radii

Combine L = σAT⁴ with A = 4πR² to get L = 4πσR²T⁴, so R = √(L / 4πσT⁴).

Worked example 6. A red super giant has L = 2.0 × 10⁵ L☉ and surface temperature 3500 K. With L☉ = 3.83 × 10²⁶ W and σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴:

L = 2.0 × 10⁵ × 3.83 × 10²⁶ = 7.66 × 10³¹ W
R² = 7.66 × 10³¹ / (4π × 5.67 × 10⁻⁸ × 3500⁴) = 7.16 × 10²³ m²
R  = 8.46 × 10¹¹ m, about 1220 times the solar radius 6.96 × 10⁸ m

Worked example 7: ratio method. With R/R☉ = √(L/L☉) × (T☉/T)² and T☉ taken as 5800 K, a star with L = 2500 L☉ and T = 9500 K has R/R☉ = 50 × 0.373 = 18.6. Rounding T☉ makes this differ slightly from the absolute method, so use one method throughout a question.

What a spectrum tells you

• Surface temperature. A star radiates close to a black body, so the wavelength of peak emission gives the temperature through λmax T = 2.9 × 10⁻³ m K. A peak at 3.2 × 10⁻⁷ m gives T = 9.1 × 10³ K.
• Composition. Cooler gas in the outer layers absorbs light at wavelengths that match the energy-level transitions of the elements present. Dark lines in the spectrum identify those elements. Line patterns for atoms are explained in atomic structure and spectra.

Common errors

• Reversing the temperature axis: hot is on the left.
• Writing d = p, or using p in degrees. Convert to arc-seconds first.
• Treating the parallax angle as the full shift rather than half.
• Using T² instead of T⁴, or forgetting the square on R.
• Giving a mass defect in u as an energy. Multiply by 931.5 to get MeV.
• Saying a star is stable because “fusion stops gravity”. The correct statement is a balance of forces.

Official syllabus

Source: IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. This page is an independent paraphrase and is not an IB publication.