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

These notes condense E.5, Fusion and stars, from the IB Physics guide, first assessment 2025, for Standard Level and Higher Level. E.5 has no additional higher level content. The study guide has the full explanations. When you are ready to practise, use the practice questions.

Related pages: IB DP Physics, the checklist, binding energy and the free diagnostics.

Key ideas in one place

• Stability: outward pressure (gas and radiation) balances inward gravity at every layer. Any change is corrected, so a star is steady while it has fuel.
• Energy source: fusion of light nuclei. The products have less mass than the reactants, and E = Δm c².
• Conditions: high temperature to overcome electric repulsion, high density so that collisions are frequent.
• Mass controls the story: more mass means hotter core, higher luminosity, shorter life, and a more violent end.
• HR diagram: luminosity up, temperature across, with temperature increasing to the left.
• Distance: d (pc) = 1 / p (arc-second).
• Size: R from L and T using L = 4πσR²T⁴.
• Spectrum: peak wavelength gives temperature; absorption lines give composition.

Formulas and data

Quantity	Relationship or value
Energy released	E = Δm c²
Mass-energy in u	1 u is equivalent to 931.5 MeV c⁻²
Energy unit	1 MeV = 1.602 × 10⁻¹³ J
Parallax	d (pc) = 1 / p (arc-second)
Distance units	1 pc = 3.26 ly = 3.09 × 10¹⁶ m = 2.06 × 10⁵ AU
Other units	1 ly = 9.46 × 10¹⁵ m; 1 AU = 1.50 × 10¹¹ m; 1 degree = 3600 arc-seconds
Luminosity	L = σAT⁴ = 4πσR²T⁴
Stefan-Boltzmann constant	σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
Peak wavelength	λmax T = 2.9 × 10⁻³ m K
Solar values	L☉ = 3.83 × 10²⁶ W; R☉ = 6.96 × 10⁸ m

These constants are data for working. In an exam, use whatever values the question supplies.

Stability and fusion in steps

Explaining the balance in a steady star

1. Gravity pulls layers inward.
2. Hot gas and radiation push outward.
3. If the core cools, the star contracts, the core heats, and fusion speeds up.
4. If the core heats, the star expands, the core cools, and fusion slows.

Fusion energy calculation

1. Add the masses of the reactants and subtract the product mass: Δm.
2. Convert: Δm (u) × 931.5 gives MeV; or Δm (kg) × c² gives J.
3. For a rate, divide the power by the energy per reaction.
4. For hydrogen used, multiply the rate by the mass of four hydrogen nuclei.

Reminder. Four hydrogen atoms (4 × 1.007825 u) to helium-4 (4.002603 u) loses 0.028697 u, which is 26.7 MeV, about 0.71 % of the mass. The Sun fuses about 9 × 10³⁷ nuclei every second.

Conditions. Two protons at 1.0 × 10⁻¹⁵ m have a potential energy of about 1.4 MeV, while the mean thermal energy at 1.5 × 10⁷ K is only about 1.9 keV. The gap is bridged by the fastest nuclei and by quantum effects.

Evolution by mass

Stage	Low and moderate mass	High mass
Birth	Cloud collapses, fusion starts	Same, but faster and hotter
Main sequence	Long (billions of years)	Short
After core hydrogen	Red giant	Super giant
End	Outer layers shed, white dwarf remains	Supernova, neutron star or black hole

Fusion up to iron releases energy. Fusion beyond iron does not, so an iron core cannot support itself.

HR diagram: regions at a glance

Region	Where	Remember
Main sequence	Diagonal band, top left to bottom right	Hydrogen fusing in the core
Red giants	Upper right	Cool but luminous, so very large
Super giants	Top	Largest and most luminous
White dwarfs	Bottom left	Hot but faint, so small
Instability strip	Near-vertical band through the giant region	Variable brightness
Lines of constant radius	Straight on log axes (L ∝ T⁴)	Above and to the right means larger

Moving across the diagram. A star with the same luminosity but half the surface temperature has a radius four times larger, because R ∝ 1/T² when L is fixed.

Must-know distinctions

• Luminosity is the total power emitted by the star. It is not the same as the brightness you see from Earth.
• Parallax angle is half the total apparent shift, with a baseline of 1 AU.
• Parsec versus light year: the parsec is defined from the angle, the light year from the distance light travels in a year. 1 pc is about 3.26 ly.
• Mass defect versus mass lost by the star: a star converts 0.7 % of the hydrogen it fuses into energy. It does not lose all of that hydrogen.
• Temperature from the spectrum versus composition from the spectrum: peak wavelength gives temperature, absorption lines give elements.

Radius method in steps

1. Convert luminosity to watts if it is given in L☉.
2. Convert temperature to kelvin.
3. Compute R = √(L / 4πσT⁴).
4. Give the answer in metres, and in R☉ if asked.

A white dwarf with L = 1.2 × 10⁻³ L☉ and T = 12 000 K: L = 4.60 × 10²³ W, so R = 5.58 × 10⁶ m, roughly the size of a planet.

Worked reminders

Parallax to metres. p = 0.0125 arc-second gives d = 1/0.0125 = 80 pc, which is 80 × 3.26 = 261 ly or 80 × 3.09 × 10¹⁶ = 2.47 × 10¹⁸ m.

Power to reaction rate. A star radiates 2.0 × 10³⁰ W and each reaction releases 1.15 × 10⁻¹² J. The rate is 2.0 × 10³⁰ / 1.15 × 10⁻¹² = 1.7 × 10⁴² reactions per second.

Comparing two stars. Since L ∝ R²T⁴, a star with 25 times the solar radius and a surface temperature of 3800 K has L/L☉ = 25² × (3800/5800)⁴ = 115. State the value of T☉ you use, because the answer depends on it.

Writing a short explanation. For “explain why a star leaves the main sequence”, give a chain: core hydrogen is used up, fusion in the core falls, outward pressure drops below gravity, the core contracts and heats, fusion continues in a shell around it, and the outer layers expand and cool.

Quick self-test

1. State what two quantities balance in a stable star.
2. A star has p = 0.20 arc-second. Find d in pc.
3. A star is 50 pc away. Find its parallax angle.
4. Convert 8.15 ly to pc.
5. A star has L = 4.60 × 10²³ W and T = 12 000 K. Is it more likely to be a white dwarf or a red giant, and why?
6. Three helium-4 nuclei (4.002603 u each) fuse into carbon-12 (12.000000 u). Find the energy released in MeV.
7. A star’s surface temperature halves while its luminosity stays constant. By what factor does R change?
8. A star’s spectrum peaks at 4.8 × 10⁻⁷ m. Find T.
9. Give one reason very massive stars have short lives.
10. Why does fusion in a core stop producing energy once iron forms?
11. Where on the HR diagram would you find a cool, luminous star?
12. State why high density as well as high temperature is needed for fusion.

Answers

1. Outward pressure (from gas and radiation) and inward gravitational force.
2. d = 1/0.20 = 5.0 pc.
3. p = 1/50 = 0.020 arc-second.
4. 8.15 / 3.26 = 2.50 pc.
5. White dwarf: the luminosity is far below the Sun’s (about 10⁻³ L☉) while the temperature is higher than the Sun’s, so the star must be small.
6. Δm = 3 × 4.002603 − 12.000000 = 0.007809 u; E = 0.007809 × 931.5 = 7.27 MeV.
7. R ∝ 1/T² at fixed L, so the radius becomes 4 times larger.
8. T = 2.9 × 10⁻³ / 4.8 × 10⁻⁷ = 6.0 × 10³ K.
9. A larger mass means a hotter core and a much higher luminosity, so fuel is used up far faster.
10. Fusing iron would absorb energy instead of releasing it, so the core cannot be held up by fusion.
11. Upper right, in the red giant or super giant region.
12. A high density makes collisions frequent enough for the fusion rate to be large.

Where marks are usually lost

• Placing hot stars on the right of the HR diagram.
• Not converting arc-seconds correctly, or using the full shift as p.
• Forgetting that the mass defect is in u and must be multiplied by 931.5 for MeV.
• Mixing L in L☉ with σ in SI units, so the radius is off by a factor of 10²⁶ in the square root.
• Writing “gravity equals fusion” instead of gravity balanced by outward pressure.
• Dropping the fourth power on T in the radius formula.
• Describing a white dwarf as cool and large. It is hot and small.
• Giving distances in pc without checking whether the question wants ly or metres.
• Leaving out units, or quoting more significant figures than the data support.

Official syllabus

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