Free Cheatsheet · Topic E.5 · SL + HL

IB Physics HL E5 Fusion and Stars — Complete Cheatsheet

Every formula, definition and exam trap for IB Physics Topic E.5 — fusion reactions, Wien's Law, Stefan-Boltzmann, the HR diagram, stellar evolution, mass-luminosity, the Jeans criterion and standard candles. Hand-built by an IBO-experienced Singapore tutor.

Topic: Fusion & Stars (Quantum & Nuclear) Syllabus: Topic E.5 (SL + HL) Read time: ~15 minutes Last updated: Apr 2026

Topic E.5 — Fusion and Stars — closes the IB Physics syllabus and is the topic that ties nuclear physics to astrophysics. SL students must handle fusion reactions, Wien's Law, the Stefan-Boltzmann law and the HR diagram; HL students extend this to the mass-luminosity relation, the Jeans criterion for star formation and Type Ia supernovae as standard candles. Examiners love mixing several formulae into a single Paper 2 question — a star's luminosity, then its radius from $L = 4\pi R^2 \sigma T^4$, then its distance from apparent brightness.

This cheatsheet condenses every formula, ratio trick and exam trap from Topic E.5 SL + HL onto one page. Scroll to the bottom for the printable PDF, the full notes pack and the gated tutorial library used by Photon Academy students in Singapore.

Cross-reference with the official IB Physics subject page for the full syllabus.

§1 — Nuclear Fusion E.5 SL/HL

Key fusion reactions

Proton-proton (net):$4\,^{1}_{1}\text{H} \rightarrow\, ^{4}_{2}\text{He} + 2 e^+ + 2 \nu_e + 2 \gamma$
D-T fusion:$^{2}_{1}\text{H} + \,^{3}_{1}\text{H} \rightarrow\, ^{4}_{2}\text{He} + \,^{1}_{0}n + 17.6$ MeV
D-D fusion:$^{2}_{1}\text{H} + \,^{2}_{1}\text{H} \rightarrow\, ^{3}_{1}\text{H} + \,^{1}_{1}p + 4.0$ MeV
TrickMass decreases in fusion (binding energy per nucleon increases). Use $E = \Delta m\, c^2$ to get the energy released per reaction.
TrapSpecific energy of fusion (energy per kg of fuel) is higher than fission, because hydrogen nuclei are far lighter than U-235.

Conditions for fusion

  • High temperature ($\sim 10^7$ K): gives protons enough KE to overcome the Coulomb repulsion barrier.
  • High density: increases the collision frequency $\Rightarrow$ higher reaction rate.
  • Both are required — don't state just one.
NoteSun's core: $T = 1.5 \times 10^7$ K, $\rho = 1.6 \times 10^5$ kg m$^{-3}$.

§2 — Stellar Radiation Laws E.5 SL/HL

Wien's Law:$\lambda_{\max} T = 2.9 \times 10^{-3}$ m K
Stefan-Boltzmann:$L = \sigma A T^4 = 4 \pi R^2 \sigma T^4$
Apparent brightness:$b = \dfrac{L}{4 \pi d^2}$
Radius ratio:$\dfrac{R_A}{R_B} = \sqrt{\dfrac{L_A}{L_B}} \cdot \left(\dfrac{T_B}{T_A}\right)^2$
TrickFor ratio problems use $L_A/L_B = (R_A/R_B)^2 (T_A/T_B)^4$ — this avoids needing $\sigma$ and cancels almost everything.
TrapWien's law: units are m·K (NOT m/K). $\lambda_{\max} = 2.9 \times 10^{-3}/T$. Convert metres to nanometres by dividing by $10^{-9}$.
TrapTemperature axis on the HR diagram runs right to left — hot stars on the LEFT, cool stars on the right.

§3 — Hertzsprung-Russell Diagram E.5 SL/HL

Hertzsprung-Russell diagram plotting stellar luminosity against surface temperature, showing the main sequence band, red giants, and white dwarfs
HR diagram: the diagonal main-sequence band runs top-left (hot, luminous) to bottom-right (cool, dim); red giants sit upper-right, white dwarfs lower-left.

The HR diagram plots luminosity (vertical, increasing upward, in $L_\odot$) against surface temperature (horizontal, increasing to the LEFT, in K).

RegionLocation on HRNotes
Main sequenceDiagonal band, upper-left to lower-right~90% of stars; hydrogen fusion in core
Red giantsUpper rightCool, large radius, evolved low/medium-mass stars
SupergiantsTop of diagramVery luminous, very large; high-mass evolved stars
White dwarfsLower leftHot but very small radius; remnants of low-mass stars
The Sun ($\odot$)Mid-main sequence, $T \approx 5800$ K, $L = L_\odot$G-type main sequence star
NoteLines of constant radius: $L \propto R^2 T^4$, so higher-$R$ lines run parallel to the main sequence, shifted upper-right. A star plotted to the upper-right of another at the same temperature must be larger.

§4 — Stellar Evolution E.5 SL/HL

Evolution pathways

Low / medium mass ($\lesssim 8\, M_\odot$, e.g. the Sun):

Main sequence $\rightarrow$ Red Giant $\rightarrow$ Planetary Nebula $\rightarrow$ White Dwarf.

High mass ($\gtrsim 8\, M_\odot$):

Main sequence $\rightarrow$ Red Supergiant $\rightarrow$ Supernova $\rightarrow$ Neutron Star or Black Hole.

Mass limits

  • Chandrasekhar limit $\approx 1.4\, M_\odot$: maximum mass of a white dwarf, set by electron degeneracy pressure.
  • Oppenheimer-Volkoff limit $\approx 3\, M_\odot$: maximum mass of a neutron star, set by neutron degeneracy pressure.
  • Above the OV limit $\Rightarrow$ collapse to a black hole.
TrickA 2-solar-mass star $\rightarrow$ red giant $\rightarrow$ planetary nebula $\rightarrow$ white dwarf. NOT a supernova — it is below the $\sim 8 M_\odot$ threshold.
Trap"Red giant" vs "red supergiant": giants come from low-mass stars; supergiants come from high-mass stars. They have different endpoints — white dwarf vs supernova / neutron star / black hole.

§5 — HL: Mass-Luminosity & Jeans Criterion E.5 HL

Mass-luminosity relation

Main-sequence:$L \propto M^{3.5} \;\Rightarrow\; \dfrac{L}{L_\odot} = \left(\dfrac{M}{M_\odot}\right)^{3.5}$
MS lifetime:$t \propto M/L \propto M^{-2.5}$
TrickTo find mass from luminosity: $M/M_\odot = (L/L_\odot)^{1/3.5} = (L/L_\odot)^{0.286}$. High-mass stars are massively more luminous AND massively shorter-lived.

Jeans criterion

  • A gas cloud collapses under gravity when $|E_{\text{grav}}| > E_{\text{thermal}}$.
  • Jeans mass: $M_J \propto T^{3/2}\, \rho^{-1/2}$.
  • Lower temperature OR higher density $\Rightarrow$ smaller $M_J$ $\Rightarrow$ easier to collapse $\Rightarrow$ star formation.
NoteSupernovae can trigger nearby cloud collapse by compressing the cloud — increasing $\rho$ and pushing it past the Jeans threshold.

§6 — HL: Standard Candles & Parallax E.5 HL

Stellar parallax

Distance (pc):$d\,(\text{pc}) = \dfrac{1}{p\,(\text{arcsec})}$
1 parsec:$3.09 \times 10^{16}$ m $\approx 3.26$ ly
TrapParallax only works out to about 1000 pc. Beyond this, the parallax angle $p$ is too small to measure — you need a different distance method (e.g. standard candles).

Type Ia supernovae as standard candles

  • A white dwarf accretes mass from a binary companion until it exceeds the Chandrasekhar limit, then undergoes runaway thermonuclear ignition.
  • Always at $\approx 1.4\, M_\odot$ $\Rightarrow$ same peak luminosity every time.
  • $L_{\text{peak}} \approx 4 \times 10^{43}$ W $\approx 5 \times 10^9\, L_\odot$.
  • Used to measure distances to remote galaxies — the foundation of the discovery of the accelerating expansion of the universe.
TrickDistance from apparent brightness: $d = \sqrt{L/(4\pi b)}$. Know this formula and use it in both directions — given $L$ and $b$, find $d$; or given $d$ and $b$, find $L$.
TrapType Ia (white dwarf accretion) is a standard candle. Type II (core collapse of a high-mass star) is NOT a standard candle — its peak luminosity varies with progenitor mass.

Worked Example — HR Diagram & Distance

Question (HL Paper 2 style — 7 marks)

Star Sirius A has a peak emission wavelength $\lambda_{\max} = 290$ nm and an apparent brightness $b = 1.2 \times 10^{-7}$ W m$^{-2}$ as measured from Earth. Its luminosity is $L = 25\, L_\odot$ where $L_\odot = 3.83 \times 10^{26}$ W.
(a) Estimate the surface temperature of Sirius A using Wien's Law. [2]
(b) Calculate the distance to Sirius A in metres. [3]
(c) State whether Sirius A lies on the main sequence and justify briefly. [2]

Solution

  1. Wien: $T = 2.9 \times 10^{-3} / \lambda_{\max} = 2.9 \times 10^{-3} / (290 \times 10^{-9})$.  (M1)
  2. $T \approx 1.0 \times 10^4$ K (about 10 000 K — a hot blue-white star).  (A1)
  3. Convert luminosity: $L = 25 \times 3.83 \times 10^{26} = 9.58 \times 10^{27}$ W.  (M1)
  4. Apply $b = L/(4 \pi d^2) \Rightarrow d = \sqrt{L/(4\pi b)} = \sqrt{(9.58 \times 10^{27})/(4\pi \times 1.2 \times 10^{-7})}$.  (M1)
  5. $d \approx 8.0 \times 10^{16}$ m (about 2.6 pc, in good agreement with the measured 2.64 pc to Sirius).  (A1)
  6. At $T \approx 10000$ K and $L \approx 25\, L_\odot$, Sirius A plots in the upper-left region of the main sequence — luminous and hot.  (R1)
  7. It IS on the main sequence (it is a hydrogen-fusing A-type star, not a giant or white dwarf — its measured radius is consistent with $L = 4\pi R^2 \sigma T^4$ for a main-sequence value).  (A1)

Examiner's note: Two common errors. (i) Forgetting to convert $L$ from solar units to watts before using the brightness equation — keep all SI units throughout. (ii) Mixing up the temperature axis direction on the HR diagram (hot is on the LEFT) when justifying part (c). Always sketch the HR diagram in rough first to anchor your reasoning.

Common Student Questions

Why does a star need both high temperature AND high density to fuse?
High temperature gives the protons enough kinetic energy to overcome the Coulomb barrier (electrostatic repulsion) when they collide. High density gives a high collision frequency, so enough fusion events happen per second to keep the star shining. Either alone is not enough — the Sun's core sits at $\sim 1.5 \times 10^7$ K AND $\sim 1.6 \times 10^5$ kg m$^{-3}$.
Which way does temperature go on the Hertzsprung-Russell diagram?
Temperature increases to the LEFT — hot blue stars on the left, cool red stars on the right. Luminosity increases upward. The main sequence runs diagonally from upper-left (hot, luminous) to lower-right (cool, faint). Confusing the temperature axis direction is a frequent IB mark loss.
Does every star end its life as a supernova?
No. Only high-mass stars ($\gtrsim 8\, M_\odot$) end as core-collapse supernovae and leave behind a neutron star or black hole. Low / medium-mass stars like the Sun ($\lesssim 8\, M_\odot$) become red giants, shed a planetary nebula and end as white dwarfs — no supernova. A 2-solar-mass star will NOT supernova.
Why are Type Ia supernovae used as standard candles, but not Type II?
Type Ia supernovae form from a white dwarf accreting mass until it exceeds the Chandrasekhar limit ($\sim 1.4\, M_\odot$) and undergoes runaway thermonuclear ignition. Because the trigger mass is always the same, the peak luminosity is always the same ($\sim 5 \times 10^9\, L_\odot$). Type II supernovae are core-collapse events from variable-mass progenitors, so their peak luminosities vary widely — they cannot be used as standard candles.
What are the units of Wien's displacement law constant?
The Wien constant is $2.9 \times 10^{-3}$ m·K (metre-kelvin), NOT m/K. The law is $\lambda_{\max} \cdot T = 2.9 \times 10^{-3}$ m·K, so to find peak wavelength you divide: $\lambda_{\max} = 2.9 \times 10^{-3}/T$ (in metres). To convert to nanometres, divide the metre value by $10^{-9}$. Mixing up the units is the most common Wien-law mark loss.

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