Key points
- An HR diagram plots luminosity (vertical, logarithmic) against surface temperature (horizontal, decreasing to the right).
- Stefan's law L = 4πr²σT⁴ means a cool star can only be very luminous if it is very large.
- Wien's law λmax T = 2.90 × 10⁻³ m K links peak wavelength to surface temperature.
- A star leaves the main sequence when the hydrogen in its core runs out.
- A white dwarf cannot have a mass above the Chandrasekhar limit, about 1.4 solar masses.
What the Hertzsprung-Russell diagram shows
The Hertzsprung-Russell (HR) diagram is a graph that sorts stars by two properties: how much energy they radiate every second (their luminosity) and how hot their surface is. When thousands of stars are plotted, they do not scatter randomly. They fall into clear groups, and each group is a different stage in the life of a star.
Luminosity is the total power radiated by a star, measured in watts. The Sun's luminosity is about 3.85 × 10²⁶ W. Because stars vary in luminosity by a factor of a million or more in either direction, the luminosity axis is logarithmic, usually in units of the Sun's luminosity so that the Sun sits at 1.
The axes: why temperature runs backwards
There are two features of the HR diagram that catch students out:
- The vertical axis is luminosity on a logarithmic scale. Each gridline is a factor of 100 (or 10) greater than the one below.
- The horizontal axis is surface temperature, and it decreases from left to right. Hot blue stars are on the left and cool red stars are on the right. This is a historical convention and it is the same in every exam question, so always check the direction of the axis before reading a value.
Some diagrams use spectral class on the horizontal axis (O, B, A, F, G, K, M from hot to cool) or colour index. These are all ways of describing temperature, so the same left-to-right pattern holds.
The regions of the diagram
| Group | Temperature | Luminosity | What it tells you |
|---|---|---|---|
| Main sequence | Wide range | Wide range, rising with temperature | Core hydrogen fusion; about 90% of a star's life is spent here |
| Red giants | Low (about 3000 to 5000 K) | High | Very large radius (cool but luminous) |
| Supergiants | Low to high | Very high | Very large and massive stars near the end of their lives |
| White dwarfs | High at first, cooling | Low | Very small radius, about the size of the Earth |
The shape comes from Stefan's law. A red giant has a low surface temperature, so each square metre radiates little energy. It is luminous only because its surface area is huge. A white dwarf has the opposite problem: each square metre radiates a lot, but there are very few square metres. You can use this reasoning in exam answers: the position on the diagram tells you the relative size of the star.
The position on the main sequence depends mainly on mass. More massive stars are hotter and much more luminous, so they use up their hydrogen faster and spend less time on the main sequence.
Stefan's law: luminosity, radius and temperature
L = 4πr²σT⁴
- L = luminosity / W
- r = radius of the star / m
- σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
- T = surface (thermodynamic) temperature / K
This treats a star as a black body, a perfect emitter. 4πr² is the surface area of a sphere.
The law says that power radiated per square metre is σT⁴, and the star has a surface area of 4πr². Temperature has a fourth-power effect: double the temperature and the luminosity increases by a factor of 2⁴ = 16, assuming the radius stays the same. Double the radius and the luminosity increases by a factor of 4.
In practice, astronomers measure how bright a star looks (its intensity at the Earth), use its distance to get the luminosity, find its temperature from its spectrum using Wien's law, and then use Stefan's law to find the radius. That is the chain of reasoning behind many exam questions.
Wien's law: peak wavelength and temperature
λmax T = 2.90 × 10⁻³ m K
- λmax = wavelength at which the star emits the most intensity / m
- T = surface temperature / K
A hotter star has a shorter peak wavelength. This is why hot stars look blue and cool stars look red.
Wien's law lets you measure the temperature of a star that is many light-years away, using nothing but its light. Note that the curves in the diagram do not match the real spectrum of a star exactly, because stars are only approximately black bodies.
Line spectra, energy levels and what stars are made of
Light from a star carries information beyond its peak wavelength. A hot, dense body such as a star's interior produces a continuous spectrum. As this light passes through the cooler gas of the star's outer atmosphere, atoms absorb photons that match the energy gaps between their electron energy levels. The result is a continuous spectrum crossed by dark absorption lines.
- A continuous spectrum contains all wavelengths in a range.
- An emission line spectrum is bright lines on a dark background, from a hot, low-density gas.
- An absorption line spectrum is dark lines on a continuous background, from cool gas in front of a hot source.
An electron moving between energy levels E₁ and E₂ emits or absorbs a photon of energy ΔE = hf = hc/λ. Each element has its own pattern of energy levels, so each has its own pattern of lines. Astronomers match the dark lines in a star's spectrum to those measured for elements in the laboratory to find out what the star's atmosphere contains.
Stellar evolution: from nebula to remnant
Formation. A cloud of gas and dust, mostly hydrogen, is pulled together by gravity. As it collapses, gravitational potential energy is transferred to thermal energy, so the temperature rises and the cloud forms a protostar. When the core reaches a temperature of the order of 10⁷ K, hydrogen nuclei can overcome their electrostatic repulsion and fuse into helium. Fusion begins and the star joins the main sequence.
Main sequence. The star is stable because the outward pressure of the hot gas and radiation balances the inward pull of gravity. The star remains in this equilibrium while hydrogen fusion in the core releases energy.
Low-mass stars (like the Sun). When the hydrogen in the core is used up, the core contracts and heats up, and hydrogen fusion continues in a shell around it. The outer layers expand and cool: the star becomes a red giant. Eventually the outer layers drift away as a planetary nebula (the name is historical and has nothing to do with planets). The core that is left is a white dwarf: a hot, dense object about the size of the Earth that no longer fuses and slowly cools. It is held up by electron degeneracy pressure, a quantum effect that does not depend on temperature.
The Chandrasekhar limit. Electron degeneracy pressure can only support a white dwarf up to about 1.4 solar masses. A remnant core above this limit cannot become a stable white dwarf.
High-mass stars. A high-mass star becomes a red supergiant and can fuse heavier elements in stages, up to iron. Iron nuclei have the highest binding energy per nucleon, so fusing them does not release energy (see binding energy and mass defect). The core collapses in a fraction of a second and the outer layers are blown off in a supernova, one of the most energetic events in the universe. What remains of the core is a neutron star, a ball of neutrons supported by neutron degeneracy pressure, or, if the remnant is massive enough, a black hole: an object so dense that nothing, not even light, can escape from within its event horizon.
Worked examples
Surface temperature from a spectrum
The radiation from a star has a peak intensity at a wavelength of 500 nm. Use Wien's law (λmax T = 2.90 × 10⁻³ m K) to calculate the surface temperature of the star.
- Convert: 500 nm = 500 × 10⁻⁹ = 5.00 × 10⁻⁷ m
- T = 2.90 × 10⁻³ / λmax
- T = 2.90 × 10⁻³ / 5.00 × 10⁻⁷ = 5800 K
Answer 5.8 × 10³ K
Luminosity of a red giant
A red giant has a surface temperature of 3500 K and a radius of 8.0 × 10¹⁰ m. (a) Calculate its luminosity. (b) The Sun's luminosity is 3.85 × 10²⁶ W. How many times more luminous is the red giant?
- L = 4πr²σT⁴
- r² = (8.0 × 10¹⁰)² = 6.4 × 10²¹ m²
- T⁴ = 3500⁴ = 1.50 × 10¹⁴ K⁴
- L = 4π × 6.4 × 10²¹ × 5.67 × 10⁻⁸ × 1.50 × 10¹⁴ = 6.8 × 10²⁹ W
- Ratio = 6.84 × 10²⁹ / 3.85 × 10²⁶ = 1777
Answer (a) 6.8 × 10²⁹ W (b) about 1.8 × 10³ times
A white dwarf
A white dwarf has a radius of 6.5 × 10⁶ m and a surface temperature of 1.2 × 10⁴ K. Calculate its luminosity and explain why it is found at the bottom left of the HR diagram.
- L = 4πr²σT⁴ = 4π × (6.5 × 10⁶)² × 5.67 × 10⁻⁸ × (1.2 × 10⁴)⁴
- = 4π × 4.225 × 10¹³ × 5.67 × 10⁻⁸ × 2.07 × 10¹⁶
- = 6.2 × 10²³ W
- Compared with the Sun this is about 0.0016 times as luminous
- The temperature is high (left) and the luminosity is very low (bottom) because the radius is tiny
Answer L = 6.2 × 10²³ W. It is hot but has a very small surface area, so it is dim and appears at the bottom left.
Comparing two stars
Star A has twice the surface temperature of star B and three times the radius. Calculate the ratio of the luminosity of A to B.
- L ∝ r²T⁴
- Ratio = (rA/rB)² × (TA/TB)⁴
- = 3² × 2⁴ = 9 × 16 = 144
Answer LA is 144 times LB
Wavelength of a spectral line
Two energy levels of a hydrogen atom are −3.40 eV and −1.51 eV. Calculate the wavelength of the photon emitted when an electron falls from the higher to the lower level. (1 eV = 1.60 × 10⁻¹⁹ J)
- Energy gap ΔE = −1.51 − (−3.40) = 1.89 eV
- In joules: 1.89 × 1.60 × 10⁻¹⁹ = 3.024 × 10⁻¹⁹ J
- λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 3.024 × 10⁻¹⁹ = 6.58 × 10⁻⁷ m
Answer 6.6 × 10⁻⁷ m (658 nm, which is red light)
Common mistakes
How OCR A examines the HR diagram
The topic appears as a mix of recall, calculation and extended explanation:
- Annotating or reading an HR diagram, and describing how a star moves across it during its life.
- Calculations using Stefan's law and Wien's law, often combined, or ratio questions comparing two stars.
- Describing stellar evolution for a star of given mass, including the Chandrasekhar limit.
- Explaining line spectra and energy levels, and calculating photon energies or wavelengths.
Synoptic links are common. The explanation of fusion uses the ideas in binding energy and mass defect, the idea of a star in equilibrium needs gravitational fields, and distances to stars lead into Hubble's law and redshift. The Module 5 hub lists every topic in this module.
Practice questions
The peak wavelength of the radiation from a star is 290 nm. Calculate its surface temperature and say what colour of light you would expect it to look.3 marks
Show the answer
Answer: 1.0 × 10⁴ K. The peak is in the ultraviolet to blue, so the star looks blue-white.
T = 2.90 × 10⁻³ / 2.90 × 10⁻⁷ = 1.0 × 10⁴ K.
The Sun has a luminosity of 3.85 × 10²⁶ W and a surface temperature of 5800 K. Calculate the radius of the Sun.4 marks
Show the answer
Answer: 6.9 × 10⁸ m
r² = L / (4πσT⁴) = 3.85 × 10²⁶ / (4π × 5.67 × 10⁻⁸ × 5800⁴) = 3.85 × 10²⁶ / 8.06 × 10¹⁷ = 4.78 × 10¹⁷ m². r = √(4.78 × 10¹⁷) = 6.91 × 10⁸ m.
A star has a surface temperature of 3000 K and is 10⁴ times as luminous as the Sun. Estimate its radius in terms of the Sun's radius, taking the Sun's temperature as 5800 K.4 marks
Show the answer
Answer: About 370 times the radius of the Sun. It is a red supergiant or giant.
L ∝ r²T⁴, so r ∝ √L / T². r/r(Sun) = √(10⁴) × (5800/3000)² = 100 × 3.74 = 374.
A dying star leaves a remnant core of mass 1.2 solar masses. A different star leaves a core of mass 2.0 solar masses. State what each is likely to become.2 marks
Show the answer
Answer: The 1.2 solar mass core is below the Chandrasekhar limit (about 1.4 solar masses) and becomes a white dwarf. The 2.0 solar mass core is above it and cannot be a white dwarf, so it becomes a neutron star (or a black hole if it is massive enough).
Compare each mass with 1.4 solar masses.
Explain why the spectrum of a star shows dark lines.3 marks
Show the answer
Answer: Light from the hot interior has a continuous spectrum. Atoms in the cooler outer atmosphere absorb photons whose energy matches the gap between two energy levels (hf = ΔE). These wavelengths are missing from the light that emerges, so dark lines appear.
Link: continuous spectrum, absorption of specific photon energies, missing wavelengths.
Next steps
Now that you know how stars are classified, see how light from far-away galaxies is used to measure the whole universe in Hubble's law and redshift. For the force that holds stars together, read gravitational fields. You can also practise this topic with step-by-step marking, which is free to start.
Frequently asked questions
Why is the temperature axis on the HR diagram backwards?
It is a historical convention: the diagram was drawn with the hottest stars (spectral class O) on the left and the coolest on the right. It stays that way in every textbook and exam, so always check which way the axis runs before reading a value.
What does the main sequence on the HR diagram show?
The main sequence is the diagonal band of stars that are fusing hydrogen into helium in their cores. A star's position on it depends mainly on its mass: massive stars are hot and luminous, and low-mass stars are cool and dim.
Why are red giants so luminous if they are cool?
Luminosity depends on both surface temperature and surface area (L = 4πr²σT⁴). A red giant has a low surface temperature, but its radius is enormous, so its surface area is huge and the total power it radiates is large.
What is the Chandrasekhar limit?
It is the maximum mass of a stable white dwarf, about 1.4 solar masses. Above it, electron degeneracy pressure cannot support the star against gravity, so the core collapses further, to a neutron star or a black hole.
What is the difference between Wien's law and Stefan's law?
Wien's law (λmax T = 2.90 × 10⁻³ m K) gives the wavelength at which a hot object radiates most strongly, so it tells you the temperature. Stefan's law (L = 4πr²σT⁴) gives the total power it radiates, so it links luminosity to radius and temperature.
Related guides
Written for the OCR A Level Physics A (H556) specification. Questions are original and written in exam style; they are not taken from OCR papers. Spotted a mistake? Tell us.