Stellar Classification

The second half of the Edexcel gravitational fields and astrophysics topic moves from gravity and orbits into the physics of stars and the universe. Stellar classification is the starting point — understanding how stars are categorised by their observable properties, particularly their spectra, temperatures, and luminosities.

How We Study Stars

We cannot visit stars, so everything we know comes from analysing the light (electromagnetic radiation) they emit. The three key observables are:

Luminosity — the total power output of a star (in watts)
Surface temperature — determined from the star's spectrum or colour
Spectrum — the pattern of absorption lines that reveals the star's chemical composition

Black Body Radiation and Wien's Law

Stars behave approximately as black body radiators — objects that absorb and emit radiation at all wavelengths. A black body spectrum has a characteristic shape: intensity rises to a peak and then falls off, with the peak wavelength depending on temperature.

Wien's displacement law relates the peak wavelength λ_max to the surface temperature T:

$\lambda_{\text{max}} = \frac{2.898 \times 10^{-3}}{T}$

where λ_max is in metres and T is in kelvins.

Star type	Surface temp (K)	λ_max (nm)	Peak colour	Observed colour
O-type	40,000	72	UV	Blue
B-type	20,000	145	UV	Blue-white
A-type	10,000	290	UV/violet	White
F-type	7,500	387	Violet	Yellow-white
G-type (Sun)	5,800	500	Green	Yellow
K-type	4,500	644	Orange	Orange
M-type	3,000	966	Infrared	Red

Common exam mistake: The Sun's peak wavelength is in the green part of the spectrum (≈500 nm), but the Sun appears yellow/white because it emits strongly across all visible wavelengths. Do not confuse peak wavelength with perceived colour.

Worked Example: Temperature from Peak Wavelength

A star has a peak emission wavelength of 350 nm. What is its surface temperature?

$T = \frac{2.898 \times 10^{-3}}{350 \times 10^{-9}} = \frac{2.898 \times 10^{-3}}{3.50 \times 10^{-7}} = 8280 \text{ K}$

This is an A-type star (white).

Worked Example: Peak Wavelength of Betelgeuse

Betelgeuse (a red supergiant) has a surface temperature of about 3,500 K.

$\lambda_{\text{max}} = \frac{2.898 \times 10^{-3}}{3500} = 8.28 \times 10^{-7} \text{ m} = 828 \text{ nm}$

This is in the near-infrared — beyond the visible red limit of about 700 nm. Betelgeuse appears red because the tail of its emission curve extends into visible red wavelengths.

Stefan's Law (Stefan-Boltzmann Law)

The luminosity of a star — its total power output — is given by Stefan's law:

$L = \sigma A T^4 = 4\pi r^2 \sigma T^4$

where:

L is the luminosity (W)
σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ W m⁻² K⁻⁴)
A = 4πr² is the surface area of the star
r is the stellar radius (m)
T is the surface temperature (K)

The T⁴ dependence is dramatic: doubling the temperature increases the luminosity by a factor of 2⁴ = 16 (for the same radius).

Worked Example: The Sun's Luminosity

The Sun has radius r = 6.96 × 10⁸ m and surface temperature T = 5,800 K.

$L = 4\pi (6.96 \times 10^8)^2 \times 5.67 \times 10^{-8} \times 5800^4$

$= 4\pi \times 4.844 \times 10^{17} \times 5.67 \times 10^{-8} \times 1.132 \times 10^{15}$

$= 6.087 \times 10^{18} \times 6.42 \times 10^{7} = 3.91 \times 10^{26} \text{ W}$

The accepted value is L☉ = 3.85 × 10²⁶ W — close to our calculation.

Worked Example: Comparing Two Stars

Star A has twice the radius and twice the temperature of Star B. Compare their luminosities.

$\frac{L_A}{L_B} = \frac{4\pi r_A^2 \sigma T_A^4}{4\pi r_B^2 \sigma T_B^4} = \left(\frac{r_A}{r_B}\right)^2 \times \left(\frac{T_A}{T_B}\right)^4 = 2^2 \times 2^4 = 4 \times 16 = 64$

Star A is 64 times more luminous. The temperature effect dominates overwhelmingly.

Spectral Classification: OBAFGKM

Stars are classified into spectral types based on their absorption spectra. The main sequence, from hottest to coolest, is:

O – B – A – F – G – K – M

The traditional mnemonic is: Oh Be A Fine Girl/Guy, Kiss Me

Each spectral class is divided into subclasses 0–9 (e.g. the Sun is G2, meaning it is near the hot end of class G).

Spectral class	Temperature range (K)	Colour	Example star
O	>30,000	Blue	Naos (ζ Puppis)
B	10,000–30,000	Blue-white	Rigel
A	7,500–10,000	White	Sirius
F	6,000–7,500	Yellow-white	Procyon
G	5,200–6,000	Yellow	Sun
K	3,700–5,200	Orange	Arcturus
M	2,400–3,700	Red	Betelgeuse

The spectral class is determined by the absorption lines present in the star's spectrum, which depend on the surface temperature. For example, O-type stars are so hot that most hydrogen is ionised, so hydrogen absorption lines are weak. In A-type stars (like Sirius), the temperature is ideal for hydrogen absorption, so the Balmer lines are strongest. In cool M-type stars, the temperature allows molecules (like TiO) to form, producing complex band spectra.

Stellar Luminosity Classes

Stars of the same spectral type can have vastly different luminosities if they differ in size. Luminosity classes distinguish these:

Luminosity class	Type	Example
Ia	Bright supergiant	Rigel
Ib	Supergiant	Betelgeuse
II	Bright giant	—
III	Giant	Arcturus
IV	Subgiant	Procyon
V	Main sequence (dwarf)	Sun

The Sun's full classification is G2V — spectral type G2, luminosity class V (main sequence).

Measuring Stellar Distance

Two key methods for measuring how far away stars are:

Trigonometric Parallax

For nearby stars (within ~100 parsecs), we measure the tiny shift in apparent position as Earth orbits the Sun. The parallax angle p (in arcseconds) is related to distance d (in parsecs) by:

$d = \frac{1}{p}$

One parsec (pc) = 3.09 × 10¹⁶ m = 3.26 light-years.

Standard Candles

For more distant objects, we use objects of known luminosity (standard candles). Comparing the known luminosity with the observed brightness (intensity) gives the distance via the inverse square law for intensity:

$I = \frac{L}{4\pi d^2}$

Rearranging:

$d = \sqrt{\frac{L}{4\pi I}}$

Type Ia supernovae and Cepheid variable stars are the most important standard candles in astronomy.

Worked Example: Distance from Parallax

A star has a parallax angle of 0.025 arcseconds. What is its distance in parsecs and metres?

$d = \frac{1}{0.025} = 40 \text{ pc}$

$d = 40 \times 3.09 \times 10^{16} = 1.24 \times 10^{18} \text{ m}$

Worked Example: Distance from Luminosity and Intensity

A star has luminosity L = 2.0 × 10²⁸ W and the observed intensity at Earth is I = 3.5 × 10⁻⁹ W m⁻².

$d = \sqrt{\frac{2.0 \times 10^{28}}{4\pi \times 3.5 \times 10^{-9}}} = \sqrt{\frac{2.0 \times 10^{28}}{4.398 \times 10^{-8}}} = \sqrt{4.55 \times 10^{35}} = 6.74 \times 10^{17} \text{ m}$

Converting: d = 6.74 × 10¹⁷ / 3.09 × 10¹⁶ = 21.8 pc.

Summary

Wien's law: λ_max = 2.898 × 10⁻³/T links peak wavelength to temperature
Stefan's law: L = 4πr²σT⁴ links luminosity to radius and temperature
Spectral classes OBAFGKM range from hottest (O) to coolest (M)
Luminosity classes I–V distinguish giants/supergiants from main sequence stars
Parallax gives distance for nearby stars: d = 1/p (in parsecs)
Standard candles and I = L/(4πd²) give distance for more remote objects

A-Level Deep Dive: Stellar Classification

Spec mapping

Edexcel 9PH0 specification Topic 12 — Space addresses stellar classification within the wider treatment of astrophysics and cosmology (refer to the official specification document for exact wording). The relevant sub-strands cover the use of black-body radiation laws — specifically Wien's displacement law $\lambda_{max} T = 2.898 \times 10^{-3}\,\text{m·K}$ and the Stefan-Boltzmann law $L = 4\pi R^2 \sigma T^4$ — together with the OBAFGKM spectral sequence, the Hertzsprung-Russell diagram, and methods for measuring stellar distance (parallax, intensity-luminosity inversion). Topic 12 is examined in 9PH0 Paper 2 (Advanced Physics II) alongside thermodynamics, nuclear and particle physics. The Edexcel A-Level Physics formula booklet does provide $\lambda_{max} T = 2.898 \times 10^{-3}\,\text{m·K}$ and $L = 4\pi R^2 \sigma T^4$ , but you must remember the OBAFGKM ordering and the qualitative shape of black-body spectra independently.

Worked example with full mark scheme

Question (8 marks):

A distant star is observed to have a peak spectral emission at wavelength $\lambda_{max} = 414\,\text{nm}$ . Independent measurement gives its luminosity as $L = 4.0 \times 10^{28}\,\text{W}$ .

(a) Calculate the surface temperature of the star and state, with reasoning, its likely spectral class on the OBAFGKM sequence. (4)

(b) Calculate the radius of the star, giving your answer in metres. Take $\sigma = 5.67 \times 10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ . (4)

Solution with mark scheme:

(a) Step 1 — apply Wien's displacement law.

$T = \dfrac{2.898 \times 10^{-3}}{\lambda_{\text{max}}} = \dfrac{2.898 \times 10^{-3}}{414 \times 10^{-9}}$

M1 — correct rearrangement of $\lambda_{max} T = 2.898 \times 10^{-3}$ with $\lambda_{max}$ converted from nm to metres. Common error: using 414 directly without converting to $4.14 \times 10^{-7}\,\text{m}$ , which yields a temperature $10^9\,\text{K}$ too small.

A1 — $T = 7000\,\text{K}$ (to 2 s.f.).

Step 2 — assign spectral class.

The OBAFGKM sequence runs hottest to coolest: O (>30 000 K), B (10 000–30 000 K), A (7500–10 000 K), F (6000–7500 K), G (5200–6000 K), K (3700–5200 K), M (<3700 K).

B1 — class identified as F-type (6000–7500 K bracket; the boundary with A is 7500 K, so 7000 K is securely F).

B1 — reasoning explicitly cites the temperature bracket. Stating "F because the colour is yellow-white" without temperature reasoning typically scores only one of the two B marks.

(b) Step 1 — rearrange the Stefan-Boltzmann law for radius.

$L = 4\pi R^2 \sigma T^4 \implies R = \sqrt{\dfrac{L}{4\pi \sigma T^4}}$

M1 — correct algebraic rearrangement to make $R$ the subject.

Step 2 — substitute numerical values.

$R = \sqrt{\dfrac{4.0 \times 10^{28}}{4\pi (5.67 \times 10^{-8})(7000)^4}}$

The denominator: $4\pi \times 5.67 \times 10^{-8} \times 2.401 \times 10^{15} \approx 1.71 \times 10^{9}$ .

M1 — correct substitution including $T^4$ (a frequent slip is forgetting the fourth power and using $T$ ).

Step 3 — evaluate.

$R^2 = \dfrac{4.0 \times 10^{28}}{1.71 \times 10^{9}} \approx 2.34 \times 10^{19}\,\text{m}^2$ $R \approx 4.84 \times 10^{9}\,\text{m}$

A1 — $R \approx 4.8 \times 10^{9}\,\text{m}$ (to 2 s.f.). For context, this is roughly seven times the radius of the Sun ( $R_\odot \approx 6.96 \times 10^{8}\,\text{m}$ ), suggesting an evolved star slightly off the main sequence.

Total: 8 marks (M2 A2 B2 M1 A1 split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): Two stars X and Y are observed.

Star X has $\lambda_{max} = 290\,\text{nm}$ and the same radius as the Sun.
Star Y has $\lambda_{max} = 580\,\text{nm}$ and the same radius as the Sun.

(a) Determine the ratio $T_X / T_Y$ of their surface temperatures. (2)

(b) Hence determine the ratio $L_X / L_Y$ of their luminosities and identify which star is intrinsically brighter. (4)

Mark scheme decomposition by AO:

(a)

M1 (AO2) — recognising that $T \propto 1/\lambda_{max}$ from Wien's law, so $T_X / T_Y = \lambda_Y / \lambda_X = 580/290$ .
A1 (AO1) — $T_X / T_Y = 2$ .

(b)

M1 (AO2) — applying Stefan-Boltzmann $L = 4\pi R^2 \sigma T^4$ with equal radii, so $L_X / L_Y = (T_X / T_Y)^4$ .
M1 (AO1) — substituting: $(2)^4 = 16$ .
A1 (AO1) — $L_X / L_Y = 16$ .
B1 (AO3) — explicit conclusion that Star X is 16 times more luminous than Star Y because $L \propto T^4$ at fixed radius — a small temperature difference produces a large luminosity difference.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. Edexcel 9PH0 Paper 2 typically gives ratio questions to test that candidates can manipulate proportional reasoning without needing a calculator-heavy substitution.

Synoptic links

Stellar classification connects to:

Hertzsprung-Russell diagram (Topic 12, next lesson): every spectral class corresponds to a vertical band on the H-R diagram. Plotting luminosity vs spectral class places stars on the main sequence, the giant branch, or the white dwarf region — turning a one-dimensional classification into a two-dimensional map of stellar evolution.
Black-body radiation (Topic 12): Wien's displacement law and the Stefan-Boltzmann law are not specific to stars — they describe any object in thermal equilibrium with its radiation field. The same physics underlies cosmic microwave background measurements ( $T \approx 2.7\,\text{K}$ , peak in microwaves) and incandescent-bulb design.
Gravitational fields (Topic 12, this course): stellar luminosity, mass and radius are linked through hydrostatic equilibrium — gravity pulls the star inward, radiation pressure pushes outward. The OBAFGKM sequence on the main sequence is essentially a mass sequence (O-type are most massive, M-type least), and mass determines surface gravity $g = GM/R^2$ .
Doppler effect (Topic 12): redshift of stellar absorption lines reveals radial velocity. Combined with luminosity-distance methods, this builds the cosmic distance ladder and supports Hubble's law $v = H_0 d$ .
Nuclear physics (Topic 11): the surface temperature you measure is set ultimately by the nuclear-fusion power generated in the core. Hotter, more massive stars (O, B classes) burn hydrogen via the CNO cycle; cooler stars (G and below) use the proton-proton chain.

Mark-scheme literacy

Stellar classification questions on 9PH0 split AO marks as follows: