Free Pearson Edexcel International A Level Physics lesson

Starlight, temperature and luminosity

Interpret spectra and separate emitted power from received intensity.

What you’ll learn

  • Read the wavelength spectrum’s peak and area as different pieces of information.
  • Combine the full emitting area with the fourth power of absolute temperature.
  • Keep intrinsic luminosity, received intensity and detector power distinct.

A thermal spectrum shifts as it warms

An ideal black body absorbs all incident electromagnetic radiation. In thermal equilibrium it emits a continuous spectrum determined by temperature. A curve of emitted power per area per wavelength against wavelength has a broad peak; the area under the complete curve represents total emitted power per area. A star is often approximated as a black body, with absorption lines superimposed.

Raising temperature increases emitted power and moves the wavelength peak to shorter wavelength. Wien’s law is λmaxT = 2.898 × 10⁻³ m K. Use metres for peak wavelength and kelvin for temperature. A visible colour is only a qualitative guide because eyes and instruments sample limited wavelengths and intervening matter can alter the received spectrum.

The peak depends on the horizontal spectral variable: a wavelength-density curve and a frequency-density curve have different peak locations and cannot be converted simply by substituting c/λmax. In this unit, apply the stated wavelength form of Wien’s law. Temperature estimates also depend on whether the observed continuum is sufficiently close to the assumed thermal model.

λmaxT = 2.898 × 10⁻³ m K area under a wavelength spectrum ∝ total emitted power per area

Symbols and units

  • λmax is the wavelength at the maximum of a wavelength-density thermal spectrum, in m; 1 nm = 10⁻⁹ m.
  • T is absolute surface temperature in K; the Wien constant is 2.898 × 10⁻³ m K.
  • For a spectrum plotted as emitted power per area per wavelength, the vertical quantity has units W m⁻² per m of wavelength; integrating over wavelength gives W m⁻².

How the equation explains the idea

  1. Use the peak of the wavelength spectrum and convert 500 nm to 5.00 × 10⁻⁷ m. This law does not use a frequency-spectrum peak or a colour description alone.
  2. Divide λmaxT = 2.898 × 10⁻³ m K by λmax: T = (2.898 × 10⁻³)/(5.00 × 10⁻⁷) = 5796 K, approximately 5800 K.
  3. With the Wien constant fixed, T₂/T₁ = λmax,1/λmax,2. A shorter wavelength peak therefore indicates a higher model temperature.
  4. The peak position estimates temperature, while the full spectral area gives emitted power per surface area. Total luminosity also requires emitting surface area and a justified thermal model.
A warmer black body has a shorter-wavelength peak

Conditions. Graph-generation model (extension): Planck wavelength spectra Mλ = 2πhc²/{λ⁵[exp(hc/(λkT)) − 1]}, with h = 6.62607015 × 10⁻³⁴ J s, c = 299792458 m s⁻¹ and k = 1.380649 × 10⁻²³ J K⁻¹. Perfect black bodies; power per emitting area per wavelength, not total source luminosity. The plotted window is 0.10–3.00 μm. This formula explains the computed curves and is not an additional Unit 5 memorization requirement.

Wavelength / (10⁻⁶ m); Spectral exitance / (10¹³ W m⁻³).

Wien peaks are approximately 0.966, 0.644 and 0.483 μm at 3000, 4500 and 6000 K. Temperature increases both peak height and total emission. The area over the entire wavelength spectrum, including the unplotted tail, is σT⁴; the area of this finite window is only part of it.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

A peak in visible light

  1. λmax = 500 nm = 5.00 × 10⁻⁷ m.
  2. T = 2.898 × 10⁻³/(5.00 × 10⁻⁷).
  3. T = 5796 K ≈ 5800 K.

Shorter wavelength peaks correspond to higher temperature.

Keep this idea. Read the wavelength spectrum’s peak and area as different pieces of information.

Temperature and radius determine emitted power

The Stefan–Boltzmann law gives total black-body luminosity L = σAT⁴, where σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. For a spherical star, emitting surface area is 4πR², not its projected disc area πR². This law uses absolute temperature and integrates radiation over all wavelengths.

At fixed area, doubling temperature multiplies luminosity by sixteen. At fixed temperature, doubling radius multiplies luminosity by four. Combining a temperature estimate from the wavelength peak with a luminosity estimate can therefore constrain radius. A small hot white dwarf can have low total luminosity despite a large power emitted per square metre.

Real surfaces may have emissivity below one; a stated grey-body model inserts a factor ε. For net exchange with a large surrounding environment at temperature Ts, a simple model uses εσA(T⁴ − Ts⁴), not only the object’s gross emission. Stellar estimates also carry strong temperature sensitivity: approximate fractional uncertainty in L includes four times the fractional uncertainty in T.

L = σAT⁴ = 4πR²σT⁴ spherical black-body model

Symbols and units

  • L is total emitted luminosity in W, not latent heat; A is emitting surface area in m², not oscillation amplitude or nuclear activity.
  • R is stellar radius in m and a sphere has A = 4πR²; T is absolute temperature in K.
  • σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ is the Stefan–Boltzmann constant.
  • Subscripts A and B in LA/LB label the two stars; they are not the area symbol A. ε, if used for a grey surface, is dimensionless emissivity.

How the equation explains the idea

  1. A spherical black body radiates from its whole surface, so substitute A = 4πR² into L = σAT⁴ to obtain L = 4πR²σT⁴.
  2. Divide the two stars’ equations. Common 4πσ cancels, leaving LA/LB = (RA/RB)²(TA/TB)⁴ for the same black-body approximation.
  3. Insert the example ratios: LA/LB = 2² × 1.5⁴ = 4 × 5.0625 = 20.25. The radius change contributes a factor of four and temperature contributes a factor of 5.0625.
  4. This is total emitted power, not the intensity at Earth. Net exchange with surroundings would also subtract their incident radiation under an appropriate exchange model.
Temperature and radius jointly determine luminosity

Conditions. Spherical black bodies with L = 4πR²σT⁴. Reference L₀ = 4πR₀²σT₀⁴, using the same σ for every curve. Radius is held fixed within each series; temperature is in kelvin before forming T/T₀.

Temperature ratio T/T₀; Luminosity ratio L/L₀.

At fixed radius, doubling temperature multiplies luminosity by 16. A star at 2T₀ with radius R₀/4 has L = L₀ because the sixteenfold temperature effect cancels its sixteenfold smaller area. The curves compare intrinsic luminosity, not received intensity at different distances.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Compare two stars

  1. Star A has twice the radius and 1.5 times the temperature of star B.
  2. LA/LB = 2² × 1.5⁴.
  3. LA/LB = 20.25.

A modest temperature ratio can strongly change luminosity.

Keep this idea. Combine the full emitting area with the fourth power of absolute temperature.

Spread luminosity over an expanding sphere

Luminosity is total emitted power; received intensity is power per detector area. For an isotropic source with no intervening absorption, its power crosses spheres of area 4πd², giving I = L/(4πd²). The distance d is from the source to the observer, not the star’s physical radius.

Doubling distance reduces intensity by four while leaving the source luminosity unchanged. A detector with effective collecting area a receives approximately P = Ia when it faces the source normally and the beam intensity is uniform across it. A measured detector band need not contain the full emission, so bolometric luminosity estimates require an appropriate spectral correction.

Dust absorption, non-isotropic emission, detector calibration and uncertain distance can bias the inference. If extinction is ignored, a source appears fainter and a known-luminosity distance estimate becomes too large. When distance is independently known, use L = 4πd²I and propagate the squared distance sensitivity rather than treating a measured intensity as intrinsic brightness.

I = L/(4πd²) L = 4πd²I detector power ≈ I × collecting area

Symbols and units

  • I is received intensity (irradiance) in W m⁻² in this lesson, not electric current; L is total source luminosity in W.
  • d is source-to-observer distance in m, distinct from the source’s physical radius.
  • a is effective collecting area in m² and P is received detector power in W. Here a is area, not acceleration.
  • The inverse-square relation assumes isotropic emission with no intervening absorption; L and I must refer to a consistent wavelength range.

How the equation explains the idea

  1. The same luminosity crosses each surrounding sphere. Divide its power by area 4πd² to obtain I = L/(4πd²).
  2. Multiply by 4πd² to recover source power: L = 4πd²I. For the example, d² = (1.0 × 10¹⁷)² = 1.0 × 10³⁴ m².
  3. Therefore L = 4π × 10³⁴ × 2.0 × 10⁻⁹ = 2.51 × 10²⁶ W. The area units cancel the per-area intensity units.
  4. A normally oriented small detector receives P ≈ Ia, not the entire L. Absorption or incomplete spectral coverage must be corrected before treating the measured intensity as the full unobscured signal.

Worked example

Find a model luminosity

  1. I = 2.0 × 10⁻⁹ W m⁻² and d = 1.0 × 10¹⁷ m.
  2. L = 4π × 10³⁴ × 2.0 × 10⁻⁹.
  3. L = 2.51 × 10²⁶ W.

Observed intensity becomes luminosity only with a distance and emission model.

Keep this idea. Keep intrinsic luminosity, received intensity and detector power distinct.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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