The big idea: A star glows like a black body — a perfect radiator. Its light carries two clues about the star.
The colour of the peak tells you how hot the surface is — that is Wien's law.
The total power it pours out depends on its size and temperature — that is the Stefan-Boltzmann law.
New words, plainly: Black body = an ideal object that absorbs all radiation falling on it and glows with a characteristic spectrum set only by its temperature. A star is a good approximation.
Peak wavelength λ_{max} = the wavelength at which the star radiates most intensely — the top of its curve.
Luminosity L = the total power the star radiates, in watts (W). (Brightness as seen from Earth is something else — that fades with distance.)
Wien's law — the COLOUR
- Looks at where the peak is (the colour)
- Hotter star → peak at a shorter wavelength (bluer)
- Gives you the surface temperature T
Stefan-Boltzmann — the POWER
- Looks at the total power radiated (luminosity L)
- Depends on the area and on T⁴ — a tiny T rise is huge
- Links luminosity, radius and temperature together
Hotter = bluer AND brighter: As a star gets hotter two things happen at once on its curve:
- the peak slides to a shorter wavelength (it looks bluer) — Wien, - the whole curve rises (it radiates far more power) — Stefan-Boltzmann.
So a blue star is both hotter and, size for size, much more luminous than a red one.
Measure where the spectrum peaks, λmax, and Wien's law gives you the star's surface temperature straight away. Shorter peak ⇒ hotter star.
- peak wavelength — where the star's radiation is most intense (m)
- surface (absolute) temperature of the star (K)
- Wien's constant, 2.9 × 10⁻³ m K (given)
Watch the units: Put λ_{max} in metres (a wavelength of 480 nm is 480 × 10⁻⁹ m) and T comes out in kelvin (K).
Rearrange the given law for whichever you need:
- temperature from the peak: T = 2.9 × 10⁻³ ÷ λmax - peak from the temperature: λmax = 2.9 × 10⁻³ ÷ T
IB-style question — surface temperature from the spectrum peak
A star's black-body spectrum peaks at a wavelength of λmax = 480 nm. Determine the surface temperature of the star. (Wien's constant = 2.9 × 10⁻³ m K.)
Solution
- Write the given Wien's law first — peak wavelength × temperature is a constant:
- Rearrange to make the temperature the subject:
- Convert the peak to metres (480 nm = 480 × 10⁻⁹ m) and substitute:
- Work it out — keep the unit:
Final answer
T = 2.9 × 10⁻³ ÷ (480 × 10⁻⁹) ≈ 6000 K — a yellow-white star, much like the Sun.
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The Stefan-Boltzmann law gives the total power a star radiates. For a star the radiating area is the surface of a sphere, A = 4πR², so the law connects luminosity, radius and temperature in one equation.
- luminosity — the total power the star radiates (W)
- Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (given)
- surface area of the star; for a sphere A = 4πR² (m²)
- surface (absolute) temperature of the star (K)
The T⁴ is the key: Because the temperature is raised to the fourth power, it dominates. Double the temperature and the luminosity jumps by 2⁴ = 16 times (radius unchanged).
The radius only enters as R² (double R → 4× the luminosity).
For comparing two stars, use the proportional form L ∝ R² T⁴ — the constants cancel.
| If you change… | What happens to luminosity L |
|---|---|
| Double the radius R (T fixed) | Area is 4πR², so L goes up 4× (L ∝ R²) |
| Double the temperature T (R fixed) | L goes up 16× — because of the T⁴ |
| Double both R and T | L goes up 4 × 16 = 64× |
Comparing two stars — the ratio trick: When a question gives one star's values and asks about a second, divide the two Stefan-Boltzmann equations so σ and 4π cancel:
LB ÷ LA = (RB ÷ RA)² × (TB ÷ TA)⁴
Then rearrange for the radius ratio: RB ÷ RA = √(LB ÷ LA) ÷ (TB ÷ TA)².
IB-style question — radius of a second star from luminosity and temperature
Star Y has 4 times the luminosity and 2 times the surface temperature of Star X. Determine the ratio of Star Y's radius to Star X's radius.
Solution
- Use the given with A = 4πR², so L ∝ R²T⁴. Write the ratio of the two stars (σ and 4π cancel):
- Rearrange to make the radius ratio the subject:
- Substitute LY/LX = 4 and TY/TX = 2:
- Work it out:
Final answer
RY/RX = √4 ÷ 2² = 2 ÷ 4 = 0.5 — Star Y is half the radius of Star X, yet far more luminous because of its higher temperature.
How this is tested: The two laws appear in both papers, almost always together:
- Paper 1A: a one-step MCQ — read off which star is hotter from its peak, or find a radius ratio of two stars from their luminosities and temperatures (L ∝ R² T⁴). - Paper 2: show that the surface temperature is about a stated value from the spectrum peak (Wien), then estimate the radius by combining the laws (often via b = L/4πd² to get L first).
Classic trap: forgetting the T⁴. When you take the ratio, the temperature ratio is raised to the fourth power and the radius ratio to the second — and λmax must be in metres for Wien.
IB-style question — surface temperatures of two stars
Star X has a black-body peak at λmax = 580 nm and Star Y has a peak at λmax = 290 nm. (a) Determine the surface temperature of each star. (Wien's constant = 2.9 × 10⁻³ m K.)
Solution
- Star X. Use the given Wien's law, rearranged for temperature:
K. - Star Y. Same law with its shorter peak:
K. - Read it off. Star Y's peak is half the wavelength, so its temperature is twice Star X's — the shorter (bluer) peak means the hotter star.
Final answer
TX ≈ 5000 K and TY ≈ 10000 K. Star Y is twice as hot as Star X.
IB-style question — radius ratio of the two stars
Star Y (from part a) is also 4 times as luminous as Star X. (b) Determine the ratio of Star Y's radius to Star X's radius.
Solution
- Use the given with A = 4πR², so L ∝ R²T⁴. Write the ratio of the two stars (σ and 4π cancel):
- Rearrange for the radius ratio:
- From part (a) the temperature ratio is TY/TX = 2, and LY/LX = 4. Substitute:
- Work it out:
Final answer
RY/RX = √4 ÷ 2² = 0.5. The hotter star Y is only half the radius of X, but its high temperature still makes it more luminous.