2.1.15Tools and Inquiry — IB Physics skills
Wien’s Displacement Law — IB Physics SL Revision Notes & Topical Questions
5 min read 5 parts Lesson 28 of 33 Unit quiz
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 5The Law
Hot objects radiate over a spread of wavelengths, and the most intense wavelength depends on temperature.
Statement
- Statement The peak wavelength of the radiation emitted by a black body is inversely proportional to its absolute temperature.
Equation
- Equation λmax T = 2.90 × 10⁻³ m K, where T is the absolute temperature in kelvin.
Black body
- Black body An ideal surface that absorbs all radiation falling on it and emits the maximum possible radiation for its temperature.
Stars
- Stars Stars are good approximations to black bodies, so the law gives an estimate of their surface temperature.
Part 2 of 5Reading Spectra
Intensity-wavelength graphs for different temperatures show the pattern clearly.
1
Each curve has a single maximum, and the wavelength at the maximum is λmax.
2
Raising the temperature moves the peak to a shorter wavelength.
3
The whole curve also rises, because a hotter body radiates more power per unit area.
4
Cooler stars have peaks in the red or infrared and look red, while hotter stars peak in blue or ultraviolet.
5
Measuring the peak of a star's spectrum and using the law gives its surface temperature.
Part 3 of 5Applications and Limits
Applications and Limits
Sun
- Sun With a surface of about 5800 K the peak is near 500 nm, in the visible part of the spectrum.
Human body
- Human body A body at about 310 K peaks in the infrared at roughly 9.4 μm, which thermal cameras detect.
Limitations
- Limitations Real objects are not perfect black bodies, so the temperature found is approximate.
Link to power
- Link to power Total power radiated depends on area and on T⁴, as given by the Stefan-Boltzmann law.
| Object | Temperature (K) | λmax |
|---|---|---|
| Human body | 310 | 9.4 × 10⁻⁶ m |
| Sun's surface | 5800 | 5.0 × 10⁻⁷ m |
| Hot star | 10 000 | 2.9 × 10⁻⁷ m |
Part 4 of 5Worked Example
Question
- The spectrum of a star has a peak at a wavelength of 500 nm.
- Estimate its surface temperature.
- Convert 500 nm to metres: 500 × 10⁻⁹ = 5.00 × 10⁻⁷ m.
- RearrangeT = 2.90 × 10⁻³ / λmax.
- T = 2.90 × 10⁻³ / 5.00 × 10⁻⁷ = 5800 K.
Answer The surface temperature of the star is about 5800 K.
Part 5 of 5Exam Checklist & Tips
2.1.15 – Key Points for the Exam
- Black body A perfect absorber and emitter of radiation.
- Peak wavelength The wavelength with the greatest radiated intensity.
- Wien's displacement law λmax T equals a constant, 2.90 × 10⁻³ m K.
- Surface temperature The temperature of the visible outer layer of a star.
Common Mistakes
- Using degrees Celsius instead of kelvin.
- Forgetting to convert nanometres to metres before using the law.
- Stating that hotter objects have longer peak wavelengths.
- Mixing up Wien's law, which gives the peak, with the Stefan-Boltzmann law for total power.
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