AP Physics 2- 15.4 Blackbody Radiation- Study Notes- New Syllabus
AP Physics 2- 15.4 Blackbody Radiation – Study Notes
AP Physics 2- 15.4 Blackbody Radiation – Study Notes – per latest Syllabus.
Key Concepts:
- Blackbody Radiation
- Wien’s Displacement Law
- Stefan-Boltzmann Law
Blackbody Radiation
Blackbody radiation is the electromagnetic radiation emitted by an object solely due to its temperature. A blackbody is an idealized perfect absorber of radiation, which also emits radiation with a characteristic spectrum dependent only on its temperature.
Physical Concept
- Any object with temperature \(\mathrm{T > 0 \, K}\) emits radiation.
- The radiation spectrum is continuous and spans a range of wavelengths.
- The emitted radiation depends on the object’s temperature, not its material.
- The intensity of radiation increases with temperature and the peak wavelength shifts toward shorter wavelengths for higher temperatures.
Wien’s Displacement Law
The wavelength at which the radiation intensity is maximum (\(\mathrm{\lambda_{max}}\)) is inversely proportional to the temperature:
$ \mathrm{\lambda_{max} T = b}, \quad b \approx 2.898 \times 10^{-3} \, \text{m·K} $
- \(\mathrm{\lambda_{max}}\) decreases as \(\mathrm{T}\) increases (hotter bodies emit shorter wavelength light).
- Explains why the Sun emits visible light while cooler objects emit infrared radiation.
Stefan-Boltzmann Law
The total power emitted per unit area of a blackbody is proportional to the fourth power of its absolute temperature:
$ \mathrm{P = \sigma T^4}, \quad \sigma \approx 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 $
- \(\mathrm{P}\) is the total radiated power per unit area.
- Explains why hotter objects radiate much more energy than cooler ones.
Key Features
- Emission spectrum depends only on temperature, not material composition.
- Peak wavelength shifts to shorter wavelengths as temperature increases.
- Total emitted power increases rapidly with temperature (\(\mathrm{T^4}\)).
- Forms the basis for understanding stellar radiation and thermal emission.
Example :
Find the wavelength of maximum emission for a blackbody at \(T = 6000 \, K\) (approximate surface temperature of the Sun).
▶️ Answer/Explanation
Step 1: Apply Wien’s law: \(\mathrm{\lambda_{max} = \dfrac{b}{T} = \dfrac{2.898 \times 10^{-3}}{6000} \approx 4.83 \times 10^{-7} \, m}\)
Step 2: Convert to nm: \(\mathrm{\lambda_{max} \approx 483 \, nm}\) (visible blue-green light)
Example :
Calculate the total power emitted per unit area by a blackbody at \(T = 500 \, K\).
▶️ Answer/Explanation
Step 1: Use Stefan-Boltzmann law: \(\mathrm{P = \sigma T^4 = 5.67 \times 10^{-8} \cdot (500)^4}\)
Step 2: Calculate: \(\mathrm{500^4 = 6.25 \times 10^{10}}\) \(\mathrm{P = 5.67 \times 10^{-8} \cdot 6.25 \times 10^{10} \approx 3544 \, W/m^2}\)
Step 3: Interpretation: A blackbody at 500 K emits approximately 3544 W/m² of power per unit area.