Classification by temperature, black-body radiation

1. Stefan’s law and Wien’s displacement law:

  • ⇒ Stefan’s Law:

  • Stefan’s Law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the power or flux) is proportional to the fourth power of the black body’s temperature:
  • [math] P = σAT^4 [/math]
  • where:
    – T is the temperature of the black body (in Kelvin, K)
    – σ is the Stefan-Boltzmann constant (approximately [math] 5.67 \times 10^{-8} W/m^2 K^4 [/math])
    – The total power radiated by a star is called its luminosity, L.
  • This law describes how much energy is emitted by an object at a given temperature.

  • ⇒ Wien’s Displacement Law:

  • Wien’s Displacement Law states that the wavelength at which a black body emits the most energy (its peak wavelength, ) is inversely proportional to its temperature:
  • [math] λ_{max} = \frac{\text{constant}}{T} [/math]
  • where:
    – λ_{max} is the peak wavelength (in meters, m)
    – Constant is Wien’s displacement constant (approximately 2.898 \times 10^{-3} mK [/math])
    – T is the temperature of the black body (in Kelvin, K)
  • Figure 1 Wien’s displacement law to estimate black-body temperature
  • This law describes how the wavelength of the emitted energy changes with temperature.

2.Black-body radiation:

  • Black body radiation is the thermal radiation emitted by an idealized perfect absorber of electromagnetic radiation, known as a black body. Black bodies absorb all incident radiation, without reflecting or transmitting any of it.
  • The characteristics of black body radiation are:
    – Spectrum: Continuous and smooth, with a peak wavelength that shifts to shorter wavelengths as temperature increases (Wien’s Displacement Law).
    – Intensity: Increases with temperature (Stefan’s Law).
    – Dependence on temperature: Radiation increases with the fourth power of temperature ([math]T^4[/math]).
    – Universality: All objects emit black body radiation, regardless of their composition or structure.
  • Black body radiation has many applications:
    – Thermodynamics: Understanding heat transfer and thermal equilibrium.
    – Radiation heat transfer: Calculating energy exchange between objects.
    – Cosmology: Studying the cosmic microwave background radiation (CMB), the oldest light in the universe.
    – Climate science: Modeling Earth’s energy balance and greenhouse effect.
  • Materials science: Analyzing thermal properties and radiation patterns of materials.
  • Some examples of black bodies include:
    – Stars (approximate black bodies)
    – Incandescent light bulbs
    – Radiators
    – The cosmic microwave background radiation (CMB)

  • ⇒ General shape of black-body curves:

  • Black-body radiation has a characteristic wavelength spectrum, which depends only on the absolute temperature of the body. The spectrum peaks at a wavelength that shifts to shorter wavelengths at higher temperatures. Figure 6 shows some spectra for black bodies at different temperatures.
  • The curves in Figure 6 show these two important trends.
    – As the body gets hotter, more radiation is emitted. The total power emitted by the body is proportional to the area under the graph. So, you can see that at 6000 K a black body radiates much more energy than the body at 4000 K
  • Figure 2 Some spectra for black bodies at different temperatures
  • The intensity of radiation peaks at a shorter wavelength at higher temperatures. You can see that the peak wavelength corresponds to red light when the temperature of the black body is 4000K. At 5000K the peak corresponds to green-yellow light. At 6000K the peak corresponds to blue-green light.
  • Example:
    Peak wavelength
    The surface temperature of the Sun is 5780K and its radius is [math]7 \times 10^5 km [/math] Calculate the peak wavelength of the radiations emitted.
  • Given data:
    Surface temperature of the Sun [math]= T = 5780K [/math]
    Radius of the Sun [math] = R = 7 \times 10^5 km [/math]

     

    Find data:
    The peak wavelength of the radiations [math] = λ_{max} = ? [/math]

    Formula:

  • [math]  \lambda_{\text{max}} = \frac{\text{constant}}{T}[/math]
  • Solution:
  • [math] \lambda_{\text{max}} = \frac{\text{constant}}{T} \\
    \lambda_{\text{max}} = \frac{2.898 \times 10^{-3}}{5780} \, \text{m} \\
    \lambda_{\text{max}} = 5 \times 10^{-7} \, \text{m} \\
    \lambda_{\text{max}} = 500 \, \text{nm} [/math]
  • This wavelength is in the blue-green of the visible spectrum.

3. Use of Stefan’s law to compare the power output, temperature and size of stars

    • ⇒ Giant stars

    • The constellation of Orion has two very luminous stars.
    • Rigel is a blue giant with a surface temperature of about 11800K and a radius of [math] 54 \times 10^6 km. [/math]
    • Betelgeuse is a red giant with a surface temperature of about 3300 K and a radius of [math] 7.7 \times 10^8 km. [/math]

    • ⇒ Rigel

    • Given data:

    • Stefan-Boltzmann constant [math]= σ = 5.7 \times 10^{-8} W/m^2 K^4 [/math]
    • Temperature = T= 11800 K
    • Radius of Rigel [math] = R = 5.4 \times 10^{10} m [/math]
    • Find data:
    • Power of Rigel = P=?
    • Formula:
    • [math] P = σAT^4 [/math]
    • [math] \text{Area of Rigel} = A = 4π R^2 [/math]
    • [math] P = σ(4πR^2)T^4 [/math]
    • Solution:
    • [math] \begin{align*}
      P &= \sigma(4\pi R^2)T^4 \\
      P &= (5.7 \times 10^{-8})(4\pi (5.4 \times 10^{10})^2 (11800)^4) \\
      P &= 4 \times 10^{31} \, \text{W} \\
      P &= 10^5 \, \text{times more luminous than the Sun}
      \end{align*} [/math]

⇒Betelgeuse:

  • Given data:
  • Stefan-Boltzmann constant [math]= σ = 5.7 \times 10^{-8} W/m^2 K^4 [/math]
  • Temperature = T= 3300 K
  • Radius of Rigel [math] = R = 7.7 \times 10^{11} m [/math]
  • Find data:
  • Power of Rigel = P=?
  • Formula:
  • [math] P = σAT^4 [/math]
  • [math] \text{Area of Rigel} = A = 4π R^2 [/math]
  • [math] P = σ(4πR^2)T^4 [/math]
  • Solution:
  • [math] \begin{align*}
    P &= \sigma(4\pi R^2)T^4 \\
    P &= (5.7 \times 10^{-8})(4\pi (7.7 \times 10^{11})^2 (3300)^4) \\
    P &= 5 \times 10^{31} \, \text{W} \\
    P &= 1.3 \times 10^5 \, \text{times more luminous than the Sun}
    \end{align*} [/math]
  • Although Betelgeuse is a relatively cool star, its radius is over 1000 times larger than of the Sun. It is because its surface area is so larger that Betelgeuse is one of the most luminous stars in the sky.
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