Respuesta :
C) Infrared.
Explanation
The wavelength of the "primary" (a.k.a. "peak") outgoing radiation from the earth depends on the surface temperature of the earth.
[tex]\displaystyle \lambda_\text{peak} = \frac{0.00289}{T}[/tex],
where
- [tex]\lambda_\text{peak}[/tex] is the peak wavelength (in meters) of the outgoing radiation, and
- [tex]T[/tex] is the absolute temperature of the earth (the one in degrees Kelvins.)
The average temperature on the earth is around 15 °C (around 60 °F.)
Convert to degrees Kelvins:
[tex]T = 15\;\textdegree\text{C} = (15 + 273.15)\;\text{K} = 288.15\;\text{K}[/tex].
Peak wavelength:
[tex]\displaystyle \lambda_\text{peak} = \frac{0.00289}{288.15} = 1.00\times 10^{-5} \;\text{m} = 1000\;\text{nm}[/tex].
What kind of electromagnetic radiation is that? Refer to a chart of the EM spectrum.
Radiations:
- [tex]\lambda < 400\;\text{nm}[/tex]: ultra-violet (UV) or radiations that are more energetic than UV;
- [tex]400\;\text{nm}<\lambda<700\;\text{nm}[/tex]: visible light;
- [tex]\lambda > 700\;\text{nm}[/tex]: infrared (IR) and less energetic radiations.
[tex]\lambda \approx 1000\;\text{nm}[/tex] is longer than the wavelength of visible light. The radiation is infrared (IR.)