Question

What is the primary outgoing radiation put off by the earth?

A. ultra-violet (UV)

B. visible light

C. infrared (IR)

D. all of the above
Please help ASAP!!! :(

Answer 1

The answer is C. If you need any help let me know. :)

Answer 2

C) Infrared.

Step-by-step explanation

The wavelength of the "primary" (a.k.a. "peak") outgoing radiation from the earth depends on the surface temperature of the earth.

`\displaystyle \lambda_\text{peak} = (0.00289)/(T)`,

where

• 
  `\lambda_\text{peak}` is the peak wavelength (in meters) of the outgoing radiation, and
• 
  `T` 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:

`T = 15\;\textdegree\text{C} = (15 + 273.15)\;\text{K} = 288.15\;\text{K}`.

Peak wavelength:

`\displaystyle \lambda_\text{peak} = (0.00289)/(288.15) = 1.00* 10^(-5) \;\text{m} = 1000\;\text{nm}`.

What kind of electromagnetic radiation is that? Refer to a chart of the EM spectrum.

Radiations:

• 
  `\lambda < 400\;\text{nm}`: ultra-violet (UV) or radiations that are more energetic than UV;
• 
  `400\;\text{nm}<\lambda<700\;\text{nm}`: visible light;
• 
  `\lambda > 700\;\text{nm}`: infrared (IR) and less energetic radiations.

`\lambda \approx 1000\;\text{nm}` is longer than the wavelength of visible light. The radiation is infrared (IR.)