Question

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

Answer 1

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.

$\displaystyle \lambda_\text{peak} = \frac{0.00289}{T}$,

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:

$\displaystyle \lambda_\text{peak} = \frac{0.00289}{288.15} = 1.00\times 10^{-5} \;\text{m} = 1000\;\text{nm}$.

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;
• $400\;\text{nm}$: 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.)

Answer 2

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