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Chapter 11: Problem 84

Surface Temperature of Venus. Access the Active Integrated Media Module "Wien's Law" in Chapter 5 of the Universe Web site or eBook. (a) Using the Wien's Law calculator, determine Venus's approximate temperature if it emits blackbody radiation with a peak wavelength of \(3866 \mathrm{~nm}\). (b) By trial and error, find the wavelength of maximum emission for a surface temperature of \(733 \mathrm{~K}\) (for present-day Venus) and a surface temperature of \(833 \mathrm{~K}\) (as it might be in the event of a global catastrophe that released more greenhouse gases into Venus's atmosphere). In what part of the electromagnetic spectrum do these wavelengths lie?

Short Answer

Expert verified
The temperature of Venus when emitting with a peak wavelength of 3866 nm is approximately 750K. For surface temperatures of 733K and 833K, the wavelengths of maximum blackbody radiation emission are in the infrared region of the electromagnetic spectrum.

Step by step solution

01

Applying Wien's Law

Wien's law can be expressed as \(T = \frac{b}{\lambda_{max}}\) where \(T\) is the temperature in Kelvin, \(\lambda_{max}\) is the peak wavelength and \(b\) is Wien's constant, approximately \(2.9 \times 10^{-3} m.K\). In part (a), we're given \(\lambda_{max} = 3866 nm = 3.866 \times 10^{-6} m\). Substituting these values in the formula we get: \( T = \frac{2.9 \times 10^{-3} m.K}{3.866 \times 10^{-6} m} \)
02

Computing the Temperature

The result from step 1 gives the temperature in Kelvin. The computation is straightforward: \(T \approx 749.87 K\)
03

Applying Wien's Law for Given Temperatures

In part (b), we need to apply Wien's law in a reversed manner to find the peak wavelength. For a temperature of 733K and 833K, we apply the formula: \(\lambda_{max} = \frac{b}{T}\) and perform the calculations to get the peak wavelengths for these temperatures.
04

Computing the Peak Wavelengths

Using the formula from step 3, we compute the peak wavelengths for 733K and 833K: \(\lambda_{max}(733K) = \frac{2.9 \times 10^{-3} m.K}{733K}\) and \(\lambda_{max}(833K) = \frac{2.9 \times 10^{-3} m.K}{833K}\)
05

Identifying Electromagnetic Spectrum Regions

With the calculated wavelengths for the two temperatures, one to see in which part of the electromagnetic spectrum these wavelengths are. The boundaries of the infrared region are from about 700 nm (0.7 µm) to 1 mm (1000 µm), implying both temperatures peak in the infrared region.

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Most popular questions from this chapter

Although the Viking Lander 1 and Viking Lander 2 landing sites are \(6500 \mathrm{~km}\) apart and have different geologic histories, the chemical compositions of the dust at both sites are nearly identical. (a) What does this suggest about the ability of the Martian winds to transport dust particles? (b) Would you expect that larger particles such as pebbles would also have identical chemical compositions at the two Viking Lander sites? Why or why not?

Describe the apparent motion of the Sun during a "day" on Venus relative to (a) the horizon and (b) the background stars. (Assume that you can see through the cloud cover.)

A full moon on Earth is bright enough to cast shadows. As seen from the Martian surface, would you expect a full Phobos or full Deimos to cast shadows? Why or why not?

Explain why Mercury does not have a substantial atmosphere.

Why was it difficult to determine Venus's surface temperature from Earth? How was this finally determined?
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