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Chapter 8: Problem 39

(a) Figure 8-18c shows how astronomers determine that the planet of HD 209458 has a surface temperature of \(1130 \mathrm{~K}\). Treating the planet as a blackbody, calculate the wavelength at which it emits most strongly. (b) The star HD 209458 itself has a surface temperature of \(6030 \mathrm{~K}\). Calculate its wavelength of maximum emission, assuming it to be a blackbody. (c) If a high-resolution telescope were to be used in an attempt to record an image of the planet orbiting HD 209458, would it be better for the telescope to use visible or infrared light? Explain your reasoning.

Short Answer

Expert verified
The wavelength of maximum emission for the planet is \(2.56 \, \mu m \) and for the star is \(480 \, nm \). For imaging the planet, it would be better to use a telescope equipped to observe in the infrared.

Step by step solution

01

Calculate the Wavelength for the Planet

We'll first apply Wien's Law to the planet. Since we know the temperature is \(1130 \, K \), we substitute it into the formula to find the most strongly emitted wavelength: \( \lambda_{max,planet} = \frac{2.898 \times 10^{-3} \, m \cdot K}{1130 \, K} \).
02

Calculate the Wavelength for the Star

Let's do the same for the star, which has a temperature of \(6030 \, K\). By substituting this temperature into Wien's Law, we find: \( \lambda_{max,star} = \frac{2.898 \times 10^{-3} \, m \cdot K}{6030 \, K} \).
03

Answer the Best Telescope for Observation

To decide if we should use visible or infrared light to view the planet, we need to understand where each type of light falls on the electromagnetic spectrum. Visible light typically corresponds to wavelengths between 400 nm and 700 nm, and infrared light corresponds to wavelengths longer than 700 nm. Considering that the wavelength corresponding to the strongest emission from the planet is longer than 700 nm, the telescope would be better equipped to image the planet in the infrared.

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