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Chapter 34: Problem 39

Treating the Sun as a \(5800-\mathrm{K}\) blackbody, compare its UV radiance at 200 nm with its visible radiance at its 500 -nm peak wavelength.

Short Answer

Expert verified
The UV radiance at 200 nm and Visible radiance at 500 nm can be calculated using Planck's Law. The comparison between the two is given as a ratio of the UV radiance to the visible radiance.

Step by step solution

01

Identify the Variables

The temperature of the Sun as a blackbody is \(T = 5800 K\). The UV wavelength we're focusing on is \(位1 = 200 nm = 200 \times 10^-9 m\) and the visible radiance wavelength is \(位2 = 500 nm = 500 \times 10^-9 m\). We're tasked to compare the UV radiance with the visible radiance.
02

Apply Planck's Law

Planck's Law states that the spectral radiance of electromagnetic radiation from a black body in thermal equilibrium is given by \(B位(T) = \frac{2hc^2}{位^5}\frac{1}{e^{hc/(位KT)}-1}\) where \(h\) is Planck constant, \(c\) is speed of light, \(K\) is the Boltzmann constant \(T\) is the temperature and \(B位(T)\) is the spectral radiance.
03

Calculate UV radiance at 200 nm

Substitute the values for \(h, K, c, 位1\) and \(T\) into Planck's Law to find the UV radiance at 200 nm.
04

Calculate Visible radiance at 500 nm

Similarly, substitute the values for \(h, K, c, 位2\) and \(T\) into Planck's Law to find the visible radiance at 500 nm.
05

Compare the Values

By comparing both values calculated in step 3 and step 4, it can be determined how the UV radiance at 200 nm compares with the visible radiance at the peak 500 nm wavelength. This comparison will be stated as a ratio of UV radiance to visible radiance.

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