Question 40.2: Cosmic Background Radiation As we learned in Section 34-5, r......

INTERPRET and ANTICIPATE
This problem gives us a chance to use Planck’s formula I_\lambda(\lambda, T)=\frac{2 \pi h c^2}{\lambda^5\left(e^{h c / \lambda k_{ B } T}-1\right)}( Eq .40 .5). It is always helpful to plot complicated functions, as we are doing here.

SOLVE
Substitute all the constants into Planck’s formula, as well as the given temperature. A spreadsheet program is helpful. A few values have been provided so that you can check your own results.

Plotting spectral intensity on the vertical axis in milliWatts per meter cubed and wavelength on the horizontal axis in millimeters (Fig. 40.8) produces a curve that looks very much like the black-body curves in Figure 40.3.

CHECK AND THINK
To check our work, let’s use Wien’s law (Eq. 40.1) to find the peak wavelength for a black body with a temperature of 2.725 K.

The peak wavelength we just found is consistent with our graph, which you can see peaks at just a little more than 1 mm. Further, we learned in Section 34-5 that the cosmic background radiation is observed in the microwave part of the electromagnetic spectrum. Compare Figure 40.8 to Table 34.2, and you will see that our entire graph falls in the microwave part of the spectrum. Finally, we note that cosmologists measure the background radiation, and then fit the spectrum to Planck’s function to arrive at the corresponding black-body temperature of 2.725 K. (This is the reverse of what we did in this example.) Often, this temperature is referred as the temperature of the Universe, which may be a little misleading because not everything in the Universe is at the same temperature.