This tells us that 5.5% of the 1-keV radiation from the center of the Galaxy would reach the earth if our assumptions are correct. At higher photon energies, the cross section is less, and the Galaxy quickly becomes quite transparent; at lower energies, the absorption rapidly becomes greater.

The problem can be turned around to find the photon energy at which the photon flux will be attenuated to e-1 = 0.368 of its initial value as it travels from the galactic center to the sun. Adopt the values of nH and r in (45), and Fpj ^p0 = 0.368, and then solve (44) for oeff. Then enter Fig. 7b to find the photon energy corresponding to oeff. The result is E ~ 1.5 keV. At this photon energy, 37% of the photons would survive the trip from the galactic center to the earth.

This calculation serves to fix one point on the 37% curve of Fig. 8 which gives propagation distances (in terms of the column density NH) that can be reached at different photon energies. In other words, the 37% curve tells us that, at E = 1.5keV,

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