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Ch.18 - Chemistry of the Environment
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All textbooksBrown 14th EditionCh.18 - Chemistry of the EnvironmentProblem 19b

Chapter 18, Problem 19b

(b) Use the energy requirements of these two pro- cesses to explain why photodissociation of oxygen is more important than photoionization of oxygen at altitudes below about 90 km.

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Hi everyone for this problem, We're told at altitudes below 80 km. Photo ionization of oxygen becomes more important than photo dissociation of oxygen is this statement true or false. So we want to know whether it's true or false if photo ionization is more important than photo dissociation at altitudes below 80 km. So the first thing we want to do is to compare the two. So let's start off by talking about photo dissociation. Okay, photo dissociation is bond breaking, and this requires killing joules per mole of energy. On the other hand, photo ionization, this is ejecting an electron and this requires 205 kg joules per mole of energy. So, as we can see, photo dissociation has a lower energy than photo ionization. Okay, so with energy and wavelength they have an inverse relationship. Okay, so this should say lower energy. So because photo dissociation has lower energy, it means it's going to require longer wavelengths and I'm going to put a lambda here for wavelength. On the other hand, photo ionization because it has higher energy, it's going to require shorter wavelengths. Okay, so let's consider our altitude given here At lower elevations, shorter wavelengths have already been absorbed. So longer wavelengths can pass through. So below 80 km is going to require longer wavelengths. So, because of this photo dissociation is going to dominate. Okay, so that is going to make this statement false. So we're gonna go ahead and write false because photo dissociation is what's going to dominate and that is going to be the answer to the statement. The statement is false. Okay, Because photo dissociation dominates. That's the end of this problem. I hope this was helpful.
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