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Photodissociation is a process where a molecule absorbs a photon and then dissociates into smaller fragments or atoms. It occurs when the energy of the absorbed photon exceeds the bond energy of the molecule. For example, O2 absorbing a photon and breaking into O + O. Photoionization, on the other hand, is a process where an atom or a molecule absorbs a photon and loses an electron, resulting in a positive ion. It occurs when the energy of the absorbed photon is greater than the ionization potential or the energy required to remove an electron. For example, O2 absorbing a photon and becoming O2+ + e-.

In order to understand which process is more important at lower altitudes, we need to compare the energy requirements for them. The energy required for photodissociation of oxygen (O2) can be represented as: E_dissociation = \(h\nu\) Where \(h\) is the Planck's constant, and \(\nu\) is the frequency of the photon. The energy required for photoionization of oxygen can be represented as: E_ionization = IP Where IP is the ionization potential of oxygen.

The ionization potential of oxygen is greater than the bond energy of oxygen, thus requiring a higher-energy photon for photoionization to occur compared to photodissociation. At altitudes below 90 km, the atmosphere filters out much of the higher-energy photons that are required for photoionization. As a result, the photons with sufficient energy for photodissociation are more abundant, leading to a greater rate of photodissociation compared to photoionization. Additionally, there is a higher concentration of oxygen molecules at lower altitudes, which further increases the likelihood of photodissociation. Therefore, photodissociation of oxygen is more important than photoionization of oxygen at altitudes below about 90 km due to the abundance of lower-energy photons and higher concentration of oxygen molecules at these altitudes.