(a) Distinguish between photodissociation and photoionization. (b) Use the energy requirements of these two processes to explain why photodissociation of oxygen is more important than photoionization of oxygen at altitudes below about \(90 \mathrm{~km}\).

Photodissociation is the process where a molecule absorbs a photon of light and breaks apart into smaller fragments. This can be represented as: \[AB + h\nu \rightarrow A + B\] Where \(AB\) is the molecule, \(h\nu\) represents the energy of the photon, and \(A\) and \(B\) are the resulting molecular fragments. On the other hand, photoionization is the process where an atom or molecule absorbs a photon and loses an electron. The general equation for this process can be represented as: \[X + h\nu \rightarrow X^{+} + e^-\] Here, \(X\) is the atom or molecule, \(h\nu\) represents the energy of the photon, \(X^{+}\) is the resulting ion, and \(e^-\) represents the released electron.

To explain the importance of photodissociation of oxygen at altitudes below \(90 \mathrm{~km}\), we need to compare the energy requirements of photodissociation and photoionization. For oxygen, the photodissociation process can be represented as: \[O_2 + h\nu \rightarrow O + O\] The energy required for this process is around \(5.12 \mathrm{~eV}\) per photon. On the other hand, the photoionization of oxygen involves: \[O_2 + h\nu \rightarrow O_2^{+} + e^-\] The energy required for oxygen photoionization is more significant at approximately \(12.08 \mathrm{~eV}\) per photon.

Since photodissociation of oxygen (\(O_2\)) requires less energy than photoionization (\(5.12 \mathrm{~eV}\) versus \(12.08 \mathrm{~eV}\)), it is more likely to occur at low altitudes where the energy of incoming photons is lower. Furthermore, when considering the Earth's atmosphere's composition, there is less ozone (O3) at altitudes below \(90 \mathrm{~km}\), which means there is less absorption of ultraviolet light by ozone. Consequently, there is a higher level of ultraviolet light available to cause photodissociation of oxygen molecules. This is why photodissociation is more important than photoionization at altitudes below about \(90 \mathrm{~km}\). Photodissociation leads to more effective dissociation of oxygen molecules at lower altitudes due to lower energy requirements and a greater availability of ultraviolet light to drive this process.