(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 km.

Short Answer

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(a) Photodissociation is the process where a molecule absorbs a photon and breaks apart into its constituent atoms or smaller molecules, while photoionization is a process in which an atom or molecule absorbs a photon and loses one or more electrons, becoming ionized. (b) Photodissociation of oxygen is more important than photoionization at altitudes below 90 km because the energy requirement for photodissociation (5.12 eV) is lower than that for photoionization (13.62 eV). At these lower altitudes, the concentration of oxygen molecules is higher, and the solar radiation is more likely to have the appropriate energy to cause photodissociation of molecular oxygen, rather than photoionization. Additionally, the absorption of high-energy photons by ozone and other molecules at higher altitudes filters out photons with sufficient energy for photoionization, further increasing the importance of photodissociation at lower altitudes.

Step by step solution

01

Define Photodissociation and Photoionization

Photodissociation and photoionization are both processes in which a molecule or atom interacts with a photon, leading to different outcomes. Photodissociation is the process in which a molecule absorbs a photon and subsequently breaks apart into its constituent atoms or smaller molecules. This process occurs when the absorbed photon has enough energy to break the chemical bonds that hold the molecule together. The general formula for this process can be written as: \(A_2 + h\nu \rightarrow A + A\) where \(A_2\) is the diatomic molecule, \(h\) is the Planck constant, \(\nu\) is the frequency of the photon, and \(A\) are the resulting separate atoms. Photoionization is a similar process in which an atom or molecule absorbs a photon and subsequently loses one or more electrons, thus becoming ionized. This will occur when the energy of the absorbed photon is higher than the ionization energy of the molecule or atom. The general formula for this process can be written as: \[A + h\nu \rightarrow A^+ + e^-\] where \(A\) is the atom or molecule, \(h\) is the Planck constant, \(\nu\) is the frequency of the photon, \(A^+\) is the ionized atom or molecule, and \(e^-\) is the ejected electron.
02

Compare the Energy Requirements

The energy requirements for photodissociation and photoionization depend on the specific molecule or atom and the energy of the photon. The photodissociation energy for molecular oxygen (\(O_2\)) is approximately 5.12 eV, while the ionization energy of atomic oxygen (\(O\)) is about 13.62 eV.
03

Photodissociation versus Photoionization at Lower Altitudes

At altitudes below 90 km, the solar radiation includes photons with a wide range of energies. Due to the energy requirements discussed in Step 2, it is more likely that oxygen molecules will encounter photons with enough energy to undergo photodissociation than photoionization. Moreover, at these lower altitudes, the concentration of oxygen molecules is higher, which further increases the probability of photodissociation occurring. Additionally, the absorption of high-energy photons by ozone (\(O_3\)) and other molecules at higher altitudes filters out photons with sufficient energy for photoionization. As a result, the photons that reach lower altitudes below 90 km are more likely to have the appropriate energy to cause photodissociation of molecular oxygen rather than photoionization. In conclusion, the photodissociation of oxygen is more important than photoionization at altitudes below 90 km due to the lower energy requirements for photodissociation, the higher concentration of oxygen molecules, and the selective filtering of high-energy photons by other atmospheric constituents.

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