Problem 19 - Chemistry The Central Science

Question

(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 \(90 \mathrm{~km}\).

Step-by-Step Solution

Answer

(a) Photodissociation is the process where a molecule absorbs a photon of light and breaks down into smaller fragments, while photoionization is when an atom or molecule absorbs a photon and loses an electron, forming a positively charged ion. (b) At altitudes below 90 km, photodissociation of oxygen is more important than photoionization due to its lower energy requirement in the UV light range (~ \(240 \mathrm{~nm} \)) compared to photoionization in the EUV light range (~ \(20 \mathrm{~nm} \)). Moreover, the Earth's atmosphere absorbs the majority of high-energy EUV photons and has a higher proportion of neutral molecules like oxygen, making photodissociation more likely to occur at these lower altitudes.

Step 1: Defining Photodissociation and Photoionization

Photodissociation is a process in which a molecule absorbs a photon of light and breaks down into smaller fragments. On the other hand, photoionization is the process where an atom or molecule absorbs a photon of light and loses an electron, forming a positively charged ion.

Step 2: Energy Requirements of Photodissociation and Photoionization

For a molecule to undergo photodissociation or photoionization, it must absorb a photon that has enough energy to break the chemical bond between its atoms or ionize it by removing an electron. The energy requirement for photodissociation of oxygen (\(O_2\)) is lower than that of photoionization. Specifically, photodissociation requires an energy in the ultraviolet (UV) light range (~ \(240 \mathrm{~nm} \)), while photoionization demands a much higher energy, in the extreme ultraviolet (EUV) light range (~ \(20 \mathrm{~nm} \)).

Step 3: Explaining the Importance of Photodissociation at Lower Altitudes

At altitudes below 90 km, the Earth's atmosphere consists of a higher proportion of neutral molecules like oxygen, which makes photodissociation more likely to occur. Moreover, the atmosphere absorbs the majority of the high-energy EUV photons that would cause photoionization before they can reach these lower altitudes. This absorption is primarily due to the interaction of EUV photons with atomic and molecular nitrogen in the atmosphere. As a result, there are fewer EUV photons available for photoionization of oxygen, making photodissociation the more significant process at altitudes below 90 km. In summary, the primary reason behind the importance of photodissociation over photoionization at altitudes below 90 km is the lower energy requirement for photodissociation and the decreasing availability of high-energy EUV photons at these altitudes due to atmospheric absorption.