Problem 32 - Chemistry: The Central Science in SI Units, Global Edition

Question

Write balanced equations for each of the following nuclear reactions: \((\mathbf{a}){ }_{92}^{238} \mathrm{U}(\mathrm{n}, \gamma){ }^{239} \mathrm{U},(\mathbf{b}){ }_{82}^{16} \mathrm{O}(\mathrm{p}, \alpha){ }^{13} \mathrm{~N},\) (c) \({ }_{8}^{18} \mathrm{O}\left(\mathrm{n}, \beta^{-}\right){ }^{19} \mathrm{~F}\).

Step-by-Step Solution

Answer

The short answer for the balanced nuclear reactions is: a) \(_{92}^{238}\mathrm{U}(\mathrm{n}, \gamma){ }^{239}\mathrm{U}\) b) \(_{8}^{16}\mathrm{O}(\mathrm{p}, \alpha){ }^{13}\mathrm{N}\) c) \(_{8}^{18}\mathrm{O}\left(\mathrm{n}, \beta^{-}\right){ }^{19}\mathrm{F}\)

Step 1: Reaction a: Balancing Uranium capture of a neutron

We have the following nuclear reaction: \[_{92}^{238}\mathrm{U}(\mathrm{n}, \gamma){ }^{239}\mathrm{U}\] Steps to balance the reaction: 1. Start with the original reaction: \(_{92}^{238}\mathrm{U}(\mathrm{n}, \gamma)\). 2. Add the neutron to the uranium nucleus: \[_{92}^{238}\mathrm{U} + _{0}^{1}\mathrm{n}\]. 3. Apply the conservation of atomic numbers: \(92 + 0 = 92\). 4. Apply the conservation of mass numbers: \(238 + 1 = 239\). 5. Write the balanced reaction as: \[_{92}^{238}\mathrm{U} + _{0}^{1}\mathrm{n} \rightarrow { }^{239}\mathrm{U}\]. The balanced equation for uranium capturing a neutron is: \[_{92}^{238}\mathrm{U}(\mathrm{n}, \gamma){ }^{239}\mathrm{U}\]

Step 2: Reaction b: Balancing Oxygen proton capture

We have the following nuclear reaction: \[_{8}^{16}\mathrm{O}(\mathrm{p}, \alpha){ }^{13}\mathrm{N}\] Steps to balance the reaction: 1. Start with the original reaction: \(_{8}^{16}\mathrm{O}(\mathrm{p}, \alpha)\). 2. Add the proton to the oxygen nucleus: \[_{8}^{16}\mathrm{O} + _{1}^{1}\mathrm{p}\]. 3. Apply the conservation of atomic numbers: \(8 + 1 = 9\). 4. Apply the conservation of mass numbers: \(16 + 1 = 17\). 5. Write the balanced reaction as: \[_{8}^{16}\mathrm{O} + _{1}^{1}\mathrm{p} \rightarrow { }_{9}^{17}\mathrm{X} + _{2}^{4}\mathrm\nuclide{\alpha}\] 6. Identify the nuclide with atomic number 9: \(!_9\nuclide{F}\). 7. Finally, substitute the final term: \[_{8}^{16}\mathrm{O} + _{1}^{1}\mathrm{p} \rightarrow { }^{13}\mathrm{N} + _{2}^{4}\mathrm\nuclide{\alpha}\] The balanced equation for oxygen capturing a proton is: \[_{8}^{16}\mathrm{O}(\mathrm{p}, \alpha){ }^{13}\mathrm{N}\]

Step 3: Reaction c: Balancing Oxygen beta decay

We have the following nuclear reaction: \[_{8}^{18}\mathrm{O}\left(\mathrm{n}, \beta^{-}\right){ }^{19}\mathrm{F}\] Steps to balance the reaction: 1. Start with the original reaction: \(_{8}^{18}\mathrm{O}\left(\mathrm{n}, \beta^{-}\right)\). 2. Apply the conservation of atomic numbers: \(8 + 1 = 9\). 3. Apply the conservation of mass numbers: \(18 + 0 = 18\). 4. Write the balanced reaction as: \[_{8}^{18}\mathrm{O} \rightarrow { }_{9}^{18}\mathrm{X} + _{-1}^{0}\mathrm\nuclide{e}\] 5. Identify the nuclide with atomic number 9: \(!_9\nuclide{F}\). 6. Finally, substitute the final term: \[_{8}^{18}\mathrm{O} \rightarrow { }^{19}\mathrm{F} + _{-1}^{0}\mathrm\nuclide{e}\] The balanced equation for oxygen beta decay is: \[_{8}^{18}\mathrm{O}\left(\mathrm{n}, \beta^{-}\right){ }^{19}\mathrm{F}\]

Key Concepts

balanced nuclear equations

Nuclear reactions involve changes in an atom's nucleus, leading to the formation of new elements. Just like chemical equations, these reactions need to be balanced to respect the conservation laws. In nuclear equations, both the conservation of mass number and atomic number must be upheld. For instance, when a nuclear reaction occurs, the total number of protons and neutrons (mass number) as well as the total number of protons (atomic number) must be identical on both sides of the equation. This ensures that the nuclear reaction is accurately represented, and each participating particle is accounted for correctly. A balanced nuclear equation provides a detailed report of the nuclear process without leaving out any particles involved—an essential aspect for detailed nuclear reaction analysis.

conservation of atomic numbers

In nuclear chemistry, the conservation of atomic numbers is a key principle. This principle dictates that the sum of atomic numbers (protons in the nucleus) of the reactants must equal the sum of atomic numbers of the products. For example, if a neutron is captured by a Uranium isotope, the atomic number of Uranium remains unchanged because a neutron has no charge. It highlights the rule that though particles like neutrons or neutrinos may change during reactions, the number of protons in the nucleus remains constant unless a conversion such as beta decay occurs. This conservation helps chemists and physicists predict the products of nuclear reactions.

conservation of mass numbers

The conservation of mass numbers ensures that the total number of nucleons (protons and neutrons) remains constant throughout a nuclear reaction. When adjusting nuclear equations, scientists tally up these numbers on both sides of the reaction to ensure they remain the same. In a nuclear reaction, such as when Oxygen captures a proton, the mass numbers are added and reviewed: the protons and neutrons from both the original oxygen nucleus and the incoming proton. This meticulous accounting guarantees that the nuclear reaction's equation is complete and properly balanced, preserving physical consistency.

neutron capture

Neutron capture is a type of nuclear reaction where a nucleus absorbs a neutron, leading to the formation of a heavier isotope. In the reaction \(_{92}^{238}\mathrm{U}(\mathrm{n}, \gamma) { }^{239}\mathrm{U}\), a **neutron** (\(_0^1\mathrm{n}\)) is added to the **Uranium nucleus** (\(_{92}^{238}\mathrm{U}\)). Neutron capture is significant in the creation of heavy elements in stars and has vital applications in nuclear reactors. Here, it often leads to subsequent radioactive decay, helping maintain a chain reaction. Understanding this process is crucial for nuclear power generation and synthesis of isotopes for medical and industrial use.

proton capture

In a proton capture reaction, a nucleus absorbs a proton, resulting in the formation of a new isotope. This process can alter both the atomic and mass numbers of an element. For example, in the equation \(_{8}^{16}\mathrm{O}(\mathrm{p}, \alpha) { }^{13}\mathrm{N}\), Oxygen captures a proton (\(_1^1\mathrm{p}\)). The atomic number increases, signifying a change from Oxygen to a new element as a proton adds another positive charge to the nucleus. This type of reaction is important for generating certain isotopes and understanding stellar nucleosynthesis, where new elements are created inside stars.

beta decay

Beta decay is a form of radioactive decay where a neutron in a radioactive nucleus transforms into a proton, emitting an electron (beta particle) and an antineutrino. In nuclear reactions such as \(_{8}^{18}\mathrm{O}\left(\mathrm{n}, \beta^{-}\right) { }^{19}\mathrm{F}\), the resulting change leads to the creation of a different element from the parent nucleus. During beta decay, the atomic number increases by one while the mass number remains unchanged. This property allows one element to convert into another, altering the nucleus's characteristics significantly. Understanding beta decay is fundamental for fields like nuclear physics, as it helps explain various natural and synthetic processes.