Question

Complete each of the following bombardment reactions: a. \(?+{ }_{18}^{40} \mathrm{Ar} \longrightarrow{ }_{19}^{43} \mathrm{~K}+{ }_{1}^{1} \mathrm{H}\) b. \({ }_{0} 1 n+{ }_{92}^{238} \mathrm{U} \longrightarrow ?\) c. \({ }_{0}^{1} n+? \longrightarrow{ }_{6}^{14} \mathrm{C}+{ }_{1}^{1} \mathrm{H}\) \(\mathbf{d} . ?+{ }_{28}^{64} \mathrm{Ni} \longrightarrow{ }_{111}^{272} \mathrm{Rg}+{ }_{0}^{1} n\)

Short answer

a) \({}_{2}^{4}\text{He}\)b) \({}_{92}^{239}\text{U}\)c) \({}_{5}^{13}\text{B}\)d) \({}_{83}^{209}\text{Bi}\)

Step-by-step solution

Understanding Bombardment Reactions

A bombardment reaction involves a target nucleus being struck by a projectile particle, resulting in the formation of new products.

Part (a) - Identifying the missing reactant

Write the equation for the reaction and balance it. For the reaction: \text{Missing Particle} + \text{\({}_{18}^{40}\text{Ar}\)} \rightarrow \text{\({}_{19}^{43}\text{K}\)} + \text{\({}_{1}^{1}\text{H}\)} Use the conservation of mass and atomic numbers to find the missing particle. Conservation of mass: \(x + 40 = 43 + 1\) leads to \(x = 4\). Conservation of atomic number: \( y + 18 = 19 + 1\) leads to \( y = 2\). Thus, the missing particle is \text{\({}_{2}^{4}\text{He}\)}.

Part (b) - Identifying the final product

Write the equation for the reaction: \text{\({}_{0}^{1}\text{n}\)} + \text{\({}_{92}^{238}\text{U}\)} \rightarrow \text{Product}. Use the conservation of mass and atomic numbers to find the product. Conservation of mass: \(1+238 = 239\). Conservation of atomic number: \(0+92 = 92\). Thus, the product is \text{\({}_{92}^{239}\text{U}\)}.

Part (c) - Identifying the missing reactant

Write the equation for the reaction and balance it. \text{\({}_{0}^{1}\text{n}\)} + \text{Missing Particle} \rightarrow \text{\({}_{6}^{14}\text{C}\)} + \text{\({}_{1}^{1}\text{H}\)}.Use conservation of mass and atomic numbers to find the missing particle. Conservation of mass: \(1+x = 14+1\) leads to \(x=14\). Conservation of atomic number: \(0+y = 6+1\) leads to \(y=5\). Thus, the missing particle is \text{\({}_{5}^{13}\text{B}\)}.

Part (d) - Identifying the missing reactant

Write the equation for the reaction and balance it. \text{Missing Particle} + \text{\({}_{28}^{64}\text{Ni}\)} \rightarrow \text{\({}_{111}^{272}\text{Rg}\)} + \text{\({}_{0}^{1}\text{n}\)}. Use the conservation of mass and atomic numbers to find the missing particle. Conservation of mass: \(x + 64 = 272 + 1\) leads to \(x=209\). Conservation of atomic number: \(y+28 = 111 + 0\) leads to \(y=83\). Thus, the missing particle is \text{\({}_{83}^{209}\text{Bi}\)}.

Key concepts

conservation of mass

In nuclear reactions, the total mass on the reactant side must equal the total mass on the product side. This principle is known as the conservation of mass.
When we balance nuclear equations, we use this concept to determine unknown particles.
For example, in the reaction: \(\text{Missing Particle} + {}_{18}^{40}\text{Ar} \rightarrow {}_{19}^{43}\text{K} + {}_{1}^{1}\text{H} \) The total mass on the left is the sum of the missing particle's mass and the argon nucleus's mass—(\text{x + 40}). On the right, we sum the masses of potassium and hydrogen—(43 and 1).
To find the missing particle's mass (x), we set up the equation: \(\text{x + 40 = 43 + 1} \) Solving for x, we get: \(\text{x = 4} \)
This shows how the conservation of mass helps us find unknown particles in nuclear reactions.

conservation of atomic numbers

The conservation of atomic numbers in nuclear reactions is as crucial as conserving mass. The sum of atomic numbers before and after a reaction must be equal. This helps balance equations and determine unknown particles.
Consider the reaction: \(\text{Missing Particle} + {}_{18}^{40}\text{Ar} \rightarrow {}_{19}^{43}\text{K} + {}_{1}^{1}\text{H} \) The atomic numbers are balanced by adding the atomic numbers of the reactants and products. \( \text{y + 18 = 19 + 1} \) By solving for y, the missing particle's atomic number: \( \text{y = 2} \)
So, the missing particle here is \(\text{{}_{2}^{4}\text{He}}\text{, or helium-4}.\)
Through this process, we ensure the conservation of atomic numbers in nuclear reactions.

nuclear reactions

Nuclear reactions involve changes in the nuclei of atoms, leading to the formation of new elements. These reactions can release or absorb large amounts of energy.
Key types of nuclear reactions include fission, fusion, and bombardment reactions.
In a bombardment reaction, a target nucleus is struck by a smaller particle like a neutron, proton, or alpha particle, resulting in new nuclear products.
For instance, if a uranium-238 nucleus is bombarded by a neutron, the product nucleus is uranium-239: \({}_{0}^{1}\text{n} + {}_{92}^{238}\text{U} \rightarrow {}_{92}^{239}\text{U} \)
Balancing these reactions relies on the conservation principles of mass and atomic numbers.
Understanding these principles helps us predict the outcomes of bombardment reactions and understand the process of element formation in nuclear chemistry.
These reactions have applications in power generation and medical treatments.