Problem 73 - Chemistry The Central Science

Question

Nuclear scientists have synthesized approximately 1600 nuclei not known in nature. More might be discovered with heavyion bombardment using high-energy particle accelerators. Complete and balance the following reactions, which involve heavy-ion bombardments: (a) \({ }_{3}^{6} \mathrm{Li}+{ }_{28}^{56} \mathrm{Ni} \longrightarrow\) ? (b) \({ }_{20}^{40} \mathrm{Ca}+{ }_{96}^{248} \mathrm{Cm} \longrightarrow{ }_{62}^{147} \mathrm{Sm}+?\) (c) \({ }_{38}^{88} \mathrm{Sr}+{ }_{36}^{84} \mathrm{Kr} \longrightarrow{ }_{46}^{116} \mathrm{Pd}+?\) (d) \({ }_{20}^{40} \mathrm{Ca}+{ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{30}^{70} \mathrm{Zn}+4{ }_{0}^{1} \mathrm{n}+2 ?\)

Step-by-Step Solution

Answer

The short version of the answer is: (a) ${}_{3}^{6}\mathrm{Li} + {}_{28}^{56}\mathrm{Ni} \longrightarrow {}_{28}^{62}\mathrm{Ni}$ (b) ${}_{20}^{40}\mathrm{Ca} + {}_{96}^{248}\mathrm{Cm} \longrightarrow {}_{62}^{147}\mathrm{Sm} + {}_{54}^{141}\mathrm{Xe}$ (c) ${}_{38}^{88}\mathrm{Sr} + {}_{36}^{84}\mathrm{Kr} \longrightarrow {}_{46}^{116}\mathrm{Pd} + {}_{28}^{56}\mathrm{Ni}$ (d) ${}_{20}^{40}\mathrm{Ca} + {}_{92}^{238}\mathrm{U} \longrightarrow {}_{30}^{70}\mathrm{Zn} + 4{ }_{0}^{1}\mathrm{n} + 2{ }_{41}^{102}\mathrm{Nb}$

Step 1: Balancing Reaction (a)

In this reaction, we have: \[ { }_{3}^{6} \mathrm{Li} + { }_{28}^{56} \mathrm{Ni} \longrightarrow ? \] To balance the reaction, we must add a nucleus on the right side of the equation such that the total atomic number (number of protons) and mass number (total number of protons and neutrons) are conserved. Atomic number balance: \(3 + 28 = 3 + x_A\), where \(x_A\) is the atomic number of the missing nucleus. This means that \(x_A = 28\). Mass number balance: \(6 + 56 = y_m + x_m\), where \(y_m\) and \(x_m\) are the mass numbers of the products. If we assume that there is only one nucleus produced, then \(y_m = 0\) and \(x_m = 62\). Thus, the balanced reaction is: \[ { }_{3}^{6}\mathrm{Li} + { }_{28}^{56}\mathrm{Ni} \longrightarrow { }_{28}^{62}\mathrm{Ni} \]

Step 2: Balancing Reaction (b)

In this reaction, we have: \[ { }_{20}^{40} \mathrm{Ca} + { }_{96}^{248} \mathrm{Cm} \longrightarrow { }_{62}^{147} \mathrm{Sm} + ? \] Atomic number balance: \(20 + 96 = 62 + x_A\), which means that \(x_A = 54\). Mass number balance: \(40 + 248 = 147 + x_m\), which means that \(x_m = 141\). Thus, the balanced reaction is: \[ { }_{20}^{40}\mathrm{Ca} + { }_{96}^{248}\mathrm{Cm} \longrightarrow{ }_{62}^{147}\mathrm{Sm} + { }_{54}^{141}\mathrm{Xe} \]

Step 3: Balancing Reaction (c)

In this reaction, we have: \[ { }_{38}^{88} \mathrm{Sr} + { }_{36}^{84} \mathrm{Kr} \longrightarrow { }_{46}^{116} \mathrm{Pd} + ? \] Atomic number balance: \(38 + 36 = 46 + x_A\), which means that \(x_A = 28\). Mass number balance: \(88 + 84 = 116 + x_m\), which means that \(x_m = 56\). Thus, the balanced reaction is: \[ { }_{38}^{88}\mathrm{Sr} + { }_{36}^{84}\mathrm{Kr} \longrightarrow{ }_{46}^{116}\mathrm{Pd} + { }_{28}^{56}\mathrm{Ni} \]

Step 4: Balancing Reaction (d)

In this reaction, we have: \[ { }_{20}^{40} \mathrm{Ca}+{ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{30}^{70} \mathrm{Zn}+4{ }_{0}^{1} \mathrm{n}+2 ? \] Atomic number balance: \(20 + 92 = 30 + 0 + 2x_A\), which means that \(x_A = 41\). Mass number balance: \(40 + 238 = 70 + 4 + 2x_m\), which means that \(x_m = 102\). Thus, the balanced reaction is: \[ { }_{20}^{40}\mathrm{Ca} + { }_{92}^{238}\mathrm{U} \longrightarrow{ }_{30}^{70}\mathrm{Zn} + 4{ }_{0}^{1}\mathrm{n} + 2{ }_{41}^{102}\mathrm{Nb} \]

Key Concepts

Heavy-Ion Bombardment

When it comes to creating new elements or isotopes, heavy-ion bombardment plays a crucial role. This process involves firing a heavy ion, which is an atom with many protons and neutrons, such as lithium or uranium, at a target nucleus at very high speeds. To achieve the necessary speeds, scientists use particle accelerators, which are machines capable of propelling charged particles to immense velocities. Upon collision, the heavy ion can fuse with the target nucleus, potentially forming a new, heavier element.

For students tackling nuclear reaction balancing, understanding the concept of heavy-ion bombardment is essential. It's the starting point of a reaction, providing the projectile that interacts with the target. The process is governed by the complicated dynamics of nuclear forces and requires precise calculations to predict the outcomes of such high-energy collisions.

Particle Accelerators

Particle accelerators are sophisticated pieces of equipment that propel charged particles, such as protons, ions, or electrons, close to the speed of light. They work using a combination of electric and magnetic fields to accelerate the particles along a designated path, often in a circular or straight line trajectory.

In the context of nuclear reactions, particularly heavy-ion bombardment, particle accelerators generate the necessary energy to overcome the repulsive electrostatic forces between positively charged nuclei. The resulting highly energetic collisions can form new isotopes and elements not found in nature. For students, understanding the role of particle accelerators is fundamental in grasping why certain nuclear reactions can only occur under laboratory conditions and not naturally on Earth.

Atomic Number Conservation

The atomic number is perhaps the most defining characteristic of an element. It is equal to the number of protons in an atom's nucleus and is denoted by the letter 'Z'. In any nuclear reaction, including heavy-ion bombardments, the law of conservation of atomic number must be upheld, meaning the total number of protons before and after the reaction must be the same.

For example, when balancing part (a) of the exercise, the combined atomic numbers on the left side of the equation should equal the atomic numbers on the right side. It is crucial for students to check this balance in order to correctly identify the products of a nuclear reaction. Correct application of this principle leads to a better understanding of nuclear stability and the formation of new elements.

Mass Number Conservation

Alongside the atomic number, the mass number (denoted by 'A') is another important aspect that must be conserved in a nuclear reaction. It represents the total number of protons and neutrons in the nucleus. The conservation of mass number ensures that the sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.

In the exercise solutions, a common step is determining the mass number of the unknown product by subtracting the known mass numbers from the total mass numbers of the reactants. This calculation is vital for completing and balancing nuclear reactions. Students should get into the practice of accounting for all the particles involved in a reaction, including any free neutrons, to ensure mass number conservation.