Question

\text { Write the balanced nuclear equation for the process summarized as }{ }_{13}^{27} \mathrm{Al}(\mathrm{n}, \alpha)_{11}^{24} \mathrm{Na} \text {. }SOLUTION Analyze We must go from the condensed descriptive form of the reaction to the balanced nuclear equation. Plan We arrive at the balanced equation by writing \(\mathrm{n}\) and \(\alpha\), each with its associated subscripts and superscripts. Solve The \(\mathrm{n}\) is the abbreviation for a neutron \(\left({ }_{0}^{1} \mathrm{n}\right)\) and \(\alpha\) represents an alpha particle ( \(\left.{ }_{2}^{4} \mathrm{He}\right)\). The neutron is the bombarding particle, and the alpha particle is a product. Therefore, the nuclear equation is $$ { }_{13}^{27} \mathrm{Al}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{11}^{24} \mathrm{Na}+{ }_{2}^{4} \mathrm{He} \text { or } \quad{ }_{13}^{27} \mathrm{Al}+\mathrm{n} \longrightarrow{ }_{11}^{24} \mathrm{Na}+\alpha $$

Short answer

The balanced nuclear equation for the given reaction is: $$ _{13}^{27} \mathrm{Al} + _{0}^{1} \mathrm{n} \longrightarrow _{11}^{24} \mathrm{Na} + _{2}^{4} \mathrm{He} $$

Step-by-step solution

Identify the given elements and particles

We are given the following elements and particles in the reaction: - Aluminum isotope: \(_{13}^{27} \mathrm{Al}\) - Neutron: \(\mathrm{n}\) - Alpha particle (\(\alpha\)): A Helium nucleus \(_{2}^{4}\mathrm{He}\) - Sodium isotope: \(_{11}^{24}\mathrm{Na}\) 2.

Write the nuclear equation in the provided format

To balance the nuclear equation, place the reacting particles on the left side of the equation and the products on the right side. $$ _{13}^{27} \mathrm{Al} + \mathrm{n} \longrightarrow _{11}^{24} \mathrm{Na} + \alpha $$ 3.

Replace the neutron and alpha particle with their respective notations

We write the neutron and alpha particle with their corresponding subscripts and superscripts, which are \(_{0}^{1}\mathrm{n}\) and \(_{2}^{4}\mathrm{He}\), respectively. Now, the balanced nuclear equation is: $$ _{13}^{27} \mathrm{Al} + _{0}^{1} \mathrm{n} \longrightarrow _{11}^{24} \mathrm{Na} + _{2}^{4} \mathrm{He} $$

Key concepts

Nuclear Transmutation

Nuclear transmutation is the process of one chemical element or isotope being transformed into another. This transformation happens at the nuclear level, meaning that it's not about chemical reactions involving electrons, but changes in the nucleus itself. In the realm of nuclear physics, transmutation can be initiated by bombarding a stable nucleus with particles. These particles could include neutrons, protons, or even other nuclei.

When a nucleus changes into another, it involves a change in the atomic number (the number of protons) which leads to the formation of a completely different element. For example, in our problem, when a neutron hits an aluminum nucleus, it results in the creation of sodium and an alpha particle. This is a classic nuclear transmutation process.

Balancing Nuclear Equations

When balancing nuclear equations, it's important to ensure that the total number of protons and neutrons is the same on both sides of the equation. The key to balancing these equations lies in the conservation laws:

• Conservation of Mass Number: The sum of the mass numbers (total protons and neutrons) of the reactants must equal the sum of the mass numbers of the products.
• Conservation of Atomic Number: The sum of the atomic numbers (protons) of the reactants must equal the sum of the atomic numbers of the products.

In our exercise, the reaction starts with aluminum \[_{13}^{27} \mathrm{Al}\] and a neutron \[_{0}^{1}\mathrm{n}\], leading to sodium \[_{11}^{24}\mathrm{Na}\] and an alpha particle \[_{2}^{4}\mathrm{He}\]. By adding the subscript and superscript numbers in both sides of the equation, we ensure the equation is balanced, thus adhering to these conservation principles.

Isotopes and Isotopic Notation

Isotopes are forms of the same element that have the same number of protons but differ in the number of neutrons. This difference in neutron number gives each isotope a different mass. In isotopic notation, you typically see the element's symbol along with two numbers:

• A superscript indicating the atomic mass number (sum of protons and neutrons).
• A subscript representing the atomic number (number of protons).

For example, the aluminum isotope in our exercise is written as \[_{13}^{27}\mathrm{Al}\], where '13' is the atomic number and '27' is the atomic mass number. This notation helps in quickly identifying isotopes and is crucial when dealing with nuclear reactions as it provides a snapshot of the nucleus's composition.

Alpha Particles in Nuclear Reactions

Alpha particles are a type of ionizing radiation made up of two protons and two neutrons, identical to the nucleus of a helium atom. In nuclear reactions, these particles are significant as they are a common product or bombarding particle.

Alpha decay, for instance, is a type of radioactive decay where a parent nucleus emits an alpha particle reducing its mass number by four and atomic number by two. In our exercise, the alpha particle \(_{2}^{4}\mathrm{He}\) is a product. This means that during the reaction, aluminum underwent a change that released this helium nucleus.

Due to their comparatively massive size and charge, alpha particles have a low penetration depth, making them less likely to travel far. However, they can still cause significant ionization along their path, which is why they are important in nuclear physics and radiation protection contexts.