A nuclear equation is a symbolic representation of a change that occurs within an atom’s nucleus, such as radioactive decay or transmutation. These reactions involve the transformation of one element into another by altering the number of protons and neutrons in the nucleus. Unlike chemical equations, which only involve the rearrangement of electrons, nuclear equations represent fundamental changes to the atomic core. For these equations to be physically accurate, they must be mathematically balanced, reflecting the fundamental conservation laws of physics.
Understanding Nuclear Notation and Particles
The standard way to represent any atomic nucleus, or nuclide, is through the notation \({}^{A}_{Z}X\). The symbol \(X\) represents the chemical element, which is determined by the atomic number, \(Z\), found in the lower left position. The atomic number specifies the number of protons in the nucleus, which defines the element’s identity. The mass number, \(A\), is located in the upper left position and represents the total count of nucleons—both protons and neutrons—within the nucleus.
Several subatomic particles are frequently involved in these transformations, each having a specific notation. An alpha particle, identical to a helium nucleus, is written as \({}^{4}_{2}He\) or \(\alpha\), indicating two protons and two neutrons. A beta particle is a high-energy electron symbolized as \({}^{0}_{-1}e\) or \(\beta\), having negligible mass and a charge of negative one.
A positron, the anti-particle of an electron, is represented by \({}^{0}_{+1}e\) and carries a positive one charge with no effective mass number. The neutron, a neutral particle, is written as \({}^{1}_{0}n\), showing a mass number of one and an atomic number of zero. A gamma ray, a form of high-energy electromagnetic radiation, is written as \({}^{0}_{0}\gamma\), as it possesses neither mass nor charge.
The Conservation Rules for Balancing
All nuclear equations are governed by two conservation rules, which are the basis for balancing the reaction. The first is the Conservation of Mass Number, which dictates that the sum of the mass numbers (\(A\) values, the top numbers) on the reactant side must equal the sum of the mass numbers on the product side. This means the total number of protons and neutrons remains constant throughout the reaction.
The second rule is the Conservation of Atomic Number, which states that the sum of the atomic numbers (\(Z\) values, the bottom numbers) for the reactants must equal the sum of the atomic numbers for the products. Since the atomic number represents the total charge in the nucleus, this rule also ensures the conservation of charge. Applying these two conservation rules allows any unknown mass number or atomic number in a nuclear equation to be determined.
Solving Common Types of Nuclear Equations
The process for solving for an unknown particle or nuclide, often represented by \({}^{A}_{Z}X\), involves the direct application of the conservation rules. First, sum all the known mass numbers (\(A\)) on both sides of the equation. Second, perform the same summation for the known atomic numbers (\(Z\)) on both sides. Finally, use the two conservation rules to set up simple algebraic equations to find the missing \(A\) and \(Z\) values; the \(Z\) value will identify the element.
Consider an alpha decay example, where Uranium-238 (\({}^{238}_{92}U\)) decays by emitting an alpha particle (\({}^{4}_{2}He\)). The incomplete equation is \({}^{238}_{92}U \rightarrow {}^{A}_{Z}X + {}^{4}_{2}He\). To find the missing mass number, \(A\), the conservation of mass number is used: \(238 = A + 4\). Solving this reveals that \(A = 234\). Next, the atomic number, \(Z\), is determined using the conservation of atomic number: \(92 = Z + 2\).
Solving this equation gives \(Z = 90\). By finding the element with an atomic number of 90 on the periodic table, the missing nuclide is identified as Thorium-234 (\({}^{234}_{90}Th\)). The complete, balanced nuclear equation is therefore \({}^{238}_{92}U \rightarrow {}^{234}_{90}Th + {}^{4}_{2}He\).
A beta decay problem requires careful attention to the negative value of the beta particle’s atomic number. For instance, in the beta decay of Iron-59 (\({}^{59}_{26}Fe\)), the incomplete reaction is \({}^{59}_{26}Fe \rightarrow {}^{A}_{Z}X + {}^{0}_{-1}e\). Mass number conservation is \(59 = A + 0\), which immediately shows that \(A = 59\).
The conservation of atomic number is where the negative value must be included: \(26 = Z + (-1)\). To solve for \(Z\), adding 1 to both sides results in \(Z = 27\). Consulting the periodic table shows that the element with an atomic number of 27 is Cobalt, making the resulting nuclide Cobalt-59 (\({}^{59}_{27}Co\)).