Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \({ }^{68} \mathrm{Ga}\) (electron capture) c. \({ }^{212} \mathrm{Fr}(\alpha)\) b. \(^{62} \mathrm{Cu}\) (positron) d. \({ }^{129} \mathrm{Sb}(\beta)\)

Electron capture is a process by which an unstable atom increases its stability. During electron capture, an inner orbital electron is drawn into the nucleus where it combines with a proton to form a neutron and a neutrino, the latter of which is ejected from the atom. This reaction is written as:
\[\begin{equation}_{Z}^{A}X + e^- \to_{Z-1}^{A}Y + \text{neutrino}\end{equation}\]For example, when gallium-68 undergoes electron capture, the equation looks like:
\[\begin{equation}_{31}^{68}\mathrm{Ga} + e^- \to_{30}^{68}\mathrm{Zn} + \text{neutrino}\end{equation}\]Through this process, gallium-68 transforms into zinc-68. To understand this concept better, know that the capture of the electron causes a conversion in the nucleus: a proton becomes a neutron, which reduces the atomic number by one while the mass number remains unchanged.

Positron emission, also known as beta-plus decay, occurs when a proton inside a nucleus is transformed into a neutron and a positron is created and emitted. Positrons are the antimatter equivalent of electrons. The equation representing this decay is quite similar to electron capture but in reverse:
\[\begin{equation}_{Z}^{A}X \to_{Z-1}^{A}Y + e^+\end{equation}\]Using copper-62 as an example, the positron emission process is written as:
\[\begin{equation}_{29}^{62}\mathrm{Cu} \to_{28}^{62}\mathrm{Ni} + e^+\end{equation}\]In this representative equation, copper-62 decays into nickel-62 with the emission of a positron. This means that the atomic number decreases by one (reflecting the loss of a proton), but the mass number stays constant, as the total number of nucleons (protons and neutrons) does not change.

Alpha decay is a type of radioactive decay in which an unstable nucleus ejects an alpha particle, which is essentially a helium-4 nucleus made up of two protons and two neutrons. As a result, the original atom loses two protons and two neutrons, leading to a decrease in both the atomic and mass numbers. The general equation is:
\[\begin{equation}_{Z}^{A}X \to_{Z-2}^{A-4}Y + _2^4\mathrm{He}\end{equation}\]When considering the alpha decay of francium-212, the equation is expressed as:
\[\begin{equation}_{87}^{212}\mathrm{Fr} \to_{85}^{208}\mathrm{At} + _2^4\mathrm{He}\end{equation}\]Here, an alpha particle is expelled from francium-212, leaving behind astatine-208. This process significantly alters the identity of the original nucleus, decreasing its atomic number by two and its mass number by four.

Beta decay is a common mode of radioactive decay in which a neutron is converted into a proton within the nucleus, releasing an electron (beta particle) in the process. The atomic number thus increases by one, but the mass number remains unchanged because the neutron and proton have similar masses. The general formula for beta decay is:
\[\begin{equation}_{Z}^{A}X \to_{Z+1}^{A}Y + e^-\end{equation}\]For instance, antimony-129 undergoing beta decay transforms according to:
\[\begin{equation}_{51}^{129}\mathrm{Sb} \to_{52}^{129}\mathrm{Te} + e^-\end{equation}\]This results in the production of tellurium-129 and an electron. It's essential to remember that beta decay increases the atomic number because a new proton is formed in the nucleus, while the ejected electron is the beta particle that gives this process its name.