Question

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. \({ }^{129} \mathrm{Sb}(\beta)\)

Short answer

a. Electron Capture in \({ }^{68} \mathrm{Ga}\): \[{}^{68}_{31} \mathrm{Ga} + e^- \rightarrow {}^{68}_{30} \mathrm{Zn}\] b. Positron Decay in \({ }^{62} \mathrm{Cu}\): \[{}^{62}_{29} \mathrm{Cu} \rightarrow {}^{62}_{30} \mathrm{Zn} + e^+\] c. Alpha Decay in \({ }^{212} \mathrm{Fr}\): \[{}^{212}_{87} \mathrm{Fr} \rightarrow {}^{208}_{85} \mathrm{At} + {}^{4}_{2}\mathrm{He}\] d. Beta Decay in \({ }^{129} \mathrm{Sb}\): \[{}^{129}_{51} \mathrm{Sb} \rightarrow {}^{129}_{52} \mathrm{Te} + e^-\]

Step-by-step solution

a. Electron Capture in \({ }^{68} \mathrm{Ga}\)

Electron capture occurs when a proton in the nucleus captures an electron and converts into a neutron. The atomic number decreases by one, while the mass number remains the same. The general equation for electron capture is: \[{}^A_Z X + e^- \rightarrow {}^A_{Z-1} Y\] So for Ga-68, the equation will be: \[{}^{68}_{31} \mathrm{Ga} + e^- \rightarrow {}^{68}_{30} \mathrm{Zn}\]

b. Positron Decay in \({ }^{62} \mathrm{Cu}\)

Positron decay occurs when a neutron in the nucleus gets converted to a proton, and a positron is released. The atomic number increases by one, while the mass number remains the same. The general equation for positron decay is: \[{}^A_Z X \rightarrow {}^A_{Z+1} Y + e^+\] For Cu-62, the equation will be: \[{}^{62}_{29} \mathrm{Cu} \rightarrow {}^{62}_{30} \mathrm{Zn} + e^+\]

c. Alpha Decay in \({ }^{212} \mathrm{Fr}\)

Alpha decay occurs when a nucleus emits an alpha particle (consisting of 2 protons and 2 neutrons). The atomic number decreases by two, and the mass number decreases by four. The general equation for alpha decay is: \[{}^A_Z X \rightarrow {}^{A-4}_{Z-2} Y + {}^{4}_{2}\mathrm{He}\] For Fr-212, the equation will be: \[{}^{212}_{87} \mathrm{Fr} \rightarrow {}^{208}_{85} \mathrm{At} + {}^{4}_{2}\mathrm{He}\]

d. Beta Decay in \({ }^{129} \mathrm{Sb}\)

Beta decay occurs when a neutron in the nucleus gets converted to a proton, and a beta particle (an electron) is released. The atomic number increases by one, while the mass number remains the same. The general equation for beta decay is: \[{}^A_Z X \rightarrow {}^A_{Z+1} Y + e^-\] For Sb-129, the equation will be: \[{}^{129}_{51} \mathrm{Sb} \rightarrow {}^{129}_{52} \mathrm{Te} + e^-\]

Key concepts

Electron Capture

Electron capture is a form of radioactive decay where an atom's nucleus absorbs an orbiting electron, which causes a proton to convert into a neutron. This process results in the decrease of the atomic number by one, while the atomic mass number remains unchanged. It's important to note that during electron capture, no particles are emitted, differing it from other radioactive decay types where particles are released.

For instance, in the decay of gallium-68 gallium-68, which has an atomic number of 31, an electron is captured by the nucleus resulting in the production of zinc-68 with an atomic number of 30. The notation for this reaction is: gallium-68gallium-68: Zn

Positron Decay

Positron decay is a process where a neutron in an atom's nucleus is transformed into a proton and a positron (the antimatter counterpart of the electron). This reaction increases the atomic number by one but keeps the mass number stable. Positrons are positively charged particles that are identical to electrons except for their charge.

For example, when copper-62 undergoes positron decay, it changes into zinc-62. Here, a neutron becomes a proton, and a positron is expelled from the nucleus. The general equation with reference to copper-62 is as follows: Zn + e+.

Alpha Decay

In alpha decay, an atomic nucleus releases an alpha particle, which is composed of two protons and two neutrons, commonly represented as a helium-4 nucleus. This causes the original atom to lose both atomic mass and atomic number. Specifically, the atomic mass decreases by four units, and the atomic number decreases by two.

The case of francium-212 illustrates alpha decay. When francium-212 decays, it emits an alpha particle and transforms into astatine-208. The reaction can be written as follows: At + He.

Beta Decay

Beta decay occurs when a neutron in the nucleus of an atom is converted into a proton, resulting in the emission of a beta particle, which is essentially an electron. This transforms the original atom to a new element with an increased atomic number by one, while the mass number stays the same.

For antimony-129 (Sb-129), the process entails the conversion of a neutron to a proton and the ejection of an electron. As a result, the antimony atom becomes an atom of tellurium-129 (Te-129). The beta decay equation for Sb-129 is: Te + e-.