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.)

$$\begin{gathered} \mathrm{a}.{}^{68}\mathrm{Ga}(\mathrm{electron}\mathrm{capture}) \\ \mathrm{b}.{}^{62}\mathrm{Cu}\left(\mathrm{Positron}\right) \\ \mathrm{c}.{}^{212}\mathrm{Fr}\left(\mathrm{\alpha }\right) \\ \mathrm{d}.{}^{129}\mathrm{Sb}\left(\mathrm{\beta }\right) \end{gathered}$$

Short answer

$$\begin{gathered} \mathrm{a}.{}_{30}{}^{68}\mathrm{Ga}+{}_{-1}{}^0\mathrm{e}\to {}_{30}{}^{68}\mathrm{Zn} \\ \mathrm{b}.{}_{29}{}^{62}\mathrm{Cu}\to {}_{+1}{}^0\mathrm{e}+{}_{28}{}^{62}\mathrm{Ni} \\ \mathrm{c}.{}_{87}{}^{212}\mathrm{Fr}\to {}_2{}^4\mathrm{He}+{}_{85}{}^{208}\mathrm{At} \\ \mathrm{d}.{}_{51}{}^{129}\mathrm{Sb}\to {}_{+1}{}^0\mathrm{e}+{}_{52}{}^{129}\mathrm{Te} \end{gathered}$$

Step-by-step solution

Equation for electron capture of 68Ga

In the process of an electron capture, an electron is captured, and it results in a proton decaying into a neutron. The complete equation becomes:

${}_{30}{}^{68}\mathrm{Ga}+{}_{-1}{}^0\mathrm{e}\to {}_{30}{}^{68}\mathrm{Zn}$

Equation for positron production of 62Cu

The product nuclide results in an atomic number of less than one unit and a positron particle is also emitted. The equation for a positron production can be written as:

${}_{29}{}^{62}\mathrm{Cu}\to {}_{+1}{}^0\mathrm{e}+{}_{28}{}^{62}\mathrm{Ni}$

Equation for the alpha decay of 212Fr

In an alpha decay, the ejection of a He from the nucleus of parent nuclide takes place. The product nucleotide has 2 units less in atomic number and 4 units less in mass number. The complete equation becomes:

${}_{87}{}^{212}\mathrm{Fr}\to {}_2{}^4\mathrm{He}+{}_{85}{}^{208}\mathrm{At}$

Equation for the beta decay of 129Sb

In a beta decay, neutrons end up decaying into a proton. The atomic number is increased, and one electron is ejected. The complete equation becomes:

${}_{51}{}^{129}\mathrm{Sb}\to {}_{+1}{}^0\mathrm{e}+{}_{52}{}^{129}\mathrm{Te}$