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. ${}_1{}^3\mathrm{He}$(β)

b. ${}_3{}^8\mathrm{Li}$(β followed by α)

c. ${}_4{}^7\mathrm{Be}$(electron capture)

d. ${}_5{}^8\mathrm{B}$(positron)

Short answer

$$\begin{gathered} \text{a}.{}_1{}^3\mathrm{He}\to {}_2{}^3\mathrm{He}+{}_{-1}{}^0\mathrm{e} \\ \mathrm{b}.{}_3{}^8\mathrm{Li}\to 2{}_2{}^4\mathrm{He} \\ \mathrm{c}.{}_4{}^7\mathrm{Be}+{}_{-1}{}^0\mathrm{e}\to {}_3{}^7\mathrm{Li} \\ \mathrm{d}.{}_5{}^8\mathrm{B}\to {}_{-1}{}^0\mathrm{e}+{}_4{}^8\mathrm{Be} \end{gathered}$$

Step-by-step solution

Radioactive decays

The process by which an atomic nucleus that is unstable in nature loses energy through an emission is called a radioactive decay. There are five types of decay:

• Alpha decay
• Beta decay
• Gamma decay
• Positron decay
• Electron capture

Equation describing the radioactive beta decay of He13

In a beta decay, the neutrons end up decaying into a proton. The atomic number is increased, and one electron is ejected. Summing up the information we have

${}_1{}^3\mathrm{He}\to {}_2{}^3\mathrm{He}+{}_{-1}{}^0\mathrm{e}$

Equation describing the radioactive beta decay followed by alpha decay of Li38 

To obtain the first product, write the equation for the beta decay:

${}_3{}^8\mathrm{Li}\to {}_4{}^8\mathrm{Be}+{}_{-1}{}^0\mathrm{e}$

Two helium atoms are ejected in the alpha decay. The equation for the successive alpha decay is shown below.

${}_4{}^8\mathrm{Be}\to 2{}_2{}^4\mathrm{He}$

Now, write the complete equation:

${}_3{}^8\mathrm{Li}\to 2{}_2{}^4\mathrm{He}$

Equation describing the electron capture of Be47 

In an electron capture, the nucleus of the parent nuclide will absorb an electron, which will cause an electron to decay into a neutron. Summing up the information, we have:

${}_4{}^7\mathrm{Be}+{}_{-1}{}^0\mathrm{e}\to {}_3{}^7\mathrm{Li}$

Equation describing the positron decay of B58

In the radioactive decay of a positron, an electron is ejected. Writing the equation, we have:

${}_5{}^8\mathrm{B}\to {}_{-1}{}^0\mathrm{e}+{}_4{}^8\mathrm{Be}$