Question

A certain stable nuclide, after absorbing a neutron, emits an electron, and the new nuclide splits spontaneously into two alpha particles. Identify the nuclide.

Short answer

The nuclide is ${}_3{}^7\mathrm{Li}\left(\mathrm{Lithium}\right)$.

Step-by-step solution

Write the given data

A stable nuclide absorbs a neutron, and then emits an electron and the new nuclide splits into two alpha particles.

Determine the concept of decay conservation law  

The decay of a radioactive nuclide is only possible on the conservation of the electron-lepton number of the parent nuclei or the nuclei that react to give daughter nuclei or the nuclei products after the reaction.

Calculate the value of the unknown nuclide

Let, the unknown nuclide be represented as ${}_Z{}^{A}X$, where, Aand Zare its massnumber and atomic number, respectively.

Thus, the reaction equation can be written as follows:

${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}+{}_0{}^1\mathrm{n}\to {}_{-1}{}^0\mathrm{e}+2{}_2{}^4\mathrm{He}$

Now, according to conservation of charge, the atomic number of the nuclide can be found to be

$$\begin{gathered} Z+0=-1+\left(2\times 2\right) \\ Z=3 \end{gathered}$$

Now, conservation of mass number yields the mass number of the nuclide as follows:

$$\begin{gathered} A+1=0+\left(2\times 4\right) \\ A=7 \end{gathered}$$

Since, from appendix G the lithium has atomic number 3, so the nuclide must be ${}_3{}^7\mathrm{Li}$.