91Ó°ÊÓ

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

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.

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:

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

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

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

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

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

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