In: Chemistry
The radioactive isotope, \({ }_{85}^{213}\) At has atomic number \(83(Z=83)\). The number of neutrons \(=213-85=128\)
For this element, the atomic number is greater than \(83 .\) Hence, to come into the belt of stability, the atomic number should be decreased.
The atomic number and mass number of an element decreases when it undergoes alpha decay. When an alpha particle is ejected, the atomic number decreases by 2 units and the mass number decreases by 4 units.
The alpha decay of \({ }_{85}^{213}\) At results \({ }_{83}^{209} \mathrm{Bi}\)
$$ { }_{85}^{213} \mathrm{At} \rightarrow{ }_{83}^{209} \mathrm{Bi}+{ }_{2}^{4} \mathrm{He} $$
Therefore, the radioactive isotope, \({ }_{85}^{213}\) At undergoes alpha decay to convert into a stable nuclide.