In a disintegration series, the radionuclide undergoes two successive -decays followed by one -decay — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

The parent nuclide is ${}_{90}^{234}Th$. Each $\beta^-$ -decay increases the atomic number by $1$ without changing the mass number. After two $\beta^-$ -decays, the nuclide becomes ${}_{92}^{234}U$. Next, it undergoes an $\alpha$ -decay, which decreases the mass number by $4$ and atomic number by $2$. The resulting nuclide will have an atomic number of $92 - 2 = 90$ and a mass number of $234 - 4 = 230$. The final nuclide is ${}_{90}^{230}Th$.