In the natural disintegration series, Thorium-234 () eventually decays to the stable isotope Lead-20 — Nuclear Chemistry and Radioactivity Chemistry Question
Question
In the natural disintegration series, Thorium-234 (${}_{90}^{234}Th$) eventually decays to the stable isotope Lead-206 (${}_{82}^{206}Pb$). What is the total combined number of $\alpha$ and $\beta^-$ particles emitted during this complete decay pathway?
💡 Solution & Explanation
The change in mass number $\Delta A = 234 - 206 = 28$. Since only $\alpha$ -particles change the mass number (by 4), the number of $\alpha$ -particles is $n_\alpha = 28 / 4 = 7$. The emission of 7 $\alpha$ -particles decreases the atomic number by $7 \times 2 = 14$. The expected atomic number without $\beta$ emissions would be $90 - 14 = 76$. The actual final atomic number is $82$. The difference is $82 - 76 = 6$, which implies $6$ $\beta^-$ -particles must be emitted (since each $\beta^-$ increases $Z$ by 1). Total particles emitted = $7 (\alpha) + 6 (\beta) = 13$.