In the Uranium series, decays through a succession of steps to finally yield the stable isotope . Ca — Nuclear Chemistry and Radioactivity Chemistry Question
Question
In the Uranium series, ${}_{92}^{238}U$ decays through a succession of steps to finally yield the stable isotope ${}_{82}^{206}Pb$. Calculate the exact total number of $\alpha$ and $\beta^-$ particles emitted during this complete decay process.
Answer: A
💡 Solution & Explanation
$\Delta A = 238 - 206 = 32$. Number of $\alpha$ -particles = $32 / 4 = 8$. Expected $Z$ after $8\alpha$ = $92 - 16 = 76$. Actual final $Z = 82$. Difference = $82 - 76 = 6$. Since each $\beta$ increases $Z$ by 1, there are $6\beta^-$ particles.
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