Question

In a decay ${}_{92}^{238}U$ is obtained in the end to ${}_{82}^{206}Pb$. How many $\alpha$ and $\beta -particles$ must have been emitted?

• A. $8\beta$ and $8\alpha$
• B. $6\beta$ and $6\alpha$
• C. $6\beta$ and $8\alpha$
• D. $6\alpha$ and $8\beta$

Answer 1

Answer: (C)

$6\beta$ and $8\alpha$

${}_{92}^{238}U{\stackrel{-8\alpha }{\to }}_{76}^{206}X{\stackrel{-6\beta }{\to }}_{82}^{206}Pb$

Thus, $6\beta$ and $8\alpha -particles$ must have been emitted in the given decay series.

Number of $\alpha -$ particles emitted $=\frac{238-206}{4}=8$

Number of $\beta -$ particles emitted $=(2\times 8)-(92-82)=6$