Question

By the successive disintegration of ${}_{92}{U}^{238}$, the final product obtained is ${}_{82}P{b}^{206}$, then how many number of $\alpha$ and $\beta$ particles are emitted?

• A. $6$ and $8$
• B. $8$ and $6$
• C. $12$ and $6$
• D. $8$ and $12$

Answer 1

Answer: (B)

$8$ and $6$

The number of $\alpha -particles$
$=\frac{\text{change in mass number}}{4}$
$=\frac{238-206}{4}$
$n_1=\frac{32}{4}=8$
Now, number of $\beta -particles{n}_2=82-(92-2n_1)$
$=82-(92-2\times 8)$
$=82-76=6$.