Question

What is the respective number of $\alpha$and $\beta$ particles emitted in the following radioactive decay $90X^{200} \rightarrow_{80}Y^{168}$

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

Answer 1

Answer: (D)

8 and 6

Answer 2

$n_{\alpha} = \frac{A - A'}{4} = \frac{200 - 168}{4} = 8$

$n_{\beta} = 2n_{a} - Z + Z' = 2 \times 8 - 90 + 80 = 6$