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What is the respective number of $α and β$ particles emitted the following radioactive decay?
${}_{90}X^{200} \to {}_{80}Y^{168}$

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6 and 8

8 and 8

6 and 6

8 and 6

answer is D.

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Detailed Solution

$n_{α} = \frac{A - A'}{4} = \frac{200 - 168}{4} = 8 n_{β} = 2 n_{α} - Z + Z' = 2 x 8 - 90 + 80 = 6$

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