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

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6 and 6

6 and 8

8 and 8

8 and 6

answer is D.

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Detailed Solution

$_{90} X^{200} \to_{80} Y^{168} + n_{12} {He}^{4} + n_{2}$

$- 1 β^{0}$

Mass No. balance

$200 = 168 + n_{1} 4$

$\Rightarrow n_{1} = 8$

At. No. balance

$90 = 80 + n_{1} (2) + n_{2} (- 1)$

$\Rightarrow n_{2} = 6$

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