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A radio active isotope $_{88} R a^{236}$ decays in series by emitting $α$ and $β^{-}$ particles into an isotope $_{83} X^{224} .$ Then number of $α$ and $β^{-}$ particles emitted by $_{88} R a^{236}$ are respectively

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2 and 2

3 and 2

3 and 1

4 and 3

answer is C.

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Detailed Solution

If x be number of $α$ - particles and y be the number of $β^{-}$ particles,
then 224 + 4x = 236 $\Rightarrow$ x = 3
Also, 88 – 2x + y = 83 $\Rightarrow$ y = 1

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