A) \[3\alpha \] and \[3\beta \]
B) \[5\alpha \] and \[3\beta \]
C) \[3\alpha \] and \[5\beta \]
D) \[5\alpha \] and \[5\beta \]
Correct Answer: A
Solution :
\[_{92}{{X}^{232}}{{\to }_{89}}{{Y}^{220}}+{{x}_{2}}H{{e}^{4}}+{{y}_{-1}}{{e}^{o}}\] no. of \[\alpha \]-particles = \[\frac{232-220}{4}=3\] no. of \[\beta \]-particles = 89 ?[92 ? 2 × 3] = 3.
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