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When ${}_{90}{Th}^{228}$ transforms to ${}_{83}{Bi}^{212}$ , then the number of the emitted $α and β -$ particles is, respectively

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detailed solution

Correct option is D

nα=228-2124=4 and nβ=2 x 4-90 +83=1

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When Th 228 90 transforms to Bi 212 83 , then the number of the emitted α and β-particles is, respectively

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When ${}_{90}{Th}^{228}$ transforms to ${}_{83}{Bi}^{212}$ , then the number of the emitted $α and β -$ particles is, respectively