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A nucleus $_{n} X^{m}$ emits one $α$ and one $β$ particle. The resulting nucleus is

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nXm − 4

n − 2Ym − 4

n − 4Zm − 4

n − 1Zm − 4

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detailed solution

Correct option is D

Decrease in mass no. =4Decrease in charge no. =2−1=1 After 1 α is emited, n − 2Ym − 4 After 1 β is emitted n − 2+1Ym − 4=n − 1Ym − 4

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A nucleus nXm emits one α and one β particle. The resulting nucleus is

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A nucleus $_{n} X^{m}$ emits one $α$ and one $β$ particle. The resulting nucleus is