A radioactive nucleus {}_{Z}^{A}X undergoes s | vedclass

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(C) $1$. In the first step,${}_{Z}^{A}X \rightarrow {}_{Z-1}B$,the atomic number decreases by $1$. This corresponds to $\beta^{+}$ decay (positron emi

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$\beta^{+}, \alpha, \beta^{-}$

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(C) $1$. In the first step,${}_{Z}^{A}X \rightarrow {}_{Z-1}B$,the atomic number decreases by $1$. This corresponds to $\beta^{+}$ decay (positron emission).$2$. In the second step,${}_{Z-1}B \rightarrow {}_{Z-3}C$,the atomic number decreases by $2$. This corresponds to $\alpha$ decay.$3$. In the third step,${}_{Z-3}C \rightarrow {}_{Z-2}D$,the atomic number increases by $1$. This corresponds to $\beta^{-}$ decay (electron emission).$4$. Therefore,the sequence of decay particles is $\beta^{+}, \alpha, \beta^{-}$.

$A$ radioactive nucleus ${}_{Z}^{A}X$ undergoes spontaneous decay in the sequence ${}_{Z}^{A}X \rightarrow {}_{Z-1}B \rightarrow {}_{Z-3}C \rightarrow {}_{Z-2}D$,where $Z$ is the atomic number of element $X$. The possible decay particles in the sequence are:

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