In the given reaction {}_{Z}^{A}X to {}_{Z+1} | vedclass

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(B) $1$. In the first step,${}_{Z}^{A}X 	o {}_{Z+1}^{A}Y$,the atomic number increases by $1$ while the mass number remains constant. This corresponds

Vedclass · Accepted Answer

$\beta, \alpha, \gamma$

Vedclass · Answer

(B) $1$. In the first step,${}_{Z}^{A}X 	o {}_{Z+1}^{A}Y$,the atomic number increases by $1$ while the mass number remains constant. This corresponds to the emission of a $\beta$-particle $({}_{-1}^{0}e)$.$2$. In the second step,${}_{Z+1}^{A}Y 	o {}_{Z-1}^{A-4}K$,the atomic number decreases by $2$ and the mass number decreases by $4$. This corresponds to the emission of an $\alpha$-particle $({}_{2}^{4}He)$.$3$. In the third step,${}_{Z-1}^{A-4}K 	o {}_{Z-1}^{A-4}K$,there is no change in the atomic number or mass number,which indicates the emission of a $\gamma$-ray $({}_{0}^{0}\gamma)$.$4$. Therefore,the sequence of emissions is $\beta, \alpha, \gamma$.

In the given reaction ${}_{Z}^{A}X \to {}_{Z+1}^{A}Y \to {}_{Z-1}^{A-4}K \to {}_{Z-1}^{A-4}K$,the radioactive radiations emitted in the sequence are:

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