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(B) $1$. In the first step,${ }_z X^A \rightarrow{ }_{z+1} Y^A$,the atomic number increases by $1$ while the mass number remains the same. This corres

Vedclass · Accepted Answer

$\beta, \alpha, \gamma$

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

In the given reaction sequence ${ }_z X^A \rightarrow{ }_{z+1} Y^A \rightarrow{ }_{z-1} K^{A-4} \rightarrow{ }_{z-1} K^{A-4}$,identify the radioactive radiations emitted in order.

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