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

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Question

(B) The given reaction sequence is: $_Z{X^A} 	o _{Z+1}{Y^A} 	o _{Z-1}{K^{A-4}} 	o _{Z-1}{K^{A-4}}$.$1$. In the first step,$_Z{X^A} 	o _{Z+1}{Y^A}$

Vedclass · Accepted Answer

$\beta, \alpha, \gamma$

Vedclass · Answer

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

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

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