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Q.
In radioactive reaction
https://cdn.mathpix.com/snip/images/f2F5c0qmJRZxp0vCFecVMy4Lt8GhR0O6j84XFXJfXug.original.fullsize.png
Successive emission of particles is
NTA AbhyasNTA Abhyas 2022
Solution:
Given the radioactive reaction,
${ }_{Z}^{A} X \rightarrow{ }_{ Z +1}^{ A _{1}^{ A } X _{1}} \rightarrow{ }_{ Z +2}^{ A } X _{2} \rightarrow{ }_{ Z }^{ A -4} X _{3} \rightarrow{ }_{ Z +1}^{ A -4} X _{4}$
In the first step, the proton number increases by $1$ and hence it is $\beta ^{-}$ decay.
In the second step, the same as first step, so $\beta ^{-}$ decay.
In the third step, the proton number reduces by $2$ and the nucleon number reduces by $4$ , hence it is an $\alpha $ decay.
In the fourth step, it is the same as the first step, so $\beta ^{-}$ decay.
