_{35}X^{88} is an unstable isotope,which deca | vedclass

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(C) The decay process is $_{35}X^{88}$ $\xrightarrow{I} Y$ $\xrightarrow{II} _{32}Z^{84}$.Step $I$: $_{35}X^{88} \rightarrow Y + _{1}e^{0}$ (Positron

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$Y$ and $Z$ may be isodiaphers.

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(C) The decay process is $_{35}X^{88}$ $\xrightarrow{I} Y$ $\xrightarrow{II} _{32}Z^{84}$.Step $I$: $_{35}X^{88} \rightarrow Y + _{1}e^{0}$ (Positron emission) gives $Y = _{34}Y^{88}$.Step $II$: $_{34}Y^{88} \rightarrow _{32}Z^{84} + _{2}He^{4}$ ($\alpha$ decay).Alternatively,if $I$ is $\alpha$ decay: $_{35}X^{88} \rightarrow _{33}Y^{84} + _{2}He^{4}$,then $II$ is $\beta^-$ decay: $_{33}Y^{84} \rightarrow _{32}Z^{84} + _{-1}e^{0}$.Checking isodiaphers: Isodiaphers have the same $(N-Z)$ value.For $Y$ $(_{34}Y^{88})$: $N-Z = (88-34) - 34 = 54 - 34 = 20$.For $Z$ $(_{32}Z^{84})$: $N-Z = (84-32) - 32 = 52 - 32 = 20$.Since both have $N-Z = 20$,$Y$ and $Z$ are isodiaphers.

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