In the radioactive disintegration series,for | vedclass

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(D) $1$. The initial nucleus is Uranium-$238$: ${}_{92}^{238}U$.$2$. When an alpha particle $({}_{2}^{4}He)$ is emitted,the atomic number decreases by

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$91, 234$

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(D) $1$. The initial nucleus is Uranium-$238$: ${}_{92}^{238}U$.$2$. When an alpha particle $({}_{2}^{4}He)$ is emitted,the atomic number decreases by $2$ and the mass number decreases by $4$. Thus,the intermediate nucleus $X$ is ${}_{90}^{234}X$.$3$. When a beta particle $({}_{-1}^{0}e)$ is emitted from $X$,the atomic number increases by $1$ while the mass number remains unchanged. Thus,the final nucleus $Y$ is ${}_{91}^{234}Y$.$4$. Comparing this with ${}_{Z}^{A}Y$,we get $Z = 91$ and $A = 234$.

In the radioactive disintegration series,for the process $_{92}^{238}U \xrightarrow{\alpha} X \xrightarrow{\beta^-} {}_{Z}^{A}Y$,the values of $Z$ and $A$ respectively are:

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