In the disintegration series_{92}^{238}U xrig | vedclass

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(D) The given disintegration series is $_{92}^{238}U \xrightarrow{\alpha} X \xrightarrow{\beta} _{Z}^{A}Y$.Step $1$: Alpha $(\alpha)$ decay of $_{92}^

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$91, 234$

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(D) The given disintegration series is $_{92}^{238}U \xrightarrow{\alpha} X \xrightarrow{\beta} _{Z}^{A}Y$.Step $1$: Alpha $(\alpha)$ decay of $_{92}^{238}U$:An alpha particle is a helium nucleus $_{2}^{4}He$. When $_{92}^{238}U$ emits an $\alpha$ particle, the atomic number decreases by $2$ and the mass number decreases by $4$.$_{92}^{238}U 	o _{90}^{234}X + _{2}^{4}He$.So, $X$ is $_{90}^{234}Th$.Step $2$: Beta $(\beta)$ decay of $_{90}^{234}X$:A beta particle is an electron $_{ - 1}^{0}e$. When $_{90}^{234}X$ emits a $\beta$ particle, the atomic number increases by $1$ and the mass number remains unchanged.$_{90}^{234}X 	o _{91}^{234}Y + _{ - 1}^{0}e$.Comparing this with $_{Z}^{A}Y$, we get $Z = 91$ and $A = 234$.

In the disintegration series
$_{92}^{238}U \xrightarrow{\alpha} X \xrightarrow{\beta} _{Z}^{A}Y$
The values of $Z$ and $A$ respectively will be:

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In the disintegration series$_{92}^{238}U \xrightarrow{\alpha} X \xrightarrow{\beta} _{Z}^{A}Y$The values of $Z$ and $A$ respectively will be: