In the given nuclear reaction,A, B, C, D, E r | vedclass

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(A) $1$. In an alpha decay,the mass number decreases by $4$ and the atomic number decreases by $2$. Thus,$_{92}U^{238} \xrightarrow{\alpha} _{90}Th^{2

Vedclass · Accepted Answer

$A = 234, B = 90, C = 234, D = 91, E = \beta$

Vedclass · Answer

(A) $1$. In an alpha decay,the mass number decreases by $4$ and the atomic number decreases by $2$. Thus,$_{92}U^{238} \xrightarrow{\alpha} _{90}Th^{234}$. Comparing this with $_{B}Th^{A}$,we get $A = 234$ and $B = 90$.$2$. In a beta decay,the mass number remains the same and the atomic number increases by $1$. Thus,$_{90}Th^{234} \xrightarrow{\beta} _{91}Pa^{234}$. Comparing this with $_{D}Pa^{C}$,we get $C = 234$ and $D = 91$.$3$. Finally,$_{91}Pa^{234} \xrightarrow{E} _{92}U^{234}$. Since the atomic number increases by $1$ and the mass number remains unchanged,$E$ must be a $\beta$ decay (emission of an electron).

In the given nuclear reaction,$A, B, C, D, E$ represent:
$_{92}U^{238} \xrightarrow{\alpha} _{B}Th^{A} \xrightarrow{\beta} _{D}Pa^{C} \xrightarrow{E} _{92}U^{234}$

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In the given nuclear reaction,$A, B, C, D, E$ represent:$_{92}U^{238} \xrightarrow{\alpha} _{B}Th^{A} \xrightarrow{\beta} _{D}Pa^{C} \xrightarrow{E} _{92}U^{234}$