Question

In the given nuclear reaction $A , B , C , D , E$ represent:
${ }_{9} U ^{238} \stackrel{\alpha}{\longrightarrow }{ }_{ B } Z ^{ A } \stackrel{\beta}{\longrightarrow }_{ D } Y ^{ C } \stackrel{e}{\longrightarrow }{ }_{92} X ^{234}$

• A. $A=234, B=90, C=234, D=91, e=\beta$
• B. $A=234, B=90, C=238, D=94, e=\alpha$
• C. $A=238, B=93, C=234, D=91, e=\beta$
• D. $A=234, B=90, C=234, D=93, e=\alpha$

Answer 1

Answer: (A)

The process is
${ }_{92} U ^{238} \stackrel{\alpha}{\longrightarrow }_{90} Z ^{234} \stackrel{\beta}{\longrightarrow }_{\text {। }} Y ^{234} \stackrel{\text {e}}{\longrightarrow }_{92} U ^{234}$
$\therefore A=234, B=90, C=234 . D=91 . e=\beta$