Question

In the given nuclear reaction, A, B, C, D, E represents
${}_{92}{U}^{238}\stackrel{\alpha }{⟶}{}_{B}T{h}^A\stackrel{\beta }{⟶}{}_{D}P{a}^C\stackrel{E}{⟶}{}_{92}{U}^{234}$

• 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

The correct option is A $A=234,B=90,C=234,D=91,E=\beta$

${}_{92}{U}^{238}\stackrel{\alpha }{⟶}{}_{90}T{h}^{234}\stackrel{\beta }{⟶}{}_{91}P{a}^{234}$
$\stackrel{E{(}_{-1}{\beta }^0)}{⟶}92U^{234}$