Question

In the given nuclear reaction A, B, C, D, E represents $92U^{238}{\overset{\quad\alpha\quad}{\rightarrow}}_{B}Th^{A}{\overset{\quad\beta\quad}{\rightarrow}}_{D}Pa^{C}{\overset{\quad E\quad}{\rightarrow}}_{92}U^{234}$

• 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

Answer: (A)

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

Answer 2

$92U^{238}{\overset{\quad\alpha\quad}{\rightarrow}}_{90}Th^{234}{\overset{\quad\beta\quad}{\rightarrow}}_{91}Pa^{234}$

${\overset{\quad E(_{- 1}\beta^{o})\quad}{\rightarrow}}_{92}U^{234}$.