Hạt nhân urani (_{92}^{238} extrm{U}) phân rã theo chuỗi phóng xạ (_{92}^{238} extrm{U}overset{alpha }{ ightarrow}Th)

* Hướng dẫn giải

Ta có: 
\(_{92}^{238}\textrm{U}\overset{\alpha }{\rightarrow}Th\overset{\beta ^{-}}{\rightarrow}Pa\overset{\beta ^{-}}{\rightarrow}X\)
         \((1)\hspace{20}(2)\hspace{19}(3)\)
\((1):\, _{92}^{238}\textrm{U}\rightarrow _{2}^{4}\textrm{He}+_{90}^{234}\textrm{Th}\)
\((2):\, _{90}^{234}\textrm{Th}\rightarrow _{-1}^{0}\textrm{e}+\, _{91}^{234}\textrm{Pa}\)
\((3):\, _{91}^{234}\textrm{Pa}\rightarrow _{-1}^{0}\textrm{e}+_{92}^{234}\textrm{X}\Rightarrow\) X gồm \(\left\{\begin{matrix} 92p\\ 142n \end{matrix}\right.\)
_{\(\Rightarrow\)} Chọn câu A.