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Câu hỏi: Cho phương trình phóng xạ ${ }_{84}^{210} \mathrm{Po} \rightarrow{ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}+{ }_{2}^{4} \mathrm{He}$. Hạt nhân ${ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}$ là
A. ${ }_{86}^{214} \mathrm{Rn}$.
B. ${ }_{82}^{205} \mathrm{~Pb}$.
C. ${ }_{82}^{206} \mathrm{~Pb}$.
D. ${ }_{82}^{207} \mathrm{~Pb}$.
A. ${ }_{86}^{214} \mathrm{Rn}$.
B. ${ }_{82}^{205} \mathrm{~Pb}$.
C. ${ }_{82}^{206} \mathrm{~Pb}$.
D. ${ }_{82}^{207} \mathrm{~Pb}$.
$\left\{ \begin{aligned}
& 210=A+4 \\
& 84=Z+2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& A=206 \\
& Z=82 \\
\end{aligned} \right.$.
& 210=A+4 \\
& 84=Z+2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& A=206 \\
& Z=82 \\
\end{aligned} \right.$.
Đáp án C.