Hat nhân \(_{{Z_1}}^{{A_1}}X\) phóng xạ

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{m_Y} = {m_{con}} = \frac{{{A_2}}}{{{A_1}}}\left( {1 - {2^{\frac{{ - t}}{T}}}} \right){m_0}\\
{m_X} = m = {m_0}{.2^{\frac{{ - t}}{T}}}
\end{array} \right.\\
 \Rightarrow \frac{{{m_Y}}}{{{m_X}}} = \frac{{\frac{{{A_2}}}{{{A_1}}}\left( {1 - {2^{\frac{{ - t}}{T}}}} \right)}}{{{2^{\frac{{ - t}}{T}}}}} = \frac{{{A_2}}}{{{A_1}}}\left( {{2^{\frac{t}{T}}} - 1} \right)\frac{{{m_Y}}}{{{m_X}}} = 7\frac{{{A_2}}}{{{A_1}}}
\end{array}\)