Gửi 6 năm trước
Hướng dẫn giải
Bài toán hạt nhân con: \(X \rightarrow Y+\alpha\)
+) \(\dfrac{N_Y}{N_X}=e^{\lambda t}-1=2^{- \dfrac{t}{T}}-1\)
+) \(\dfrac{m_Y}{m_X}=\dfrac{N_Y}{N_X}. \dfrac{A_Y}{A_X}=\left (2^{- \dfrac{t}{T}}-1 \right )\dfrac{A_Y}{A_X}\)
+) Nếu:
\(\left\{\begin{matrix}t_1 \rightarrow \dfrac{N_Y}{N_X}=k\\ t_2=t_1+nT \rightarrow \dfrac{N_Y}{N_X}=k'\end{matrix}\right.\Rightarrow k'=2^n.k+2^n-1\)
+) \(t=nT \Leftrightarrow \left\{\begin{matrix}N=\dfrac{N_0}{2^n};m=\dfrac{m_0}{2^n}\\ \Delta N=\left ( 1-\dfrac{1}{2^n} \right )N_0\\ \Delta m=\left ( 1-\dfrac{1}{2^n} \right )m_0\end{matrix}\right.\)
Hướng dẫn giải
Bài toán hạt nhân con: \(X \rightarrow Y+\alpha\)
+) \(\dfrac{N_Y}{N_X}=e^{\lambda t}-1=2^{- \dfrac{t}{T}}-1\)
+) \(\dfrac{m_Y}{m_X}=\dfrac{N_Y}{N_X}. \dfrac{A_Y}{A_X}=\left (2^{- \dfrac{t}{T}}-1 \right )\dfrac{A_Y}{A_X}\)
+) Nếu:
\(\left\{\begin{matrix}t_1 \rightarrow \dfrac{N_Y}{N_X}=k\\ t_2=t_1+nT \rightarrow \dfrac{N_Y}{N_X}=k'\end{matrix}\right.\Rightarrow k'=2^n.k+2^n-1\)
+) \(t=nT \Leftrightarrow \left\{\begin{matrix}N=\dfrac{N_0}{2^n};m=\dfrac{m_0}{2^n}\\ \Delta N=\left ( 1-\dfrac{1}{2^n} \right )N_0\\ \Delta m=\left ( 1-\dfrac{1}{2^n} \right )m_0\end{matrix}\right.\)
Gửi 6 năm trước