3340.sagews

%latex
\textbf{Q5}\\
To work out the amount of 129-I nuclei after nuclei-synthesis we need to factor in a constant rate of creation and a rate of decay which is proportional to the number of nuclei. The rate of 129-I creation would be:\\
$\frac{dN(t)_{129}}{dt}=R-\lambda N_{129}$\\
Rearranging this gives:\\
$\frac{dN_{129}}{N_{129}}=R\frac{dN_{129}}{N_{129}}\frac{dt}{dN_{129}}-\lambda dt$\\
Now we substitute $\frac{dN(t)_{129}}{dt}=R-\lambda N_{129}$ and $\frac{dN_{129}}{N_{129}}$.\\
$ln(N_{129})=R ln(N_{129}) (R-\lambda N_{129})-\lambda t$\\
Exponentiation gives:\\
$N_{129}=e^{R ln(N_{129}) (R-\lambda N_{129})}e^{-\lambda t}$\\
$N_{129}=e^{ln((N_{129})^{R(R-\lambda N_{129})}}e^{-\lambda t}$\\
$N_{129}=N_{129}^{R(R-\lambda N_{129})}e^{-\lambda t}$\\
$N_{129}^{R(R-\lambda N_{129})-1}=e^{\lambda t}$\\
$N_{129}=(e^{\lambda t})^{\frac{1}{R(R-\lambda N_{129})-1}}$\\
So over the period $\tau$ we have:\\
$N_{129}=(e^{\lambda \tau})^{\frac{1}{R(R-\lambda N_{129})-1}}$\\
$N_{127}=R\tau$\\
For the next period we have normal decay for $N_{129}$:\\
$N_{129}(t)=N(\tau) e^{-\lambda t}$\\
$N_{127}(t)=R\tau$\\
These describe the amount of each isotope of iodine that is incorporated into the meteorite. Since all the $129-I$ has decayed the expression for the 129-I at creation of the meteorite is the same as the amount of 129-Xe that would exist now. So the ratio is:\\
$\frac{[^{129}Xe]}{[^{127}I]}=\frac{N_{129}(t)}{N_{127}(t)}=\frac{N(\tau) e^{-\lambda t}}{ R\tau}=\frac{(e^{\lambda \tau})^{\frac{1}{R(R-\lambda N_{129})-1}} e^{-\lambda t}}{ R\tau}$\\
Or if we assume nucleosynthesis was instant we get a simpler ratio of :\\
$\frac{[^{129}Xe]}{[^{127}I]}=\frac{N_{129}(t)}{N_{127}(t)}=\frac{N(\tau) e^{-\lambda t}}{ R\tau}=\frac{R\tau e^{-\lambda (t+\tau)}}{ R\tau}=e^{-\lambda (t+\tau)}$