We can use the following formula in order to solve the questions:
[tex]N(t)=N_o(\frac{1}{2})^{\frac{t}{t_{1/2}}}[/tex]
Where:
[tex]\begin{gathered} N(t)=Remaining_{\text{ }}quantity_{\text{ }}after_{\text{ }}time_{\text{ }}t \\ t_{1/2}=half-life=1599 \\ t=time(in_{\text{ }}years) \\ N_o=Initial_{\text{ }}quantity \end{gathered}[/tex]
(1)
[tex]\begin{gathered} t=1300 \\ t_{1/2}=1599 \\ N(1300)=5g \\ so: \\ 5=N_o(0.5)^{\frac{1300}{1599}} \\ N_o=\frac{5}{(0.5)^{\frac{1300}{1599}}} \\ N_o\approx8.784 \end{gathered}[/tex]
(2)
Using the initial quatity calculated previously:
[tex]\begin{gathered} t=13000 \\ N(13000)=8.784(0.5)^{\frac{13000}{1599}} \\ N(13000)=0.031 \end{gathered}[/tex]
Answers:
For (1): 8.784
For (2): 0.031
