Using the following formula:
[tex]\begin{gathered} N(t)=N_o(0.5)^{\frac{t}{t_{1/2}}} \\ where: \\ N(t)=Remaining_{\text{ }}quantity_{\text{ }}after_{\text{ }}time_{\text{ }}t \\ N_o=Initial_{\text{ }}quantity \\ t=time_{\text{ }}in_{\text{ }}years \\ t_{1/2}=half-life=24100 \end{gathered}[/tex]
(1)
[tex]\begin{gathered} t=2000 \\ t_{1/2}=24100 \\ N(2000)=5g \\ so: \\ 5=N_o(0.5)^{\frac{2000}{24100}} \\ N_o=\frac{5}{(0.5)^{\frac{2000}{24100}}} \\ N_o\approx5.296 \end{gathered}[/tex]
(2)
Using the initial quantity calculated previously:
[tex]\begin{gathered} t=20000 \\ N(20000)=5.296(0.5)^{\frac{20000}{24100}} \\ N(20000)=2.979 \end{gathered}[/tex]
Answers:
For (1): 5.296
For (2): 2.979
