A sample of $600 \mathrm{~g}$ of radioactive lead-210 decays to polonium-210 according to the function $A(t)=600 e^{-0.032 t}$, where $t$ is time in years. Find the amount of radioactive lead remaining after (a) $3 \mathrm{yr}$, (b) $8 \mathrm{yr}$, (c) $20 \mathrm{yr}$. (d) Find the half-life.
Answer
Final Answer: The amount of radioactive lead remaining after (a) 3 years is approximately \(\boxed{545.08 \, g}\), (b) 8 years is approximately \(\boxed{464.49 \, g}\), (c) 20 years is approximately \(\boxed{316.38 \, g}\). The half-life is approximately \(\boxed{21.66 \, yr}\).
Step 1 ：Given the function \(A(t)=600 e^{-0.032 t}\), where \(t\) is time in years and \(A(t)\) is the amount of radioactive lead remaining.
Step 2 ：For part (a), substitute \(t = 3\) into the function to find the amount of radioactive lead remaining after 3 years: \(A(3) = 600 e^{-0.032 \times 3} \approx 545.08 \, g\).
Step 3 ：For part (b), substitute \(t = 8\) into the function to find the amount of radioactive lead remaining after 8 years: \(A(8) = 600 e^{-0.032 \times 8} \approx 464.49 \, g\).
Step 4 ：For part (c), substitute \(t = 20\) into the function to find the amount of radioactive lead remaining after 20 years: \(A(20) = 600 e^{-0.032 \times 20} \approx 316.38 \, g\).
Step 5 ：For part (d), the half-life is the time it takes for half of the substance to decay. This can be found by setting \(A(t)\) to half of the initial amount and solving for \(t\): \(300 = 600 e^{-0.032 t}\). Solving this equation gives \(t \approx 21.66 \, yr\).
Step 6 ：Final Answer: The amount of radioactive lead remaining after (a) 3 years is approximately \(\boxed{545.08 \, g}\), (b) 8 years is approximately \(\boxed{464.49 \, g}\), (c) 20 years is approximately \(\boxed{316.38 \, g}\). The half-life is approximately \(\boxed{21.66 \, yr}\).
