Final answer:
The amount of lead-210 remaining after 3, 8, and 10 years are approximately 367.5g, 229.2g, and 185.0g, respectively. To calculate the half-life, we solve for t1/2 in the equation 250 = 500e^-0.032t1/2, which gives us a half-life of approximately 21.65 years.
Step-by-step explanation:
The student's question deals with the topic of radioactive decay and the calculation of the remaining amount of a radioactive substance after a certain period of time using an exponential decay function. To answer the question, we apply the given decay function A(t)=500e^-0.032t to find the amount of lead-210 remaining after different time periods.
(a) After 3 years: A(3) = 500e^-0.032×3 ≈ 367.5g
(b) After 8 years: A(8) = 500e^-0.032×8 ≈ 229.2g
(c) After 10 years: A(10) = 500e^-0.032×10 ≈ 185.0g
For the half-life (d), we use the fact that after one half-life, half of the original substance remains, so we solve the equation 250 = 500e^-0.032t1/2 for t1/2.
250 = 500e^-0.032t1/2
1/2 = e^-0.032t1/2
ln(1/2) = -0.032t1/2
t1/2 = ln(1/2) / -0.032 ≈ 21.65 years
The half-life of lead-210 is approximately 21.65 years.