207k views
0 votes
A sample of 500 g of radioactive​ lead-210 decays to​ polonium-210 according to the function A(t)=500e^-0.032t ​, where t is time in years. Find the amount of radioactive lead remaining after ​(a) ​3yr, ​(b) 8​yr, ​(c) 10 yr. ​(d) Find the​ half-life.

1 Answer

2 votes

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.

User Garry Wong
by
8.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories