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A sample of 600 g of radioactive lead-210 decays to polonium-210 according to the function A(t) = 600º 0.0321

, Where t
is time in years. Find the amount of radioactive lead remaining after (a) 3 yr, (b) 8 yr, (c) 15уг. (d) Find the half-life.

User UncleAdam
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Final answer:

The amount of remaining radioactive lead can be calculated using the given function. After 3 years, approximately 573.43 grams of lead remains. After 8 years, approximately 306.45 grams remains, and after 15 years, approximately 164.15 grams remains. The half-life of the radioactive lead is approximately 21.6 years.

Step-by-step explanation:

The radioactive decay of lead-210 to polonium-210 can be described by the function A(t) = 600 * 0.0321^t, where t is the time in years. To find the amount of radioactive lead remaining after a certain time, we substitute the value of t into the function and calculate the result.

(a) To find the amount remaining after 3 years, we substitute t=3 into the function: A(3) = 600 * 0.0321^3 = 600 * 0.0321^3 = 573.43 grams.

(b) To find the amount remaining after 8 years, we substitute t=8 into the function: A(8) = 600 * 0.0321^8 = 306.45 grams.

(c) To find the amount remaining after 15 years, we substitute t=15 into the function: A(15) = 600 * 0.0321^15 = 164.15 grams.

(d) The half-life can be found by finding the time it takes for the initial amount to reduce to half its value. Let's set A(t) = 300 (half of 600) and solve for t: 300 = 600 * 0.0321^t. Taking the logarithm of both sides gives: ln(300/600) = t * ln(0.0321), which simplifies to t = ln(300/600) / ln(0.0321) = 21.6 years.

User Richard Toth
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