Problem 18 - College Algebra Essentials

Question

The half-life of the radioactive element plutonium-239 is \(25,000\) years. If 16 grams of plutonium- 239 are initially present, how many grams are present after \(25,000\) years? \(50,000\) years? \(75,000\) years? \(100,000\) years? \(125,000\) years?

Step-by-Step Solution

Answer

The amount of Plutonium-239 remaining will be 8 grams after 25,000 years, 4 grams after 50,000 years, 2 grams after 75,000 years, 1 gram after 100,000 years and 0.5 grams after 125,000 years.

Step 1: Determination of half-life periods

Initially, determine the number of half-life periods for each given time. This is done by dividing the given time by the half-life. Here, the half-life is 25,000 years.

Step 2: Calculation after 25,000 years

Since 25,000 years is equal to one half-life period, half of the initial amount of plutonium-239 will remain. So, 16 grams / 2 = 8 grams.

Step 3: Calculation after 50,000 years

Since 50,000 years is equal to two half-life periods, the remaining amount of plutonium-239 will be halved twice. So, 16 grams / 2 / 2 = 4 grams.

Step 4: Calculation after 75,000 years

Since 75,000 years is equal to three half-life periods, the remaining amount of plutonium-239 will be halved three times. So, 16 grams / 2 / 2 / 2 = 2 grams.

Step 5: Calculation after 100,000 years

Since 100,000 years is equal to four half-life periods, the remaining amount of plutonium-239 will be halved four times. So, 16 grams / 2 / 2 / 2 / 2 = 1 gram.

Step 6: Calculation after 125,000 years

Since 125,000 years is equal to five half-life periods, the remaining amount of plutonium-239 will be halved five times. So, 16 grams / 2 / 2 / 2 / 2 / 2 = 0.5 grams.