University students, studying the activity of a particular radioactive isotope which had a half-life of 12.0 hours. If the original activity of the sample was 448 kBq, what would the activity be 3.00 days later?
Three days is six 12-hour intervals, which is six half-lives.
Multiply the original activity by (1/2)^6 = 1/64
448/64 =__ kBq
To find the activity of the radioactive isotope 3.00 days later, we need to consider its half-life and the decay equation. Here's how you can calculate it:
Step 1: Convert the half-life from hours to days.
Since the given half-life is 12.0 hours, divide it by 24 to convert it to days.
Half-life in days = 12.0 hours / 24 = 0.5 days.
Step 2: Determine the number of half-life periods.
To find the number of half-life periods that have passed in 3.00 days, divide 3.00 days by the half-life in days.
Number of half-life periods = 3.00 days / 0.5 days = 6 half-life periods.
Step 3: Calculate the remaining activity.
To calculate the remaining activity of the sample 3.00 days later, we can use the decay formula:
Remaining Activity = Initial Activity × (1/2)^(number of half-life periods).
Let's plug in the values:
Remaining Activity = 448 kBq × (1/2)^(6).
Step 4: Evaluate the equation and find the answer.
Using a calculator, evaluate the expression:
Remaining Activity ≈ 448 kBq × 0.015625 ≈ 7.000 kBq.
Therefore, the activity of the radioactive isotope 3.00 days later would be approximately 7.000 kBq.