Final answer:
Using the half-life of 20.4 minutes for C-116, the time interval required for the sample's activity to decrease to 38.0% of the original value is approximately 28.6 minutes.
Step-by-step explanation:
To calculate the time interval required for the activity of a sample of the isotope C-116 to decrease to 38.0% of its original value, we'll use the concept of half-life. The half-life is the time it takes for half of a radioactive substance to decay. Given a half-life of 20.4 minutes, we want to find out how long it takes for the sample to reach 38.0% of its initial activity.
To solve this, we can use the formula for exponential decay: A(t) = A0 × (1/2)t/T, where A(t) is the activity at time t, A0 is the initial activity, and T is the half-life.
We need to solve for t when A(t)/A0 = 0.38. The equation becomes: 0.38 = (1/2)t/20.4. Taking the natural logarithm of both sides, we get the equation: ln(0.38) = (t/20.4) × ln(0.5). Solving for t, we find the time interval needed.
t = 20.4 × (ln(0.38) / ln(0.5))
Using a calculator, t ≈ 20.4 × (-0.9678 / -0.6931) ≈ 28.6 minutes
Therefore, it will take approximately 28.6 minutes for the activity of the C-116 isotope to decrease to 38.0% of its original value.