Question

The activity of a radioactive sample is measured as 9750 counts per minute at \( t = 0 \) and as 975 counts per minute at \( t = 5 \) minutes. The decay constant is approximately:

• A. 0.922 per minute
• B. 0.691 per minute
• C. 0.461 per minute
• D. 0.230 per minute

Hint:

The decay constant \( \lambda \) can be calculated from the ratio of initial and final activities using the exponential decay law.

Answer 1

The Correct Option is C

Step 1: The decay of a radioactive sample follows the exponential decay law: \[ A(t) = A_0 e^{-\lambda t}, \] where \( A(t) \) is the activity at time \( t \), \( A_0 \) is the initial activity, and \( \lambda \) is the decay constant.
Step 2: Using the given data \( A_0 = 9750 \, \text{counts/min} \) and \( A(5) = 975 \, \text{counts/min} \), we can solve for \( \lambda \). \[ \frac{975}{9750} = e^{-\lambda \times 5} \quad \Rightarrow \quad \lambda \approx 0.461 \, \text{per minute}. \]
Final Answer: \[ \boxed{0.461 \, \text{per minute}} \]