Step 1: Understanding the Concept:
This problem involves radioactive decay and the concept of half-life. The activity (or radiation level) of a radioactive sample decreases exponentially with time. We need to find the time it takes for the activity to reduce to the safe level.
Step 2: Key Formula or Approach:
The formula for radioactive decay relates the initial activity (\(A_0\)), the activity at time \(t\) (\(A(t)\)), and the half-life (\(T_{1/2}\)):
\[ A(t) = A_0 \left( \frac{1}{2} \right)^{t/T_{1/2}} \]
Alternatively, we can write \(\frac{A(t)}{A_0} = \left( \frac{1}{2} \right)^n\), where \(n\) is the number of half-lives that have passed, and \(t = n \times T_{1/2}\).
Step 3: Detailed Explanation:
Let the safe radiation level be \(A_{safe}\).
The initial radiation level is given as 32 times the safe level, so \(A_0 = 32 \times A_{safe}\).
The final radiation level, when the material can be handled safely, is \(A(t) = A_{safe}\).
The half-life is \(T_{1/2} = 2.5\) hours.
We need to find the time \(t\).
Using the decay formula:
\[ \frac{A(t)}{A_0} = \frac{A_{safe}}{32 \times A_{safe}} = \frac{1}{32} \]
We set this equal to the decay factor:
\[ \left( \frac{1}{2} \right)^n = \frac{1}{32} \]
Since \(32 = 2^5\), we can write:
\[ \left( \frac{1}{2} \right)^n = \frac{1}{2^5} = \left( \frac{1}{2} \right)^5 \]
By comparing the exponents, we find that the number of half-lives that must pass is \(n=5\).
Now, we can calculate the total time \(t\):
\[ t = n \times T_{1/2} = 5 \times 2.5 \text{ hours} = 12.5 \text{ hours} \]
Step 4: Final Answer:
The material can be handled safely after 12.5 hours. Therefore, option (D) is correct.