The half-life of `Tc^(99)` is 6.0 hr. The total residual activity in a patient 30 hr after receiving an injection containing `Tc^(99)` must be more than 0.01 `muC_(i)`. What is the maximum activity `("in"muC_(i))` that the sample injected can have?

Options

A. 0.16
B. 0.32
C. 0.64
D. 0.08

Answer

To solve the problem, we need to determine the maximum initial activity of the injected sample of Technetium-99 (Tc-99) based on its half-life and the residual activity after a certain time.

### Step-by-Step Solution:

1. **Identify the Half-Life**:
The half-life of Tc-99 is given as 6.0 hours.

2. **Determine the Total Time**:
The total time after the injection is given as 30 hours.

3. **Calculate the Number of Half-Lives**:
To find out how many half-lives fit into the total time, we use the formula:
\[ n = \frac{\text{Total Time}}{\text{Half-Life}} = \frac{30 \text{ hours}}{6 \text{ hours}} = 5 \]
So, there are 5 half-lives in 30 hours.

4. **Understand the Activity Decay Formula**:
The activity \( A \) at any time can be calculated using the formula:
\[ A = A_0 \left( \frac{1}{2} \right)^n \]
where \( A_0 \) is the initial activity and \( n \) is the number of half-lives.

5. **Set Up the Equation**:
We know that the residual activity after 30 hours must be more than 0.01 µCi:
\[ A > 0.01 \text{ µCi} \]
Substituting the formula for activity:
\[ A_0 \left( \frac{1}{2} \right)^5 > 0.01 \]

6. **Calculate \( \left( \frac{1}{2} \right)^5 \)**:
Calculate \( \left( \frac{1}{2} \right)^5 \):
\[ \left( \frac{1}{2} \right)^5 = \frac{1}{32} \]

7. **Rearranging the Inequality**:
Substitute this back into the inequality:
\[ A_0 \cdot \frac{1}{32} > 0.01 \]
To find \( A_0 \), multiply both sides by 32:
\[ A_0 > 0.01 \cdot 32 \]

8. **Calculate the Maximum Initial Activity**:
Calculate the right side:
\[ A_0 > 0.32 \text{ µCi} \]

Thus, the maximum initial activity that the sample injected can have is greater than 0.32 µCi.

### Final Answer:
The maximum activity \( A_0 \) that the sample injected can have is **0.32 µCi**.

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