The radioisotpe, N-13,has a half-life of 10.0 minutes is used to image organs in the body. If an injected sample has an activity of 40 microcuries `(40,muCi)`, what is its activity after 25 minutes in the body?

Explanation

To solve the problem of determining the activity of the radioisotope N-13 after 25 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Initial activity \( R_0 = 40 \, \mu Ci \)
- Half-life \( t_{1/2} = 10 \, \text{minutes} \)
- Time elapsed \( t = 25 \, \text{minutes} \)

2. **Determine the Rate Constant \( k \):**
The rate constant \( k \) can be calculated using the half-life formula:
\[ k = \frac{\ln 2}{t_{1/2}} \]
Substituting the half-life value:
\[ k = \frac{\ln 2}{10} \approx \frac{0.693}{10} = 0.0693 \, \text{min}^{-1} \]

3. **Use the First-Order Decay Formula:**
The activity after time \( t \) can be calculated using the first-order decay formula:
\[ R = R_0 \cdot e^{-kt} \]
Substituting the values we have:
\[ R = 40 \cdot e^{-0.0693 \cdot 25} \]

4. **Calculate the Exponent:**
First, calculate \( -kt \):
\[ -kt = -0.0693 \cdot 25 = -1.7325 \]

5. **Calculate \( e^{-kt} \):**
Now, calculate \( e^{-1.7325} \):
\[ e^{-1.7325} \approx 0.176 \]

6. **Calculate the Final Activity \( R \):**
Now substitute back to find \( R \):
\[ R = 40 \cdot 0.176 \approx 7.04 \, \mu Ci \]

### Final Answer:
The activity of the N-13 after 25 minutes is approximately \( 7.04 \, \mu Ci \).