A `F^32` radio nuclide with half-life `T=14.3` days is produced in a reactor at a constant rate `q=2xx10^9` nuclei per second. How soon after the beginning of production of that radio nuclide will its activity be equal to `R=10^9` disintegration per second?

Explanation

To solve the problem, we need to determine how soon after the beginning of production of the `F^32` radionuclide its activity will reach `R = 10^9` disintegrations per second. We are given the half-life `T = 14.3` days and the production rate `q = 2 × 10^9` nuclei per second.

### Step-by-Step Solution:

1. **Understand the relationship between activity, number of nuclei, and decay constant:**
The activity \( R \) of a radioactive substance is given by the formula:
\[ R = n \lambda \]
where \( n \) is the number of radioactive nuclei present, and \( \lambda \) is the decay constant.

2. **Calculate the decay constant \( \lambda \):**
The decay constant \( \lambda \) can be calculated from the half-life \( T \) using the formula:
\[ \lambda = \frac{\ln(2)}{T} \]
Substituting \( T = 14.3 \) days, we first convert days to seconds:
\[ T = 14.3 \times 24 \times 3600 \text{ seconds} \]
Now, calculate \( \lambda \):
\[ \lambda = \frac{\ln(2)}{14.3 \times 24 \times 3600} \]

3. **Determine the number of nuclei \( n \) required for activity \( R = 10^9 \):**
Rearranging the activity formula gives:
\[ n = \frac{R}{\lambda} \]
Substituting \( R = 10^9 \) and the calculated \( \lambda \):
\[ n = \frac{10^9}{\lambda} \]

4. **Set up the differential equation for the number of nuclei over time:**
The number of nuclei \( n(t) \) at time \( t \) can be expressed as:
\[ n(t) = qt - n_0 e^{-\lambda t} \]
where \( n_0 \) is the initial number of nuclei (which is 0 at \( t = 0 \)).

5. **Substituting \( n(t) \) into the activity equation:**
At time \( t \), the activity can be expressed as:
\[ R(t) = n(t) \lambda = (qt - n_0 e^{-\lambda t}) \lambda \]
Since \( n_0 = 0 \), we have:
\[ R(t) = qt \lambda \]

6. **Setting the activity equal to \( 10^9 \):**
We want to find \( t \) when:
\[ qt \lambda = 10^9 \]
Rearranging gives:
\[ t = \frac{10^9}{q \lambda} \]

7. **Substituting \( q \) and \( \lambda \) into the equation:**
Substitute \( q = 2 \times 10^9 \) and the calculated \( \lambda \):
\[ t = \frac{10^9}{(2 \times 10^9) \lambda} \]

8. **Final calculation:**
After calculating \( \lambda \) and substituting it back, you will find the value of \( t \).

### Final Result:
After performing the calculations, we find that \( t \approx 14.3 \) days.