A `.^(32)P` radionuclide with half - life T = 14.3 days is produced in a reactor at a constant rate `q=2.7xx10^(9)` nuclei per second. How soon after the beginning of production of that nuclide will its activity be equal to `A=1.0xx10^(9)dis.//s`?

Options

A. 9.5 days
B. 8 days
C. 7.5 days
D. 6 days

Answer

To solve the problem, we need to determine how long it will take for the activity of the radionuclide \(^{32}P\) to reach \(A = 1.0 \times 10^9\) disintegrations per second, given that it is produced at a constant rate \(q = 2.7 \times 10^9\) nuclei per second and has a half-life of \(T = 14.3\) days.

### Step-by-Step Solution:

1. **Calculate the decay constant (\(\lambda\))**:
The decay constant \(\lambda\) can be calculated using the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
where \(T_{1/2}\) is the half-life in seconds. First, convert the half-life from days to seconds:
\[ T_{1/2} = 14.3 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} = 1230720 \text{ seconds} \]
Now, calculate \(\lambda\):
\[ \lambda = \frac{\ln(2)}{1230720} \approx 5.64 \times 10^{-7} \text{ s}^{-1} \]

2. **Determine the number of nuclei (\(n\)) required for the desired activity**:
The activity \(A\) is related to the number of nuclei \(n\) and the decay constant \(\lambda\) by:
\[ A = n \lambda \]
Rearranging gives:
\[ n = \frac{A}{\lambda} \]
Substituting the values:
\[ n = \frac{1.0 \times 10^9}{5.64 \times 10^{-7}} \approx 1.77 \times 10^{15} \text{ nuclei} \]

3. **Set up the differential equation for the number of nuclei**:
The change in the number of nuclei over time can be described by the equation:
\[ \frac{dn}{dt} = q - n \lambda \]
where \(q\) is the production rate.

4. **Integrate the equation**:
Rearranging gives:
\[ \frac{dn}{q - n \lambda} = dt \]
Integrating both sides from \(0\) to \(n\) and \(0\) to \(t\):
\[ \int_{0}^{n} \frac{1}{q - n \lambda} dn = \int_{0}^{t} dt \]
The left side integrates to:
\[ -\frac{1}{\lambda} \ln(q - n \lambda) \bigg|_{0}^{n} = t \]

5. **Substituting limits**:
\[ -\frac{1}{\lambda} \left( \ln(q - n \lambda) - \ln(q) \right) = t \]
Simplifying gives:
\[ t = -\frac{1}{\lambda} \ln\left(\frac{q - n \lambda}{q}\right) \]

6. **Substituting the values**:
Substitute \(q = 2.7 \times 10^9\), \(n = 1.77 \times 10^{15}\), and \(\lambda = 5.64 \times 10^{-7}\):
\[ t = -\frac{1}{5.64 \times 10^{-7}} \ln\left(\frac{2.7 \times 10^9 - 1.77 \times 10^{15} \times 5.64 \times 10^{-7}}{2.7 \times 10^9}\right) \]

7. **Calculate the final time**:
After calculating the logarithm and the time, you will find that:
\[ t \approx 9.5 \text{ days} \]

### Final Answer:
The time after which the activity of the radionuclide \(^{32}P\) will be equal to \(1.0 \times 10^9\) disintegrations per second is approximately **9.5 days**.