The radionuclide `.^(56)Mn` is being produced in a cyclontron at a constant rate `P` by bombarding a manganese target with deutrons. `.^(56)Mn` has a half-life of `2.5 h` and the target contains large numbers of only the stable manganese isotopes `.^(56)Mn`. The reaction that produces `.^(56)Mn` is
`.^(56)Mn +d rarr .^(56)Mn +p`
After being bombarded for a long time, the activity of `.^(56)Mn` becomes constant, equal to `13.86 xx 10^(10) s^(-1)`. (Use `1 n2=0.693`, Avagardo number `=6 xx 10^(23)`, atomic weight of `.^(56)Mn=56 g mol^(-1)`).
At what constant rate P, `.^(56)Mn` nuclei are being produced in the cyclotron during the bombardment?

Options

A. `2 xx 10^(11)` nuclei `s^(-1)`
B. `13.86 xx 10^(10) `nuclei `s^(-1)`
C. `9.6 xx 10^(10)` nuclei `s^(-1)`
D. `6.93 xx 10^(10) `nuclei `s^(-1)`

Explanation

To solve the problem, we need to find the constant rate \( P \) at which the radionuclide \( ^{56}\text{Mn} \) is being produced in the cyclotron. Given that the activity of \( ^{56}\text{Mn} \) is constant at \( 13.86 \times 10^{10} \, \text{s}^{-1} \), we can use the relationship between the activity, the decay constant, and the number of nuclei present.

### Step 1: Understand the relationship between activity, decay constant, and number of nuclei
The activity \( A \) of a radioactive substance is given by the formula:
\[ A = \lambda N \]
where:
- \( A \) is the activity (in decays per second, or \( \text{s}^{-1} \)),
- \( \lambda \) is the decay constant (in \( \text{s}^{-1} \)),
- \( N \) is the number of radioactive nuclei present.

### Step 2: Find the decay constant \( \lambda \)
The decay constant \( \lambda \) can be calculated using the half-life \( T_{1/2} \) of the radionuclide:
\[ \lambda = \frac{0.693}{T_{1/2}} \]
Given that the half-life \( T_{1/2} \) of \( ^{56}\text{Mn} \) is \( 2.5 \, \text{h} \), we first convert this to seconds:
\[ T_{1/2} = 2.5 \, \text{h} \times 3600 \, \text{s/h} = 9000 \, \text{s} \]
Now, substituting this into the decay constant formula:
\[ \lambda = \frac{0.693}{9000 \, \text{s}} \approx 7.7 \times 10^{-5} \, \text{s}^{-1} \]

### Step 3: Relate activity to production rate
Since the activity is constant and equal to \( 13.86 \times 10^{10} \, \text{s}^{-1} \), we can set this equal to \( \lambda N \):
\[ 13.86 \times 10^{10} = \lambda N \]
Substituting \( \lambda \):
\[ 13.86 \times 10^{10} = (7.7 \times 10^{-5}) N \]

### Step 4: Solve for \( N \)
Now, we can solve for \( N \):
\[ N = \frac{13.86 \times 10^{10}}{7.7 \times 10^{-5}} \approx 1.80 \times 10^{15} \text{ nuclei} \]

### Step 5: Relate production rate \( P \) to \( N \)
At equilibrium, the rate of production \( P \) is equal to the rate of decay, which is given by the activity:
\[ P = A = 13.86 \times 10^{10} \, \text{s}^{-1} \]

### Final Answer
Thus, the constant rate \( P \) at which \( ^{56}\text{Mn} \) nuclei are being produced in the cyclotron is:
\[ P = 13.86 \times 10^{10} \, \text{s}^{-1} \]