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 `l n2=0.693`, Avagadro number` =6 xx 10^(23)`, atomic weight of `.^(56)Mn`=`56 g mol^(-1))`.
After the activity of `.^(56)Mn` becomes constant, number of `.^(56)Mn` nuclei present in the target is equal to .

Options

A. `5xx10^(11)`
B. `20xx10^(11)`
C. `1.2xx10^(14)`
D. `1.8xx10^(15)`

Explanation

To solve the problem, we need to find the number of nuclei of the radionuclide \(^{56}\text{Mn}\) present in the target after the activity becomes constant. We will use the relationship between the activity, decay constant, and the number of nuclei.

### Step-by-step Solution:

1. **Understanding the Equilibrium Condition**:
At equilibrium, the rate of production of \(^{56}\text{Mn}\) (denoted as \(P\)) is equal to the rate of decay of \(^{56}\text{Mn}\). This can be expressed mathematically as:
\[ P = \lambda N \]
where \(N\) is the number of nuclei and \(\lambda\) is the decay constant.

2. **Finding the Decay Constant**:
The decay constant \(\lambda\) is related to the half-life (\(t_{1/2}\)) of the radionuclide by the formula:
\[ \lambda = \frac{\ln 2}{t_{1/2}} \]
Given that the half-life of \(^{56}\text{Mn}\) is \(2.5\) hours, we first convert this into seconds:
\[ t_{1/2} = 2.5 \text{ hours} = 2.5 \times 3600 \text{ seconds} = 9000 \text{ seconds} \]
Now, substituting this value into the decay constant formula:
\[ \lambda = \frac{0.693}{9000} \approx 7.7 \times 10^{-5} \text{ s}^{-1} \]

3. **Substituting the Activity**:
The activity \(A\) (which is given as \(13.86 \times 10^{10} \text{ s}^{-1}\)) can be substituted into the equilibrium equation:
\[ N = \frac{P}{\lambda} \]
Therefore:
\[ N = \frac{13.86 \times 10^{10}}{7.7 \times 10^{-5}} \]

4. **Calculating the Number of Nuclei**:
Performing the calculation:
\[ N \approx \frac{13.86 \times 10^{10}}{7.7 \times 10^{-5}} \approx 1.80 \times 10^{15} \]

5. **Final Answer**:
Thus, the number of \(^{56}\text{Mn}\) nuclei present in the target after the activity becomes constant is approximately:
\[ N \approx 1.8 \times 10^{15} \]