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^(2)`, atomic weight of `.^(56)Mn=56 g mol^(-1)`).
After a long time bombardment, number `.^(56)Mn` nuclei present in the target depends upon.
(i) number of Mn nuclei present at the start of the process.
(ii) half life of Mn.
(iii) constant rate of production P

Options

A. All (i), (ii), and (iii) are correct.
B. Only (i) and (ii) are correct.
C. Only (ii) and (iii) are correct.
D. Only (i) and (iii) are correct.

Explanation

To solve the problem, we need to analyze the situation where the radionuclide \(\text{Mn}^{56}\) is produced at a constant rate \(P\) and has a half-life of \(2.5\) hours. The activity of \(\text{Mn}^{56}\) becomes constant after a long time of bombardment. We will derive the relationship between the number of \(\text{Mn}^{56}\) nuclei present in the target and the factors mentioned in the question.

### Step-by-Step Solution

1. **Understanding the Activity**:
The activity \(A\) of a radioactive substance is given by the equation:
\[ A = \lambda N \]
where \(N\) is the number of radioactive nuclei present and \(\lambda\) is the decay constant.

2. **Decay Constant**:
The decay constant \(\lambda\) is related to the half-life \(T_{1/2}\) by the formula:
\[ \lambda = \frac{\ln 2}{T_{1/2}} \]
Given that the half-life \(T_{1/2} = 2.5 \text{ hours} = 2.5 \times 3600 \text{ seconds} = 9000 \text{ seconds}\), we can calculate \(\lambda\):
\[ \lambda = \frac{0.693}{9000} \approx 7.7 \times 10^{-5} \text{ s}^{-1} \]

3. **Equilibrium Condition**:
After a long time of bombardment, the rate of production of \(\text{Mn}^{56}\) becomes equal to the rate of decay:
\[ P = \lambda N \]
Rearranging this gives:
\[ N = \frac{P}{\lambda} \]

4. **Substituting for \(\lambda\)**:
Substituting the expression for \(\lambda\) into the equation for \(N\):
\[ N = \frac{P}{\frac{\ln 2}{T_{1/2}}} = \frac{P \cdot T_{1/2}}{\ln 2} \]

5. **Conclusion**:
From the equation \(N = \frac{P \cdot T_{1/2}}{\ln 2}\), we can see that \(N\) depends on:
- The constant rate of production \(P\) (directly proportional).
- The half-life \(T_{1/2}\) (directly proportional).
- It does not depend on the initial number of \(\text{Mn}^{56}\) nuclei present.

### Final Answer:
The number of \(\text{Mn}^{56}\) nuclei present in the target depends upon:
- (ii) half-life of \(\text{Mn}\).
- (iii) constant rate of production \(P\).