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.

Options

A. a. All (a), (b) and (c ) are correct
B. b. only (a) and (b) are correct
C. c. only (b) and (c )are correct
D. d. only (a) and (c ) are correct

Answer

To solve the problem regarding the production of the radionuclide \(\text{Mn}^{56}\) in a cyclotron, we will follow these steps:

### Step 1: Understand the relationship between production and decay
After a long time of bombardment, the activity (rate of decay) of the radionuclide becomes constant. This means that the rate of production \(P\) of \(\text{Mn}^{56}\) is equal to the rate of decay of \(\text{Mn}^{56}\).

### Step 2: Write the equation for the rate of decay
The rate of decay can be expressed using the decay constant \(\lambda\) and the number of nuclei \(N\):
\[ \text{Rate of decay} = \lambda N \]
where \(N\) is the number of \(\text{Mn}^{56}\) nuclei present.

### Step 3: Relate the decay constant to half-life
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 \(t_{1/2}\) of \(\text{Mn}^{56}\) is 2.5 hours, we need to convert this into seconds for consistency in units:
\[ t_{1/2} = 2.5 \, \text{hours} = 2.5 \times 3600 \, \text{seconds} = 9000 \, \text{seconds} \]

### Step 4: Substitute \(\lambda\) into the decay equation
Substituting \(\lambda\) into the decay equation gives:
\[ \text{Rate of decay} = \frac{\ln 2}{t_{1/2}} N \]
Thus, we can write:
\[ P = \frac{\ln 2}{t_{1/2}} N \]

### Step 5: Solve for \(N\)
Rearranging the equation to solve for \(N\):
\[ N = \frac{P \cdot t_{1/2}}{\ln 2} \]

### Step 6: Identify the dependencies
From the equation \(N = \frac{P \cdot t_{1/2}}{\ln 2}\), we can see that the number of \(\text{Mn}^{56}\) nuclei present in the target depends on:
- The constant rate of production \(P\)
- The half-life \(t_{1/2}\)

### Conclusion
Thus, the number of \(\text{Mn}^{56}\) nuclei present in the target after a long time of bombardment depends on the constant rate of production \(P\) and the half-life \(t_{1/2}\).