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)`).
At what constant rate P, `.^(56)Mn` nuclei are being produced in the cyclontron during the bombardment?

Options

A. `2xx10^(11) "nuclei"//s`
B. `13.86xx10^(10) "nuclei"//s`
C. `9.6xx10^(10) "nuclei"//s`
D. `6.93xx10^(10) "nuclei"//s`

Answer

To find the constant rate \( P \) at which \( ^{56}Mn \) nuclei are being produced in the cyclotron, we can use the relationship between the activity of a radioactive substance and its decay constant. Here’s a step-by-step solution:

### 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),
- \( \lambda \) is the decay constant (in \( s^{-1} \)),
- \( N \) is the number of radioactive nuclei present.

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

### Step 3: Use the constant activity to find \( N \)
The problem states that the activity \( A \) of \( ^{56}Mn \) is \( 13.86 \times 10^{10} \, s^{-1} \). We can rearrange the activity formula to solve for \( N \):
\[ N = \frac{A}{\lambda} \]
Substituting the values we have:
\[ N = \frac{13.86 \times 10^{10} \, s^{-1}}{7.7 \times 10^{-5} \, s^{-1}} \approx 1.80 \times 10^{15} \, \text{nuclei} \]

### Step 4: Relate the production rate \( P \) to the decay rate
At equilibrium, the rate of production \( P \) of \( ^{56}Mn \) is equal to the rate of decay, which is equal to the activity \( A \):
\[ P = A = 13.86 \times 10^{10} \, s^{-1} \]

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

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