A radionuclide with half - life 1620 s is produced in a reactor at a constant rate of 1000 nuclei per second. During each decay energy, 200 MeV is released. If the production of radionuclides started at t = 0, then the rate of release of energy at t = 3240 s is

Options

A. `1.5xx10^(5)" MeV s"^(-1)`
B. `1.5xx10^(2)" MeV s"^(-1)`
C. `2.5xx10^(2)" MeV s"^(-1)`
D. `3.5xx10^(5)" MeV s"^(-1)`

Answer

To solve the problem step by step, we need to determine the rate of energy release from the decay of a radionuclide produced at a constant rate.

### Step 1: Identify the parameters
- Half-life (\( t_{1/2} \)) = 1620 s
- Production rate (\( \alpha \)) = 1000 nuclei/s
- Energy released per decay = 200 MeV
- Time (\( t \)) = 3240 s

### Step 2: Calculate the decay constant (\( \lambda \))
The decay constant \( \lambda \) can be calculated using the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Substituting the half-life:
\[ \lambda = \frac{0.693}{1620 \, \text{s}} \approx 0.000427 \, \text{s}^{-1} \]

### Step 3: Determine the number of radionuclides at time \( t \)
The number of radionuclides \( N(t) \) at time \( t \) can be expressed as:
\[ N(t) = \frac{\alpha}{\lambda} \left(1 - e^{-\lambda t}\right) \]
Substituting \( \alpha = 1000 \, \text{nuclei/s} \) and \( t = 3240 \, \text{s} \):
\[ N(3240) = \frac{1000}{0.000427} \left(1 - e^{-0.000427 \times 3240}\right) \]
Calculating \( e^{-0.000427 \times 3240} \):
\[ e^{-1.38} \approx 0.251 \]
Thus,
\[ N(3240) = \frac{1000}{0.000427} \left(1 - 0.251\right) \approx \frac{1000}{0.000427} \times 0.749 \approx 1754.4 \, \text{nuclei} \]

### Step 4: Calculate the decay rate at time \( t \)
The decay rate \( R(t) \) is given by:
\[ R(t) = \lambda N(t) \]
Substituting the values:
\[ R(3240) = 0.000427 \times 1754.4 \approx 0.749 \, \text{decays/s} \]

### Step 5: Calculate the rate of energy release
The rate of energy release \( P \) can be calculated as:
\[ P = R(t) \times \text{Energy per decay} \]
Substituting the values:
\[ P = 0.749 \, \text{decays/s} \times 200 \, \text{MeV} \approx 149.8 \, \text{MeV/s} \]

### Step 6: Convert MeV to a more standard unit (if necessary)
To express this in terms of Joules (optional):
\[ 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \]
Thus,
\[ P \approx 149.8 \times 1.6 \times 10^{-13} \approx 2.3968 \times 10^{-11} \, \text{J/s} \]
However, since the question asks for the answer in MeV/s, we keep it as:
\[ P \approx 149.8 \, \text{MeV/s} \approx 1.5 \times 10^5 \, \text{MeV/s} \]

### Final Answer:
The rate of release of energy at \( t = 3240 \, \text{s} \) is approximately \( 1.5 \times 10^5 \, \text{MeV/s} \).

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