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Step-by-Step Solution
Step 1: Identify Given Data
• Mass of the fuel used: 2 kg of $\,^{235}\text{U}$
• Energy released per fission: 200 MeV
• Avogadro number: $N_A = 6.023 \times 10^{26}\,\text{atoms per kmol}$
• 1 eV = $1.6 \times 10^{-19}\,\text{J}$
• Time of fuel usage: 30 days
Step 2: Calculate the Number of Atoms of $\,^{235}\text{U}$ in 2 kg
First, find how many moles (in kmol) of $\,^{235}\text{U}$ are in 2 kg. One kmole of $\,^{235}\text{U}$ has a mass of 235 kg. Hence, the number of kmoles in 2 kg is
$\displaystyle \text{Number of kmoles} \;=\; \frac{2}{235}.$
Next, multiply by Avogadro's number $N_A$ (in atoms per kmol) to find the total number of atoms:
$\displaystyle \text{Number of atoms} \;=\; \biggl(\frac{2}{235}\biggr) \;\times\; 6.023 \times 10^{26}.$
Step 3: Compute Total Energy Released
Each fission releases 200 MeV. We first compute the total number of fissions (equal to the total number of atoms) and multiply by the energy released per atom (200 MeV):
$\displaystyle E_{\text{total (in eV)}} = \biggl(\frac{2}{235} \times 6.023 \times 10^{26}\biggr) \times 200 \times 10^6 \;\text{eV}.$
To convert this energy from eV to joules, we multiply by $1.6 \times 10^{-19}\,\text{J/eV}$:
$\displaystyle E_{\text{total (in J)}} = \biggl(\frac{2}{235} \times 6.023 \times 10^{26}\biggr) \times 200 \times 10^6 \times 1.6 \times 10^{-19}\,\text{J}.$
Step 4: Convert 30 Days into Seconds
The power is the total energy divided by the time over which it is released. Convert 30 days into seconds:
$\displaystyle 30\,\text{days} \;=\; 30 \times 24 \times 3600 \;\text{seconds}.$
Step 5: Calculate the Power Output
Power = (Total energy) / (Total time). Substituting the values:
$\displaystyle P \;=\; \frac{\bigl(\frac{2}{235} \times 6.023 \times 10^{26}\bigr) \times 200 \times 10^6 \times 1.6 \times 10^{-19}}{30 \times 24 \times 3600} \;\text{watts}.$
Evaluating this expression gives a value close to:
$\displaystyle P \approx 63 \times 10^6\,\text{W} \;=\; 63\,\text{MW}.$
Since we are seeking an approximate answer, it is typically rounded to:
$\displaystyle 60\,\text{MW}.$
Step 6: Final Answer
The power output of the reactor is closest to 60 MW.
