AIIMS 2007 Physics Nuclear Reactors MCQ Question

To calculate the power output of a $^{235}\text{U}$ reactor, we need to follow a series of steps involving conversions from mass to the number of atoms, the energy produced per fission event, and finally the power output over time.

Step 1: Calculate the number of moles of $^{235}\text{U}$

First, we need to determine the number of moles in 2 kg of $^{235}\text{U}$. The molar mass of $^{235}\text{U}$ is approximately 235 g/mol.

The number of moles $n$ can be calculated using the formula:

$$n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$$

Converting 2 kg to grams:

$$2 \text{ kg} = 2000 \text{ g}$$

Now, we can find the number of moles:

$$n = \frac{2000 \text{ g}}{235 \text{ g/mol}} \approx 8.51 \text{ mol}$$

Step 2: Calculate the number of fissions

Next, we convert the number of moles to the number of atoms using Avogadro's number, $N_A = 6 \times 10^{23} \text{ atoms/mol}$:

$$\text{Number of atoms} = n \times N_A = 8.51 \text{ mol} \times 6 \times 10^{23} \text{ atoms/mol} \approx 5.11 \times 10^{24} \text{ atoms}$$

Step 3: Calculate total energy produced

Each fission releases 185 MeV. First, we need to convert this energy into joules. The conversion factor is:

$$1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$$

Thus,

$$185 \text{ MeV} = 185 \times 10^{6} \text{ eV} = 185 \times 10^{6} \times 1.6 \times 10^{-19} \text{ J} \approx 2.96 \times 10^{-11} \text{ J}$$

Next, we calculate the total energy produced by all fissions:

$$\text{Total Energy} = \text{Number of fissions} \times \text{Energy per fission}$$

Since each atom undergoes one fission, we have:

$$\text{Total Energy} = 5.11 \times 10^{24} \text{ atoms} \times 2.96 \times 10^{-11} \text{ J} \approx 1.51 \times 10^{14} \text{ J}$$

Step 4: Calculate power output

Power is defined as energy per unit time. Given that the reactor uses up the fuel in 30 days, we need to convert this time into seconds:

$$30 \text{ days} = 30 \times 24 \times 60 \times 60 \text{ s} = 2,592,000 \text{ s}$$

Now we can calculate the power output $P$:

$$P = \frac{\text{Total Energy}}{\text{Time}} = \frac{1.51 \times 10^{14} \text{ J}}{2,592,000 \text{ s}} \approx 58.3 \text{ MW}$$

Conclusion

Thus, the power output of the $^{235}\text{U}$ reactor is approximately 58.3 MW, which corresponds to option C.

Explanation of Other Options

• A) 56.3 MW: This option is lower than the calculated value, possibly due to an underestimation of the energy produced or the time over which it was measured.
• B) 60.3 MW: This option is higher, which could arise from an overestimation of energy produced or fissions.
• D) 54.3 MW: Similar to option A, this is also lower and likely results from incorrect calculations or assumptions in energy conversion or time.

Thus, the correct answer is clearly option C (58.3 MW).