Question:
In a nuclear reactor, \(^{235}\text{U}\) undergoes fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 yr, then find the total mass of uranium required.
In a nuclear reactor, \(^{235}\text{U}\) undergoes fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 yr, then find the total mass of uranium required.
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1 year ≈ \(3.15 \times 10^7\) seconds. Avogadro's number \(N_A = 6.02 \times 10^{23}\).
Updated On: Apr 20, 2026
- \(36.5 \times 10^3\text{ kg}\)
- \(36 \times 10^2\text{ kg}\)
- \(39.5 \times 10^3\text{ kg}\)
- \(38.2 \times 10^3\text{ kg}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Energy produced = Power × Time. Efficiency = Output/Input.
Step 2: Detailed Explanation:
Output energy = \(1000 \times 10^6 \times 10 \times 3.15 \times 10^7 = 3.15 \times 10^{17}\text{ J}\).
Input energy = \(\frac{3.15 \times 10^{17}}{0.1} = 3.15 \times 10^{18}\text{ J}\).
Energy per fission = \(200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11}\text{ J}\).
Number of fissions = \(\frac{3.15 \times 10^{18}}{3.2 \times 10^{-11}} = 9.84 \times 10^{28}\).
Mass of U = \(\frac{9.84 \times 10^{28}}{6.02 \times 10^{26}} \times 235 = 163.5 \times 235 = 38422\text{ kg} \approx 38.4 \times 10^3\text{ kg}\).
Closest is \(38.2 \times 10^3\text{ kg}\).
Step 3: Final Answer:
\(\boxed{38.2 \times 10^3\text{ kg}}\)
Energy produced = Power × Time. Efficiency = Output/Input.
Step 2: Detailed Explanation:
Output energy = \(1000 \times 10^6 \times 10 \times 3.15 \times 10^7 = 3.15 \times 10^{17}\text{ J}\).
Input energy = \(\frac{3.15 \times 10^{17}}{0.1} = 3.15 \times 10^{18}\text{ J}\).
Energy per fission = \(200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11}\text{ J}\).
Number of fissions = \(\frac{3.15 \times 10^{18}}{3.2 \times 10^{-11}} = 9.84 \times 10^{28}\).
Mass of U = \(\frac{9.84 \times 10^{28}}{6.02 \times 10^{26}} \times 235 = 163.5 \times 235 = 38422\text{ kg} \approx 38.4 \times 10^3\text{ kg}\).
Closest is \(38.2 \times 10^3\text{ kg}\).
Step 3: Final Answer:
\(\boxed{38.2 \times 10^3\text{ kg}}\)
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