CUET · PHYSICS · PYQ PAPER 2025
A nucleus with mass number 240 and binding energy per nucleon 7.6 MeV splits into two fragments each of mass number 120 with binding energy per nucleon 8.5 MeV. The amount of energy released in this process will be :
- A \(0.9 MeV\)
- B \(17 MeV\)
- C \(216 MeV\)
- D \(2040 MeV\)
Answer & Solution
Correct Answer
(C) \(216 MeV\)
Step-by-step Solution
Detailed explanation
\(BE_{initial} = 240 \times 7.6 \text{ MeV} = 1824 \text{ MeV}\) \(BE_{final} = 2 \times (120 \times 8.5 \text{ MeV}) = 2040 \text{ MeV}\) \(E_{released} = BE_{final} - BE_{initial} = 2040 \text{ MeV} - 1824 \text{ MeV} = 216 \text{ MeV}\)
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