Q88 / 90 · JEE Main 2024 (27 Jan Shift 1) · Physics

In a nuclear fission process, a high mass nuclide (A≈236) with binding energy 7.6 MeV/Nucleon dissociated into two middle mass nuclides (A≈118), having binding energy of 8.6 MeV/Nucleon. The energy released in the process would be _______ MeV.

Correct answer: 236

Solution

The formula to calculate the energy released in the given process can be written as

Q=BEProduct -BEReactant    ...1

From equation (1), it follows that

Q=2(118)(8.6)-236(7.6) MeV=236 MeV

Hence, the correct answer is 236.

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