JEE Main 2022 · 24 June · morning · Single correct · +4 / −1

Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments 'B' and 'C' of mass numbers 105 and 115. The binding energy of nucleons in 'B' and 'C' is 6.4 MeV per nucleon. The energy Q released per fission will be :

  1. (a) 0.8 MeV
  2. (b) 275 MeV
  3. (c) 220 MeV
  4. (d) 176 MeV
Answer and solution

Answer (d) 176 MeV

Solution

Physics sir · 3 steps

Energy released is the binding energy gained: total BE of the fragments minus total BE of the parent.

  1. Why BE matters. A more tightly bound system has less mass. When nucleus A splits into the more tightly bound B and C, the extra binding energy is released as Q.

    Q = (BE_B + BE_C) - BE_A

  2. Total binding energies. Parent: BE_A = 220 × 5.6 = 1232 MeV. Fragments: BE_B = 105 × 6.4 = 672 MeV and BE_C = 115 × 6.4 = 736 MeV, together 1408 MeV.
  3. So. Q = 1408 - 1232 = 176 MeV. As a check, 105 + 115 = 220 nucleons each gain 0.8 MeV, and 220 × 0.8 = 176. The answer is (d).

The idea to keep. In fission or fusion, Q equals the total binding energy after minus the total before. Multiply each BE per nucleon by its mass number before you subtract.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 110 times in JEE Main since 2020. This question is from 2022.