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JEE Main 2026 (4 Apr, Shift 1) · Nuclear Fission & Fusion

Nuclear Fission & Fusion question from JEE Main 2026 (4 Apr, Shift 1)

Q3JEE Main 2026 · 4 APR · SHIFT 1Medium

Nuclear Fission & Fusion

Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass numbers 3,4 and 10 are 5.6 MeV,7.4 MeV5.6\text{ MeV}, 7.4\text{ MeV} and 6.1 MeV, respectively, then in the process, ΔMc2=‾\Delta \text{Mc}^{2} = \underline{\qquad} MeV.

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Answer: (c)

  1. BE of reactants =2×3×5.6+4×7.4=33.6+29.6=63.2 MeV= 2\times3\times5.6 + 4\times7.4 = 33.6 + 29.6 = 63.2\text{ MeV}.
  2. BE of product =10×6.1=61.0 MeV= 10\times6.1 = 61.0\text{ MeV}.
  3. ΔMc2=∣61.0−63.2∣=2.2 MeV\Delta Mc^2 = |61.0 - 63.2| = 2.2\text{ MeV} (energy must be supplied).
  4. So the answer is (c) 2.22.2.
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