Binding Energy
Given that the mass defect of a helium-4 nucleus is about 0.0304 u, an examiner asks candidates to find the total binding energy of the nucleus in mega-electronvolts.
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About This Question
- Subject
- physics
- Chapter
- atoms and nuclei
- Topic
- binding energy
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
28.3 MeV
Binding energy, as described in NCERT, is the energy released when free nucleons combine into a nucleus, and by mass-energy equivalence it equals the mass defect converted to energy, EB=Δm×931.5 MeV per u. The nucleus is bound precisely because assembling it lowers the total mass. Substituting the given defect, EB=0.0304×931.5≈28.3 MeV. The value 931.5 MeV is wrong because that is the energy of a full mass unit, not of the small 0.0304 u defect. The value 7.1 MeV is wrong because that is the binding energy per nucleon (28.3 divided by 4), not the total. The value 3.4 MeV is wrong because it corresponds to only a fraction of the defect and matches no correct step in the conversion. As stated in NCERT Class 12, Chapter 13 (Nuclei), a larger total binding energy means a more tightly held nucleus, and this energy would have to be supplied to break the nucleus back into free nucleons. Helium-4 is notably stable for such a light nucleus, which is why alpha particles are emitted as intact units in radioactive decay rather than as separate nucleons. A plausibility check: dividing 28.3 MeV among four nucleons gives about 7 MeV each, comfortably within the typical 7 to 9 MeV per-nucleon range seen across the periodic table, confirming the result.
This medium difficulty physics question is from the chapter atoms and nuclei, covering the topic of binding energy. It appeared in the 2025 exam.
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