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ICSE Class 10 Answered

Find the energy released during the nuclear reaction: 1H1 + 3Li7 → 2He4 + 2He4 Given that Mass of lithium = 7.0160 a.m.u. Mass of 2He4 = 4.0026 a.m.u. Proton = 1.0078 a.m.u.
Asked by Topperlearning User | 18 Sep, 2017, 04:50: PM
answered-by-expert Expert Answer

Mass of the reactant nuclei = 7.0160 + 1.0078 = 8.0238 a.m.u.

Mass of the product nuclei = 4.0026 +4.0026 = 8.0052 a.m.u.

Mass defect = Δm = 8.0238 - 8.0052 = 0.0186 a.m.u.

According to the mass-energy equivalence relation this loss in mass is released in form of energy. Thus,

Energy released = 0.0186 a.m.u. × 931.5 MeV

-------------------= 17.326 MeV

Answered by | 18 Sep, 2017, 06:50: PM

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