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In a certain star, three alpha particles undergo fusion in a single reaction to form 126C nucleus. Calculate the energy released in this reaction in MeV |
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Answer» Mass of three alpha particles = 3 x 4.002604 u = 12.007812 u Mass of 122C nucleus = 12.000000 u Mass defect = 12.007812 - 12.000000 Δm = 0.007812 u Energy released during the fusion = Δm x 931.5 MeV = 0.007812 x 931.5 MeV = 7.2768 MeV |
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