1.

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

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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