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Question

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.
Given: m(42He)=4.002604 u and m(126C)=12.000000 u

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Solution

Mass of 3 alpha particles =3×4.002604 u
=12.007812 u.
Mass of 1 6C12atom=12.007812 u
Mass Defect =m=12.00781212.000000=0.007812
Energy released =0.007812×931.5=7.2768 MeV.

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