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)=4.002604u and m(126C)=12.000000u
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Solution
Mass of 3 alpha particles =3×4.002604u =12.007812u. Mass of 1 6C12atom=12.007812u Mass Defect =△m=12.007812−12.000000=0.007812 Energy released =0.007812×931.5=7.2768MeV.