In a stellar nuclear reaction, three alpha particles ( ₂ He ^4 ) fuse to form a single carbon nucleus ( ₆ C ¹² ). The total energy released in this reaction is 7.2 MeV . If the binding energy per nucleon of the carbon-12 nucleus is 7.7 MeV , the binding energy per nucleon of an alpha particle is :

Step-by-step solution

Let the binding energy per nucleon of an alpha particle be x . Total binding energy of the product (carbon-12 nucleus) is E_C = 12 7.7 = 92.4 MeV . Total binding energy of the reactants (three alpha particles) is E_ = 3 (4 x) = 12x . The energy released ( Q ) in the fusion reaction is given by: Q = E_C - E_ 7.2 = 92.4 - 12x 12x = 92.4 - 7.2 = 85.2 x = 85.2 12 = 7.1 MeV . Answer: 7.1 MeV