In a nuclear fusion reaction, three identical nuclei of an element X (mass number A = 10 ) fuse to form a single heavier nucleus Y . The binding energy per nucleon for X is 5 MeV and for Y is 8 MeV . The mass defect of this fusion reaction is: (Take 1 MeV = 1.6 10⁻¹³ J and c = 3 10^8 m/s )

Step-by-step solution

The fusion reaction is: 3X Y Total binding energy of the reactants: BE_ reactants = 3 (A_X BE/A _X) BE_ reactants = 3 (10 5 MeV ) = 150 MeV Since the mass number is conserved, A_Y = 3 10 = 30 . Total binding energy of the product: BE_ product = A_Y BE/A _Y BE_ product = 30 8 MeV = 240 MeV Energy released ( Q -value) in the reaction: Q = BE_ product - BE_ reactants Q = 240 MeV - 150 MeV = 90 MeV Converting the energy to Joules: E = 90 1.6 10⁻¹³ J = 144 10⁻¹³ J Using mass-energy equivalence E = m c^2 : m = E c^2 = 14