In a nuclear reaction ₄¹⁰ Be + ₁^1 p ₅¹⁰ ~B + ₀^1 n , what is the total energy released if the binding energy of beryllium -9 is 58.2 MeV and that of boron- 10 is 64.7 MeV ?

Step-by-step solution

The nuclear reaction is given by ₄¹⁰ Be + ₁¹ p ₅¹⁰ B + ₀¹ n . The energy released in a nuclear reaction is given by the difference in the total binding energy of the products and the reactants. Q = BE ( products ) - BE ( reactants ) . The binding energy of ₁¹ p (proton) is 0 MeV and the binding energy of ₀¹ n (neutron) is 0 MeV. The binding energy of ₄¹⁰ Be is given as 58.2 MeV and the binding energy of ₅¹⁰ B is 64.7 MeV. Q = [ BE ( ₅¹⁰ B ) + BE ( ₀¹ n )] - [ BE ( ₄¹⁰ Be ) + BE ( ₁¹ p )] Q = [64.7 + 0] - [58.2 + 0]