Show by calculation which process produces more energy per gram of material reacting. fission of U-235: \(\quad{ }^{235} \mathrm{U}_{22} \mathrm{U}+{ }_{0} n \longrightarrow{ }_{40}^{94} \mathrm{Zr}+{ }_{58}^{140} \mathrm{Ce}+6_{-1}{ }^{0} e+2{ }_{0}^{1} n\) fusion of deuterium: \(\quad{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{1}^{3} \mathrm{H}+{ }_{1}^{1} \mathrm{H}\) Nuclear masses for Ce-140 and Zr-94 are \(139.8734\) and 93.8841, respectively.

For each reaction, we need to find the mass defect (∆m), which is the difference between the total mass of the reactants and the total mass of the products. For fission of U-235: ∆m = mass of reactants - mass of products Reactants: U-235 mass = 235 u Neutron mass = 1 u Total mass of reactants = 236 u Products: Zr-94 mass = 93.8841 u Ce-140 mass = 139.8734 u 6 electrons mass ≈ 0 u (negligible) 2 neutrons mass = 2 u Total mass of products = 235.7575 u ∆m = 236u - 235.7575u = 0.2425u For fusion of deuterium: ∆m = mass of reactants - mass of products Reactants: 2 Deuterium mass = 2 * 2 u = 4 u Products: Tritium mass = 3 u Hydrogen mass = 1 u Total mass of products = 4 u ∆m = 4u - 4u = 0u (There is no mass defect in this case)

Using Einstein's mass-energy equivalence relation, E=mc², we can calculate the energy produced per reaction: For fission of U-235: Δm = 0.2425 u 1 u = 931.5 MeV/c² ΔE = (0.2425 u)(931.5 MeV/c²/u) = 226.03 MeV For fusion of deuterium: Δm = 0 u ΔE = (0 u)(931.5 MeV/c²/u) = 0 MeV

Calculate the energy per gram by dividing the energy produced by the reactants with their molar masses. For fission of U-235: Energy per mole of reactants = 226.03 MeV / 1 mol of U-235 Molar mass of U-235 = 235 g/mol Energy per gram of U-235 = (226.03 MeV / 1 mol of U-235) / (235 g/mol) = 0.9614 MeV/g For fusion of deuterium: Since the energy produced by the fusion process is zero, there is no energy produced per gram of reactants.

Comparing the energy produced per gram of material for the two processes: Fission of U-235: 0.9614 MeV/g Fusion of Deuterium: 0 MeV/g The fission of U-235 produces more energy per gram of material reacting, as compared to the fusion of Deuterium.