Calculate the energy released/absorbed (in MeV ) in the nuclear reaction: _1^1 H +_1^3 H →_1^2 H +_1^2 H; Given: m (_1^1 H )=1.007825 μ; m (_1^2 H…

Mass of reactants $\displaystyle =(1.007825+3.016049) \mathrm{u}$ $\displaystyle =4.023874 \mathrm{u}$ Mass of product $\displaystyle =2 \times 2.014102 \mathrm{u}$ $\displaystyle =4.028204 \mathrm{u}$ Mass defect, $\displaystyle \Delta m=4.023874 \mathrm{u}-4.028204 \mathrm{u}$ $\displaystyle =-0.00433 \mathrm{u}$ As the mass defect is negative, energy is absorbed. Energy absorbed, $\displaystyle E=0.00433 \times 931.5 \mathrm{MeV}$ $\displaystyle =4.03 \mathrm{MeV}$