How much energy must be imparted to the proton so as to start the reaction ${ }_{\text{3}}{\text{Li}}^{\text{7}}+{}_{\text{1}}{}^{ }\mathrm{H}^{\text{1}}\to {}_4{}^{ }\text{Be}^{\text{7}}\text{+}{}_0{}^{ }\text{n}^{\text{1}}$$(\mathrm{Given} \mathrm{that} \mathrm{mass} \mathrm{of}{ }_{\text{3}}{\text{Li}}^{\text{7}}= 7.016 \mathrm{amu} ,{}_{\text{1}}{}^{ }\mathrm{H}^{\text{1}}= 1.00783 \mathrm{amu}, {}_4{}^{ }\text{Be}^{\text{7}}=7.01693\mathrm{amu}\mathrm{and}{}_0{}^{ }\text{n}^{\text{1}}=1.00866\mathrm{amu},$$1\mathrm{amu}=931.5\mathrm{MeV}/{\mathrm{C}}^2)$

 

Mass of reactants = (7.01600+1.00783)amu
Mass of products = (7.01693+1.00866)amu
Therefore mass defect = (8.02383–8.02559)amu
                                               = –0.00176amu

$\mathrm{The}\mathrm{Q}–\mathrm{value}\mathrm{of}\mathrm{reaction},\mathrm{Q}=-0.00176\times 931.5\mathrm{MeV}$

                                                         $=-1.64\mathrm{Mev}$

            According to conservation law of momentum, the energy supplied in the form of 

$\mathrm{K}.\mathrm{E}.\mathrm{of}\mathrm{proton}=-\left(1+\frac{\mathrm{m}}{\mathrm{M}}\right)\mathrm{Q}=-\left(1+\frac{1}{7}\right)\left(-1.64\right)\mathrm{MeV}$

                                                             = 1.8742 MeV