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)$

 

detailed solution

Correct option is C

Mass of reactants = (7.01600+1.00783)amuMass of products = (7.01693+1.00866)amuTherefore mass defect = (8.02383–8.02559)amu                                               = –0.00176amuThe Q–value of reaction,Q = −0.00176×931.5MeV                                                         =−1.64Mev            According to conservation law of momentum, the energy supplied in the form of  K.E. of proton =−1+mMQ=−1+17−1.64MeV                                                             = 1.8742 MeV