• # question_answer ${{\,}_{12}}{{A}^{27}}+{{\,}_{2}}H{{e}^{4}}\to {{\,}_{14}}S{{i}^{30}}+{{\,}_{1}}{{H}^{1}}+Q$Mass of${{\,}_{13}}A{{l}^{27}}=26.9815$amu and mass  of ${{\,}_{14}}S{{i}^{30}}=29.9738.$The Q is equal to A)  2.329 MeV                 B)  3.298 MeVC)  12.98 MeV                 D)  5.478 MeV

The sum of the masses of product = 29.9738 + 1.0078 = 30.9816 amu. The sum of the masses of reactants = 26.9815 + 4.0026 = 30.9841 amu. Difference in their masses $(\Delta \Mu )$ = 30.9816 - 30.9841 = -0.0025 amu. Here the value of Q is +ve. So $Q=+0.0025\times 931.5$ = +2.329 MeV Hence, the correct option is [a].