${}^7\mathrm{Li}_3$ fuses with a proton according to the nuclear reaction given below :  ${ }_1^1\mathrm{p}{+}^7{\mathrm{Li}}_3^7{\to }^4\mathrm{He}{+}^4\mathrm{He}$.  Given that the atomic masses of ${ }_1^1\mathrm{H},{ }_2^4\mathrm{He}{\text{ and }}^7{\mathrm{Li}}_3$ are 1.007825 u, 4.002603 u and 7.016004 u respectively, where $\mathrm{u}=931.5\mathrm{MeV}/{\mathrm{c}}^2$ , then the Q – value of the reaction is

The total mass of the initial particles ${\mathrm{m}}_{\mathrm{i}}=1.007825+7.016004=8.023829\mathrm{u}$

total mass of final particles ${\mathrm{m}}_{\mathrm{f}}=2\times 4.002603=8.005206\mathrm{u}$.

Difference between the initial and final mass of particles $={\mathrm{m}}_{\mathrm{i}}-{\mathrm{m}}_{\mathrm{f}}=8.023829-8.005206=0.018623\mathrm{u}$. The Q value is given by $\mathrm{Q}=\left(\mathrm{\Delta m}\right){\mathrm{c}}^2=0.018623\times 931.5=17.35\mathrm{MeV}$