Combining the reactions, it can be written that
Li63+n10→He42+H31
H21+H31→He42+n10
_____________________ \mathrm{Li}63+H21\rightarrow 2(\mathrm{He}42)
Hence, the energy released in process can be calculated as follows
Q====Δmc2[M(Li)+M(H21)−2×M(He42)]×931.5MeV[6.01690+2.01471−2×4.00388]×931.5MeV22.22MeV