In the nuclear reaction, 

${}_1\mathrm{H}^2+{}_1\mathrm{H}^2\to {}_2\mathrm{He}^3+{}_0\mathrm{n}^1$

If the mass of the deuterium atom = 2.014741 anu, mass of ${}_2\mathrm{He}^3$ atom = 3.016977 amu and mass of neutron = 1.008987 amu, then the Q value of the reaction is nearly.

$\mathrm{Q}=\left(\sum _{ }{\mathrm{B}}_{\mathrm{r}}-\sum _{ }{\mathrm{B}}_{\mathrm{p}}\right){\mathrm{C}}^2$

Where $\sum _{ }{\mathrm{B}}_{\mathrm{r}}$ = sum of the masses of reactants

and  $\sum _{ }{\mathrm{B}}_p$=sum of the masses of the products

$\sum _{ }{\mathrm{B}}_{\mathrm{r}}=2 \times 2.014741 \mathrm{amu}$

$\sum _{ }{\mathrm{B}}_r=(4.029482) \mathrm{amu}$

$\sum _{ }{\mathrm{B}}_{\mathrm{p}}=(3.016977+1.008987) \mathrm{amu} =4.025964 \mathrm{amu}$

$\sum _{ }{\mathrm{B}}_{\mathrm{r}}-\sum _{ }{\mathrm{B}}_{\mathrm{p}}=(4.029482-4.025964) \mathrm{amu} =0.003518 \mathrm{amu}$

Decrease in mass appears as equivalent energy

$\therefore$  Q = 0.003518 x 931 MeV

    = 3.27 MeV