In the nuclear reaction H 2 1 +H 2 1 →He 3 2 +n 1 0 If the mass of the deuterium atom = 2.014741 anu, mass of He 3 2  atom = 3.016977 amu and mass of neutron = 1.008987 amu, then the Q value of the reaction is nearly.

Q=∑ Br-∑ BpC2Where ∑ Br = sum of the masses of reactantsand  ∑ Bp=sum of the masses of the products∑ Br=2 x 2.014741 amu∑ Bp=(3.016977) amu∑ Bp=(3.016977+1.008987) amu        =4.025964 amu∑ Br+∑ Bp=(4.029482-4.025964) amu                          =0.003518 amuDecrease in mass appears as equivalent energy∴  Q = 0.003518 x 931 MeV    = 3.27 MeV