Assuming that four protons combine to form a helium nucleus and two positrons each of mass 0 . 000549 a . m . u . , calculate the energy released. Given, the mass of 1 H 1 = 1 .007825 a . m . u . and mas of 2 He 4 = 4 . 002603 a . m . u .

Step-by-step solution

The nuclear fusion reaction can be written as: 1 H 1 + 1 H 1 + 1 H 1 + 1 H 1 → 2 He 4 + 2   1 e 0 + Q If m N   1 H 1 and m N 2 He 4 represent the masses of 1 H 1 and 2 He 4 nuclei respectively, then Q = 14   m N 1 H 1 - m N 2 He 4 - 2 m e × 931.5   MeV .....(i) Where m e represents the mass of the positron 1 e 0 . Here, m N   2 He 4   4.002603   a . m . u . ; m e = 0.0005499 a . m . u . Substituting for m N 1 He 1 ,   m N 2 He 4 and m e in the equation (i), we have