Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV )

Step-by-step solution

Q = ∑ B . E .   of   products -   ∑ B . E .   of   reactants The equation for two deuterons combining to form Helium nucleus is given as H 2 + 1 H 2 → 2 H 4   Energy   released = Q = 4 B . E . 2 H 4   - 4 B . E . 1 H 2    = 4 × 7 − 4 × 1.1 = 28 − 4.4 The energy released in this process is = 23.6   MeV