The binding energy per nucleon of deuterium and helium nuclei are 1 . 1 MeV and 7 . 0 MeV respectively. When two deuterium nuclei fuse together to form a helium nucleus, the energy released in the fusion is

Step-by-step solution

The fusion reaction is as given below: 1 H 2 + 1 H 2 → 2 H e 4 + Q The binding energy of the reacting nuclei = 1.1 × 2 + 1.1 × 2 = 4.4 M e V The binding energy of the product nuclei = 7.0 × 4 = 28.0 M e V Hence, the energy released in the fusion reaction, Q = 28.0 - 4.4 = 23.6 M e V .