In a nuclear reaction 2 deuteron nuclei combine to form a helium nucleus. The energy released in MeV will be: (Given mass of deuteron =2.01355 ~amu . and mass of helium nucleus =4.0028 ~amu .

Step-by-step solution

The nuclear reaction is given by: 2₁²H ₂⁴He + Q . The mass of the reactants is 2 2.01355 amu = 4.02710 amu . The mass of the product is 4.0028 amu . The mass defect m is calculated as: m = (4.02710 - 4.0028) amu = 0.0243 amu . The energy released Q is given by Q = m 931.5 MeV/amu . Q = 0.0243 931.5 MeV 22.62545 MeV . Rounding to the nearest provided option, the energy released is 22.62 MeV . Answer: 22.62 MeV