Q 12-13-135JEE MainJEE Main 2017 (8 Apr)Easy
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):
Answer: (A) $23.6$ MeV
Energy released = binding energy of the products − binding energy of the reactants:
$$Q = 4(7.0) - 2\times2(1.1) = 28.0 - 4.4 = 23.6\ \text{MeV}$$
Solution by Sreeraj P, M.Sc Physics