Q. The nuclear reaction $\ ^{2}H + \ ^{2}H \rightarrow \ ^{4}He$ (mass of deuteron $= 2.0141\ a.m.u.$ and mass of $He = 4.0024\ a.m.u.$) is
A) Fusion reaction releasing $24\ MeV$ energy
B) Fusion reaction absorbing $24\ MeV$ energy
C) Fission reaction releasing $0.0258\ MeV$ energy
D) Fission reaction absorbing $0.0258\ MeV$ energy
Explanation
(a) Total mass of reactants
$= (2.0141) \times 2 = 4.0282\ amu$
Total mass of products $= 4.0024\ amu$
Mass defect $= 4.0282\ amu - 4.0024\ amu$
$= 0.0258\ amu$
$\therefore$ Energy released $E = 931 \times 0.0258 = 24\ MeV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.