Q. The binding energy per nucleon of deuterium and helium atom is $1.1\ MeV$ and $7.0\ MeV$. If two deuterium nuclei fuse to form helium atom, the energy released is
A) $19.2\ MeV$
B) $23.6\ MeV$
C) $26.9\ MeV$
D) $13.9\ MeV$
Explanation
(b) $\ _{1}H^{2} +_{1}H^{2} \rightarrow_{2}He^{4} +$ energy
Binding energy of a $(\ _{1}H^{2})$ deuterium nuclei
$= 2 \times 1.1 = 2.2\ MeV$
Total binding energy of two deuterium nuclei
$= 2.2 \times 2 = 4.4\ MeV$
Binding energy of a $(\ _{2}H^{4})$ nuclei $= 4 \times 7 = 28\ MeV$
So, energy released in fusion $= 28 - 4.4 = 23.6\ MeV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.