Q. Energy released in fusion of $1\ kg$ of deuterium nuclei
A) $8 \times 10^{13}J$
B) $6 \times 10^{27}J$
C) $2 \times 10^{7}kwh$
D) $8 \times 10^{23}MeV$
Explanation
(d) Fusion reaction of deuterium is
$\ _{1}H^{2} + \ _{1}H^{2} \rightarrow \ _{2}He^{3} + \ _{0}n^{1} + 3.27\ MeV$
So $E = \frac{6.02 \times 10^{23} \times 10^{3} \times 3.27 \times 1.6 \times 10^{- 13}}{2 \times 2} = 7.8 \times 10^{13}J$
$= 8 \times 10^{13}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.