Q. The mass defect in a particular nuclear reaction is $0.3\ grams$. The amount of energy liberated in kilowatt hours is (Velocity of light $= 3 \times 10^{8}m/s$)
A) $1.5 \times 10^{6}$
B) $2.5 \times 10^{6}$
C) $3 \times 10^{6}$
D) $7.5 \times 10^{6}$
Explanation
(d) $E = \mathrm{\Delta}mc^{2} \Rightarrow E = \frac{0.3}{1000} \times \left( 3 \times 10^{8} \right)^{2} = 2.7 \times 10^{13}J$
$= \frac{2.7 \times 10^{13}}{3.6 \times 10^{6}} = 7.5 \times 10^{6}kWh$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.