Q. The energy in $MeV$ is released due to transformation of $1\ kg$ mass completely into energy $(c = 3 \times 10^{8}m/s)$
A) $7.625 \times 10\ MeV$
B) $10.5 \times 10^{29}\ MeV$
C) $2.8 \times 10^{- 28}\ MeV$
D) $5.625 \times 10^{29}\ MeV$
Explanation
(d) $E = \mathrm{\Delta}mc^{2} = 1 \times \left( 3 \times 10^{8} \right)^{2} = 9 \times 10^{16}J$
$\Rightarrow E = \frac{9 \times 10^{16}}{1.6 \times 10^{- 19}} = 5.625 \times 10^{35}eV = 5.625 \times 10^{29}MeV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.