An atom bomb of mass 1 kg explodes, releasing \( 9\times10^{13}\,\mathrm{J} \) of energy. What percentage of the mass is converted into energy?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.1%
By Einstein’s mass–energy equivalence \( E=mc^2 \), the mass converted into the released energy is \( m=E/c^2 \), with \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \).
\( \begin{aligned} m &= \frac{E}{c^2} \\ &= \frac{9\times10^{13}\,\mathrm{J}}{(3\times10^{8}\,\mathrm{m\,s^{-1}})^2} \\ &= \frac{9\times10^{13}}{9\times10^{16}}\,\mathrm{kg} \\ &= 1\times10^{-3}\,\mathrm{kg} \end{aligned} \)
\( \begin{aligned} \text{Percentage converted} &= \frac{1\times10^{-3}\,\mathrm{kg}}{1\,\mathrm{kg}}\times100\% \\ &= 0.1\% \end{aligned} \)