A) \[{{10}^{-15}}\]J
B) 1 J
C) \[{{10}^{-12}}\]J
D) \[{{P}_{0}}\]J
Correct Answer: D
Solution :
\[E=m{{c}^{2}}=1\times {{(3\times {{10}^{8}})}^{2}}=9\times {{10}^{16}}\approx {{10}^{17}}J\]
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