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Q.

When $U^{235}$ undergoes fission, 0.1% of the original mass is converted into energy. The energy released by an atom bomb with 10kg of $U^{235}$ is

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a

$9 \times 10^{17} J$

b

2 MeV

c

20 MeV

d

$9 \times 10^{14} J$

answer is D.

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Detailed Solution

$Δ M = 0.1 % M$ = $10^{- 3} (10 kg) = 10^{- 2} kg$

Energy released = $Δ {MC}^{2}$

= $10^{- 2} {(3 \times 10^{8})}^{2}$ = $9 \times 10^{14} J$

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