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What is the total energy, in MeV, released if all the atoms in 1 kg of pure $Pu_{94}^{239}$undergo fission when the average energy released per fission is 180 MeV?
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The correct answer is:
4.536 × 1026MeV
Correct Option is : (D)
4.536 × 1026MeV
4.536 × 1026MeV
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