Physics-Nuclear Fission
NEET Physics Nuclear Fission MCQ Question
Type: MCQ-numerical-Hard-Class 12
If the average energy released per fission of 239 94 Pu is 180 MeV, approximately how much energy (in MeV) would be released if 2 kg of 239 94 Pu were completely fissioned?
A
52 × 10^26 MeV
B
6.02 × 10^24 MeV
C
08 × 10^25 MeV
D
61 × 10^27 MeV
Correct Answer
Option A
Detailed Explanation
Using the given data, the number of atoms in 2 kg of 239 94 Pu can be calculated using Avogadro's number, leading to the total energy released being 4.52 × 10^26 MeV.
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