Physics-Nuclear Fission
NEET Physics Nuclear Fission MCQ Question
Type: MCQ-numerical-Medium-Class 12
If 1 kg of pure 239 94 Pu undergoes fission, how much energy is released in MeV?
A
52 × 10^26 MeV
B
08 × 10^24 MeV
C
60 × 10^23 MeV
D
7.22 × 10^25 MeV
Correct Answer
Option A
Detailed Explanation
The energy released per fission is 180 MeV. The number of atoms in 1 kg of 239 94 Pu is given by (1000 g / 239 g/mol) × 6.023 × 10^23 atoms/mol. Multiplying this by 180 MeV gives the total energy released.
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