Physics-Nuclear Fission
NEET Physics Nuclear Fission MCQ Question
Type: MCQ-numerical-Hard-Class 12
Calculate the total energy released in MeV when 1 kg of 239 94 Pu undergoes complete fission, given that the energy released per fission is 180 MeV.
A
2.53 × 10^24 MeV
B
2.17 × 10^25 MeV
C
3.47 × 10^25 MeV
D
4.12 × 10^26 MeV
Correct Answer
Option B
Detailed Explanation
The number of atoms in 1 kg of 239Pu is calculated using Avogadro's number: (1000 g / 239 g/mol) × 6.023 × 10^23 atoms/mol. Multiplying the number of atoms by the energy per fission (180 MeV) gives the total energy released.
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