NEET Physics Nuclear Fission MCQ Question
If 1 kg of pure 239Pu undergoes fission, how much energy in joules is released, given the average energy released per fission is 180 MeV?
7.24 × 10^13 J
8.64 × 10^13 J
6.84 × 10^13 J
5.92 × 10^13 J
Correct Answer
Detailed Explanation
To find the total energy released, calculate the number of atoms in 1 kg of 239Pu using Avogadro's number and multiply by the energy per fission (180 MeV, converted to joules using 1 MeV = 1.6 × 10^-13 J).
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