Energy released per fission is 200MeV.
The number of atoms n in 120g:
n=240120×6×1023atoms.
Total energy released, E=n×200×106
=6×1025MeV
The energy released per fission of nucleus of X240 is 200MeV. The energy released if all the atoms in 120g of pure X240 undergo fission is _____×1025MeV.
(Given NA=6×1023)
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