Atoms & Nuclei

JEE Main level, numerical answer, medium

224 of 239

Question 224 of 239

The energy released per fission of nucleus of X240X240 is 200MeV200MeV. The energy released if all the atoms in 120g120g of pure X240X240 undergo fission is _____×1025MeV\times {10}^{25}MeV.
(Given NA=6×1023{N}_{A}=6\times {10}^{23})

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Answer: 6

Energy released per fission is 200MeV200MeV.
The number of atoms nn in 120g120g:
n=120240×6×1023atoms.n=\dfrac{120}{240}\times 6\times {10}^{23}\text{atoms}.
Total energy released, E=n×200×106E=n\times 200\times {10}^{6}
=6×1025MeV=6\times {10}^{25}MeV

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