Q.

The energy released per fission of nucleus of  240X is 200MeV . The energy released if all the atoms in 120 g of pure  240X undergo fission is _____ ×1025MeV Given NA=6×1023

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answer is 6.

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Detailed Solution

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No .of atoms =120240×6×1023 =3×1023

Energy released if all the atoms in 120 gm of pure X  240 nuclei is =3×1023×200MeV =6×1025MeV

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