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The energy released per fission of nucleus of $^{240} X is 200 MeV$ . The energy released if all the atoms in 120 g of pure $^{240} X$ undergo fission is _____ $\times 10^{25} MeV \cdot (Given N_{A} = 6 \times 10^{23})$

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answer is 6.

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Detailed Solution

No .of atoms $= \frac{120}{240} \times 6 \times 10^{23} = 3 \times 10^{23}$

Energy released if all the atoms in 120 gm of pure ${}^{240}X$ nuclei is $= 3 \times 10^{23} \times 200 MeV$ $= 6 \times 10^{25} MeV$

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