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

Option: 1

6


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

no.of atoms =\frac{120}{240}\times 6\times 10^{23}

=3\times 10^{23}

Energy rebased =200\times 3\times 10^{23}

                                     =6\times 10^{25}

Posted by

Rishi

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