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The fission properties of .94^239Pu are ...

The fission properties of `._94^239Pu` are very similar to those of `._92^235U`. The average energy released per fission is 180 MeV . How much energy, in MeV , is released if all the atoms in 1 kg of pure `._94^239Pu` undergo fission ?

Text Solution

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Number of nuclei in 1 kg of `._94^239Pu = (6.023xx10^23)/239xx1000.0=2.52 xx 10^24`
Energy released per fission = 180 MeV
`therefore` Total energy released = `2.52xx10^24 xx180` MeV
`=4.54xx10^26` MeV
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