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

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

A

`4.53xx10^26` MeV

B

`2.21xx10^14` MeV

C

`1xx10^13` MeV

D

`6.33xx10^24` MeV

Text Solution

Verified by Experts

The correct Answer is:
A
Number of atoms in 1 kg of pure `.^239Pu=(6.023xx10^23)/239xx1000=2.52xx10^24`
As average energy released per fission is 180 MeV
`therefore ` Total energy released = `2.52xx10^24 xx 180` MeV
`=4.53xx10^26` MeV
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