AP EAMCET Engineering 2011 Solved Paper

Energy released in the fission of one nucleus =200MeV
=200×1.6×10−13J
=3.2×10−11J
Energy to be released =1000J
3.2×10−11J energy is released due to number of nucleus fissioned =1
∴ 1000J energy is released due to number of nucleus fissioned=

1×1000

3.2×10−11

=3.125×1013 nucleus