AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

The number of nucleus in 100gm of 92U235 is calculated as,
N=

100

M

×NA
N=

100

235

×(6.022×1023 )
N=2.56×1023 nucleus
The energy released in the fission of one nucleus is,
E′0=E0
E′0=1.88×106×1.6×10−19J
E′0=300.8 ×10−13J
The energy released in N nucleus fission is calculated as,
E=E′0N
E=300.8×10−13×2.56×1023
E=7.71×1012J