During the fission of U-235, energy of the order of 180 MeV is generated per nucleus fissioned . The amount of energy released by the fission of 0.235 g of U-235 is

Options

A. `1.73 xx 10^(7)` KJ
B. `1.08 xx 10^(23)` KJ
C. `1.73 xx 10^(16)` KJ
D. `1.08 xx 10^(7)` KJ

Answer

A

Explanation

No. of atoms present in 0.235 g of U-235
`=(0.235xx6.02xx10^(23))/(235)=6.02xx10^(20)`
Energy produced per nucleus = 180 MeV
Total energy produced
`=6.02 xx 10^(20) xx 180 MeV`
`=1.08 xx 10^(23) MeV`
`=(1.08 xx 10^(23))xx(1.602xx10^(-16))KJ`
`=1.73 xx 10^(7) KJ`.