During the fission of `""^(235) U` , 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 `""^(235)U` is :

Options

A. `1.73 xx 10^(7) kJ`
B. `1.08 xx 10^(25)` kJ
C. `1.73 xx 10^(10) J`
D. `1.08 xx 10^(7) kJ`

Answer

A, C

Explanation

1 nuclide `to` 180 Me V , `((0.235)/(235) xx N_(0)) to Q = ?`
`Q = (0.235 xx 6 xx10^(23) xx 180)/(235 xx 1) = 1080 xx 10^(20) MeV`
`Q = 1.080 xx 10^(23) Me V = 1.080 xx 10^(23) xx 10^(6) xx 1.602 xx (10^(-19))/(1000) KJ`
= `(1.73 xx 10^(23 + 6 - 19 - 3)) = 1.73 xx 10^(7) KJ`