1.

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 :

Answer»

`1.73 xx 10^(7) kJ`
`1.08 xx 10^(25)` kJ
`1.73 xx 10^(10) J`
`1.08 xx 10^(7) kJ`

Solution :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`


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