1.

In nuclear fission, 0.01% mass is converted into energy. The energy released by the fission of 100 kg mass will be:

Answer»

`9 xx 10^(15) j`
`9 xx 10^(11) KJ`
`9 xx 10^(17) j`
`9 xx 10^(13) J`

Solution :`Delta m = 100 xx (0.01)/(100) = 0.01 kg`
`E = Delta mc^(2) = 0.01 xx (3 xx 10^(8))^(2) J`
`= 9 xx 10^(14) J = 9 xx 10^(11) kJ`


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