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Calculate the energy released in MeV in the following nuclear reaction : `._(92)^(238)Urarr._(90)^(234)Th+._(2)^(4)He+Q ["Mass of "._(92)^(238)U=238.05079 u` Mass of `._(90)^(238)Th=234.043630 u` Massof `._(2)^(4)He=4.002600 u` `1u = 931.5 MeV//c^(2)]`

Answer» Given nuclear reaction is `._(92)^(238)U rarr ._(90)^(234)Th + ._(2)^(4)He+Q`
Mass defect `=M_(U)-M_(Th)-M_(He)`
`=238.05079-234.043636-4.002600=0.00456 u`.
Energy released `=(0.00456 u) xx (431.5 MeV//C^(2))=4.25 MeV`.


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