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Question
The atomic mass of Uranium `""_92^238"U"` is 238.0508 u, while that of Thorium `""_90^234"Th"` is 234.0436u, and that of Helium `""_2^4"He"` "is 4.0026u. Alpha decay converts `""_92^238"U"` into `""_92^234"Th"` as, shown below:
`""_92^238"U" -> ( ""_90^234"Th" + ""_2^4"He" + "Energy" )`
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Solution
The given nuclear reaction is
`""_92^238"U" -> ""_90^234"Th" + ""_2^4"He" + "Energy"`
To find the energy released in the above reaction, we have to find the mass object Δm.
Total mass of the reactants = 238.0508 u
Total mass of the products = 234.0436 + 4.0026
Total mass of the products = 238.0462 u
∴ Δm = 238.0508 - 238.0462
∴ Δm = 0.0046 u
∴ Energy released = 0.0046 x 931.0 MeV
Energy released = 4.28 MeV
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