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 Worked
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The radioisotope thorium 228 is an alpha emitter. Calculate the energy Q released when one such nucleus decays. Mass of Th 228 = 228.02875u Mass of Ra 224 = 224.02022u Mass of alpha particle = 4.00260u
Mass defect (Dm) = mass of Th228 – mass of Ra224 – mass of alpha particle = 228.02875u – 224.02022u – 4.00260u = 0.005926u
But 1u = 931.3 MeV (from data sheet) Therefore Mass defect is equivalent to 0.005926 x 931.3 = 5.5188838 MeV However we can only quote to 4 significant figures (as this is the least number given) therefore the answer is Q = 5.519 MeV |
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