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Find the Q-value and the kinetic energy of the emitted `alpha` - particle in the `alpha` - decay of `._(88)^(226)Ra` Given m `(._(88)^(226)Ra)=226.02540 u`, `m (._(86)^(222)Rn)=222.01750 u`, `m (._(86)^(222)Rn)=220.01137 u, m (._(84)^(216)Po)` `= 216.00189 u`. |
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Answer» `._(88)Ra^(226)rarr ._(85)Rn^(222)+._(2)He^(4)` Q value `[m(._(88)Ra^(226))-m(._(86)Rn^(222))-m_(alpha)]xx931.5 MeV` `= [226.02540-222.01750-4.00260]xx931.5 MeV` `Q = 0.0053xx931.5 MeV = 4.94 MeV` K.E of a particle `= ((A-4)Q)/(A)=(226-4)/(226)xx4.94 = 4.85 MeV` |
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