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Find the Q-value and the kinetic energy of the emitted alpha - particle in the alpha - decay of ._(86)^(220)Rn. 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» SOLUTION :PROCEEDING as above, in CASE of `._(86)RN^(220)` Q = 6.41 MeV K.E of a particle `= ((A-4)Q)/(A) = ((220-4))/(220)xx6.41 = 6.29 MeV` |
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