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Under certain circumstances, a nucleus can decay by emitting a particle more massive than an alpha - particle. Consider the following decay processes : ._(88)^(223)Ra to ._(82)^(209)Pb + ._(6)^(14)C ._(88)^(223)Ra to ._(86)^(219)Pb+ ._(2)^(4)He. Calculate the Q - value for these decays and determine that both are energetically allowed. |
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Answer» Solution :i) For the decay process `._(88)Ra^(223)to ._(82)Pb^(209)+ ._(6)C^(14)+Q` mass defect, `Delta M` = mass of `Ra^(223)` - (mass of `Pb^(209)` + mass of `C^(14)`) `= 223.01850 - (208.98107+14.00324)` = 0.03419 U `Q = 0.03419 XX 931 MeV = 31.83 MeV` ii) For the decay process `._(88)Ra^(223)to ._(86)Rn^(219)+ ._(2)He^(4)+Q` mass defect, `Delta M` = mass of `Ra^(223)` - (mass of `Rn^(219)+` mass of `He^(4)`) `= 223.01850 - (219.00948 + 4.00260)` `= 0.00642 u` `therefore Q = 0.00642xx931 MeV = 5.98 MeV` As Q values are positive in both the cases, therefore both the decays are energetically possible. |
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