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The radionuclide .^(11)C decays according to ._(6)^(11)C to ._(5)^(11)B+ e^(+)+v : T_(1//2)=20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values , m(._(6)^(11)C)=11.011434 u and m(._(6)^(11)B)=11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted. |
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Answer» Solution :Mass defect in the given reaction is `Delta m = m(._(6)C^(11))` `= [m (._(5)B^(11))+Me]` This is in terms of NUCLEAR masses. If we EXPRESS the Q VALUE interms of atomic masses we have to subtract `6m_(e)` from atomic mass of carbon and `5 m_(e)` from that of boron to GET the corresponding nuclear masses Therefore, we have `Delta m = [m(._(6)C^(11))-6 m_(e)-m(._(5)B^(11))+5m_(e)-m_(e)]` `=[m(._(6)C^(1))-m(._(5)B^(11))-2m_(e)]` ` = [11.011434-11.009305-2xx0.000548]u` `= 0.001033 u` As, 1u = 931 MeV `Q = 0.001033xx931 MeV = 0.961 MeV` Which is the maximum energy of emitted POSITION. |
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