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The radionuclide " "^(11)C decays according to " "_(6)^(11)C (to) " "_(5)^(11)B + e^(+) +nu : T_(1/2) = 20.3min The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m(" "_(6)^(11)C) =11.011434u and m(" "_(6)^(11)B) = 11.009305u, Calculate Q and compare it with the maximum energy of the positron emitted. |
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Answer» Solution :Q-value=[`" "^(m)N(" "^(11)C)-{m_(N)(" "^(11)B) + me} ]u XX 931.5 MeV = [{m(" "_(6)^(11)C) -6m_(e)}-{m(" "_(5)^(11)B)-5 m_(e) + m_(e)}]u xx 931.5 MeV=[ m(" "_(6)^(11)C) - m(" "_(5)^(11)B) - 2m_(e)]u xx 931.5 MeV= [11.011434 -11.009305 - 2 xx 0.000548] xx 931.5 MeV = 0.961 MeV` The maximum energy of the positron (0.960 MeV) is ALMOST equal to Q-value. |
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