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Q.13.13  The radionuclide ^{11}C decays according to

                   _{6}^{11}\textrm{C}\rightarrow 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}\textrm{C})=11.011434\; u and m(_{6}^{11}\textrm{B})=11.009305\; u
                  calculate Q and compare it with the maximum energy of the positron emitted.

Answers (1)

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If we use atomic masses

\\\Delta m=m(_{6}^{11}\textrm{C})-m(_{5}^{11}\textrm{B})-2m_{e}\\ \Delta m=11.011434-11.009305-2\times 0.000548\\ \Delta m=0.001033u

Q-value= 0.001033\times931.5=0.9622 MeV which is comparable with a maximum energy of the emitted positron.

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