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Exercises · 13.13

Q.The radionuclide 11C^{11}\text{C} decays according to
[!FORMULA] 611C→511B+e++ν : T1/2=20.3 min^{11}_{6}\text{C} \rightarrow {}^{11}_{5}\text{B} + e^{+} + \nu \ : \ T_{1/2} = 20.3\ \text{min}
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values: m(611C)=11.011434m(^{11}_{6}\text{C}) = 11.011434 u and m(511B)=11.009305m(^{11}_{5}\text{B}) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.

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Because atomic masses include electrons, a β+\beta^+ decay's Q-value needs a correction of 2mec22m_ec^2 (one for the created positron, one because the parent atom carries one more bound electron than the daughter). The computed Q≈0.961Q\approx0.961 MeV agrees almost exactly with the stated maximum positron energy of 0.960 MeV.

For β+\beta^{+} decay, ZAX→Z−1AY+e++ν^{A}_{Z}X \rightarrow {}^{A}_{Z-1}Y + e^{+} + \nu, the parent's neutral atomic mass includes ZZ electrons while the daughter's includes only Z−1Z-1. Since a positron is also produced, using ATOMIC masses requires:

Q=[M(ZAX)−M(Z−1AY)−2me]c2Q = \left[M(^{A}_{Z}X) - M(^{A}_{Z-1}Y) - 2m_e\right]c^2

Substituting the given data (me=0.000548m_e = 0.000548 u):

Q=[11.011434−11.009305−2(0.000548)]×931.5 MeVQ = \left[11.011434 - 11.009305 - 2(0.000548)\right] \times 931.5\ \text{MeV}

=[0.002129−0.001096]×931.5= \left[0.002129 - 0.001096\right] \times 931.5

=0.001033×931.5=0.9622 MeV= 0.001033 \times 931.5 = 0.9622\ \text{MeV}

This Q-value represents the total energy shared between the emitted positron and the neutrino (plus a tiny, negligible recoil energy of the B-11 daughter nucleus). The MAXIMUM kinetic energy of the positron occurs when the neutrino carries away essentially zero energy, so: …

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