The radionuclide 11C decays according to
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
m (116C) = 11.011434 u and m (115B) = 11.009305 u,
calculate Q and compare it with the maximum energy of the positron emitted.
Important: We must consider electron mass in β decays, this mass is no more negligible.
The nuclear reaction is given by :
If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case of 11Cand 5 me in the case of 11B one excess electron, one electron is already there so in the final equation there will be 2 electrons.
Hence Q value for this reaction is given by = [11.011434-(11.009305 + 2 × me)] × c2
We know, me = 0.000548 u
Q = [11.011434-(11.009305 + 2 × 0.000548)] × c2
= 0.001033 u × c2
= 0.962 MeV
Hence the Q value is comparable with the maximum energy of the positron emitted.