Question

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.

Answer 1

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.