The radionuclide `._6C^(11)` decays according to `._6C^(11)to ._5B^(11) +e^(+) +v`: half life =20.3min. The maximum energy of the emitted positron is `0.960 MeV`. Given the mass values
`m(._6C^(11))=11.011434u, m(._6B^(11))=11.009305u`
Calculate Q and compare it with maximum energy of positron emitted.

Explanation

Given, radioactive `""^(11)C` decay equation
`""_(8)C^(11) to ""_(5)B^(11) + e^(+) +v +Q`, the Q-value of this decay process is given by
`Q = Deltam xx c^(2)`
where `Deltam` is the mass defect.
As the mass of positron (`e^(+)`) is very small, we need to consider nuclear masses of `""_(6)C^(11)` and `""_(5)B^(11)` to calculate mass-defect.
Now, if we express the Q-value in terms of nuclear masses we have to subtract mass of 6 electrons `(6m_(e))` from the atomic mass of carbon and 5me from that of boron’s atomic mass to get the corresponding nuclear masses. Therefore, we have
`Q =(m(""_(11)C) - 6m_(e)-m(""_(11)B) + 5m_(e) - m_(e)c^(2)`
`=(m(""^(11)C)-m(""^(11)B)-2m_(e)c^(2))`
`=0.001033 xx 931.5 = 0.961` MeV
This energy is comparable to the actual maximum energy o f the emitted positron.