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

`""_(6)^(11) C to ""_(5)^(11)B + e^(+) + v + Q`
`Q = [m_N (""_(6)^(11)C) - m ""_(5)^(11)B - m_e] c^2`
where the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add `6m_e` in case of `""_(6)^(11)C` and `5m_e` in case of `_""_(5)^(11)B`. Hence
`Q = [m (""_(6)^(11)C) - m ""_(5)^(11)B - 2m ] c^2 ` (Note `m_e` has been doubled)
Using given masses, `Q = 0.961 MeV`
`Q = E_d+ E_e+ E_n`
The daughter nucleus is too heavy compared to `e^+` and v, so it carries negligible energy `(Ed ~~ 0)`. If the kinetic energy `(E_v)` carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q, hence maximum `E_e ~~ Q`).