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

To solve the problem, we need to calculate the Q value of the decay process of the radionuclide \( _6C^{11} \) to \( _5B^{11} \) plus a positron and a neutrino. The Q value represents the energy released during the decay, which can be compared to the maximum energy of the emitted positron.

### Step-by-Step Solution:

1. **Identify the Reaction:**
The decay reaction is given as:
\[
_6C^{11} \rightarrow _5B^{11} + e^+ + \nu
...