The radionuclide $^{11} C$ decays according to
$_{6}^{11} C \rightarrow_{5}^{11} B + e^{+} + \nu: \quad T_{1/2} = 20.3 \; min$
The maximum energy of the emitted positron is $0.960 \; MeV$. Given the mass values:
$m(_{6}^{11} C) = 11.011434 \; u$ and $m(_{5}^{11} B) = 11.009305 \; u$
Calculate $Q$ and compare it with the maximum energy of the positron emitted.

(A) The given nuclear reaction is:
$_{6}^{11} C \rightarrow_{5}^{11} B + e^{+} + \nu$
The $Q$-value of the decay is given by the difference in mass between the parent nucleus and the products:
$Q = [m(_{6}^{11} C) - m(_{5}^{11} B) - 2m_{e}]c^{2}$
Here,$m(_{6}^{11} C)$ and $m(_{5}^{11} B)$ are atomic masses,and $m_{e} = 0.000548 \; u$ is the mass of the positron.
Substituting the values:
$Q = [11.011434 - 11.009305 - 2(0.000548)] \; u \times c^{2}$
$Q = [0.002129 - 0.001096] \; u \times c^{2}$
$Q = 0.001033 \; u \times c^{2}$
Using the conversion $1 \; u = 931.5 \; MeV/c^{2}$:
$Q = 0.001033 \times 931.5 \; MeV \approx 0.962 \; MeV$
The calculated $Q$-value $(0.962 \; MeV)$ is very close to the maximum energy of the emitted positron $(0.960 \; MeV)$. This is because the neutrino carries away a negligible amount of energy in this specific decay mode.