Determine the maximum energy that a positron can possess during the...

The $\beta^{+}$ decay equation is given by:

^{11}_{6}\text{C} \rightarrow ^{11}_{5}\text{B} + e^{+} + \nu_{e}$<br><br>The Q-value of the reaction is:<br><br>$Q = [m_{N}(^{11}\text{C}) - m_{N}(^{11}\text{B}) - m_e]c^2$<br><br>Expressing nuclear masses in terms of atomic masses:<br><br>$m_{N}(^{11}\text{C}) = m(^{11}\text{C}) - 6m_e$<br><br>$m_{N}(^{11}\text{B}) = m(^{11}\text{B}) - 5m_e$<br><br>Substituting these into the Q-value equation:<br><br>$Q = [m(^{11}\text{C}) - 6m_e - (m(^{11}\text{B}) - 5m_e) - m_e]c^2$<br><br>$Q = [m(^{11}\text{C}) - m(^{11}\text{B}) - 2m_e]c^2$<br><br>Given values:<br><br>$m(^{11}\text{C}) = 11.0114 \text{ u}$<br><br>$m(^{11}\text{B}) = 11.0093 \text{ u}$<br><br>$m_e = 0.0005486 \text{ u}$<br><br>Calculating the mass defect:<br><br>$\Delta m = 11.0114 - 11.0093 - 2(0.0005486)$<br><br>$\Delta m = 0.0021 - 0.0010972 = 0.0010028 \text{ u}$<br><br>Using the conversion factor$1 \text{ u} = 931 \text{ MeV}/c^2$:<br><br>$Q = 0.0010028 \times 931 \text{ MeV}$<br><br>$Q = 0.9336 \text{ MeV} = 933.6 \text{ keV}$<br><br>The maximum kinetic energy of the positron is equal to the Q-value of the reaction, which is$933.6 \text{ keV}$.<br><br>Answer:$933.6 \text{ keV}