Determine the maximum energy that a positron can possess during the ⁺ -decay of ¹¹ C into ¹¹ B . The atomic masses of ¹¹ C and ¹¹ B are 11.0114 u and 11.0093 u , respectively.

Step-by-step solution

The ⁺ decay equation is given by: ¹¹₆ C ¹¹₅ B + e⁺ + _ e The Q-value of the reaction is: Q = [m_ N (¹¹ C ) - m_ N (¹¹ B ) - m_e]c^2 Expressing nuclear masses in terms of atomic masses: m_ N (¹¹ C ) = m(¹¹ C ) - 6m_e m_ N (¹¹ B ) = m(¹¹ B ) - 5m_e Substituting these into the Q-value equation: Q = [m(¹¹ C ) - 6m_e - (m(¹¹ B ) - 5m_e) - m_e]c^2 Q = [m(¹¹ C ) - m(¹¹ B ) - 2m_e]c^2 Given values: m(¹¹ C ) = 11.0114 u m(¹¹ B ) = 11.0093 u m_e = 0.0005486 u Calculating the mass defect: m = 11.0114 - 11.0093 - 2(0.0005486)