The mass defect of the nuclear reaction: _5B^8 -> _4Be^8 + e^+ is — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

**Step 1: Write the beta-plus decay.** $$\ce{^8_5B -> ^8_4Be + e^+ + \nu_e}$$ **Step 2: Express mass defect using nuclear masses.** The nuclear mass defect is: $$\Delta m = m_N(\ce{^8B}) - m_N(\ce{^8Be}) - m_e$$ (The positron $e^+$ has the same mass as an electron $m_e$.) **Step 3: Convert nuclear masses to atomic masses.** $$m_N(\ce{^8B}) = M(\ce{^8B}) - 5\,m_e$$ $$m_N(\ce{^8Be}) = M(\ce{^8Be}) - 4\,m_e$$ **Step 4: Substitute and simplify.** $$\Delta m = \bigl[M(\ce{^8B}) - 5m_e\bigr] - \bigl[M(\ce{^8Be}) - 4m_e\bigr] - m_e$$ $$= M(\ce{^8B}) - M(\ce{^8Be}) - 5m_e + 4m_e - m_e$$ $$\boxed{\Delta m = M(\ce{^8B}) - M(\ce{^8Be}) - 2\,m_e}$$ **Key point:** Unlike beta-minus decay, beta-plus decay requires subtracting $2m_e$ because the parent has one fewer atomic electron (Z decreases by 1) AND the emitted positron accounts for one more $m_e$. **Answer: D — $\Delta m = M(\ce{^8B}) - M(\ce{^8Be}) - 2m_e$**