The mass defect of nuclear reaction: _4Be^10 -> _5B^10 + e^- is — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

**Step 1: Write the beta-minus decay.** $$\ce{^{10}_4Be -> ^{10}_5B + e^- + \bar{\nu}_e}$$ **Step 2: Express mass defect using nuclear masses.** The nuclear mass defect (energy released) is: $$\Delta m = m_N(\ce{^{10}Be}) - m_N(\ce{^{10}B}) - m_e$$ **Step 3: Convert nuclear masses to atomic masses.** Atomic mass includes the electron masses of the orbital electrons: $$m_N(\ce{^{10}Be}) = M(\ce{^{10}Be}) - 4\,m_e$$ $$m_N(\ce{^{10}B}) = M(\ce{^{10}B}) - 5\,m_e$$ **Step 4: Substitute and simplify.** $$\Delta m = \bigl[M(\ce{^{10}Be}) - 4m_e\bigr] - \bigl[M(\ce{^{10}B}) - 5m_e\bigr] - m_e$$ $$= M(\ce{^{10}Be}) - M(\ce{^{10}B}) - 4m_e + 5m_e - m_e$$ $$\boxed{\Delta m = M(\ce{^{10}Be}) - M(\ce{^{10}B})}$$ The electron masses cancel exactly, so the mass defect equals simply the difference in atomic masses. **Answer: A — $\Delta m = M(\ce{^{10}Be}) - M(\ce{^{10}B})$**