## **22.1 MASS DEFECT** In nuclear physics, we observe that the total mass of an atomic nucleus is surprisingly less than the combined mass of its individual protons and neutrons if they were separate. This difference in mass is called the **mass defect**. This section will define mass defect, show its simple mathematical formula, and explain why this small missing mass is incredibly important for understanding nuclear energy. ### **1. Definition of Mass Defect** * **Common Observation vs. Nuclear Physics:** Unlike macroscopic objects where the total mass is the sum of its constituents (e.g., 10 balls in a bag), the total mass of an atomic nucleus is *less* than the sum of the masses of its individual constituent particles (protons and neutrons, collectively called nucleons). * **Formal Definition:** The **mass defect** ($\Delta m$) is defined as the difference between the mass of the nucleus ($m_{\text{nucleus}}$) and the sum of the masses of its constituent particles (protons and neutrons). * **Example:** Carbon-12 has a mass of exactly 12 u, which is less than the sum of the masses of its six protons and six neutrons if calculated individually. Similarly, the mass of deuterium (${}^2\text{H}$) is less than the sum of the masses of one proton and one neutron. ### **2. Mathematical Formulation of Mass Defect** * For a nucleus with a specific number of protons ($Z$) and neutrons ($A-Z$): * $m_p$: mass of a proton * $m_n$: mass of a neutron * $m_{\text{nucleus}}$: mass of the actual nucleus * **General Equation for Mass Defect:** $$ \begin{equation*} \Delta m=\left[(A-Z) m_{n}+Z m_{p}\right]-m_{\text{nucleus }} \end{equation*} $$ * **Specific Case (e.g., Deuterium ${}^2\text{H}$):** (where A=2, Z=1, so A-Z=1) $$ \Delta m=\left(m_{n}+m_{p}\right)-m_{\text{nucleus }} $$ ### **3. Atomic Mass Unit (amu)** * **Purpose:** Used to express masses of atoms and nuclei in nuclear physics. * **Definition:** One twelfth of the mass of a carbon-12 atom. * **Conversion:** $1 \mathrm{amu} = 1 \mathrm{u} = 1.66 \times 10^{-27} \mathrm{~kg}$ ### **4. Example Problem: Mass Defect in Helium-4** * **Problem:** Find the mass defect in the formation of Helium-4 nucleus ($^4\text{He}$). * Given: $m_{\text{He-4}} = 4.002603 \mathrm{u}$ * $m_p = 1.007276 \mathrm{u}$ * $m_n = 1.008665 \mathrm{u}$ * **Solution Steps:** 1. For Helium-4, $A=4$ and $Z=2$. 2. Number of protons ($Z$) = 2. 3. Number of neutrons ($A-Z$) = $4-2=2$. 4. Apply the formula: $\Delta m = [(A-Z) m_n + Z m_p] - m_{\text{nucleus}}$ 5. $\Delta m = [2(1.008665) + 2(1.007276) - 4.002603] \mathrm{u}$ 6. $\Delta m = [2.017330 + 2.014552 - 4.002603] \mathrm{u}$ 7. $\Delta m = [4.031882 - 4.002603] \mathrm{u}$ 8. **Result:** $\Delta \mathrm{m}=0.029279 \mathrm{u}$ ### **5. Significance and Binding Energy** The mass defect represents the mass converted into binding energy that holds the nucleus together, according to Einstein's mass-energy equivalence ($E=\Delta mc^2$). This energy is released when nucleons combine to form a nucleus. For more on how atoms are structured at a discrete level, see .