# 22.2 BINDING ENERGY Following our discussion on mass defect, we now turn to **binding energy**, a concept directly linked to the "missing mass" within a nucleus. This energy is the powerful glue that holds a nucleus together, overcoming the immense repulsive forces between protons. This section will define binding energy, explain its connection to Einstein's famous mass-energy equivalence, and introduce the crucial idea of binding energy per nucleon, which helps us understand why some nuclei are more stable than others. We will also explore the binding energy curve, a vital tool for predicting whether nuclear fission or fusion will release energy. ### 1. Binding Energy Definition * **Binding Energy** is defined as *the minimum energy required to break an isolated nucleus into its constituent particles*. * Conversely, it is also the energy released during the formation of a nucleus. ### 2. Mass Defect ($\Delta m$) and Einstein's Mass-Energy Equivalence * The formation of a nucleus involves a "missing mass," known as the **mass defect**. * This mass defect is converted into energy, which is the binding energy. * Einstein's famous equation relates energy ($E$) and mass ($\Delta m$): $$ E=\Delta m c^{2} $$ * $E$: Energy required to break the nucleus or energy released during its formation. * $\Delta m$: Mass defect (in kg). * $c$: Speed of light (approx. $3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$). * This relation signifies that mass and energy are inter-convertible quantities. * **Calculation of Mass Defect**: For a nucleus with mass $m_{\text{nucleus}}$, $Z$ protons, and $N$ neutrons: $$ \Delta m = (N \cdot m_n + Z \cdot m_p) - m_{\text{nucleus}} $$ where $m_n$ is the mass of a neutron and $m_p$ is the mass of a proton. ### 3. Units and Conversions * In $E=\Delta m c^2$, mass ($\Delta m$) must be in kilograms (kg), and energy ($E$) will be in Joules (J). * **Atomic Mass Unit (amu)**: * $1 \text{ amu} = 1.66 \times 10^{-27} \text{ kg}$ * **Electron-Volt (eV)**: A common energy unit in nuclear physics. * $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$ * **Conversion of 1 amu to Energy Units**: * $1 \text{ amu} = 1.494 \times 10^{-10} \text{ J}$ * $1 \text{ amu} = 931.5 \text{ MeV}$ (Mega electron-Volts) * This conversion is very convenient for expressing nuclear masses in energy units. ### 4. Binding Energy per Nucleon (Packing Fraction) * The absolute value of binding energy does not fully explain nuclear stability. * **Binding Energy per Nucleon** (also called **packing fraction**, $f$) is a more important measure of stability. * It is the binding energy divided by the total number of nucleons (mass number, A). $$ f=\frac{E}{A}=\frac{\left((A-Z) m_{n}+Z m_{p}-m_{\text {nucleus }}\right) c^{2}}{A} $$ * **Significance**: The greater the binding energy per nucleon, the greater the stability of the nucleus. * The rest energy of a bound nucleus is less than the combined rest energy of its separated nucleons. ### 5. Binding Energy Curve * The **binding energy curve** graphically represents the variation of binding energy per nucleon with the mass number (A). * **Shape and Trends (Fig. 22.2 (b))**: * **Light Nuclei (A=1 to approx. A=20)**: Binding energy per nucleon generally increases. * **Medium-Sized Nuclei (A=50-100)**: The curve peaks around $A=56$ (e.g., Iron-56, ${ }^{56} \mathrm{Fe}$). This region represents the *most stable nuclei* or the "valley of stability." * **Heavy Nuclei (A > 100)**: Binding energy per nucleon gradually decreases. This is due to increasing repulsive electrostatic forces between protons, which are not fully balanced by the strong nuclear force. * **Implications for Nuclear Reactions**: * **Fission**: Heavy nuclei (e.g., Uranium-235) have lower binding energy per nucleon and are less stable. They tend to undergo fission, splitting into more stable, medium-sized nuclei and releasing energy. * **Fusion**: Light nuclei (e.g., Hydrogen isotopes) have lower binding energy per nucleon. They can fuse to form more stable, heavier nuclei (e.g., Helium), releasing significant energy. This is because the product nucleus has a higher binding energy per nucleon than the reactants. | Particle | Mass (kg) | Mass (amu) | MeV | | :--- | :--- | :--- | :--- | | Electron | $9.1164 \times 10^{-31}$ | 0.0005485 | 0.511 | | Proton | $1.672 \times 10^{-27}$ | 1.007276 | 938.27 | | Neutron | $1.675 \times 10^{-27}$ | 1.008665 | 939.57 | | Hydrogen | $1.672 \times 10^{-27}$ | 1.007825 | 938.78 | | Deuterium | $3.344 \times 10^{-27}$ | 2.014102 | 1876.12 | | Tritium | $5.0083 \times 10^{-27}$ | 3.016049 | 2809.43 | | Helium | $6.646 \times 10^{-27}$ | 4.002603 | 3728.40 | | Carbon-12 | $1.992 \times 10^{-26}$ | 12.000000 | 11177.9 | | Table 22. 2: Binding energy of some nuclei in MeV | | | | :---: | :---: | :---: | | Nucleus | Binding energy | Binding energy
per nucleon | | ${ }^{2} \mathrm{H}$ | 2.23 | 1.12 | | ${ }^{4} \mathrm{He}$ | 28.3 | 7.07 | | ${ }^{12} \mathrm{C}$ | 89.1 | 7.4 | | ${ }^{56} \mathrm{Fe}$ | 486.24 | 8.68 | | ${ }^{206} \mathrm{~Pb}$ | 1622 | 7.87 | | ${ }^{235} \mathrm{U}$ | 1786 | 7.6 | ## Possible Questions/Answer - **Q:** What is the binding energy of Helium-4, given its mass is 4.002603 amu and the sum of its constituent protons (2) and neutrons (2) is 4.031882 amu? **A:** 1. **Mass Defect ($\Delta m$):** $\Delta m = (4.031882 - 4.002603) \text{ amu} = 0.029279 \text{ amu}$ 2. **Convert $\Delta m$ to kg:** $\Delta m = (0.029279 \times 1.66 \times 10^{-27}) \text{ kg} = 4.860314 \times 10^{-29} \text{ kg}$ 3. **Calculate Binding Energy ($E$) using $E = \Delta m c^2$:** $E = (4.860314 \times 10^{-29} \text{ kg}) (3 \times 10^8 \text{ m/s})^2 = 4.374283 \times 10^{-12} \text{ J}$ - **Q:** Calculate the binding energy of a tritium nucleus (mass = 3.016049 u) in joules and electron-volts, given proton mass ($m_p$) = 1.007276 u and neutron mass ($m_n$) = 1.008665 u. **A:** 1. **Identify Components:** Tritium ($^3$H) has 1 proton (Z=1) and 2 neutrons (N=A-Z=3-1=2). 2. **Calculate Mass Defect ($\Delta m$):** $\Delta m = (N \cdot m_n + Z \cdot m_p) - m_{\text{tritium}}$ $\Delta m = (2 \times 1.008665 \text{ u} + 1 \times 1.007276 \text{ u}) - 3.016049 \text{ u}$ $\Delta m = (2.017330 + 1.007276) \text{ u} - 3.016049 \text{ u}$ $\Delta m = 3.024606 \text{ u} - 3.016049 \text{ u} = 0.008557 \text{ amu}$ 3. **Convert $\Delta m$ to kg:** $\Delta m = (0.008557 \times 1.66 \times 10^{-27}) \text{ kg} = 1.420 \times 10^{-29} \text{ kg}$ 4. **Calculate Binding Energy ($E$) in Joules:** $E = \Delta m c^2 = (1.420 \times 10^{-29} \text{ kg}) (3 \times 10^8 \text{ m/s})^2 = 1.278 \times 10^{-12} \text{ J}$ 5. **Convert Binding Energy to electron-volts:** $E = 1.278 \times 10^{-12} \text{ J} \times \frac{1 \text{ eV}}{1.60 \times 10^{-19} \text{ J}} = 7.99 \times 10^6 \text{ eV}$ - **Q:** Find the packing fraction for Carbon-12 ($^{12}$C) nucleus, given its mass is 12.000 u, proton mass is 1.007276 u, and neutron mass is 1.008665 u. **A:** 1. **Identify Components:** Carbon-12 has 6 protons (Z=6) and 6 neutrons (N=A-Z=12-6=6). 2. **Calculate Mass Defect ($\Delta m$):** $\Delta m = (N \cdot m_n + Z \cdot m_p) - m_{\text{C-12}}$ $\Delta m = (6 \times 1.008665 \text{ u} + 6 \times 1.007276 \text{ u}) - 12.000 \text{ u}$ $\Delta m = (6.051990 + 6.043656) \text{ u} - 12.000 \text{ u}$ $\Delta m = 12.095646 \text{ u} - 12.000 \text{ u} = 0.095646 \text{ u}$ 3. **Convert Binding Energy ($E$) to MeV:** Using $1 \text{ u} = 931.5 \text{ MeV}$: $E = 0.095646 \text{ u} \times 931.5 \text{ MeV/u} = 89.1 \text{ MeV}$ 4. **Calculate Packing Fraction ($f$):** $f = \frac{E}{A} = \frac{89.1 \text{ MeV}}{12} = 7.4 \text{ MeV/nucleon}$ | Concept/Formula | Description | | :-------------- | :---------- | | **Binding Energy** | Minimum energy to break a nucleus. | | **Mass Defect ($\Delta m$)** | Missing mass converted to binding energy. | | $E = \Delta m c^2$ | Einstein's mass-energy equivalence. | | $\Delta m = (N m_n + Z m_p) - m_{\text{nucleus}}$ | Mass defect calculation. | | $1 \text{ amu}$ | $1.66 \times 10^{-27} \text{ kg}$ or $931.5 \text{ MeV}$ | | $1 \text{ eV}$ | $1.60 \times 10^{-19} \text{ J}$ | | **Binding Energy per Nucleon ($f$)** | $f = E/A$; measure of nuclear stability. | | **Binding Energy Curve** | Graph of $f$ vs. $A$; peaks at $A \approx 56$. | The concept of binding energy and its variation across different nuclei is fundamental to understanding nuclear stability, radioactivity, and energy generation in nuclear power plants and stars. ## References (Derived from FBISE textbook)