Mass-Energy Equivalence and Nuclear Binding Energy

Introduction to Mass-Energy Equivalence

  • Historical Context:     * Before Albert Einstein's theory, mass and energy were viewed as two distinct and independent entities.     * Mass: A fundamental property of matter that every substance possesses.     * Energy: The capacity to perform work. It was considered to be associated either with a body's position or its motion, but not its intrinsic mass.     * Conservation Laws: The law of energy conservation and the law of mass conservation were treated as independent principles. The total mass of the universe and the total energy of the universe were believed to be separately constant.

  • Einstein's Theory:     * Through the special theory of relativity, Einstein demonstrated that mass and energy are interrelated.     * Every substance possesses energy simply due to the existence of its mass.     * These independent conservation laws were unified into a single law: the conservation of (mass + energy).

Einstein's Mass-Energy Relation

  • The Equivalence Formula: If a substance loses a mass amount Δm\Delta m, an equivalent amount of energy ΔE\Delta E is produced:     ΔE=(Δm)c2\Delta E = (\Delta m) c^2     where cc is the speed of light in a vacuum (3.0×108m/s3.0 \times 10^8\,m/s).

  • Magnitude of Energy:     * According to this relation, a mass of 1kg1\,kg is equivalent to an enormous amount of energy: 9×1016J9 \times 10^{16}\,J.     * In kilowatt-hours, this equals 2.5×1010kWh2.5 \times 10^{10}\,kWh.

  • Experimental Observability:     * If energy ΔE\Delta E is added to matter, its mass increases by Δm=ΔEc2\Delta m = \frac{\Delta E}{c^2}.     * Because the value of cc is so high, mass changes in macroscopic processes (like heating a substance or compressing a spring) are too small to be measured even by the most sensitive balances.

  • Nuclear Context:     * Mass-energy conversion is significant at the nuclear level, such as in nuclear fission, where mass decrease results in the release of equivalent energy.     * The Sun: The sun continuously loses mass, which we receive on Earth as radiated energy.

  • Key Distinction: In this relation, "mass" does not mean "matter." While matter remains the same, the mass of a particle can increase with its velocity.

Pair Production and Pair Annihilation

  • Pair Production:     * Definition: The conversion of energy directly into mass. This occurs when an energetic γ\gamma-ray photon falls on a heavy substance.     * Process: The γ\gamma-ray is absorbed by a nucleus, producing an electron (1β0_{-1}\beta^0) and a positron (+1β0_{+1}\beta^0).     * Equation: hν(γ-photon)1β0(electron)++1β0(positron)h\nu (\gamma\text{-photon}) \rightarrow _{-1}\beta^0 (\text{electron}) + _{+1}\beta^0 (\text{positron}).     * Rest-Mass Energy: Calculated as E0=m0c2E_0 = m_0 c^2. For an electron or positron (m0=9.1×1031kgm_0 = 9.1 \times 10^{-31}\,kg), the rest-mass energy is approximately 0.51MeV0.51\,MeV.     * Threshold Energy: For pair production to occur, the γ\gamma-photon must have at least 2×0.51MeV=1.02MeV2 \times 0.51\,MeV = 1.02\,MeV. If energy is lower, the photon may cause the Photoelectric effect or Compton effect instead.     * Excess Energy: Energy above 1.02MeV1.02\,MeV becomes kinetic energy for the particles. The positron typically receives more than half of the kinetic energy because the positively charged nucleus repels it (acceleration) and attracts the electron (retardation).     * Role of Heavy Nucleus: A heavy nucleus must be present to ensure that both energy and momentum are conserved during the process.

  • Pair Annihilation:     * Definition: The conversion of mass into energy. This happens when an electron and a positron come close and destroy each other.     * Process: They combine to produce two γ\gamma-photons (energy).     * Equation: 1β0(electron)++1β0(positron)=hν(γ-photon)+hν(γ-photon)_{-1}\beta^0 (\text{electron}) + _{+1}\beta^0 (\text{positron}) = h\nu (\gamma\text{-photon}) + h\nu (\gamma\text{-photon}).     * Photon Energy: Each γ\gamma-photon carries 0.51MeV0.51\,MeV of energy.     * Conservation Mechanics: Two photons are produced (rather than one) to conserve both energy and momentum.     * Spin and Directionality:         * If the spins of the electron and positron are antiparallel: Two γ\gamma-photons are produced travelling in opposite directions.         * If the spins are parallel: Three γ\gamma-photons are produced, usually inclined at 120120^{\circ} to each other.         * The probability ratio of 2-photon to 3-photon emission is approximately 1100:11100:1.

Mass Defect and Nuclear Binding Energy

  • Mass Defect (Δm\Delta m):     * The rest-mass of a stable nucleus (MNM_N) is always less than the sum of the masses of its constituent nucleons (protons and neutrons) in their free state.     * Formula: Δm=(mass of protons+mass of neutrons)mass of nucleus\Delta m = (\text{mass of protons} + \text{mass of neutrons}) - \text{mass of nucleus}.     * For an atom ZXA_ZX^A: Δm=[Zmp+(AZ)mn]MN\Delta m = [Z m_p + (A - Z) m_n] - M_N.

  • Nuclear Binding Energy (EbE_b):     * The mass that "disappears" (Δm\Delta m) reappears as energy (Δmc2\Delta m c^2) during nucleus formation. This energy binds nucleons together.     * Definition: The minimum energy required to separate the nucleons of a nucleus and place them at rest at an infinite distance apart.     * Calculated using atomic masses (m(ZXA)m(_ZX^A)) and hydrogen atom mass (mHm_H):         Eb=[ZmH+(AZ)mnm(ZXA)]c2E_b = [Z m_H + (A - Z) m_n - m(_ZX^A)] c^2     * Conversion factor: 1u×c2=931.5MeV1\,u \times c^2 = 931.5\,MeV.

  • Binding Energy per Nucleon (EbnE_{bn}):     * Determines the stability of the nucleus.     * Definition: The average energy required to remove a single nucleon from the nucleus to infinity.     * Formula: Ebn=EbAE_{bn} = \frac{E_b}{A}.     * A higher binding energy per nucleon indicates greater nuclear stability.

The Binding Energy Curve

  • Inferences from the Curve:     * Peak Stability: The curve has a broad, flat maximum between mass numbers A=50A = 50 and A=80A = 80. Nuclei in this range are the most stable, with Ebn8.5MeVE_{bn} \approx 8.5\,MeV.     * Iron (56Fe^{56}Fe): Has the maximum stability with Ebn8.8MeVE_{bn} \approx 8.8\,MeV.     * Heavier Nuclei (A > 80): EbnE_{bn} decreases gradually (e.g., 7.6MeV7.6\,MeV for 238U^{238}U). In these heavy nuclei, the binding energy is insufficient to completely overcome the Coulombian repulsion between the large number of protons. These nuclei are often radioactive (beyond 83Bi209_{83}Bi^{209}).     * Lighter Nuclei (A < 50): EbnE_{bn} decreases. Below A=20A = 20, it drops sharply (e.g., 1.1MeV1.1\,MeV for heavy hydrogen H2H^2), indicating lower stability.     * Subsidiary Peaks: Peaks at 2He4_2He^4, 6C12_6C^{12}, and 8O16_8O^{16} indicate that even-even nuclei are more stable than their neighbors.

  • Nuclear Processes:     * Nuclear Fission: When a very heavy nucleus (238U^{238}U) splits into two lighter fragments near the flat maximum, EbnE_{bn} increases by about 1MeV1\,MeV, releasing energy. This is the basis for nuclear reactors and bombs.     * Nuclear Fusion: When very light nuclei (like H2H^2) combine to form a heavier nucleus (like He4He^4), EbnE_{bn} increases significantly more than in fission, releasing vast amounts of energy. This process powers the sun and stars.

Detailed Analysis of Solved Numerical Problems

  • Category I: Mass-Energy Relation:     * Example 1: Destruction of 1.0g1.0\,g of matter.         * Δm=1.0×103kg\Delta m = 1.0 \times 10^{-3}\,kg; c=3.0×108m/sc = 3.0 \times 10^8\,m/s.         * ΔE=(1.0×103)×(3.0×108)2=9.0×1013J\Delta E = (1.0 \times 10^{-3}) \times (3.0 \times 10^8)^2 = 9.0 \times 10^{13}\,J.         * In kWh: 9.0×10133600=2.5×1010Wh=2.5×107kWh\frac{9.0 \times 10^{13}}{3600} = 2.5 \times 10^{10}\,Wh = 2.5 \times 10^7\,kWh.

  • Category II: Pair Production & Annihilation:     * Example 2: Kinetic energy in pair production from a 2.26MeV2.26\,MeV photon.         * Rest mass energy of pair = 1.02MeV1.02\,MeV.         * Excess kinetic energy = 2.261.02=1.24MeV2.26 - 1.02 = 1.24\,MeV.         * Average K.E. per particle = 0.62MeV0.62\,MeV.     * Example 3: Energy and wavelength from electron-positron annihilation.         * Energy released = 1.02MeV1.02\,MeV.         * Energy per photon (EE) = 0.51MeV=0.82×1013J0.51\,MeV = 0.82 \times 10^{-13}\,J.         * Wavelength (λ\lambda) = hcE=0.024A˚\frac{hc}{E} = 0.024\,\text{\AA}.     * Example 4: Head-on collision of 1MeV1\,MeV positron and 1MeV1\,MeV electron.         * Total energy = Rest mass energy (1.02MeV1.02\,MeV) + Total K.E. (2MeV2\,MeV) = 3.02MeV3.02\,MeV.         * Energy per photon = 1.51MeV1.51\,MeV. λ=8.2×103A˚\lambda = 8.2 \times 10^{-3}\,\text{\AA}.

  • Category III: Mass Defect & Binding Energy:     * Example 5: Reaction H2+H2He4H^2 + H^2 \rightarrow He^4.         * Mass of 2H2=4.028204u2H^2 = 4.028204\,u; Mass of He4=4.002604uHe^4 = 4.002604\,u.         * Δm=0.0256u\Delta m = 0.0256\,u. Energy released = 0.0256×931.5=23.8464MeV0.0256 \times 931.5 = 23.8464\,MeV.     * Example 8: Helium (He4He^4) nucleus.         * Δm=[2(1.007276)+2(1.008665)4.001506]=0.030376u\Delta m = [2(1.007276) + 2(1.008665) - 4.001506] = 0.030376\,u.         * B.E. = 0.030376×931=28.3MeV0.030376 \times 931 = 28.3\,MeV.     * Example 11: Fission of XX (A=240A=240, Ebn=7.6MeVE_{bn}=7.6\,MeV) into YY (A=110A=110) and ZZ (A=130A=130) with Ebn=8.5MeVE_{bn}=8.5\,MeV.         * Initial B.E. = 240×7.6=1824MeV240 \times 7.6 = 1824\,MeV.         * Final B.E. = (110×8.5)+(130×8.5)=2040MeV(110 \times 8.5) + (130 \times 8.5) = 2040\,MeV.         * Energy released (QQ) = 20401824=216MeV2040 - 1824 = 216\,MeV.

  • Category IV: Q-Value and Decay Decay:     * Example 18: β\beta-decay of 10Ne23_{10}Ne^{23}.         * Q=[m(Ne23)m(Na23)]c2=0.004696×931.5=4.374MeVQ = [m(Ne^{23}) - m(Na^{23})] c^2 = 0.004696 \times 931.5 = 4.374\,MeV.         * Max K.E. of electron = 4.374MeV4.374\,MeV (assuming antineutrino carries no energy).     * Example 21: Fission of 56Fe^{56}Fe into two equal 28Al^{28}Al.         * Δm=55.934942(27.98191)=0.02888u\Delta m = 55.93494 - 2(27.98191) = -0.02888\,u.         * Q=26.81MeVQ = -26.81\,MeV. Since QQ is negative, the reaction is not energetically possible.

Questions & Discussion

  • Q1: Pair-production means?     * Response: (a) Annihilation of a γ\gamma-ray into an electron and a positron.

  • Q2: Minimum gamma-ray energy for pair-production?     * Response: (c) 1.02MeV1.02\,MeV.

  • Q3: Binding energy per nucleon as mass number increases?     * Response: (d) First increases, then decreases.

  • Q4: Binding energy of the hydrogen nucleus?     * Response: (d) Zero (since it consists of only one proton).

  • Q5: What decides stability of a nucleus?     * Response: (b) Binding energy/nucleon.

  • Q6: Sharp peak for Helium in the B.E. curve?     * Response: (a) It is very stable.

  • Q7: B.E./nucleon being almost same for many nuclei indicates?     * Response: (d) Saturation of nuclear forces.

  • Q8: Most stable nuclei are?     * Response: (c) even-even.

  • Q10: Binding energy of isotope 8O17_{8}O^{17}?     * Response: (9mn+8mpm0)c2(9 m_n + 8 m_p - m_0)c^2.