Comprehensive Study Notes on Nuclear Physics

13.1 INTRODUCTION

• Prior chapter summary: In the previous chapter, it was discussed that in every atom, the positive charge and mass are densely concentrated in the nucleus.
• Dimensions:
  • The nucleus's dimensions are much smaller than those of the atom, specifically smaller by a factor of about 10%.
  • This implies that the volume of a nucleus is approximately $10^{-12}$ times the volume of an atom.
  • Analogy: If an atom is enlarged to the size of a classroom, the nucleus would be the size of a pinhead.
  • Despite its small size, the nucleus contains more than 99.9% of the atom's mass.
• Questions:
  • Does the nucleus have a structure similar to that of the atom?
  • What are the constituents of the nucleus?
  • How are these constituents held together?
• Topics of discussion in this chapter: Various properties of nuclei (size, mass, stability) and associated nuclear phenomena (radioactivity, fission, fusion).

13.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS

• Small atomic mass: The mass of an atom is extremely small compared to kilograms; for instance, the mass of a carbon atom (¹²C) is $1.992647 imes 10^{-26}$ kg.
• Use of atomic mass unit (u): Since kilograms are not convenient for such small masses, atomic mass unit (u) is utilized.
  • Definition: 1 u is defined as 1/12th of the mass of a carbon (¹²C) atom.
  • Calculation:
    $u = \frac{1.992647 \times 10^{-26} \text{ kg}}{12} = 1.660539 \times 10^{-27} \text{ kg}$ (Equation 13.1)
• Integer multiples approximation: {
  • The atomic masses of various elements express nearly as integral multiples of a hydrogen atom's mass, with notable exceptions.
  • Example: The atomic mass of chlorine is 35.46 u.
• Mass spectrometry: Accurate measurements of atomic masses are made using a mass spectrometer.
• Isotopes:
  • Find: Isotopes are atomic species of the same element that exhibit the same chemical properties but differ in mass.
  • Origin of name: The term "isotope" comes from Greek meaning "the same place" (referring to their position on the periodic table).
  • Composition: Practically every element consists of several isotopes with different relative abundances.
  • Example: Chlorine has two isotopes with masses 34.98 u (75.4% abundance) and 36.98 u (24.6% abundance).
  • Calculation of average atomic mass:
    $\text{Average mass of Cl} = \frac{(75.4 \times 34.98) + (24.6 \times 36.98)}{100} = 35.47 \, u$
  • A comparison of hydrogen isotopes: Hydrogen has three isotopes with masses 1.0078 u (proton), 2.0141 u (deuterium), and 3.0160 u (tritium).
  • Abundance of isotopes: The lightest isotope (proton) has a relative abundance of 99.985%.
• Proton mass:
  • The mass of a proton, $m_p = 1.00727 \, u = 1.67262 \times 10^{-27} \, kg$ (Equation 13.2).
  • This is calculated by subtracting the mass of an electron: $m_{e^-} = 0.00055 \, u$ from hydrogen's mass.
• Proton's role:
  • Protons carry positive charge, with one unit of fundamental charge and are stable.
  • Initial belief about electrons in the nucleus was ruled out based on quantum theory principles.
  • Number of electrons outside a nucleus, denoted as Z (atomic number), equals the total positive charge of the protons.

EXAMPLES

Example 13.1: Nuclear Density Calculation

• Given: Mass of iron nucleus as 55.85 u; A = 56.
• Calculation:
  1. Convert atomic mass unit to kg: $m_e = 55.85 imes 9.27 \times 10^{-26} \, kg$
  2. Nuclear density formula: $\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{9.27 \times 10^{-26}}{(4/3)(1.2 \times 10^{-15})^3 imes 56}$
  3. Result: Nuclear density = $2.29 \times 10^{17} \, kg/m^3$.
  • Comparison: This density equates to that in neutron stars, indicating a substantial level of compression similar to that of a nucleus.

13.4 MASS-ENERGY AND NUCLEAR BINDING ENERGY

13.4.1 Mass - Energy

• Einstein's theory of special relativity emphasizes the necessity of treating mass as another form of energy.
  • Prior understanding: Mass and energy were conserved separately when undergoing reactions.
  • Einstein's discovery: Mass can convert into different energy forms (e.g., kinetic energy) and vice versa.
• Mass-energy equivalence relation: $E = mc^2$ (Equation 13.6)
  • Where E represents energy, m denotes mass, and c indicates the speed of light in vacuum, $c \approx 3 \times 10^8 \, m/s$.
• Example 13.2: Energy equivalent of 1 g of substance
  • Calculation: $E = 10^3 \times c^2 = 10^3 \times (3 \times 10^8)^2 \Rightarrow E = 10^3 \times 9 \times 10^{16} = 9 \times 10^{13} \, J$
  • Interpretation: Converting 1 g of matter results in an enormous energy release.
• Verification: Experimental affirmation of Einstein's mass-energy relation arises from nuclear reactions involving nucleons, nuclei, electrons, and new particles.
  • Energy conservation law: States initial energy and final energy remain equal, incorporating mass's energy association.

13.4.2 Nuclear Binding Energy

• Mass paradox: It is anticipated that a nucleus's mass equals the total mass of its orbiting protons and neutrons. However, it is consistently found that this is not the case.
  • Example: Analyze the nucleus ¹⁶O (oxygen), consisting of 8 neutrons and 8 protons:
  1. Expected total mass computed as:
     $\text{Mass}_{\text{expected}} = 8 \times 1.00866 + 8 \times 1.00727 + 8 \times 0.00055 = 16.12744 \, u$
  2. Spectroscopically derived atomic mass: $15.99493 \, u$
  3. Therefore, the experimental mass of the nucleus becomes:
     $15.99053 \, u$ after subtracting electron mass.
  • Discrepancy: The nucleus's mass is lower than the sum of the individual constituent masses by 0.13691 u.
• Mass defect: Represented by the formula $\Delta M = [Z m + (A - Z) m_n ] - M$ (Equation 13.7) (A being the mass number, Z is atomic number).
  • Significance: This mass defect implies that energy is required to break the nucleus into individual protons and neutrons. Consequently, the energy associated with this mass defect can be demonstrated as:
    $E = \Delta M c^2$ (Equation 13.8)
• Example 13.3: Energy equivalent of an atomic mass unit
  • Determine:
  1. Express 1 u in kg:
     $1 u \approx 1.6605 \times 10^{-27} \, kg$
  2. Conversion to energy: $E = 1.6605 \times 10^{-27} \times (2.9979 \times 10^{8})^2 = 1.4924 \times 10^{-10} \, J$
     • Conversion to MeV:
       $E \approx 931.5 \, MeV \text{ or } 1 u \approx 931.5 \text{ MeV}/c^2$
  3. For ¹⁶O:
     $\Delta M = 0.13691 \, u \Rightarrow \text{Energy} = 0.13691 \times 931.5 \, MeV/c^2 \approx 127.5 \, MeV/c^2$
• Implication: The energy necessary to separate the ¹⁶O nucleus into its constituents is found to be 127.5 MeV/c².

13.5 NUCLEAR FORCE

• Fundamental Binding Forces:
  • The nuclear force is paramount, necessitating strength to overcome the repulsion between protons and satisfactorily bind neutrons and protons in the nucleus.
  • Comparative strength: The nuclear force surpasses the Coulomb force acting between protons and is far greater than gravitational forces.
• Force characteristics:
  1. The nuclear force decays sharply beyond a few femtometers in distance, leading to saturation properties in larger nuclei.
  2. This rapid decrease results in characteristics where nucleons only notably engage neighboring nucleons within a short range.
  3. Qualitative absence of stringent mathematical formulations characterizes nuclear force, unlike evident laws governing gravitation or electrostatics.

13.6 RADIOACTIVITY

• Discovery: Radioactivity was incidentally discovered by A. H. Becquerel in 1896 during studies of fluorescent materials and uranium.
• Radioactive decay: A fundamental descriptor of radioactivity is an unstable nucleus undergoing decay.
  • Types of decay:
  1. α-decay: Emission of helium nucleus (He).
  2. β-decay: Emission of either electrons or positrons (the latter being particles with opposite charge to electrons).
  3. γ-decay: Emission of high-energy photons, typically exceeding hundreds of keV.

13.7 NUCLEAR ENERGY

• Binding Energy Curve: A graphical representation of binding energy per nucleon demonstrates relative stability and energy output of nuclear processes.
  • Notable regions: Binding energy per nucleon remains roughly constant (around 8.0 MeV) for mass numbers between A = 30 and A = 170, whereas lighter and heavier nuclei exhibit lower binding energies.

13.7.1 Fission

• Nuclear fission: This critical nuclear reaction occurs when heavy isotopes (like ²³⁵U) are bombarded by neutrons, resulting in nuclei fragmenting into lighter nuclear species.
  • Specific reaction example:
  1. $²³⁵U + n \rightarrow ²³⁶U \rightarrow ¹⁴⁴Ba + ¹⁹Kr + 3n$ (Equation 13.10).
  2. Alternative paths exist leading to various intermediate nuclear fragments.
  • Energy release: The energy derived from one fission event approximates $200 \, MeV$.
  • Energy conversion methodology: Energy from fission results in kinetic energy transfer of fragmented nuclei to surrounding matter, typically apparent as heat.

13.7.2 Nuclear Fusion

• Fusion reaction: Detailing the process by which light nuclei amalgamate into heavier nuclei, releasing energy due to increased binding.
  • Example reactions: 1. $H + H \rightarrow D + e + v + 0.42 \, MeV$ (Equation 13.13a)
  1. $²H + ²H \rightarrow He + n + 3.27 \, MeV$ (Equation 13.13b)
  2. $H + ²H \rightarrow ³H + H + 4.03 \, MeV$ (Equation 13.13c)
  • Repulsive nature: Overcoming Coulombic repulsion requires significant energy (Coulomb barrier) for fusion to proceed.
  • Thermonuclear fusion: The high temperatures achieved (around $3 \times 10^8 \text{ K}$) enable fusion in stellar environments like the sun.

13.7.3 Controlled Thermonuclear Fusion

• Fusion reactors: Aim to replicate stellar fusion conditions, achieving temperatures around $10^8 ext{ K}$ to achieve plasma confinement.
• Global initiatives: Various countries, including India, pursue advancements in controlled fusion for future energy solutions.