The nucleus

• Factors determining nuclear stability:

  1. Nuclear force (attractive) between all nucleons
  2. Coulomb repulsion (repulsive) between protons
  3. Pauli exclusion principle (limiting factor)

• For light stable nuclides: N/Z ≈ 1 (equal numbers of protons and neutrons)

• For heavy stable nuclides: N/Z increases to about 1.6

• This increase occurs because:

  • Nuclear force is short-ranged (effective only up to ~1 fm)
  • Coulomb repulsion is long-ranged (affects all proton pairs)
  • As nuclei get larger, more neutrons are needed to counterbalance the increasing Coulomb repulsion without adding more protons
  • Neutrons contribute only to attractive nuclear force without adding repulsive Coulomb force

The binding energy per nucleon can be expressed as:

$\frac{B}{A} = a_1 - \frac{a_2}{A^{1/3}} - \frac{a_3 Z(Z-1)}{A^{4/3}}$

Each term represents a physical phenomenon:

1. Volume term ($a_1$):

   • Represents binding from nucleon-nucleon interactions in bulk
   • Proportional to number of nucleons (A)
   • Positive contribution to binding energy

2. Surface term ($-\frac{a_2}{A^{1/3}}$):

   • Accounts for reduced binding of nucleons at nuclear surface
   • Proportional to surface area ($R^2 \propto A^{2/3}$)
   • Negative contribution (reduces binding)

3. Coulomb term ($-\frac{a_3 Z(Z-1)}{A^{4/3}}$):

   • Represents electrostatic repulsion between protons
   • Proportional to number of proton pairs $\frac{Z(Z-1)}{2}$
   • Inversely proportional to nuclear radius ($R \propto A^{1/3}$)
   • Negative contribution (reduces binding)

This explains why mid-range nuclei (A≈50-80) have maximum binding energy per nucleon (~8.7 MeV), making them most stable.

The Paradox:

• An isolated neutron undergoes beta-minus decay: n → p + e⁻ + ν̄ (half-life ≈15 minutes)
• Yet stable nuclei contain neutrons that don't decay

Explanation:

1. Energy considerations:

   • Neutron mass (939.57 MeV/c²) exceeds proton mass (938.27 MeV/c²)
   • Isolated neutron decay has positive Q-value: ~0.78 MeV
   • In nuclei, total energy includes nuclear binding energy

2. Nuclear binding effects:

   • Neutrons in nuclei occupy specific quantum states with defined energies
   • If converting a neutron to proton would place the proton in an already filled state, Pauli exclusion principle prevents decay
   • The energy required to place proton in higher available energy state can exceed the 0.78 MeV energy gain from decay

3. N/Z ratio effects:

   • Nuclei seek optimal N/Z ratio for stability
   • Even though individual neutrons could decay energetically, the resulting nucleus would be less stable overall
   • Nuclear force "prefers" certain N/Z combinations that maximize total binding energy

This demonstrates how collective quantum effects can drastically modify individual particle behavior.

Key Characteristics of Nuclear Forces:

1. Short-range:

   • Effective only up to ~1-2 fm
   • Implies each nucleon interacts with only nearest neighbors
   • Explains constant nuclear density (volume ∝ A)

2. Strength:

   • 50-60 times stronger than electromagnetic forces at short distances
   • Overcomes Coulomb repulsion between protons within the nucleus
   • Enables nuclear binding despite electromagnetic repulsion

3. Charge independence:

   • Nearly identical for p-p, n-n, and p-n interactions
   • Explains similar behavior of protons and neutrons as nucleons
   • Leads to concept of "isospin" symmetry in nuclear physics

4. Non-central nature:

   • Force depends on nucleon spin orientations
   • Stronger when spins are parallel, weaker when anti-parallel
   • Results in nuclear energy level splitting patterns

5. Saturation property:

   • Each nucleon interacts with limited number of neighbors
   • Explains why binding energy per nucleon reaches plateau (~8.7 MeV)
   • Key to understanding why fusion and fission both release energy

Mass Excess Definition:

• The difference between the actual atomic mass (m) and the mass number (A) in atomic mass units:
• Mass excess = (m - A) u·c²

Calculation:

• Mass excess (in MeV) = 931(m - A) MeV
• Where m is the atomic mass in u and A is the mass number

Usefulness:

1. Computational Convenience:

   • Atomic masses are close to whole numbers
   • Working with small differences (mass excess) reduces computational errors
   • Avoids repeatedly writing nearly identical large numbers

2. Energy Calculations:

   • Directly relates to binding energy calculations
   • Q-values can be calculated from mass excess differences
   • Simplifies reaction energy and decay energy calculations

3. Visualization of Nuclear Stability:

   • Plotting mass excess per nucleon vs. A shows nuclear stability valley
   • Unstable nuclei "roll down" toward stability through radioactive decay
   • Helps predict decay modes based on which direction reduces mass excess

Example: Hydrogen-1 with mass 1.00783 u has mass excess of 931(1.00783-1) = 7.29 MeV

To calculate the energy released when four hydrogen-1 nuclei fuse to form helium-4:

1. Identify the key values:

   • B/A for ⁴He ≈ 7.1 MeV per nucleon
   • H-1 (proton) has no binding energy (B/A = 0)

2. Calculate total binding energy of products:

   • For ⁴He: (B/A)×A = 7.1 MeV × 4 nucleons = 28.4 MeV

3. Calculate total binding energy of reactants:

   • For four H-1: 4 × 0 MeV = 0 MeV

4. Energy released = BE(products) - BE(reactants)

   • Energy released = 28.4 MeV - 0 MeV = 28.4 MeV

This provides the theoretical maximum energy release. In practice, the actual fusion process occurs through intermediate steps (deuterium, tritium, etc.) and involves some mass converted to neutrinos that escape, yielding about 26.7 MeV of usable energy.

Note: This calculation illustrates why hydrogen fusion powers stars and why the helium abundance increased rapidly in the early universe.

• Statistical uncertainty in radioactive decay:

  • Is inherent to the quantum nature of the process
  • Cannot be eliminated by improving measurement techniques
  • Follows precise mathematical laws (Poisson statistics)
  • Is a fundamental property of nature, not a limitation of equipment

• Whereas experimental measurement error:

  • Results from equipment limitations or environmental factors
  • Can potentially be reduced with better instruments
  • May follow different statistical distributions (often Gaussian)
  • Is a practical limitation, not a fundamental one

• Example: When monitoring a radioactive sample:

  • Even with perfect detection equipment, the number of decays per minute will naturally fluctuate
  • These fluctuations follow a predictable statistical pattern based on the sample size
  • The relative magnitude of fluctuations decreases with √N (where N is the number of atoms)

• Application: In radiation safety, statistical uncertainties must be accounted for when establishing safety margins

The asymmetric distribution occurs due to:

• Nuclear shell effects that favor the formation of nuclei with "magic numbers" of nucleons
• The lower potential energy configuration of asymmetric fragments compared to symmetric ones
• Quantum mechanical effects that make fragments with mass numbers around A≈95 and A≈140 more energetically favorable

In uranium-236 fission:

• Peak probability occurs at mass numbers A≈95 and A≈140
• The probability of equal-sized fragments (A≈118) is approximately 100 times lower
• This bimodal distribution is characteristic of thermal neutron-induced fission of uranium

Note: This asymmetry demonstrates that nuclear fission is not a simple splitting in half, but a complex quantum mechanical process governed by nuclear structure effects.

Control rods in a nuclear reactor serve to:

• Absorb neutrons to control the rate of fission reactions
• Regulate power output by adjusting their insertion depth into the core
• Provide emergency shutdown capability (SCRAM) when fully inserted
• Maintain criticality at exactly k=1 during normal operation
• Compensate for fuel depletion over time

Control rods are typically made of neutron-absorbing materials such as cadmium, boron, or hafnium.

Example: When operators need to reduce reactor power, control rods are inserted deeper into the moderator, capturing more neutrons that would otherwise cause additional fission events.

Classical vs. Quantum View of Nuclear Fission

Classical Physics Prediction:

• Fission impossible without sufficient external energy to overcome the entire barrier
• Energy conservation must be strictly maintained at all times
• Heavy nuclei should be permanently stable if they don't have enough energy to overcome the fission barrier

Quantum Mechanical Reality:

• Nuclei can "tunnel" through energy barriers they classically couldn't overcome
• Allows spontaneous fission without external energy input
• Temporary apparent violation of energy conservation permitted

Explanation via Heisenberg's Uncertainty Principle

• Energy-time uncertainty relation: $\Delta E \cdot \Delta t \geq \hbar/2$
• System can temporarily "borrow" energy $\Delta E = E_2 - E_1$ to penetrate barrier
• This borrowed energy must be "returned" within time $\Delta t$

Factors Affecting Penetration Probability

1. Barrier Height: Probability decreases exponentially with higher barriers ($\propto e^{-k\Delta E}$)
2. Barrier Width: Thinner barriers are more easily penetrated
3. Intermediate State Duration: Shorter durations increase penetration probability

This quantum tunneling mechanism explains why some heavy nuclei undergo spontaneous fission while others remain stable for extremely long periods, with half-lives determined by their specific barrier characteristics.