The nucleus

Nuclear Mass Relation:

• For a nucleus $^A_Z$X with Z protons and N neutrons:
  • Expected mass if constituents were free: $Z\cdot m_p + N\cdot m_n$
  • Actual nuclear mass: $m(^A_Z\text{X})$
  • Observed: $m(^A_Z\text{X}) < Z\cdot m_p + N\cdot m_n$

Mass Difference Physical Meaning:

• The mass difference directly represents binding energy via $E = mc^2$
• Binding energy: $B = [Z\cdot m_p + N\cdot m_n - m(^A_Z\text{X})]c^2$
• This is the energy required to completely disassemble the nucleus into free nucleons
• Also equals energy released when free nucleons combine to form the nucleus
• Typically expressed in MeV (1 u = 931.478 MeV/c²)

Example calculation: For helium-4: $B = [2(1.007825) + 2(1.008665) - 4.00260]c^2 = 0.03038 \cdot 931.478 = 28.3$ MeV

This mass-energy equivalence demonstrates Einstein's $E = mc^2$ and shows that nuclear binding is an energy-intensive process.

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.

Using the activity equation: $A = A_0 \cdot 2^{-t/t_{1/2}}$

Given information:

• $A/A_0 = 0.8$ (80% remaining)
• $t = 10$ hours
• Need to find $t_{1/2}$

Step 1: Substitute values into the equation $0.8 = 2^{-10/t_{1/2}}$

Step 2: Take log (base 2) of both sides $\log_2(0.8) = -10/t_{1/2}$ $\log_2(0.8) = \frac{\ln(0.8)}{\ln(2)} = -0.322$

Step 3: Solve for $t_{1/2}$ $t_{1/2} = \frac{-10}{-0.322} = 31.1$ hours

Relationship: $t_{av} = \frac{1}{\lambda} = \frac{t_{1/2}}{0.693} \approx 1.44 \times t_{1/2}$

Physical significance:

• Average-life represents the mean lifetime of all nuclei in a sample
• It is the statistical expectation value of how long a nucleus will exist before decaying
• Average-life is always longer than half-life by a factor of 1.44
• Can be calculated as $t_{av} = \frac{\int_0^{\infty} t \lambda N e^{-\lambda t} dt}{\int_0^{\infty} \lambda N e^{-\lambda t} dt} = \frac{1}{\lambda}$

Example: If $t_{1/2} = 10$ days, then $t_{av} = 14.4$ days

Delayed neutrons:

• Neutrons emitted not during initial fission but during subsequent decay of fission fragments
• Released significantly later (seconds to minutes) than prompt neutrons
• Example: In $^{235}$U fission producing $^{137}$I, the decay chain $^{137}$I → $^{137}$Xe → $^{137}$Cs releases a neutron from $^{137}$Xe

Importance in reactor control:

• Comprise ~0.65% of all neutrons in uranium fission
• Create critical time delay that makes controlled chain reactions possible
• Allow human/mechanical control systems to respond to power changes
• Without delayed neutrons, reactors would respond too quickly for control systems
• Make negative feedback mechanisms effective for stability
• Enable use of control rods for power adjustment

If only prompt neutrons existed, nuclear reactors would be impossible to control safely.

The Lawson criterion:

• Fundamental requirement for energy-positive fusion
• Expressed as: $n\tau > 10^{14}$ s·cm$^{-3}$
• Where:
  • $n$ = particle density (particles per cm$^{3}$)
  • $\tau$ = confinement time (seconds)
  • Product $n\tau$ is called the Lawson number

For a viable fusion reactor:

• Must achieve Q > 1 (fusion energy output > energy input)
• Requires sufficient triple product of:
  • Density (10$^{20}$-10$^{22}$ particles/m$^3$)
  • Confinement time (1-10 seconds)
  • Temperature (10$^8$-10$^9$ K)
• D-T reaction is easiest, requiring lower temperatures (~10$^8$ K)
• D-D reactions require higher temperatures (~10$^9$ K)

This criterion explains why achieving practical fusion power remains challenging despite decades of research.

Barrier penetration in nuclear fission is a quantum mechanical phenomenon where:

• A heavy nucleus transitions from an initial state (E₁) to a lower-energy final state (E₃) despite needing to pass through a higher-energy intermediate state (E₂)

• This process contradicts classical physics because:

• Classical systems cannot spontaneously overcome energy barriers
• Quantum systems can "tunnel" through forbidden energy regions

• The energy difference ΔE = E₂-E₁ is called the "height of the barrier"

• The process is governed by Heisenberg's uncertainty principle: ΔE·Δt ≥ ℏ/2

• Example: In U-238, despite needing ~6 MeV to reach the intermediate deformed state, quantum tunneling allows occasional spontaneous fission with a half-life of ~10¹⁶ years

The three energy states in nuclear fission via barrier penetration represent different nuclear configurations:

1. Initial state (E₁): • Stable heavy nucleus in ground state • Roughly spherical configuration • Metastable equilibrium

2. Intermediate state (E₂): • Deformed "saddle point" configuration • Elongated nucleus with narrowing in middle • Highest energy state in the process • Represents the activation energy barrier

3. Final state (E₃): • Two separate middle-weight nuclei (fission fragments) • Lower energy than initial state • Energy difference (E₁-E₃) is released as kinetic energy of fragments and radiation

The energy difference (E₂-E₁) represents the barrier height that classically would prevent spontaneous fission but can be overcome through quantum tunneling.

Inertial confinement fusion is a technique to achieve controlled nuclear fusion by:

• Using intense laser beams directed from multiple angles onto a D-T (deuterium-tritium) pellet • Rapidly compressing and heating the fuel pellet to fusion conditions • Creating pressure and temperature high enough to overcome Coulomb repulsion

Key characteristics: • Operates in pulses rather than continuous operation • Achieves ultra-high density (10³-10⁴ times initial density) for very brief periods • Self-heating occurs as fusion-produced alpha particles remain trapped in the compressed plasma

Unlike magnetic confinement (Tokamak), which uses magnetic fields to contain rarefied plasma for longer durations, inertial confinement relies on the fuel's inertia to maintain compression during the brief fusion reaction.