Unit 24: Nuclear Physics

Discovery of the Nucleus

Rutherford's Alpha Particle Scattering Experiment

In 1911, Ernest Rutherford directed a narrow beam of alpha particles (helium nuclei, He^{2+}) at a thin gold foil. Most particles passed through with little or no deflection, but a small fraction (≈1 in 8000) were scattered at large angles, and some even bounced back toward the source.

“It was almost as incredible as if you fired a 15‑inch shell at a piece of tissue paper and it came back and hit you.” – Ernest Rutherford

From these observations Rutherford concluded that:

• The atom contains a tiny, dense, positively charged core – the nucleus.
• The nucleus occupies only a fraction (≈10^{-12}) of the atom’s volume but contains almost all its mass.
• Electrons surround this nucleus and account for the atom’s size.

Numerical Problem – Distance of Closest Approach

An alpha particle with kinetic energy K = 5 MeV is directed toward a gold nucleus (Z = 79). Estimate the distance of closest approach r_min assuming a head‑on collision.

Solution: At the point of closest approach, the initial kinetic energy is completely converted into electrostatic potential energy:

K = (1/(4πϵ₀)) * (2Ze²)/r_min

Using 1/(4πϵ₀) = 9×10⁹ N·m²/C², e = 1.6×10⁻¹⁹ C, and converting K to joules (5 MeV = 5×1.6×10⁻¹³ J = 8×10⁻¹³ J):

r_min = (9×10⁹ * 2 * 79 * (1.6×10⁻¹⁹)²) / (8×10⁻¹³)
      ≈ 4.5×10⁻¹⁴ m

Thus the alpha particle cannot penetrate closer than about 45 fm to the gold nucleus.

Nuclear Density and Structure

Key Definitions

• Mass number (A): total number of nucleons (protons + neutrons).
• Atomic number (Z): number of protons (also equals number of electrons in a neutral atom).
• Neutron number (N): N = A – Z.

Nuclear Density

Experimental scattering data show that the nuclear radius follows R = R₀ A^{1/3} with R₀ ≈ 1.2 fm. The volume of a nucleus is therefore V = (4/3)πR³ = (4/3)πR₀³ A. Since the mass of a nucleus is approximately Am_u (where m_u is the atomic mass unit), the density becomes:

ρ = (Am_u) / V ≈ (Am_u) / [(4/3)πR₀³ A] = m_u / [(4/3)πR₀³]

Notice that A cancels, giving a density that is independent of the mass number. Substituting the numbers:

ρ ≈ 1.66×10⁻²⁷ kg / [(4/3)π (1.2×10⁻¹⁵ m)³] ≈ 2.3×10¹⁷ kg/m³

This value is often quoted as ≈10¹⁷ kg/m³.

Numerical Problem – Verify Nuclear Density

Calculate the density of ^{56}_{26}Fe using R₀ = 1.2 fm and compare with the constant value.

Solution:

1. Radius: R = 1.2 fm × 56^{1/3} ≈ 1.2 × 3.83 ≈ 4.60 fm = 4.60×10⁻¹⁵ m
2. Volume: V = (4/3)πR³ ≈ (4/3)π (4.60×10⁻¹⁵)³ ≈ 4.08×10⁻⁴³ m³
3. Mass: m ≈ 56 u = 56 × 1.66×10⁻²⁷ kg ≈ 9.30×10⁻²⁶ kg
4. Density: ρ = m/V ≈ 9.30×10⁻²⁶ / 4.08×10⁻⁴³ ≈ 2.28×10¹⁷ kg/m³

The result matches the expected constant nuclear density.

Atomic Mass, Isotopes, Isobars, and Isotones

Atomic Mass Unit (u)

One atomic mass unit is defined as 1/12 the mass of a ^{12}C atom:

1 u = 1.660539×10⁻²⁷ kg

Using Einstein’s relation, this mass corresponds to an energy:

1 u c² = 931.5 MeV

Isotopes, Isobars, Isotones

Numerical Problem – Mass of an Isotope

The atomic mass of ^{235}_{92}U is 235.0439 u. Find its mass in kilograms and its rest energy in MeV.

Solution:

• Mass in kg: m = 235.0439 u × 1.660539×10⁻²⁷ kg/u ≈ 3.903×10⁻²⁵ kg
• Rest energy: E = mc² = 235.0439 u × 931.5 MeV/u ≈ 2.190×10⁵ MeV ≈ 219 GeV

Einstein’s Mass‑Energy Relation

The equivalence of mass and energy is expressed by:

E = mc²

where c = 2.998×10⁸ m/s.

Mass Defect

When nucleons bind to form a nucleus, the total mass of the separate nucleons is greater than the mass of the resulting nucleus. The difference is the mass defect:

Δm = [Z m_p + (A–Z) m_n] – m_nucleus

The energy released (or required) to break the nucleus apart is:

ΔE = Δm × c² = Δm × 931.5 MeV/u

Numerical Problem – Mass Defect of Helium‑4

Given: m_p = 1.007276 u, m_n = 1.008665 u, m(^{4}_{2}He) = 4.002603 u. Compute the mass defect and binding energy.

Solution:

• Total mass of constituents: 2 m_p + 2 m_n = 2(1.007276) + 2(1.008665) = 4.031882 u
• Mass defect: Δm = 4.031882 u – 4.002603 u = 0.029279 u
• Binding energy: BE = 0.029279 u × 931.5 MeV/u ≈ 27.2 MeV

Binding Energy, Packing Fraction, and BE per Nucleon

Binding Energy (BE)

As derived above, BE = Δm × 931.5 MeV. It quantifies the stability of a nucleus.

Binding Energy per Nucleon

BE/A (usually plotted vs. A) shows a peak around iron (A≈56), indicating greatest stability.

Packing Fraction

Defined as the mass defect per nucleon:

Packing fraction = Δm / A

A negative packing fraction indicates a bound nucleus (mass less than sum of parts).

Numerical Problem – BE per Nucleon and Packing Fraction for Uranium‑235

Atomic mass of ^{235}_{92}U = 235.0439 u. Use m_p = 1.007276 u, m_n = 1.008665 u.

Solution:

1. Total constituent mass: 92 m_p + 143 m_n = 92(1.007276) + 143(1.008665) = 236.956 u
2. Mass defect: Δm = 236.956 u – 235.0439 u = 1.9121 u
3. Binding energy: BE = 1.9121 u × 931.5 MeV/u ≈ 1781 MeV
4. BE per nucleon: BE/A = 1781 MeV / 235 ≈ 7.58 MeV/nucleon
5. Packing fraction: Δm/A = 1.9121 u / 235 ≈ 0.00814 u (or ≈7.58 MeV when multiplied by 931.5).

Creation and Annihilation

Pair Production

A high‑energy photon (gamma ray) can convert into an electron‑positron pair when its energy exceeds the combined rest mass energy of the two particles:

E_γ ≥ 2 m_e c² = 2 × 0.511 MeV = 1.022 MeV

The process occurs in the vicinity of a nucleus to conserve momentum.

Pair Annihilation

When an electron meets a positron, they annihilate, typically producing two gamma‑ray photons each of energy 0.511 MeV (if at rest):

e⁻ + e⁺ → γ + γ

Quarks and Gluons

Nucleons are not fundamental; they consist of three quarks bound by gluons:

• Proton: uud (two up quarks, one down quark)
• Neutron: udd (one up, two down)
• Gluons mediate the strong force, exchanging color charge between quarks.

Numerical Problem – Threshold Photon Energy for Pair Production

Calculate the minimum photon energy required for pair production in the field of a nucleus.

Solution: As shown above, the threshold is 2 m_e c² = 1.022 MeV. Any photon with energy ≥ 1.022 MeV can, in principle, create an electron‑positron pair.

Nuclear Fission and Fusion

Nuclear Fission

Fission is the splitting of a heavy nucleus into lighter fragments, accompanied by the release of neutrons and a large amount of energy.

Typical reaction (Uranium‑235):

^{235}_{92}U + ^1_0n → ^{141}_{56}Ba + ^{92}_{36}Kr + 3 ^1_0n + ≈200 MeV

The emitted neutrons can induce further fissions, leading to a chain reaction.

• Controlled fission**: Neutrons are absorbed by moderators (e.g., water, graphite) to sustain a steady reaction – the principle behind nuclear reactors.
• Uncontrolled fission**: Minimal neutron absorption leads to an exponential increase in fission events – the basis of atomic bombs.