Nuclei

NUCLEI

INTRODUCTION AND ATOMIC STRUCTURE

The nucleus is the dense, positively charged core at the centre of every atom. Rutherford's alpha scattering experiments established that nuclear radius is approximately 10⁴ times smaller than atomic radius, making the nucleus occupy only about 10⁻¹² of atomic volume. Despite occupying negligible volume, the nucleus contains over 99.9% of the atom's mass. If an atom were enlarged to classroom size, the nucleus would be the size of a pinhead. This chapter explores nuclear composition, binding energy, and nuclear phenomena including radioactivity, fission, and fusion.

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ATOMIC MASS UNIT AND ISOTOPE DEFINITION

**Atomic Mass Unit (u)** is defined as 1/12th of the mass of the carbon-12 (¹²C) atom. This provides a convenient scale for expressing extremely small atomic masses:

1 u = (1.992647 × 10⁻²⁶ kg) / 12 = **1.660539 × 10⁻²⁷ kg**

Atomic masses of most elements are approximately integral multiples of hydrogen mass, though exceptions exist. For example, natural chlorine has atomic mass 35.46 u, which is a weighted average of two isotopes:

³⁴Cl: mass = 34.98 u, abundance = 75.4%

³⁶Cl: mass = 36.98 u, abundance = 24.6%

Average mass = (75.4 × 34.98 + 24.6 × 36.98)/100 = 35.47 u

**Isotopes** are atoms of the same element with identical atomic number Z but different mass numbers A. They possess identical chemical properties (same electronic structure) but differ in mass due to different neutron numbers. Chlorine naturally consists of isotopic mixtures; hydrogen has three isotopes with masses 1.0078 u, 2.0141 u (deuterium), and 3.0160 u (tritium).

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COMPOSITION OF NUCLEUS

Proton: The Hydrogen Nucleus

The lightest hydrogen isotope (¹H, protium) nucleus is called a **proton**. It is a stable particle carrying one unit of positive fundamental charge (e = 1.602 × 10⁻¹⁹ C):

**Mass of proton**: mₚ = **1.00727 u = 1.67262 × 10⁻²⁷ kg**

The proton mass equals the hydrogen atom mass (1.00783 u) minus electron mass (0.00055 u).

Discovery of the Neutron

Deuterium (²₁H) and tritium (³₁H) are hydrogen isotopes containing one proton each but with mass ratios of 1:2:3 relative to protium. This discrepancy in mass indicated the presence of additional neutral matter in deuterium and tritium nuclei. In 1932, **James Chadwick** discovered the neutron by bombarding beryllium nuclei with alpha particles. The resulting neutral radiation could eject protons from light nuclei. Conservation of energy and momentum analyses ruled out photons and identified a new neutral particle—the neutron.

**Mass of neutron**: mₙ = **1.00866 u = 1.6749 × 10⁻²⁷ kg**

Unlike protons, free neutrons are unstable and decay into a proton, electron, and antineutrino (mean life ≈ 1000 s). However, neutrons are stable inside nuclei. Chadwick received the 1935 Nobel Prize in Physics for this discovery.

Nuclear Nomenclature and Composition Terms

**Atomic number (Z)**: Number of protons in the nucleus

**Neutron number (N)**: Number of neutrons in the nucleus

**Mass number (A)**: Total number of nucleons (protons + neutrons): **A = Z + N**

**Nucleon**: Generic term for either proton or neutron

Nuclear species are represented as **ᴬ_Z X**, where X is the chemical symbol. For example, **¹⁹⁷₇₉Au** (gold nucleus) contains 197 nucleons: 79 protons and 118 neutrons.

Related Nuclear Classifications

**Isobars**: Nuclides with same mass number A but different atomic numbers (e.g., ³₁H and ³₂He)

**Isotones**: Nuclides with same neutron number N but different atomic numbers (e.g., ¹⁹⁸₈₀Hg and ¹⁹⁷₇₉Au)

The isotopic composition explains why elements show fractional atomic masses. Chlorine's average mass of 35.46 u results from natural mixtures of ³⁵Cl and ³⁷Cl isotopes.

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SIZE OF THE NUCLEUS

Experimental Determination via Alpha Scattering

Rutherford's alpha scattering experiment, conducted by Geiger and Marsden on thin gold foils, revealed that alpha particles (⁴₂He nuclei, charge +2e, kinetic energy ~5.5 MeV) approach gold nuclei to distances as close as 4.0 × 10⁻¹⁴ m. The scattering followed Rutherford's Coulomb-only predictions at this energy scale, establishing that actual nuclear radius must be less than this distance of closest approach.

Using electron bombardment (which penetrates better than alpha particles), precise measurements of various nuclei established the empirical **nuclear radius formula**:

**R = R₀A^(1/3)**

where:

**R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm** (1 femtometre = 10⁻¹⁵ m)

**A** = mass number of nucleus

Crucial Insight: Constant Nuclear Density

Since R ∝ A^(1/3), nuclear volume V = (4/3)πR³ ∝ A. Therefore, nuclear density = mass/volume ∝ A/A = constant. All nuclei have approximately the same density, independent of mass number A. This remarkable property shows nuclei behave like incompressible liquid drops.

**Nuclear density** ≈ **2.3 × 10¹⁷ kg/m³**

This is vastly larger than ordinary matter density (water ≈ 10³ kg/m³) because atoms are mostly empty space. Matter in neutron stars approximates nuclear density, demonstrating extreme compression.

Worked Example: Nuclear Density Calculation

**Example**: Iron-56 nucleus (mFe = 55.85 u, A = 56). Calculate nuclear density.

**Solution**:

Convert mass: 55.85 u × 1.66054 × 10⁻²⁷ kg/u = 9.27 × 10⁻²⁶ kg

Nuclear volume: V = (4/3)π(R₀A^(1/3))³ = (4/3)π(1.2 × 10⁻¹⁵)³ × 56

V = (4/3)π × 1.728 × 10⁻⁴⁵ × 56 = 4.04 × 10⁻⁴⁴ m³

Density = (9.27 × 10⁻²⁶) / (4.04 × 10⁻⁴⁴) = **2.29 × 10¹⁷ kg/m³**

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MASS-ENERGY AND BINDING ENERGY

Einstein's Mass-Energy Equivalence

**Einstein's mass-energy relation**: **E = mc²**

Einstein's theory of special relativity unified mass and energy. Before relativity, mass and energy were conserved separately; Einstein showed mass itself is energy. Any mass m has energy equivalent mc², where c = 3 × 10⁸ m/s is the speed of light.

**Worked Example**: Energy equivalent of 1 g of substance.

E = 10⁻³ kg × (3 × 10⁸ m/s)² = 10⁻³ × 9 × 10¹⁶ = **9 × 10¹³ J**

This enormous energy release explains why even small mass defects in nuclear reactions release significant energy. Conservation of energy in nuclear reactions must account for mass-energy transformation.

Mass Defect and Nuclear Binding Energy

Precise mass spectrometry reveals that **nuclear mass is always less than the sum of masses of constituent nucleons**. For example, ¹⁶₈O nucleus:

Mass of 8 neutrons: 8 × 1.00866 u = 8.06928 u

Mass of 8 protons: 8 × 1.00727 u = 8.05816 u

Total constituent mass: 16.12744 u

Actual ¹⁶O atomic mass: 15.99493 u

Less electron mass (8 × 0.00055 u): 15.99053 u (nuclear mass)

**Mass defect**: ΔM = 16.12744 − 15.99053 = 0.13691 u

**Mass defect formula**:

**ΔM = [Zm_p + (A−Z)m_n + Zm_e] − M_atom**

Or equivalently: **ΔM = [Zm_p + (A−Z)m_n] − M_nucleus**

where M_atom is the measured atomic mass and M_nucleus is the actual nuclear mass.

Binding Energy Definition

The mass defect represents the mass converted to binding energy when nucleons combine to form the nucleus. The **binding energy** is the energy released when separated nucleons combine into a nucleus, or equivalently, the energy required to disassemble the nucleus into free nucleons:

**E_b = ΔM × c²**

**Energy conversion factor**: 1 u = 931.5 MeV/c²

**Worked Example**: Binding energy of ¹⁶₈O nucleus.

ΔM = 0.13691 u = 0.13691 × 931.5 MeV/c² = **127.5 MeV**

Therefore, 127.5 MeV must be supplied to completely disassemble the oxygen nucleus into 8 protons and 8 neutrons. Conversely, 127.5 MeV is released when these nucleons combine to form ¹⁶O.

Binding Energy Per Nucleon

A more meaningful measure of nuclear stability is **binding energy per nucleon**:

**E_bn = E_b / A**

This represents the average energy per nucleon needed for complete nuclear dissociation. Higher binding energy per nucleon indicates greater stability and stronger binding.

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BINDING ENERGY CURVE AND NUCLEAR STABILITY

Shape and Key Features of the Binding Energy Curve

The plot of binding energy per nucleon (E_bn) versus mass number (A) for known nuclei shows distinctive features:

**For intermediate-mass nuclei (30 < A < 170)**:

E_bn is approximately constant at ~8.5 MeV/nucleon

Maximum occurs at A ≈ 56 (iron-56): E_bn ≈ 8.75 MeV/nucleon

At A = 238 (uranium): E_bn ≈ 7.6 MeV/nucleon

This plateau region includes the most stable nuclei

**For light nuclei (A < 30)**:

E_bn is significantly lower (~7 MeV/nucleon for helium, ~8 MeV for carbon)

These nuclei are less tightly bound

Binding increases rapidly with A in this region

**For heavy nuclei (A > 170)**:

E_bn decreases noticeably as A increases

²³⁸U: E_bn ≈ 7.6 MeV/nucleon

Heavier nuclei are progressively less stable per nucleon

Nuclear Force and Short-Range Nature

The approximately constant E_bn for 30 < A < 170 reveals crucial properties of the **nuclear force**:

**Attractive and immensely strong**: E_bn ~ 8 MeV exceeds Coulomb energy per nucleon by orders of magnitude

**Short-range (~2 fm)**: A nucleon influences only nearby neighbors within the nuclear force range

**Saturation property**: Each nucleon can bind with maximum ~p neighbors. Adding more nucleons doesn't increase binding per nucleon because interior nucleons cannot form additional bonds beyond saturation

In large nuclei, most nucleons reside in the interior beyond the nuclear force range from the surface, making binding energy per nucleon independent of A (for 30 < A < 170).

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NUCLEAR FORCE CHARACTERISTICS

Comparison with Other Fundamental Forces

1. **Strength**: Nuclear force >> Coulomb force >> Gravitational force

Must overcome Coulomb repulsion between protons

Operates only at nuclear scales (~fm)

2. **Range**: Short-range force

Effective only within ~2 fm

Negligible beyond 3 fm

Contrasts with Coulomb (infinite range) and gravity (infinite range)

3. **Force vs. Distance Relationship**:

For r < r₀ (≈0.8 fm): **Strongly repulsive** (prevents nucleon collapse)

At r = r₀: **Potential energy minimum** (most favorable separation)

For r > r₀ (up to ~3 fm): **Attractive force** (increases magnitude as r decreases)

Beyond ~3 fm: **Negligible**

This short-range behavior creates the characteristic potential well shown in potential energy vs. separation graphs.

Charge Independence of Nuclear Force

The nuclear force operates between:

**Neutron-neutron (n-n)**: Same attractive strength

**Proton-proton (p-p)**: Same attractive strength (minus Coulomb repulsion)

**Neutron-proton (n-p)**: Same attractive strength (slightly stronger than n-n or p-p)

The force is essentially **charge-independent**, depending only on nuclear separation, not on particle charge. This explains why stable nuclei contain roughly equal numbers of protons and neutrons (symmetric nuclei) for A < 40, though heavy nuclei require more neutrons to provide additional nuclear attraction to overcome increasing Coulomb repulsion.

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IMPLICATIONS OF BINDING ENERGY CURVE

Nuclear Fission (Qualitative)

A heavy nucleus (A ≈ 240) has lower binding energy per nucleon (~7.6 MeV) compared to mid-mass nuclei (A ≈ 120, ~8.6 MeV/nucleon).

When a nucleus at A = 240 spontaneously or forcibly splits into two A ≈ 120 fragments:

Initial total binding energy: 240 × 7.6 = 1824 MeV

Final total binding energy: 2 × 120 × 8.6 = 2064 MeV

**Energy released**: 2064 − 1824 = 240 MeV

This energy release makes fission energetically favorable. The energy appears as kinetic energy of fission fragments and neutrons, forming the basis of nuclear power and weapons.

Nuclear Fusion (Qualitative)

Two light nuclei (A ≤ 10) with low binding energy per nucleon (~7 MeV/nucleon each) combine to form heavier nuclei with higher binding energy per nucleon (~8.5 MeV/nucleon).

Example: Two deuterium nuclei (A = 2) fuse to helium-4 (A = 4):

Initial binding energy: 2 × 2 × 7.0 = 28 MeV

Final binding energy: 4 × 8.8 = 35.2 MeV

**Energy released**: 35.2 − 28 = 7.2 MeV

The final system is more tightly bound, releasing energy. Fusion powers the sun and stars, and is the basis for controlled thermonuclear reactors.

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WORKED NUMERICAL PROBLEMS

**Problem 1**: Calculate the binding energy per nucleon for ⁴₂He (mass = 4.00260 u).

**Solution**:

Z = 2 (protons), N = 2 (neutrons), A = 4

Sum of constituent masses: 2(1.00727) + 2(1.00866) = 4.03186 u

Mass defect: ΔM = 4.03186 − 4.00260 = 0.02926 u

Binding energy: E_b = 0.02926 × 931.5 = 27.25 MeV

**Binding energy per nucleon**: E_bn = 27.25 / 4 = **6.81 MeV/nucleon**

**Problem 2**: For ⁵⁶₂₆Fe (mass = 55.9349 u), find total binding energy.

**Solution**:

Z = 26, N = 30, A = 56

Constituent masses: 26(1.00727) + 30(1.00866) = 56.46242 u

Mass defect: ΔM = 56.46242 − 55.9349 = 0.52752 u

**Binding energy**: E_b = 0.52752 × 931.5 = **491.3 MeV**

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EXAM-IMPORTANT POINTS SUMMARY

1. **Nuclear size formula** R = R₀A^(1/3) with R₀ = 1.2 fm shows nuclei have constant density

2. **Mass defect always exists**: actual nuclear mass < sum of nucleon masses

3. **Binding energy = ΔM × c²** is the energy holding nucleus together

4. **Binding energy per nucleon** (E_bn) is the key stability indicator

5. **E_bn curve shape** explains both fission (heavy nucleus) and fusion (light nuclei) as energy sources

6. **Nuclear force** is short-range (~2 fm), attractive, immensely strong, and charge-independent

7. **Saturation of nuclear force** explains constant E_bn for 30 < A < 170

8. Conversion factor: **1 u = 931.5 MeV/c²** is essential for all binding energy calculations

9. Iron-56 (A = 56) is the most stable nucleus with maximum binding energy per nucleon (~8.75 MeV)