Nuclear Physics Complete Guide: Atomic Nucleus, Radioactive Decay & Binding Energy Explained

House of Physics October 02, 2025
Nuclear Physics: Complete Guide to Atomic Nuclei, Radioactivity, and Nuclear Reactions

Nuclear Physics: Complete Guide to Atomic Nuclei and Radioactivity

Mastering Nuclear Structure, Radioactive Decay, Binding Energy, and Nuclear Reactions | Based on HRK Physics Chapter 54
Nuclear Physics Atomic Nucleus Radioactivity Binding Energy Nuclear Reactions Reading Time: 30 min

📋 Table of Contents

📜 Historical Background

The development of nuclear physics spans the 20th century with key discoveries:

  • J.J. Thomson (1897): Discovered electrons and proposed the "plum pudding" atomic model
  • Ernest Rutherford (1911): Conducted gold foil experiment, discovered atomic nucleus
  • Niels Bohr (1913): Proposed quantum model of the atom
  • James Chadwick (1932): Discovered the neutron
  • Enrico Fermi (1930s): Conducted pioneering work on nuclear reactions
  • Lise Meitner & Otto Hahn (1938): Discovered nuclear fission

These discoveries revolutionized our understanding of matter and led to both nuclear energy and medical applications.

Introduction to Nuclear Physics

🔬 What is Nuclear Physics?

Nuclear physics is the branch of physics that studies atomic nuclei and their constituents, interactions, and properties. The nucleus occupies only about \(10^{-15}\) of the volume of the atom but provides most of its mass as well as the force that holds it together.

There are many similarities between the study of atoms and the study of nuclei. Both systems are governed by the laws of quantum mechanics. Like atoms, nuclei have excited states that can decay to ground state through the emission of photons (gamma rays).

📝 Key Characteristics of Atomic Nuclei

  • Extremely Small: Nuclear radius is about \(10^{-15}\) m compared to atomic radius of \(10^{-10}\) m
  • Extremely Dense: Nuclear density is approximately \(2.3 \times 10^{17}\) kg/m³
  • Composed of Nucleons: Protons and neutrons bound by the strong nuclear force
  • Quantum Systems: Exhibit discrete energy levels and quantum behavior

Rutherford's Gold Foil Experiment

🧪 The Experiment That Changed Everything

In 1911, Ernest Rutherford performed his famous gold foil experiment to investigate the structure of the atom. This experiment overturned J.J. Thomson's "plum pudding" model and revealed the existence of the atomic nucleus.

⚙️ Rutherford Experiment Setup

Alpha Source
↓ α particles
Thin Gold Foil
↙ ↓ ↘ Scattered α particles
Movable Detector

Experimental Setup:

  1. A source emitted alpha particles (helium nuclei)
  2. A thin gold foil served as the target
  3. A movable detector measured scattered particles at different angles

Surprising Results:

  • Most alpha particles passed through undeflected
  • Some were deflected at small angles
  • A very few were deflected at large angles, even backwards

Rutherford famously said: "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."

💡 Rutherford's Conclusion

Rutherford concluded that atoms contain a very small, dense, positively charged central region called the nucleus, where most of the atom's mass is concentrated. This explained why:

  • Most alpha particles passed through (empty space between nuclei)
  • Some were slightly deflected (passing near nuclei)
  • A few were strongly deflected (direct collisions with nuclei)

Nuclear Properties

🔍 Understanding Nuclear Structure

Nuclei are composed of protons and neutrons, collectively called nucleons. While protons and neutrons were once considered elementary particles, we now know they are composed of quarks.

Nuclear Systematic

🧮 Nuclear Notation and Classification

Nuclear Notation

Nuclei are denoted as \( ^A_ZX \), where:

  • \( Z \) = atomic number (number of protons)
  • \( A \) = mass number (total number of nucleons)
  • \( N = A - Z \) = neutron number
  • \( X \) = chemical symbol

Nuclear Classification

  • Isotopes: Same Z, different A (same element, different mass)
  • Isotones: Same N, different Z
  • Isobars: Same A, different Z
  • Isomers: Same A and Z, different energy states

Nuclear Force

⚡ The Strong Nuclear Force

The nucleus contains positively charged protons that should repel each other due to electrostatic forces. Yet nuclei are stable. This stability is due to the strong nuclear force.

Property Electromagnetic Force Strong Nuclear Force
Range Long range (\( \propto 1/r^2 \)) Short range (\( \approx 10^{-15} \) m)
Strength Relatively weak Very strong (100× electromagnetic)
Charge Dependence Acts only on charged particles Acts on all nucleons
Saturation No saturation Shows saturation

Nuclear Radii

📏 Nuclear Size Measurements

Empirical Formula

Nuclear radii follow the empirical formula:

\[ R = R_0 A^{1/3} \]

where:

  • \( R \) = nuclear radius
  • \( R_0 \approx 1.2 \times 10^{-15} \) m = 1.2 fm
  • \( A \) = mass number

Nuclear Density

Assuming spherical nuclei:

\[ V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi R_0^3 A \]
\[ \rho = \frac{m}{V} = \frac{A m_p}{\frac{4}{3} \pi R_0^3 A} \]
\[ = \frac{3 m_p}{4 \pi R_0^3} \]
\[ \approx 2.3 \times 10^{17} \, \text{kg/m}^3 \]

This constant density shows that nuclear matter is incompressible.

Nuclear Mass and Binding Energies

⚖️ Mass Defect and Binding Energy

The mass of a nucleus is always less than the sum of the masses of its constituent protons and neutrons. This mass difference is called the mass defect.

\[ \Delta m = Z m_p + (A - Z) m_n - m_{\text{nucleus}} \]

According to Einstein's mass-energy equivalence, this mass defect corresponds to the binding energy that holds the nucleus together:

\[ E_b = \Delta m c^2 \]

Nuclear Spin and Magnetism

🔄 Nuclear Angular Momentum

Like electrons, nucleons have intrinsic angular momentum (spin). The total nuclear spin is the vector sum of the spins of all nucleons.

  • Proton spin: \( \frac{1}{2} \hbar \)
  • Neutron spin: \( \frac{1}{2} \hbar \)
  • Total nuclear spin is quantized: \( I = 0, \frac{1}{2}, 1, \frac{3}{2}, \ldots \)

Nuclear spin is responsible for nuclear magnetic moments, which are about 1000 times smaller than electron magnetic moments.

Binding Energy

🔗 What is Binding Energy?

Binding energy is the energy required to completely separate a nucleus into its constituent protons and neutrons. It represents the work that must be done against the attractive nuclear force to break the nucleus apart.

Binding Energy per Nucleon

🧮 Calculating Binding Energy

Step 1: Mass Defect

For a nucleus \( ^A_ZX \), the mass defect is:

\[ \Delta m = [Z m_p + (A - Z) m_n] - m_{\text{nucleus}} \]

Step 2: Binding Energy

\[ E_b = \Delta m c^2 \]
\[ = \{ [Z m_p + (A - Z) m_n] - m_{\text{nucleus}} \} c^2 \]

Step 3: Binding Energy per Nucleon

\[ \text{Binding Energy per Nucleon} = \frac{E_b}{A} \]

📈 Binding Energy per Nucleon Curve

[Graph: Binding energy per nucleon vs. mass number showing peak at Fe-56]

The binding energy per nucleon curve shows that:

  • Light nuclei (A < 20) have lower binding energy per nucleon
  • Medium mass nuclei (A ≈ 50-60) have maximum binding energy per nucleon
  • Heavy nuclei (A > 60) have decreasing binding energy per nucleon

Iron-56 (\( ^{56}_{26}Fe \)) has the highest binding energy per nucleon (≈ 8.8 MeV), making it the most stable nucleus.

Sample Problems

Sample Problem 1: Binding Energy of Helium-4

Calculate the binding energy and binding energy per nucleon for helium-4 (\( ^4_2He \)).

Given: \( m_p = 1.007276 \, u \), \( m_n = 1.008665 \, u \), \( m(^4_2He) = 4.001506 \, u \), \( 1 \, u = 931.5 \, \text{MeV}/c^2 \)

Mass defect:
\[ \Delta m = [Z m_p + (A - Z) m_n] - m_{\text{nucleus}} \]
\[ = [2(1.007276) + 2(1.008665)] - 4.001506 \]
\[ = [2.014552 + 2.017330] - 4.001506 \]
\[ = 4.031882 - 4.001506 \]
\[ = 0.030376 \, u \]
Binding energy:
\[ E_b = \Delta m \times 931.5 \, \text{MeV}/u \]
\[ = 0.030376 \times 931.5 \]
\[ = 28.30 \, \text{MeV} \]
Binding energy per nucleon:
\[ \frac{E_b}{A} = \frac{28.30}{4} \]
\[ = 7.075 \, \text{MeV/nucleon} \]
Sample Problem 2: Binding Energy Comparison

Compare the binding energy per nucleon for deuterium (\( ^2_1H \)) and helium-4 (\( ^4_2He \)).

Given: \( m(^2_1H) = 2.014102 \, u \), other masses as before

For deuterium (\( ^2_1H \)):
\[ \Delta m = [1(1.007276) + 1(1.008665)] - 2.014102 \]
\[ = [1.007276 + 1.008665] - 2.014102 \]
\[ = 2.015941 - 2.014102 \]
\[ = 0.001839 \, u \]
Binding energy:
\[ E_b = 0.001839 \times 931.5 \]
\[ = 1.712 \, \text{MeV} \]
Binding energy per nucleon:
\[ \frac{E_b}{A} = \frac{1.712}{2} \]
\[ = 0.856 \, \text{MeV/nucleon} \]
For helium-4 (from previous problem):
\[ \frac{E_b}{A} = 7.075 \, \text{MeV/nucleon} \]
Comparison:
Helium-4 is much more tightly bound than deuterium
This explains why fusion of deuterium to form helium releases energy

Radioactive Decay

☢️ What is Radioactivity?

Radioactivity is the spontaneous disintegration of unstable atomic nuclei through the emission of particles or electromagnetic radiation. This process transforms the original nucleus (parent) into a different nucleus (daughter).

Alpha Decay

α-Decay Process

General Alpha Decay Equation

\[ ^A_ZX \rightarrow ^{A-4}_{Z-2}Y + ^4_2He + Q \]

where Q is the disintegration energy released in the decay.

Energy Considerations

The Q-value for alpha decay is given by:

\[ Q = [m_X - (m_Y + m_\alpha)]c^2 \]

For alpha decay to occur spontaneously, Q must be positive.

Example: Uranium-238 Decay

\[ ^{238}_{92}U \rightarrow ^{234}_{90}Th + ^4_2He + 4.27 \, \text{MeV} \]

💡 Alpha Decay Characteristics

  • Occurs primarily in heavy nuclei (Z > 82)
  • Alpha particles are emitted with discrete energies
  • Alpha particles have low penetrating power (stopped by paper or skin)
  • Highly ionizing due to +2 charge and relatively large mass

Beta Decay

β-Decay Processes

Beta Minus (β⁻) Decay

\[ ^A_ZX \rightarrow ^A_{Z+1}Y + e^- + \bar{\nu}_e + Q \]

A neutron transforms into a proton, emitting an electron and an antineutrino.

Beta Plus (β⁺) Decay

\[ ^A_ZX \rightarrow ^A_{Z-1}Y + e^+ + \nu_e + Q \]

A proton transforms into a neutron, emitting a positron and a neutrino.

Electron Capture (EC)

\[ ^A_ZX + e^- \rightarrow ^A_{Z-1}Y + \nu_e + Q \]

An orbital electron is captured by the nucleus, converting a proton to a neutron.

Type Process Change in Z Change in N
β⁻ Decay n → p + e⁻ + ṽ +1 -1
β⁺ Decay p → n + e⁺ + ν -1 +1
Electron Capture p + e⁻ → n + ν -1 +1

Gamma Decay

γ-Ray Emission

Gamma decay involves the emission of high-energy photons (gamma rays) from an excited nuclear state. Unlike alpha and beta decay, gamma decay does not change the atomic number or mass number of the nucleus.

\[ ^A_ZX^* \rightarrow ^A_ZX + \gamma \]

where \( ^A_ZX^* \) represents an excited nuclear state.

💡 Gamma Decay Characteristics

  • Does not change Z or A of the nucleus
  • Involves transition between nuclear energy levels
  • Gamma rays have high penetrating power
  • Often follows alpha or beta decay when daughter nucleus is left in excited state

Laws of Radioactivity and Half-Life

📊 Radioactive Decay Law

Radioactive decay is a random process that follows an exponential decay law. The rate of decay at any time is proportional to the number of radioactive atoms present at that time.

🧮 Mathematical Description

Differential Form

\[ \frac{dN}{dt} = -\lambda N \]

where:

  • \( N \) = number of radioactive atoms
  • \( \lambda \) = decay constant
  • Negative sign indicates decrease with time

Integrated Form

\[ N = N_0 e^{-\lambda t} \]

where \( N_0 \) is the initial number of atoms at t = 0.

Half-Life

Half-life (\( T_{1/2} \)) is the time required for half of the radioactive atoms to decay:

\[ \frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \]
\[ \frac{1}{2} = e^{-\lambda T_{1/2}} \]
\[ \ln 2 = \lambda T_{1/2} \]
\[ T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \]

Activity

Activity (A) is the rate of decay:

\[ A = -\frac{dN}{dt} = \lambda N \]
\[ A = A_0 e^{-\lambda t} \]

where \( A_0 = \lambda N_0 \) is the initial activity.

Sample Problems

Sample Problem 3: Half-Life Calculation

A radioactive sample has an initial activity of 800 decays per minute. After 30 minutes, the activity is 200 decays per minute. Calculate the half-life of the sample.

Using the decay law:
\[ A = A_0 e^{-\lambda t} \]
\[ 200 = 800 e^{-\lambda (30)} \]
\[ \frac{200}{800} = e^{-30\lambda} \]
\[ \frac{1}{4} = e^{-30\lambda} \]
\[ \ln\left(\frac{1}{4}\right) = -30\lambda \]
\[ -\ln 4 = -30\lambda \]
\[ \lambda = \frac{\ln 4}{30} \]
\[ = \frac{1.386}{30} \]
\[ = 0.0462 \, \text{min}^{-1} \]
Half-life:
\[ T_{1/2} = \frac{\ln 2}{\lambda} \]
\[ = \frac{0.693}{0.0462} \]
\[ = 15 \, \text{minutes} \]
Sample Problem 4: Radioactive Dating

A piece of wood from an ancient artifact has a carbon-14 activity that is 25% of the activity found in living wood. Calculate the age of the artifact. (Half-life of carbon-14 is 5730 years)

Given:
\[ \frac{A}{A_0} = 0.25 \]
\[ T_{1/2} = 5730 \, \text{years} \]
Decay constant:
\[ \lambda = \frac{\ln 2}{T_{1/2}} \]
\[ = \frac{0.693}{5730} \]
\[ = 1.21 \times 10^{-4} \, \text{year}^{-1} \]
Using decay law:
\[ A = A_0 e^{-\lambda t} \]
\[ 0.25 = e^{-\lambda t} \]
\[ \ln(0.25) = -\lambda t \]
\[ -1.386 = -(1.21 \times 10^{-4}) t \]
\[ t = \frac{1.386}{1.21 \times 10^{-4}} \]
\[ = 11455 \, \text{years} \]

Mean Life of Radioactive Elements

⏱️ Mean Life Concept

The mean life (τ) of a radioactive substance is the average lifetime of all the radioactive atoms before they decay. It is related to the decay constant and half-life.

🧮 Mean Life Calculation

Definition

\[ \tau = \frac{1}{\lambda} \]

Relation to Half-Life

\[ \tau = \frac{1}{\lambda} = \frac{T_{1/2}}{\ln 2} \]
\[ = \frac{T_{1/2}}{0.693} \approx 1.443 T_{1/2} \]

Physical Meaning

The mean life is the time it would take for all radioactive atoms to decay if they continued to decay at the initial rate.

Units for Measuring Ionizing Radiation

📏 Radiation Measurement Units

Several units are used to measure different aspects of ionizing radiation. Understanding these units is essential for radiation protection and applications.

Unit Quantity Measured Definition
Becquerel (Bq) Activity 1 decay per second
Curie (Ci) Activity 3.7 × 10¹⁰ decays per second
Gray (Gy) Absorbed Dose 1 joule per kilogram
Rad Absorbed Dose 0.01 Gy
Sievert (Sv) Dose Equivalent Gray × quality factor
Rem Dose Equivalent 0.01 Sv

Radioactive Dating

⏳ Dating with Radioactivity

Radioactive dating uses the predictable decay of radioactive isotopes to determine the age of materials. Different isotopes are used for dating different types of materials and time scales.

🌿 Carbon-14 Dating

Used for dating organic materials up to about 50,000 years old. Based on the decay of carbon-14 (half-life: 5730 years) that is continuously produced in the atmosphere by cosmic rays.

🪨 Potassium-Argon Dating

Used for dating rocks and geological formations. Based on the decay of potassium-40 to argon-40 (half-life: 1.25 billion years). Useful for dating materials millions to billions of years old.

💎 Uranium-Lead Dating

Used for dating the oldest rocks on Earth and meteorites. Based on the decay of uranium-238 to lead-206 (half-life: 4.47 billion years) and uranium-235 to lead-207 (half-life: 704 million years).

Q-Energy in Nuclear Reactions

⚡ Q-Value of Nuclear Reactions

The Q-value of a nuclear reaction is the energy released or absorbed during the reaction. It is equal to the difference between the total mass-energy of the reactants and products.

🧮 Q-Value Calculation

General Nuclear Reaction

\[ a + X \rightarrow Y + b + Q \]

Q-Value Formula

\[ Q = [m_a + m_X - (m_Y + m_b)]c^2 \]
\[ = (\text{total initial mass} - \text{total final mass})c^2 \]

Interpretation

  • Q > 0: Exothermic reaction (releases energy)
  • Q < 0: Endothermic reaction (absorbs energy)
  • Q = 0: Elastic reaction (no energy change)
Sample Problem 5: Q-Value Calculation

Calculate the Q-value for the nuclear reaction: \( ^2_1H + ^3_1H \rightarrow ^4_2He + ^1_0n \)

Given: \( m(^2_1H) = 2.014102 \, u \), \( m(^3_1H) = 3.016049 \, u \), \( m(^4_2He) = 4.001506 \, u \), \( m_n = 1.008665 \, u \), \( 1 \, u = 931.5 \, \text{MeV}/c^2 \)

Mass defect:
\[ \Delta m = [m(^2_1H) + m(^3_1H)] - [m(^4_2He) + m_n] \]
\[ = [2.014102 + 3.016049] - [4.001506 + 1.008665] \]
\[ = 5.030151 - 5.010171 \]
\[ = 0.019980 \, u \]
Q-value:
\[ Q = \Delta m \times 931.5 \, \text{MeV}/u \]
\[ = 0.019980 \times 931.5 \]
\[ = 18.61 \, \text{MeV} \]
This is the D-T fusion reaction, which releases significant energy and is being developed for fusion power.

Frequently Asked Questions

❓ Why are some nuclei radioactive while others are stable?

The stability of a nucleus depends on the balance between the attractive strong nuclear force and the repulsive electromagnetic force between protons. Factors that determine nuclear stability include:

  • Neutron-to-Proton Ratio: Light stable nuclei have N/Z ≈ 1, while heavier stable nuclei need more neutrons (N/Z up to 1.5 for very heavy elements)
  • Magic Numbers: Nuclei with certain numbers of protons or neutrons (2, 8, 20, 28, 50, 82, 126) are particularly stable
  • Even vs. Odd Nucleon Numbers: Even-even nuclei (even Z, even N) are generally more stable than odd-odd nuclei
  • Binding Energy per Nucleon: Nuclei with higher binding energy per nucleon are more stable

Nuclei that don't meet these stability criteria will undergo radioactive decay to become more stable.

❓ What is the difference between nuclear fission and nuclear fusion?

Nuclear fission and fusion are opposite processes that both release large amounts of energy:

Aspect Nuclear Fission Nuclear Fusion
Process Heavy nucleus splits into lighter nuclei Light nuclei combine to form heavier nucleus
Typical Elements Uranium, Plutonium Hydrogen, Helium
Energy per Reaction ~200 MeV ~17.6 MeV (D-T fusion)
Energy per Nucleon ~0.9 MeV/nucleon ~3.5 MeV/nucleon
Current Applications Nuclear power plants, atomic bombs Hydrogen bombs, experimental reactors
Challenges Radioactive waste, safety Extreme temperature and pressure requirements

Both processes release energy because the binding energy per nucleon is higher for medium-mass nuclei than for very heavy or very light nuclei.

❓ How does radiation affect living organisms?

Ionizing radiation affects living organisms through several mechanisms:

  • Direct Damage: Radiation can directly break chemical bonds in DNA and other important molecules
  • Indirect Damage: Radiation creates free radicals in water that can then damage biological molecules
  • Cellular Effects: Can cause cell death, mutations, or transformation to cancerous cells

The effects depend on:

  • Dose: Higher doses cause more severe effects
  • Dose Rate: The same total dose spread over time is less harmful
  • Type of Radiation: Alpha particles are more damaging internally, while gamma rays penetrate deeper
  • Tissue Sensitivity: Rapidly dividing cells (bone marrow, gastrointestinal tract) are more sensitive

At low doses, the body can repair most damage. At high doses, radiation sickness or death can occur.

📚 Master Nuclear Physics

Understanding nuclear physics is essential for many modern technologies, from nuclear power and medical imaging to dating archaeological artifacts. Continue your exploration of the fascinating world of atomic nuclei and their applications.

Read More: Physics HRK Notes on Modern Physics

© House of Physics | HRK Physics Chapter 54: Nuclear Physics

Based on Halliday, Resnick, and Krane's "Physics" with additional insights from university physics curriculum

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