Unit 5 · Chapter 02

02Atomic and Nuclear Physics

Analyze atomic models, electron energy levels, nuclear structure, radioactive decay series, half-life calculations, and nuclear fission and fusion.

Atomic and nuclear physics explains the structure of matter, the source of stellar energy, and the principles behind nuclear power and medical imaging.

How does the structure of the atom — and the nucleus at its core — determine the energy emitted as light, the stability of matter, and the transformations we call radioactive decay?

Lesson Overview

This lesson traces our understanding of the atom from Rutherford's nuclear model — where a tiny, dense, positively charged nucleus is surrounded by orbiting electrons — to Bohr's quantised model of hydrogen, which explains why atoms emit only specific wavelengths of light. We then descend into the nucleus itself: protons and neutrons bound together by the strong nuclear force, held in place by a binding energy equal to the mass defect times c². Finally we study radioactive decay — alpha, beta, and gamma — and the statistical laws (half-life, decay constant) that govern how unstable nuclei transform over time, culminating in the enormous energy releases of fission and fusion.

Radioactive Decay Types

Summary of the four decay modes

TypeParticle emittedCharge change (Z)Mass change (A)Penetrating power
Alpha (α)₂⁴He nucleus−2−4Low — stopped by paper
Beta-minus (β⁻)e⁻ + antineutrino+10Medium — stopped by aluminium
Beta-plus (β⁺)e⁺ + neutrino−10Medium — stopped by aluminium
Gamma (γ)High-energy photon00High — reduced by lead / concrete

Bohr Model — Hydrogen Energy Levels

E_n = −13.6 / n² eV  ·  r_n = n² × 0.0529 nm

n = principal quantum number

Level nEnergy (eV)Radius (nm)Series (transitions to this level)
n = 1−13.6000.0529Lyman (UV) — all higher levels → n=1
n = 2−3.4000.2116Balmer (visible/UV) — all higher → n=2
n = 3−1.5110.4761Paschen (IR) — all higher → n=3
n = 4−0.8500.8464Brackett (IR)
n = 5−0.5441.3225Pfund (IR)
n = 6−0.3781.9044Humphreys (far IR)

Spectral line energy: ΔE = hf = E_i − E_f  (always positive; photon is emitted when electron drops to lower level)

Key Equations

Bohr energy levels

Eₙ = −13.6 / n² (eV)

Bohr orbital radius

rₙ = n² × a₀, a₀ = 0.0529 nm

Photon energy / frequency

ΔE = hf = Eᵢ − E_f

Mass number

A = Z + N

Binding energy

E_b = Δm · c²

Half-life decay

N = N₀ (½)^(t / t½)

Exponential decay law

N = N₀ e^(−λt)

Decay constant

t½ = ln 2 / λ ≈ 0.693 / λ

Worked Examples

Example 1

A hydrogen electron transitions from n = 3 to n = 1. Find the energy of the emitted photon, its frequency, and its wavelength. Identify the spectral series.

Step 1 — Energy of each level:

E₃ = −13.6 / 3² = −13.6 / 9 = −1.511 eV

E₁ = −13.6 / 1² = −13.600 eV

Step 2 — Photon energy (energy released):

ΔE = E₁ − E₃ = −13.600 − (−1.511) = −12.089 eV → |ΔE| = 12.089 eV

Convert: 12.089 eV × 1.602×10⁻¹⁹ J/eV = 1.937×10⁻¹⁸ J

Step 3 — Frequency:

f = ΔE / h = 1.937×10⁻¹⁸ / 6.626×10⁻³⁴ = 2.924×10¹⁵ Hz

Step 4 — Wavelength:

λ = c / f = 3.00×10⁸ / 2.924×10¹⁵ = 1.026×10⁻⁷ m = 102.6 nm

Step 5 — Series: transition ends at n = 1 → Lyman series (ultraviolet)

Answer:ΔE = 12.09 eV, f = 2.92×10¹⁵ Hz, λ = 102.6 nm — Lyman series (UV)
Example 2

Calculate the nuclear binding energy of helium-4 (₂⁴He). Given: mₚ = 1.00728 u, mₙ = 1.00866 u, actual mass of ⁴He = 4.00260 u. (1 u = 931.5 MeV/c²)

Step 1 — Identify constituents: Z = 2 protons, N = 2 neutrons

Step 2 — Total constituent mass:

m_constituents = 2(1.00728) + 2(1.00866)

= 2.01456 + 2.01732 = 4.03188 u

Step 3 — Mass defect:

Δm = m_constituents − m_actual = 4.03188 − 4.00260 = 0.02928 u

Step 4 — Binding energy:

E_b = Δm × 931.5 MeV/u = 0.02928 × 931.5 = 27.27 MeV

Answer:Δm = 0.02928 u → E_b = 27.27 MeV
Example 3

Write the alpha decay equation for radium-226 (₈₈²²⁶Ra) and verify conservation of mass number A and atomic number Z.

Alpha particle: ₂⁴He (A = 4, Z = 2)

Daughter nucleus: A_d = 226 − 4 = 222, Z_d = 88 − 2 = 86

Z = 86 is radon (Rn)

Decay equation:

₈₈²²⁶Ra → ₈₆²²²Rn + ₂⁴He

Check A: 226 = 222 + 4 ✓

Check Z: 88 = 86 + 2 ✓

Answer:₈₈²²⁶Ra → ₈₆²²²Rn + ₂⁴He (A and Z both conserved)
Example 4

Iodine-131 has a half-life of 8 days. A sample initially contains N₀ = 1.00×10¹² atoms. How many atoms remain after 24 days?

Step 1 — Number of half-lives elapsed:

n = t / t½ = 24 / 8 = 3 half-lives

Step 2 — Apply half-life formula:

N = N₀ × (½)³ = 1.00×10¹² × (1/8)

N = 1.25×10¹¹ atoms

Answer:N = 1.25×10¹¹ atoms remain after 24 days
Example 5

Carbon-14 has a half-life of 5730 years. (a) Find the decay constant λ. (b) Estimate the activity of a 1.00 g sample of pure ¹⁴C in decays per second.

Part (a) — Decay constant:

λ = ln 2 / t½ = 0.6931 / 5730 yr = 1.209×10⁻⁴ yr⁻¹

Part (b) — Number of atoms in 1.00 g of ¹⁴C (molar mass ≈ 14 g/mol):

N = (1.00 / 14) × 6.022×10²³ = 4.301×10²² atoms

Activity in yr⁻¹:

A = λN = 1.209×10⁻⁴ × 4.301×10²² = 5.200×10¹⁸ yr⁻¹

Convert to decays per second (1 yr ≈ 3.156×10⁷ s):

A = 5.200×10¹⁸ / 3.156×10⁷ ≈ 1.65×10¹¹ Bq

Answer:λ = 1.21×10⁻⁴ yr⁻¹; Activity ≈ 1.65×10¹¹ decays/s (Bq)

Guided Problems

Guided Problem 1

A hydrogen electron drops from n = 4 to n = 2. Calculate the energy and wavelength of the emitted photon. Which spectral series does this belong to?

Hint: Use Eₙ = −13.6/n² for both levels, find ΔE = E₂ − E₄ (take the magnitude), then λ = hc/ΔE. Transitions ending at n = 2 are the Balmer series.

Guided Problem 2

Lithium-7 (₃⁷Li) has Z = 3 protons and N = 4 neutrons. The actual atomic mass is 7.01601 u. Using mₚ = 1.00728 u and mₙ = 1.00866 u, find the mass defect and binding energy.

Hint: Calculate total constituent mass = 3mₚ + 4mₙ, subtract the actual mass to get Δm, then multiply by 931.5 MeV/u.

Guided Problem 3

Thorium-234 (₉₀²³⁴Th) undergoes beta-minus decay. Write the complete decay equation, identifying the daughter nucleus.

Hint: In β⁻ decay a neutron converts to a proton: Z increases by 1, A stays the same. Look up or deduce the element with Z = 91.

Guided Problem 4

A radioactive sample has an initial activity of 8000 Bq. After 30 minutes the activity has fallen to 1000 Bq. Find the half-life and the decay constant.

Hint: Activity is proportional to N, so A = A₀(½)^(t/t½). Solve for t½: how many halvings reduce 8000 to 1000? Then λ = ln2/t½.

Guided Problem 5

Using the exponential decay law N = N₀e^(−λt), show that when t = t½ the equation reduces to N = N₀/2, and hence derive t½ = ln2/λ.

Hint: Substitute N = N₀/2 into the exponential law and solve for t. You will need to take the natural logarithm of both sides.

Key Vocabulary

Atomic number (Z)

The number of protons in the nucleus of an atom. It defines the element and determines the chemical properties.

Example: Carbon always has Z = 6 protons.

Mass number (A)

The total number of nucleons (protons + neutrons) in the nucleus: A = Z + N.

Example: Carbon-14 has A = 14, Z = 6, N = 8.

Isotope

Atoms of the same element (same Z) with different numbers of neutrons (different A). Isotopes have identical chemical behaviour but different nuclear properties.

Example: ¹²C and ¹⁴C are both carbon but differ in neutron count.

Binding energy (E_b)

The energy required to completely separate all nucleons in a nucleus. It equals the mass defect multiplied by c²: E_b = Δm·c².

Example: ⁴He binding energy ≈ 28.3 MeV.

Mass defect (Δm)

The difference between the total mass of the individual nucleons and the actual mass of the assembled nucleus. The 'missing' mass is converted to binding energy.

Example: For ⁴He: Δm = 4.03188 u − 4.00260 u = 0.02928 u.

Radioactive decay

The spontaneous transformation of an unstable nucleus into a more stable configuration, emitting radiation (alpha, beta, or gamma) in the process.

Example: ²²⁶Ra → ²²²Rn + ⁴He (alpha decay).

Half-life (t½)

The time required for exactly half of the radioactive nuclei in a sample to decay. It is constant for a given isotope and related to λ by t½ = ln2/λ.

Example: I-131 has t½ = 8 days; after 16 days only ¼ of the original sample remains.

Nuclear fission

The splitting of a heavy nucleus (e.g. ²³⁵U) into two smaller daughter nuclei plus neutrons and energy. The released energy comes from the increase in binding energy per nucleon.

Example: ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n + ~200 MeV.

Workbook Check — Interactive Quiz

Interactive Practice — 5 Questions

1

A hydrogen electron falls from n = 2 to n = 1. What is the energy of the emitted photon?

2

Which statement correctly describes the mass defect of a nucleus?

3

Polonium-210 (₈₄²¹⁰Po) undergoes alpha decay. What is the daughter nucleus?

4

A sample of 6.4×10¹⁰ radioactive atoms has a half-life of 5 years. How many atoms remain after 15 years?

5

Which type of radiation has the greatest penetrating power?

Independent Practice

1

Calculate the wavelength of the photon emitted when a hydrogen electron transitions from n = 4 to n = 2. (R_H = 1.097 × 10⁷ m⁻¹)

2

Find the energy of the photon emitted when a hydrogen electron drops from n = 3 to n = 1. (E_n = −13.6/n² eV)

3

In a nuclear fission reaction, ²³⁵U absorbs a neutron and splits into ⁹²Kr and ¹⁴¹Ba plus neutrons. Write the balanced nuclear equation and find the number of neutrons released.

4

Calculate the mass defect and energy released in the fusion reaction: ²H + ³H → ⁴He + n. (m_²H = 2.01410 u, m_³H = 3.01605 u, m_⁴He = 4.00260 u, m_n = 1.00867 u, 1 u = 931.5 MeV)

5

★ A hydrogen atom electron is in the n = 5 state. (a) Calculate the energy of this state in eV. (b) List all possible photon wavelengths emitted as the electron cascades down to n = 1. (c) Which transitions produce visible light (400–700 nm)? (d) Calculate the ionization energy needed to remove the electron from n = 5. (E_n = −13.6/n² eV, hc = 1,240 eV·nm)

Challenge
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Common Mistakes

Using the Bohr energy formula E_n = −13.6/n² eV for atoms other than hydrogen

The Bohr model applies only to hydrogen-like atoms (one electron). For multi-electron atoms, electron-electron repulsion makes the formula invalid

Confusing mass number A with atomic mass (in amu)

Mass number A = Z + N is an integer (count of nucleons). Atomic mass in amu includes binding energy effects and is slightly less than A for stable nuclei

Forgetting that alpha decay reduces A by 4 and Z by 2, while beta-minus decay increases Z by 1 with no change in A

α decay: A→A−4, Z→Z−2. β⁻ decay: A unchanged, Z→Z+1 (neutron→proton). β⁺ decay: A unchanged, Z→Z−1. γ decay: no change in A or Z

Applying N = N₀(½)^(t/t½) with t and t½ in different units

Always use the same time unit for t and t½. If t½ = 8 days and t = 24 days, then t/t½ = 3 half-lives. Convert both to the same unit first

Thinking nuclear binding energy is the energy needed to hold the nucleus together (like glue)

Binding energy is the energy you must ADD to completely separate the nucleus into individual protons and neutrons. Higher binding energy per nucleon = more stable nucleus

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Math Tips

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Bohr energy levels: E_n = −13.6/n² eV. Ground state (n=1): −13.6 eV. First excited state (n=2): −3.4 eV. Ionization energy from ground state = 13.6 eV

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Half-life calculation: after n half-lives, N = N₀/2ⁿ = N₀×(0.5)ⁿ. After 1 t½: 50% remains. After 2: 25%. After 3: 12.5%. After 10: ~0.1%

Mass-energy: 1 amu = 931.5 MeV/c². Binding energy = Δm × 931.5 MeV/amu. Iron-56 has the highest binding energy per nucleon (~8.8 MeV) — the most stable nucleus

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Decay constant and half-life: λ = ln2/t½ = 0.693/t½. Activity A = λN = λN₀e^(−λt). At t = 0: A₀ = λN₀. Activity halves every half-life

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Nuclear reaction check: verify conservation of (1) mass number A, (2) atomic number Z, (3) charge, and (4) lepton number. If any is violated, the reaction is forbidden