Unit 5 · Lesson 2a

2aAtomic Models and Bohr's Model

Trace the evolution of atomic models from Thomson to Rutherford to Bohr, and master the quantized energy levels of hydrogen.

The Bohr model was the first successful quantum model of the atom — it correctly predicted hydrogen's spectral lines and introduced the concept of quantized energy levels that underlies all of modern chemistry and atomic physics.

Lesson Overview

Our understanding of the atom evolved through a series of models. Thomson's plum-pudding model (1904) placed electrons embedded in a positive sphere. Rutherford's gold foil experiment (1911) revealed a tiny, dense, positive nucleus. Bohr's model (1913) combined classical orbits with quantum rules: electrons occupy only specific allowed orbits with quantized angular momentum, and emit or absorb photons only when jumping between orbits. For hydrogen, the energy levels are Eₙ = −13.6/n² eV. While the Bohr model was eventually superseded by quantum mechanics, it correctly predicts hydrogen's spectral lines.

Key Concepts

Thomson Model

Plum-pudding: electrons embedded in a diffuse positive sphere (1904)

Rutherford Model

Nuclear model: tiny dense positive nucleus, electrons orbit outside (1911)

Bohr Postulates

Electrons in fixed orbits; only certain radii allowed; photon emitted/absorbed on transition

Hydrogen Energy Levels

Eₙ = −13.6/n² eV (n = 1, 2, 3, …); ground state n=1 at −13.6 eV

Bohr Radius

Smallest allowed orbit: a₀ = 5.29×10⁻¹¹ m; rₙ = n²a₀

Limitations of Bohr Model

Only works for hydrogen; cannot explain multi-electron atoms or fine structure

Example 1

Calculate the energy of the hydrogen atom in the n = 3 energy level.

Answer:Eₙ = −13.6/n² eV = −13.6/9 = −1.51 eV. The negative sign indicates the electron is bound to the nucleus.
Example 2

A hydrogen electron transitions from n = 4 to n = 2. Calculate the energy of the photon emitted.

Answer:ΔE = E₄ − E₂ = (−13.6/16) − (−13.6/4) = −0.85 − (−3.40) = 2.55 eV. The photon has energy 2.55 eV (visible light, blue-green, Balmer series).
Example 3

Describe Rutherford's gold foil experiment and explain what result was expected vs. what was observed.

Answer:Expected (Thomson model): alpha particles should pass straight through with minor deflections. Observed: most passed through, but a small fraction deflected at large angles — some even bounced back. Conclusion: the atom has a tiny, dense, positively charged nucleus containing nearly all the mass, with electrons orbiting far outside.
Example 4

What is the radius of the n = 3 orbit in hydrogen? (a₀ = 5.29×10⁻¹¹ m)

Answer:rₙ = n²a₀ = (3)²(5.29×10⁻¹¹) = 9 × 5.29×10⁻¹¹ = 4.76×10⁻¹⁰ m = 0.476 nm.
Example 5

What is the ionization energy of hydrogen from the ground state? What does ionization mean?

Answer:Ionization means removing the electron completely (n → ∞, E = 0). From ground state (n=1, E = −13.6 eV): ionization energy = 0 − (−13.6) = 13.6 eV. This is the energy needed to free the electron from the atom.
Guided Problem 1

Calculate the energy of the hydrogen atom in the n = 5 level and find the energy needed to excite it from n = 1 to n = 5.

Hint: Use Eₙ = −13.6/n² eV. The excitation energy = E₅ − E₁.

Guided Problem 2

A photon of energy 10.2 eV is absorbed by a hydrogen atom in the ground state. To which energy level does the electron jump?

Hint: Find n such that Eₙ = E₁ + 10.2 eV = −13.6 + 10.2 = −3.4 eV. Solve −13.6/n² = −3.4.

Guided Problem 3

Why did the Rutherford model of the atom have a fatal flaw according to classical electromagnetism?

Hint: A charged particle moving in a circle accelerates. What does classical EM theory say about accelerating charges?

Guided Problem 4

Calculate the radius of the n = 2 orbit in hydrogen and compare it to the ground state radius.

Hint: Use rₙ = n²a₀. How does r₂ compare to r₁?

Guided Problem 5

List two phenomena that the Bohr model successfully explains and two that it cannot explain.

Hint: Think about what the Bohr model was designed for (hydrogen spectra) and where it breaks down (multi-electron atoms, fine structure).

Key Vocabulary

Rutherford Model

The nuclear model of the atom: a tiny, dense, positively charged nucleus surrounded by electrons orbiting at relatively large distances.

Example: Rutherford's gold foil experiment showed that most of an atom is empty space, with mass concentrated in a nucleus ~10⁻¹⁵ m across.

Bohr Model

A quantum model of the hydrogen atom where electrons occupy only specific allowed circular orbits with quantized angular momentum.

Example: In the Bohr model, the hydrogen ground state has energy −13.6 eV and radius 5.29×10⁻¹¹ m.

Energy Level

A discrete, allowed energy state for an electron in an atom. For hydrogen: Eₙ = −13.6/n² eV.

Example: The n=2 level of hydrogen has energy −3.4 eV; the n=3 level has −1.51 eV.

Ionization Energy

The minimum energy required to completely remove an electron from an atom in its ground state.

Example: Hydrogen's ionization energy is 13.6 eV — the energy needed to remove the electron from n=1 to n=∞.

Interactive Practice — 5 Questions

1

Rutherford's gold foil experiment showed that:

2

In the Bohr model, the energy of hydrogen's n=2 level is:

3

When an electron in hydrogen jumps from n=3 to n=1, the atom:

4

The ionization energy of hydrogen from the ground state is:

5

A major limitation of the Bohr model is that it:

Independent Practice

1

List the three atomic models (Thomson, Rutherford, Bohr) in chronological order. For each, describe the key experimental evidence that supported or refuted it.

2

Calculate the energies of the first four hydrogen energy levels (n = 1, 2, 3, 4) in eV. Draw an energy level diagram.

3

A hydrogen electron transitions from n = 5 to n = 2. Calculate the photon energy emitted and determine whether it is in the UV, visible, or IR range.

4

Explain why the Rutherford model predicted that atoms should be unstable (electrons should spiral into the nucleus) and how Bohr's postulates resolved this problem.

5

★ Calculate the wavelength of the photon emitted when a hydrogen electron transitions from n = 3 to n = 2 (the H-alpha line). Compare your answer to the observed value of 656 nm.

Challenge
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Common Mistakes

Thinking higher n means lower energy (more negative).

Higher n means higher (less negative) energy. n=1 is the lowest energy (most negative, −13.6 eV); n=∞ is 0 eV (free electron).

Confusing emission (electron drops to lower n) with absorption (electron jumps to higher n).

Emission: electron falls to lower level → photon released. Absorption: photon absorbed → electron jumps to higher level. Photon energy = |ΔE| in both cases.

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Math Tips

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For hydrogen energy levels: Eₙ = −13.6/n² eV. Photon energy for a transition: E_photon = |E_upper − E_lower| = 13.6(1/n_lower² − 1/n_upper²) eV. Always take the absolute value — photon energy is positive.