Atoms

What this chapter is about

This chapter traces how scientists uncovered the structure of atoms. You begin with early ideas such as Thomson's model, then move to Rutherford's alpha-particle scattering experiment that revealed the tiny, dense, positively charged nucleus. The failure of classical physics to explain atomic stability and discrete spectral lines leads you to Bohr's model of the hydrogen atom, which introduces quantised orbits and energy levels.

A Class 12 student meets this chapter because it bridges electromagnetism and quantum ideas. You apply Coulomb's law, centripetal force and energy conservation inside a single atom, and you see the first use of quantisation in physics. The hydrogen spectrum becomes a test-bed: Bohr's postulates explain the Balmer, Lyman and other series with remarkable accuracy.

After studying this chapter you should be able to describe the key experiments and models historically, derive the radius, velocity and energy of Bohr orbits, calculate wavelengths of spectral lines using the Rydberg formula, and appreciate why classical mechanics fails at atomic scales.

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Key ideas

• Thomson's model (plum-pudding model): Positive charge is spread uniformly through a sphere and electrons are embedded like plums in a pudding. This model could not explain large-angle scattering of alpha particles.
• Rutherford's nuclear model: Alpha particles fired at thin gold foil showed that most pass straight through, a few deflect at large angles, and very few bounce back. Conclusion: almost all atomic mass and all positive charge sit in a nucleus far smaller than the atom itself (radius ≈ 10⁻¹⁵ m versus atomic radius ≈ 10⁻¹⁰ m).
• Impact parameter and scattering angle: A smaller impact parameter (distance of approach to nucleus) gives a larger scattering angle. The relation involves cot(θ/2).
• Bohr's postulates (hydrogen atom):
  1. Electrons move in circular orbits around the nucleus under electrostatic attraction.
  2. Only orbits for which angular momentum L = n h/(2π), with n = 1, 2, 3 …, are allowed (quantisation of angular momentum).
  3. An electron in an allowed orbit does not radiate energy; energy is emitted or absorbed only when it jumps between orbits, with ΔE = h ν.
• Bohr radius and energy levels: The radius of the nth orbit is rₙ = n² a₀, where a₀ ≈ 0.529 Å. Energy Eₙ = −13.6/n² eV. Negative energy means the electron is bound.
• Hydrogen spectrum: Transitions from higher to lower levels emit photons whose wavelengths satisfy 1/λ = R (1/n₁² − 1/n₂²), with Rydberg constant R ≈ 1.097 × 10⁷ m⁻¹. Different series arise for different final levels n₁ (Lyman → 1, Balmer → 2, Paschen → 3, etc.).
• Limitations of Bohr model: It works well for hydrogen and hydrogen-like ions but cannot explain spectra of multi-electron atoms, fine structure, or the Zeeman effect without further modification.

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Formulas and facts to remember

1. Bohr radius: a₀ = ε₀ h² / (π m e²) ≈ 5.29 × 10⁻¹¹ m.
2. Radius of nth orbit: rₙ = n² a₀ / Z (Z = atomic number; for hydrogen Z = 1).
3. Velocity in nth orbit: vₙ = (Z e²) / (2 ε₀ n h) = (c α Z) / n, where α ≈ 1/137 is the fine-structure constant.
4. Total energy of nth level: Eₙ = −(m e⁴ Z²) / (8 ε₀² h² n²) = −13.6 Z² / n² eV.
5. Rydberg formula for wavelengths: 1/λ = R Z² (1/n₁² − 1/n₂²), n₂ > n₁.
6. Frequency of emitted photon: ν = (Eₙ₂ − Eₙ₁) / h.
7. Kinetic and potential energy in orbit: K = −E (positive), U = 2 E (negative), so total E = K + U = −K.
8. Distance of closest approach (α-particle): d = 2 k Z e² / (½ m v²), giving nuclear size upper limit.

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Worked examples

Example 1 – Radius and speed of an electron in the second orbit of hydrogen

Problem: Find the radius and orbital speed of the electron when n = 2 for hydrogen.

Solution:

Radius: r₂ = n² a₀ = 4 × 0.529 Å = 2.12 Å = 2.12 × 10⁻¹⁰ m.

Speed: v₂ = v₁ / n. First find v₁ = e² / (2 ε₀ h) ≈ 2.18 × 10⁶ m/s.

So v₂ = (2.18 × 10⁶) / 2 = 1.09 × 10⁶ m/s.

Answer: Radius ≈ 2.12 Å; speed ≈ 1.09 × 10⁶ m/s.

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Example 2 – Wavelength of the first Balmer line

Problem: Calculate the wavelength of the photon emitted when an electron in hydrogen falls from n = 3 to n = 2.

Solution:

Use Rydberg formula: 1/λ = R (1/n₁² − 1/n₂²) with n₁ = 2, n₂ = 3.

1/λ = 1.097 × 10⁷ (1/4 − 1/9) = 1.097 × 10⁷ × (9 − 4)/36 = 1.097 × 10⁷ × 5/36.

1/λ = 1.524 × 10⁶ m⁻¹.

λ = 1 / (1.524 × 10⁶) = 6.56 × 10⁻⁷ m = 656 nm (red light).

Answer: λ ≈ 656 nm, which lies in the visible (Balmer) series.

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Example 3 – Energy required to ionise hydrogen from the first excited state

Problem: How much energy is needed to remove the electron completely from hydrogen when it is in the n = 2 level?

Solution:

Energy in n = 2: E₂ = −13.6 / 4 = −3.4 eV.

At infinity (ionised), E = 0. Energy required = 0 − (−3.4) = 3.4 eV.

Answer: Ionisation energy from n = 2 is 3.4 eV.

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Common mistakes

• Using positive energy values for bound states → Remember energies are negative; the electron is bound.
• Confusing impact parameter with scattering angle (thinking larger impact parameter gives larger deflection) → Larger impact parameter means weaker deflection.
• Applying Bohr's formula directly to multi-electron atoms → Bohr model is accurate only for one-electron systems (H, He⁺, Li²⁺, etc.).
• Forgetting to square n in the energy or radius formulas → rₙ ∝ n² and Eₙ ∝ 1/n².
• Mixing up emission and absorption: using n₂ < n₁ for emission → For emission n₂ > n₁; electron falls from higher to lower orbit.

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Quick revision

1. Rutherford's experiment proved the nucleus is tiny, dense and positively charged.
2. Bohr's quantisation condition: L = n h/(2π).
3. Energy levels: Eₙ = −13.6/n² eV for hydrogen.
4. Rydberg formula: 1/λ = R (1/n₁² − 1/n₂²).
5. Kinetic energy = −(total energy); potential energy = 2 × (total energy).
6. Bohr model fails for atoms with more than one electron.

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