Atoms Class 12 Notes (2026-27) — CBSE
Class 12 Physics Chapter 12 notes: Rutherford's alpha-scattering experiment, Bohr's model of the atom, energy levels, and the hydrogen spectral series.
Atoms — Class 12 Physics Notes
Chapter Snapshot
How is an atom built? Rutherford's alpha-scattering experiment revealed the tiny, dense nucleus, but his model could not explain why atoms are stable or why they emit line spectra. Bohr fixed both with his quantum postulates, giving formulas for orbit radius and energy and explaining the hydrogen spectral series.
Board relevance: part of the ~12-mark Atoms + Nuclei unit. Expect a Bohr energy-level calculation or a spectral-series question. The energy formula En = −13.6/n² eV is the single most useful thing to memorise.
Key Concepts & Definitions
Rutherford's alpha-scattering experiment
Alpha particles were fired at a thin gold foil. Observations:
- Most passed straight through → the atom is mostly empty space.
- A few were deflected through large angles → there is a concentrated positive charge.
- About 1 in 8000 bounced straight back → all the positive charge and nearly all the mass sit in a tiny central nucleus (~10⁻¹⁵ m, about 10⁴ times smaller than the atom).
Distance of closest approach — at the turning point, all the alpha particle's kinetic energy converts to potential energy:
r₀ = (1/4πε₀)(2Ze²/K)
Limitations of Rutherford's model:
1. Stability: an orbiting (accelerating) electron should continuously radiate energy, spiral inward, and collapse into the nucleus within ~10⁻⁸ s. Atoms are actually stable.
2. Spectrum: it predicts a continuous spectrum, but atoms emit sharp line spectra.
Bohr's postulates
1. Stationary orbits: electrons revolve in certain allowed circular orbits without radiating energy.
2. Quantisation of angular momentum:
mvr = nh/2π (n = 1, 2, 3 … the principal quantum number)
3. Frequency condition: energy is emitted or absorbed only when an electron jumps between orbits:
hν = E₂ − E₁
Formulas
For a hydrogen-like atom of atomic number Z:
Quantity Formula
Radius of nth orbit rn = 0.53 n²/Z Å (r ∝ n²)
Energy of nth orbit En = −13.6 Z²/n² eV
Velocity in nth orbit vn ∝ Z/n (v₁ ≈ c/137)
Transition wavelength 1/λ = RZ²(1/n₁² − 1/n₂²)
- R = 1.097 × 10⁷ m⁻¹ (Rydberg constant); n₂ n₁.
- For hydrogen (Z = 1): ground state r = 0.53 Å, E₁ = −13.6 eV; E₂ = −3.4 eV, E₃ = −1.51 eV, E₄ = −0.85 eV.
- Ionisation energy of hydrogen = +13.6 eV (energy to take the electron from n = 1 to infinity).
- Energy is negative because the electron is bound; its magnitude is the binding energy. Energy levels get closer together as n increases, converging to 0.
Hydrogen Spectral Series
Each series is defined by the level the electron falls to (n₁):
Series n₁ Region
Lyman 1 Ultraviolet
Balmer 2 Visible
Paschen 3 Infrared
Brackett 4 Infrared
Pfund 5 Far infrared
Only the Balmer series lies in the visible region — which is why it was discovered first. Within any series, the transition from n₂ = n₁ + 1 gives the longest wavelength (least energy) and the transition from n₂ = ∞ gives the series limit (shortest wavelength).
Worked Examples
Example 1 — Energy level: Find the energy of the n = 3 level of hydrogen.
E₃ = −13.6/3² = −13.6/9 = −1.51 eV.
Example 2 — Transition energy: Find the energy released when an electron falls from n = 3 to n = 2 in hydrogen.
ΔE = E₃ − E₂ = (−1.51) − (−3.4) = 1.89 eV (a Balmer line, visible red).
Example 3 — Wavelength: Find the wavelength of that photon.
λ = 1240/1.89 ≈ 656 nm (the red H-alpha line).
Example 4 — Radius: Find the radius of the n = 2 orbit of hydrogen.
r₂ = 0.53 × 2² = 0.53 × 4 = 2.12 Å.
Example 5 — Ionisation: How much energy is needed to ionise a hydrogen atom in its ground state?
From E₁ = −13.6 eV to E = 0, so 13.6 eV is required.
Important Question Patterns
1. Rutherford (2–3 marks): observations and what each shows; distance of closest approach; the two limitations of the model.
2. Bohr's postulates (3 marks): state all three; the quantisation condition mvr = nh/2π; how it explains stability and line spectra.
3. Energy levels (3 marks): En = −13.6/n² eV; energy/wavelength of a transition; ionisation energy; why energies are negative.
4. Spectral series (2–3 marks): name the series and their regions; which is visible; longest wavelength and series limit.
5. Radius/velocity (2 marks): rn = 0.53n²/Z Å; how r, v and E vary with n.
⚡ Quick Revision
- Rutherford: most α pass through (empty space), few deflected, ~1/8000 rebound (tiny dense nucleus). Fails on stability and line spectra.
- Bohr: (1) stationary non-radiating orbits, (2) mvr = nh/2π, (3) hν = E₂ − E₁.
- rn = 0.53 n²/Z Å (r ∝ n²); En = −13.6 Z²/n² eV; v ∝ Z/n.
- Hydrogen: E₁ = −13.6, E₂ = −3.4, E₃ = −1.51, E₄ = −0.85 eV. Ionisation energy = 13.6 eV.
- Energy is negative (bound electron); levels crowd together as n increases.
- 1/λ = R(1/n₁² − 1/n₂²), R = 1.097 × 10⁷ m⁻¹.
- Series: Lyman (n₁=1, UV), Balmer (n₁=2, visible), Paschen (3, IR), Brackett (4, IR), Pfund (5, far IR).
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