Atoms Class 12 Notes | CBSE Physics Chapter 12

Atoms is Chapter 12 of CBSE Class 12 Physics. This chapter traces the development of atomic models from Thomson’s plum pudding model to Rutherford’s nuclear model to Bohr’s model. You will learn Bohr’s postulates, energy levels of hydrogen atom, and spectral series.

This chapter carries 4–5 marks. Bohr’s postulates, energy level calculations, and hydrogen spectral series are most tested.

---

Key Concepts

1. Rutherford’s Nuclear Model

Alpha particle scattering experiment showed that the atom has a tiny, dense, positively charged nucleus at the centre, with electrons orbiting around it.

Limitation: An orbiting electron should continuously radiate energy and spiral into the nucleus — but atoms are stable. Classical physics couldn’t explain this.

2. Bohr’s Model of Hydrogen Atom

Postulates

1. Electrons revolve in fixed circular orbits (stationary orbits) without radiating energy
2. Quantisation: Angular momentum is quantised: L = mvr = nh/(2π), n = 1, 2, 3…
3. Energy is emitted/absorbed only when an electron jumps between orbits: E = hν = E_i − E_f

Key Formulae for Hydrogen (Z = 1)

Quantity	Formula
Radius of nth orbit	rn = 0.529 × n² Å (= n² × a₀, where a₀ = 0.529 Å)
Velocity in nth orbit	vn = 2.18 × 10⁶/n m/s
Energy of nth level	En = −13.6/n² eV

Ground state (n=1): E = −13.6 eV; First excited (n=2): E = −3.4 eV

3. Hydrogen Spectral Series

1/λ = R(1/n₁² − 1/n₂²) where R = 1.097 × 10⁷ m⁻¹ (Rydberg constant)

Series	Transition to n₁	Region
Lyman	n₁ = 1	Ultraviolet
Balmer	n₁ = 2	Visible
Paschen	n₁ = 3	Infrared
Brackett	n₁ = 4	Infrared
Pfund	n₁ = 5	Infrared

---

Solved Examples

Example 1

Find the energy of the electron in the 3rd orbit of hydrogen.

Answer: E₃ = −13.6/3² = −13.6/9 = −1.51 eV

Example 2

Find the wavelength of the first line of the Balmer series (n₂ = 3 to n₁ = 2).

Answer: 1/λ = R(1/4 − 1/9) = R(9−4)/36 = 5R/36 = (5 × 1.097 × 10⁷)/36 = 1.524 × 10⁶

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

---

Quick Revision Points

• Bohr: quantised orbits, L = nh/(2π), En = −13.6/n² eV
• rn ∝ n²; vn ∝ 1/n; En ∝ −1/n²
• Ground state: n=1, E = −13.6 eV; ionisation energy = 13.6 eV
• Lyman (UV), Balmer (visible), Paschen/Brackett/Pfund (IR)
• 1/λ = R(1/n₁² − 1/n₂²)

---

Previous: Ch 11 — Dual Nature
Next: Ch 13 — Nuclei