Atoms

Atoms

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

• Atoms consist of a tiny, dense, positively charged nucleus (radius ~10⁻¹⁵ m) surrounded by electrons revolving at distances ~10⁻¹⁰ m, leaving atoms mostly empty space (Rutherford’s α-scattering conclusion).
• Bohr’s two key postulates for hydrogen-like ions: electrons occupy stationary orbits without radiating, and angular momentum is quantised as L = m v r = n h / (2π).
• Radii scale as rₙ = n² a₀ / Z, with a₀ = 0.529 Å (Bohr radius); total energy Eₙ = −13.6 Z² / n² eV (negative = bound state).
• Spectral line wavelength: 1/λ = R Z² (1/n₁² − 1/n₂²), R ≈ 1.097 × 10⁷ m⁻¹. Series to remember: Lyman (UV, n₁=1), Balmer (visible, n₁=2), Paschen (IR, n₁=3).
• CUET pattern: straight numericals on energy/wavelength/ionisation potential, plus a one-line statement on Rutherford observations.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Thomson, Rutherford, and the Birth of the Nucleus

Thomson’s plum-pudding model assumed positive charge spread uniformly with embedded electrons. Rutherford’s gold-foil experiment (1911) fired α-particles (⁴He nuclei, kinetic energy ~7.7 MeV from ²¹⁸Po source) at a thin gold foil. The surprising observations: (i) most α-particles passed straight through → atom is mostly empty space; (ii) a few deflected through large angles → a concentrated positive charge exists inside; (iii) about 1 in 8000 bounced back → the entire positive charge and virtually all the mass reside in a volume ~10⁻⁴ the size of the atom, the nucleus. Scattering cross-section varies as dN/dθ ∝ 1/sin⁴(θ/2) — minimum distance of approach r₀ = (1/4πε₀)(2 Z e² / E_kin), an estimate that gives nuclear size.

Bohr’s Quantisation Rules

Bohr (1913) applied three postulates to hydrogen-like ions (single electron, nuclear charge +Z e):

1. Electrons revolve in circular stationary orbits without radiating; centripetal force supplied by Coulomb attraction: (mv²)/r = (1/4πε₀)(Z e² / r²)
2. Angular momentum is quantised: m v r = n h / (2π), n = 1, 2, 3, …
3. A photon of energy hν = E₂ − E₁ is emitted (or absorbed) on transition between orbits.

Derived Quantities

From postulates 1 and 2:

• vₙ = (Z e²)/(2 ε₀ n h)
• rₙ = n² a₀ / Z, a₀ = ε₀ h²/(π m e²) = 0.529 Å = 0.0529 nm
• Eₙ = −(13.6 eV) Z²/n² (kinetic = +|Eₙ|, potential = −2|Eₙ|, by virial theorem)

Higher n → larger orbit; higher Z → tighter orbit. Ionisation energy of H = 13.6 eV; of He⁺ (Z=2) = 54.4 eV; of Li²⁺ (Z=3) = 122.4 eV.

Hydrogen Spectral Series

Wavelength of emitted photon when electron falls from n₂ to n₁:

1/λ = R Z² (1/n₁² − 1/n₂²), R = 1.097 × 10⁷ m⁻¹.

Series	n₁	Region	First line (n₂)
Lyman	1	UV	121.6 nm
Balmer	2	Visible	656.3 nm (Hα, red)
Paschen	3	IR	1875 nm
Brackett	4	IR	4051 nm
Pfund	5	IR	—

Typical CUET Question Patterns

(a) Numerical: “Find wavelength emitted when electron in Li²⁺ jumps from n=4 to n=2.” (Answer: 1/λ = R·9·(1/4 − 1/16) = R·9·(3/16).) (b) Identification: pick the series given (n₁, n₂). (c) Conceptual MCQ: why is total energy negative — because the electron is bound; positive energy would mean a free electron.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Why Bohr Works (and Where It Breaks)

Bohr’s model succeeded for hydrogen because the reduced-mass correction is small (m_e/m_p ≈ 1/1836) and the system is genuinely one-electron. For He, Li, …, electron–electron repulsion makes the Hamiltonian non-separable, and the formula Eₙ = −13.6 Z²/n² eV no longer holds; instead, you get Z_eff < Z (Slater’s rules, ~Z − 0.31 for 1s electrons, etc., covered in Chemical Bonding chapters).

Three phenomena Bohr cannot explain, which you should know exist even if not solved here:

1. Fine structure — splitting of Balmer lines (e.g., Hα into a doublet) due to spin–orbit coupling and relativistic correction (Sommerfeld’s elliptical orbits extended Bohr partially; full treatment needs Dirac).
2. Zeeman effect — splitting of spectral lines in an external magnetic field, requiring vector coupling of orbital and spin angular momenta (Landé g-factor).
3. Wave-particle duality — Bohr orbits are not classical paths; they correspond to standing-wave orbits (de Broglie: 2π r = n λ, equivalent to Bohr’s quantisation).

Common Mistakes in Calculations

• Confusing Bohr radius (0.529 Å, atomic scale) with nuclear radius (~1.2 fm × A^(1/3)). The atom is ~10⁵ times larger than the nucleus.
• Using the Rydberg formula without Z² for hydrogen-like ions (He⁺, Li²⁺). Forgetting Z gives a wrong wavelength by a factor of 4 or 9.
• Writing angular momentum as mvr = nh instead of nh/(2π) — this is the most-tested recall trap in MCQs.
• Treating the orbit as decaying classically — Bohr postulates it does not radiate; energy is conserved within a stationary state.

Worked Micro-Example (CUET-style)

A hydrogen atom absorbs a photon of energy 12.09 eV. Find the final state and the series of the next emission.

Ground state E₁ = −13.6 eV. After absorption: E = −13.6 + 12.09 = −1.51 eV ≈ −13.6/n² ⇒ n² = 9 ⇒ n = 3. Electron is in n = 3 (excited state). Possible emissions: 3→1 (Lyman, λ ≈ 102.6 nm), 3→2 (Balmer, λ ≈ 656.3 nm), 2→1 (Lyman, λ ≈ 121.6 nm). All three belong to Lyman or Balmer series depending on the cascade path.

Practice Prompts

1. Numerical: The shortest wavelength of the Balmer series is 364.6 nm. Using this, estimate the Rydberg constant R for hydrogen and hence the ionisation energy of Li²⁺.
2. Conceptual: In Rutherford scattering, why does the use of thicker foil not increase the back-scattering fraction beyond a point — and what does this tell you about the number of nuclei per atom?

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