Atomic Structure

🟢 Lite — Quick Review (1h–1d)

Rapid summary of atomic models, quantum numbers, electronic configuration and key formulae for JEE Main revision.

Atomic Structure — Key Facts for JEE Main

• Dalton’s Atomic Theory (1808): All matter consists of indivisible atoms; atoms of the same element are identical; compounds form by combination of atoms in fixed ratios. Limited by inability to explain subatomic particles.
• Thomson’s Model (1897): Plum-pudding model — positive sphere with embedded electrons. Disproved by Rutherford’s gold-foil experiment (1911).
• Rutherford’s Nuclear Model: Atom has a dense positively charged nucleus (radius ~10⁻¹⁵ m) surrounded by electrons in empty space (radius ~10⁻¹⁰ m). Failed to explain continuous emission spectra and stability of atoms.
• Bohr’s Model (1913): Electrons revolve in discrete orbits (n = 1, 2, 3…) with quantised energy. Postulates: stationary states, angular momentum = nh/2π, photon emission on transition.
• Energy Levels: $E_n = -\dfrac{13.6 \text{ eV}}{n^2}$ for hydrogen atom. Ionisation energy = 13.6 eV.
• Rydberg Formula: $\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ where $R_H = 1.097 \times 10^7 \text{ m}^{-1}$.
• Spectral Series: Lyman (UV, $n_1=1$), Balmer (visible, $n_1=2$), Paschen (IR, $n_1=3$), Brackett, Pfund.
• Quantum Numbers:
  • Principal (n): 1, 2, 3… (shell: K, L, M, N…)
  • Azimuthal (l): 0 to n-1 (subshell: s, p, d, f)
  • Magnetic (m_l): -l to +l
  • Spin (m_s): +½ or -½
• Pauli’s Exclusion Principle: No two electrons in an atom can have identical all four quantum numbers.
• Hund’s Rule of Maximum Multiplicity: Electrons fill degenerate orbitals singly before pairing.
• Aufbau Principle: Orbitals fill in order of increasing energy: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s.
• de Broglie Equation: $\lambda = \dfrac{h}{mv}$ — dual nature of matter.
• Heisenberg’s Uncertainty Principle: $\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi}$ — impossible to simultaneously determine exact position and momentum.
• Photoelectric Effect: $KE_{max} = h\nu - \phi$; threshold frequency $\nu_0 = \phi/h$.
• Ionic Radii Trend: Cation < neutral atom < anion; isoelectronic series: Ne > O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺.
• Effective Nuclear Charge (Z_eff): $Z_{eff} = Z - \sigma$; increases across a period, decreases down a group.

⚡ JEE Main Exam Tips:

• Remember: 1s fills first (lowest energy). Exception: 4s fills before 3d (Aufbau rule).
• For hydrogen-like species (H, He⁺, Li²⁺), $E_n \propto Z^2/n^2$.
• Shorter wavelength → higher energy. Lyman series (UV) has shortest wavelength in hydrogen spectrum.
• Wave number $\bar{\nu} = 1/\lambda$ measured in cm⁻¹.
• Maximum electrons in shells: 2n² (K=2, L=8, M=18, N=32).

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🟡 Standard — Regular Study (2d–2mo)

Deeper understanding of atomic structure, quantum mechanical model, electronic configurations, and problem-solving strategies for JEE Main.

Atomic Structure — JEE Main Study Guide

The quantum mechanical model replaced Bohr’s planetary model, introducing wave functions (ψ) that describe electron behaviour. The Schrödinger wave equation, $H\psi = E\psi$, yields wave functions for hydrogen atom where the solutions require three quantum numbers (n, l, m_l). The fourth quantum number (m_s) arises from relativistic correction. Each orbital is characterised by a unique wave function; the probability density is given by |ψ|², which gives the electron cloud model.

Electronic Configuration Writing: Follow the order: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s². Exceptions occur for Cr ([Ar] 3d⁵ 4s¹) and Cu ([Ar] 3d¹⁰ 4s¹) due to extra stability of half-filled and fully-filled d-subshells. The atomic number of chromium is 24; its ground state configuration is [Ar] 3d⁵ 4s¹ — not [Ar] 3d⁴ 4s² — because the exchange energy stabilization in 3d⁵ exceeds the energy difference between 3d and 4s.

Solved Example: Calculate the wavelength of photon emitted when an electron in hydrogen atom falls from n=5 to n=2.

Solution: Using Rydberg formula: 1/λ = R_H(1/n₁² – 1/n₂²). With n₁=2, n₂=5: 1/λ = (1.097×10⁷ m⁻¹)(1/4 – 1/25) = (1.097×10⁷)(0.25 – 0.04) = (1.097×10⁷)(0.21) = 2.3037×10⁶ m⁻¹. Thus λ = 4.34×10⁻⁷ m = 434 nm. This falls in the visible Balmer series (blue-green region).

Solved Example: Write the quantum numbers for the 3p electron of silicon (Si, Z=14).

Solution: Silicon’s electronic configuration: 1s² 2s² 2p⁶ 3s² 3p². For a 3p electron: n=3, l=1 (p-subshell). For the two 3p electrons in different orbitals (Hund’s rule), m_l values are –1, 0, +1. m_s can be +½ or –½ for each. So one valid set: (3, 1, 0, +½).

Trends in Periodic Table:

• Atomic radius decreases across Period 2 (Li to Ne) due to increasing Z_eff.
• Ionisation energy generally increases across a period with dips at Group 13 (Boron) and Group 16 (Oxygen).
• Electron affinity becomes more negative across a period; Group 17 halogens have highest electron affinity.
• Electronegativity increases across a period and decreases down a group (F > Cl > Br > I).

Common Mistakes to Avoid:

• Confusing shell number (n) with number of electrons (2n²).
• Using n and l incorrectly for d-block elements — remember (n-1)d fills after ns.
• Forgetting that Heisenberg’s principle applies to conjugate pairs (position/momentum, energy/time).

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage of advanced atomic structure concepts, quantum mechanical derivations, many-electron systems, and detailed JEE Advanced-level problem solving.

Atomic Structure — Comprehensive JEE Notes

Quantum Mechanical Derivation of Bohr’s Quantisation: From de Broglie’s hypothesis, electron wavelength λ = h/mv. For a stable orbit, circumference must be an integer multiple of wavelength: 2πr = nλ = nh/mv → mvr = nh/2π → angular momentum is quantised. Substituting into Coulomb’s law ($F = ke^2/r^2$) and centripetal force ($mv^2/r$), we get $r_n = n^2a_0/Z$ where $a_0 = 0.529 Å$ is Bohr radius. Energy $E_n = -Z^2R_H/n^2$ with $R_H = 13.6$ eV.

Time-Independent Schrödinger Equation: For hydrogen atom: $\hat{H}\psi = \left(-\dfrac{\hbar^2}{2m}\nabla^2 - \dfrac{ke^2}{r}\right)\psi = E\psi$. Solving in spherical coordinates yields R(r), Θ(θ), Φ(φ) solutions. Radial distribution function gives probability of finding an electron at distance r, irrespective of direction. The radial probability curve for 2s shows a node (zero probability at intermediate r), contrasting with 2p which has no radial node but has a planar node.

Many-Electron Atoms and Slater’s Rules: For atoms with >1 electron, the Hamiltonian includes electron-electron repulsion terms that preclude exact solutions. Slater’s rules estimate effective nuclear charge Z_eff = Z – σ, where σ is shielding constant. For 1s electrons, σ = 0.30 per other 1s electron; for ns or np electrons, σ = 0.35 per other electron in same group + 0.85 per electron in (n-1) shell + 1.00 per electron in lower shells.

Term Symbols and Hund’s Cases: For carbon atom (1s² 2s² 2p²), the ground state term is ³P (triplet P) following Hund’s rules. The term symbol format: $^{2S+1}L_J$ where S = total spin, L = total orbital angular momentum (S, P, D, F…), J = L+S to |L–S|. For p² configuration: L = 1 (P), S = 1 (triplet), J = 0, 1, 2 → ³P₀, ³P₁, ³P₂. The ground state is ³P₀ for carbon.

Selection Rules for Spectral Transitions: Electric dipole transitions require Δl = ±1 and Δm_l = 0, ±1. Spin-forbidden transitions have lower probability. For alkaline metals (Na), the prominent yellow D-line (589 nm) corresponds to 3p → 3s transition, split into D₁ (589.0 nm) and D₂ (589.6 nm) due to spin-orbit coupling (fine structure).

Photoelectric Effect Derivation: Einstein’s equation: $KE_{max} = h\nu - \phi$. If photon of energy 4.5 eV strikes a metal with work function 2.3 eV, KE_max = 2.2 eV. Using $KE = \frac{1}{2}mv^2$, electron velocity v = $\sqrt{2KE/m}$ = $\sqrt{2(2.2 \text{ eV})/m_e}$. Converting 1 eV = 1.602×10⁻¹⁹ J: KE = 3.52×10⁻¹⁹ J. So v = $\sqrt{2(3.52×10⁻¹⁹)/(9.11×10⁻³¹)}$ = $\sqrt{7.73×10¹¹}$ = 8.79×10⁵ m/s.

Zeeman Effect: When atoms are placed in external magnetic field, spectral lines split into multiple components due to interaction of magnetic moment with field. Normal Zeeman effect: splitting into three lines (π and σ components) for transitions where Δm = 0, ±1. The energy change ΔE = m_l μ_B B where μ_B is Bohr magneton.

Advanced JEE Problems: Problem: The wavelength of the second line of the Balmer series is 4861 Å. Calculate the wavelength of the first line of the Lyman series. Answer: For Balmer n₂=3 (second line: n₁=2 to n₂=3), use Rydberg with λ = 4861 Å = 4.861×10⁻⁷ m → R_H = 1.097×10⁷ m⁻¹. For Lyman first line (n₁=1 to n₂=2), 1/λ_L = R_H(1/1² – 1/2²) = (3/4)R_H → λ_L = 4/(3R_H) = 1.21×10⁻⁷ m = 1215 Å. Compare energies: Lyman photon is ~4× more energetic.

Advanced Topic: Compton Effect: Scattering of X-ray photons by electrons: λ’ – λ = (h/m_ec)(1 – cos θ). For θ = 90°, shift = h/m_ec = 0.00243 nm (Compton wavelength). This confirms particle nature of radiation.

Revision Checklist:

• Memorise all four quantum numbers and their allowed values.
• Be able to write electron configurations for any element up to Z=36 quickly.
• Derive relationship between Rydberg constant and Bohr radius.
• Understand splitting in magnetic fields (Zeeman effect).
• Apply de Broglie and Heisenberg to numerical problems.

⚡ Pro Tip: In JEE Advanced, questions combining de Broglie, Heisenberg, and spectral series are common. Always check units (eV vs J, nm vs m) before substituting into formulae. Use the unified formula $E_n = -13.6 Z^2/n^2$ eV for hydrogen-like species.