Structure of Atom
'The atom is the basic building block of matter. But inside the atom lies a strange and fascinating world.' — Quantum Physics
1. Chapter Overview
This chapter takes you INSIDE the atom — beyond Dalton's indivisible sphere to the world of PROTONS, NEUTRONS, ELECTRONS, and the QUANTUM MECHANICAL model. You will study the EVOLUTION of atomic models (Thomson, Rutherford, Bohr), the WAVE-PARTICLE duality of matter, QUANTUM NUMBERS (the 'address' of an electron), ATOMIC ORBITALS, and ELECTRONIC CONFIGURATIONS.
2. Subatomic Particles
| Particle | Symbol | Charge (C) | Mass (kg) | Discoverer |
|---|---|---|---|---|
| Electron | e⁻ | -1.602 × 10⁻¹⁹ | 9.1 × 10⁻³¹ | J.J. Thomson (1897) |
| Proton | p⁺ | +1.602 × 10⁻¹⁹ | 1.672 × 10⁻²⁷ | E. Goldstein (1886) |
| Neutron | n⁰ | 0 | 1.675 × 10⁻²⁷ | J. Chadwick (1932) |
Atomic Number and Mass Number
- Atomic Number (Z) = Number of PROTONS in the nucleus
- Mass Number (A) = Number of PROTONS + Number of NEUTRONS
- Isotopes: Same Z, different A (same element, different number of neutrons)
3. Atomic Models — A Historical Journey
Thomson's Model (Plum Pudding Model)
- Atom as a SPHERE of positive charge with electrons EMBEDDED
- Failed to explain Rutherford's experimental results
Rutherford's Model (Gold Foil Experiment)
- Most α-particles passed straight through → atom is EMPTY space
- Some α-particles were DEFLECTED → small, dense POSITIVE nucleus
- A few BOUNCED BACK → nucleus is VERY dense
- Drawbacks: Could not explain STABILITY of atom (electron would spiral into nucleus) or atomic SPECTRA
Bohr's Model (for Hydrogen)
Postulates:
- Electrons move in STATIONARY circular orbits (shells) around nucleus
- Angular momentum is QUANTIZED: mvr = nh/2π (n = 1, 2, 3...)
- Energy is emitted/absorbed ONLY when electron JUMPS between orbits
Bohr's Radius: r_n = n² × a₀ (a₀ = 0.529 Å = Bohr radius for n = 1) Bohr's Energy: E_n = -13.6/n² eV = -2.18×10⁻¹⁸/n² J Spectral Lines: ΔE = hν = E_f — E_i
Worked Problem: Calculate the energy of the n = 3 level in hydrogen. A: E₃ = -13.6/9 = -1.51 eV.
4. Dual Nature of Matter and Radiation
Planck's Quantum Theory
- Energy is quantised: E = hν (h = 6.626 × 10⁻³⁴ J·s)
- h = Planck's constant
Photoelectric Effect (Einstein)
- Light ejects electrons from metal surface
- KE_max = hν — hν₀ (ν₀ = threshold frequency)
- Proves particle nature of light (photons)
de Broglie's Hypothesis
- Matter has wave-like properties
- λ = h/p = h/mv (de Broglie wavelength)
- EVERY moving particle has an associated wavelength
Heisenberg's Uncertainty Principle
- It is IMPOSSIBLE to simultaneously know the EXACT position and EXACT momentum of a particle
- Δx × Δp ≥ h/4π
- Fundamental LIMITATION, not a measurement issue
Worked Problem
Q: Find de Broglie wavelength of an electron moving at 2 × 10⁶ m/s. (m = 9.1 × 10⁻³¹ kg) A: λ = h/mv = 6.626×10⁻³⁴/(9.1×10⁻³¹ × 2×10⁶) = 3.64 × 10⁻¹⁰ m = 3.64 Å.
5. Quantum Mechanical Model
Quantum Numbers (The Electron's Address)
| Quantum Number | Symbol | Values | Significance |
|---|---|---|---|
| Principal | n | 1, 2, 3... | Shell size and ENERGY |
| Azimuthal | l | 0 to n-1 | Subshell SHAPE (s, p, d, f) |
| Magnetic | m_l | -l to +l | Orbital ORIENTATION |
| Spin | m_s | +½ or -½ | Electron SPIN direction |
Shapes of Orbitals
- s-orbital (l = 0): SPHERICAL
- p-orbital (l = 1): DUMBBELL shape (p_x, p_y, p_z)
- d-orbital (l = 2): CLOVERLEAF shape (5 orientations)
- f-orbital (l = 3): Complex shapes (7 orientations)
Nodes
- Radial nodes = n — l — 1
- Angular nodes = l
- Total nodes = n — 1
6. Electronic Configuration
Filling Order — Aufbau Principle
- Electrons fill orbitals in INCREASING order of energy
- Energy order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s...
- Mnemonic: (n + l) rule — lower (n + l) fills first
Pauli's Exclusion Principle
- NO two electrons in an atom can have the SAME set of all four quantum numbers
- Each orbital can hold MAXIMUM 2 electrons with OPPOSITE spins
Hund's Rule of Maximum Multiplicity
- Electrons occupy ALL orbitals of a subshell SINGLY before pairing
- Maximises TOTAL spin (most stable)
Electronic Configurations of Key Elements
| Element | Z | Configuration |
|---|---|---|
| H | 1 | 1s¹ |
| He | 2 | 1s² |
| C | 6 | 1s² 2s² 2p² |
| N | 7 | 1s² 2s² 2p³ |
| O | 8 | 1s² 2s² 2p⁴ |
| Na | 11 | 1s² 2s² 2p⁶ 3s¹ |
| Fe | 26 | 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ |
Exceptions
- Cr (Z=24): [Ar] 4s¹ 3d⁵ (NOT 4s² 3d⁴) — HALF-FILLED stability
- Cu (Z=29): [Ar] 4s¹ 3d¹⁰ (NOT 4s² 3d⁹) — FULLY-FILLED stability
7. Common Mistakes
- Bohr's model FAILS for multi-electron atoms: It works well only for H and H-like ions (He⁺, Li²⁺)
- 4s fills BEFORE 3d (Aufbau), but 3d has LOWER energy than 4s in actual atoms
- Nodes are regions of ZERO probability: Not 'minimum' probability
- m_s = ±½, NOT the values +1 and -1: These are for spin, not charge
- Heisenberg principle ≠ measurement error: It is a FUNDAMENTAL property of quantum systems
8. CBSE Exam Focus
- Bohr's model — radius, energy derivation (5-mark)
- de Broglie wavelength numericals (3-mark)
- Quantum numbers — writing sets for electrons (3-mark)
- Electronic configuration of elements (1/3-mark)
- Heisenberg uncertainty principle — numericals (3-mark)
- Photoelectric effect numericals
9. Key Formulas
- E_n = -13.6/n² eV (hydrogen)
- ΔE = hν = hc/λ
- λ = h/mv (de Broglie)
- Δx × Δp ≥ h/4π
- mvr = nh/2π (Bohr quantisation)
- R_H = 1.097 × 10⁷ m⁻¹ (Rydberg constant)
- 1/λ = R_H(1/n₁² — 1/n₂²)
10. Self-Test (5+ Q&A)
Q1: Find the wavelength of the Hα line (transition from n=3 to n=2) in the Balmer series. (R_H = 1.097 × 10⁷ m⁻¹) A: 1/λ = R_H(1/4 — 1/9) = R_H(5/36) = 1.097×10⁷ × 5/36 = 1.524×10⁶ m⁻¹. λ = 656.3 nm.
Q2: Write the four quantum numbers for the 3p¹ electron of aluminium (Z=13). A: Al: 1s² 2s² 2p⁶ 3s² 3p¹. n=3, l=1, m_l can be -1,0,+1 (typically +1), m_s = +½.
Q3: Calculate the uncertainty in position if Δv = 0.001 m/s for an electron (m = 9.1×10⁻³¹ kg). A: Δp = mΔv = 9.1×10⁻³⁴. Δx ≥ h/(4πΔp) = 6.626×10⁻³⁴/(4×3.14×9.1×10⁻³⁴) = 0.058 m.
Q4: State and explain Hund's rule with an example. A: Hund's rule: Electrons fill each orbital singly with parallel spins before pairing. Example: Carbon (1s² 2s² 2p²) — the two 2p electrons go into separate p orbitals with parallel spins.
Q5: Why is Cr's configuration [Ar]4s¹3d⁵ instead of 4s²3d⁴? A: Half-filled d⁵ configuration has EXTRA STABILITY due to exchange energy and symmetry. One electron from 4s promotes to 3d to achieve this stable configuration.
11. Conclusion
The structure of the atom has EVOLVED from Thomson's sphere to the quantum mechanical model. Bohr's model, while limited, introduces the crucial idea of QUANTISATION. The quantum mechanical model, with its orbital concept and quantum numbers, provides a COMPLETE description of electron distribution in atoms. Electronic configuration determines the CHEMICAL PROPERTIES of elements — making this chapter fundamental to understanding the periodic table and chemical bonding.
