J.J. Thomson's Model (1897) — Plum Pudding Model
- Atom is spherical and electrically neutral (+ve charge = −ve charge)
- +ve charge uniformly distributed throughout volume; electrons embedded in between
- Mass of atom also uniformly distributed inside volume
- Drawbacks: Can't explain (1) spectrum of atoms, (2) scattering of α-particles
Rutherford's Atomic Model (1911) — Nuclear Model
- α-particle scattering experiment (gold foil experiment)
- Most α-particles went straight through → atom is mostly hollow
- 1 out of 8000 deflected by 180° → tiny dense nucleus
- Entire +ve charge and almost whole mass concentrated at centre → Nucleus
- Electrons could not deflect α-particle → electrons are very light
- Radius of atom (ratom) = 10⁻¹⁰ m = 1 Å
- Radius of nucleus = 10⁻¹⁵ m = 1 fermi
- Drawbacks: Can't explain (1) line spectrum of atoms, (2) stability of atom (electron should spiral into nucleus)
Closest distance of approach (r₀): ½Mαv² = k·ze·(2e)/r₀
r₀ = 4kze²/(Mαv²) (K.E. = P.E. at closest approach)
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Bohr's Atomic Model and Postulates
Bohr's Three Postulates
- 1st Postulate: Electrons revolve in certain stable (stationary) orbits without emission of radiant energy. Each orbit has definite total energy → stationary states of atom.
- 2nd Postulate: Electron revolves only in orbits for which angular momentum L = nh/(2π) (n = 0,1,2,...). Quantization of angular momentum.
- 3rd Postulate: Electron may make transition from one orbit to another of lower energy. When it does, a photon is emitted with energy Ei − Ef = Ephoton.
Velocity of e⁻ in nth orbit: V = (e²/2ε₀h) × Z/n = 2.18×10⁶ × Z/n m/s = (c/137) × Z/n
V ∝ Z/n (V ∝ Z; V ∝ 1/n)
Radius of nth orbit: r = 0.53 × n²/Z Å = a₀n²/Z (a₀ = 0.53 Å = Bohr radius)
r ∝ n²/Z (r ∝ n²; r ∝ 1/Z)
Time period: T = 2πr/V = 1.51×10⁻¹⁶ × n³/Z² seconds (T ∝ n³/Z²)
Frequency: f = 2/3 × 10¹⁶ × Z²/n³ Hz (f ∝ Z²/n³)
Magnetic field at centre: B ∝ Z³/n⁵
Energy of Electron in nth Orbit (H-atom: Z=1)
- KE = ze²/(8πε₀r) = 13.6 Z²/n² eV (+ve)
- PE = −ze²/(4πε₀r) = −27.2 Z²/n² eV (−ve, twice KE in magnitude)
- Total Energy TE = KE + PE = −KE = PE/2 = −13.6 Z²/n² eV
- PE = −2KE = 2TE → |PE| = 2|TE| = 2|KE|
- For H-atom: TE = −13.6/n² eV (n=1: −13.6 eV; n=2: −3.4 eV; n=3: −1.51 eV; n=4: −0.85 eV; n=5: −0.54 eV)
- Ionization energy = |TE| = 13.6 Z²/n² eV
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Hydrogen Spectrum and Spectral Series
Energy of emitted photon: E = Ei − Ef = 13.6(1/nf² − 1/ni²) eV
Rydberg formula: 1/λ = R(1/nf² − 1/ni²) where R = 1.097×10⁷ m⁻¹ (Rydberg constant)
Wave number = 1/λ; Angular wave no. = 2π/λ = k
| Series | nf | ni | Region | 1st line (α) | Last line (series limit) |
|---|
| Lyman | 1 | 2,3,4... | UV ray | ni=2 → λmax, fmin | ni=∞ → λmin, fmax |
| Balmer | 2 | 3,4,5... | Visible | ni=3 → λmax | ni=∞ → λmin |
| Paschen | 3 | 4,5,6... | Infra-red | ni=4 | ni=∞ |
| Brackett | 4 | 5,6,7... | IR | ni=5 | ni=∞ |
| P-fund | 5 | 6,7,8... | IR | ni=6 | ni=∞ |
| Humphrey | 6 | 7,8,9... | Far IR | ni=7 | ni=∞ |
Number of Spectral Lines
- Between levels n₁ and n₂: N = (n₂−n₁)(n₂−n₁+1)/2
- From n levels to ground: N = n(n−1)/2
- Total from n2 to n1 levels: N = ⁿ²C₂ − ⁿ¹C₂
Drawbacks of Bohr's Model
- Couldn't explain fine structure of spectral lines
- Can't explain splitting of spectral lines in magnetic field (Zeeman effect) or electric field (Stark effect)
- Valid only for single-electron atoms (H, He⁺, Li²⁺)
- Can't explain the intensity of spectral lines
- Only considered particle nature of electron (not wave)
ΔKE = (4n/(1+n)²) × KE₀ where n = Matom/Mcollide
Max energy transferred (elastic): (ΔKE)max = 10.02 eV = 12.09 eV (for H-atom, 1st excitation)
If collision is inelastic → energy transfer matches excitation energy; otherwise elastic
Definition
- Inverse process of photoelectric effect
- Collection of photons; energy from 0 to KEmax of electrons
- Produced by sudden deceleration of high-speed electrons hitting a metal target
- Intensity ∝ tube current; Frequency ∝ tube voltage
Emax = hc/λmin = KEmax of electron = eV₀
Cutoff wavelength: λmin = hc/(eV₀) (λmin ∝ 1/V₀)
| Type of X-ray | Wavelength | Properties |
|---|
| Soft X-rays | λ = 4Å | Low penetrating power; produced at low potential difference |
| Hard X-rays | λ = 1Å | High penetrating power; produced at high potential difference |
X-ray Spectrum
- Continuous X-ray spectrum: Due to retardation of electrons; λmin ∝ 1/V₀
- Characteristic X-ray spectrum: Due to inner shell electron transitions; sharp lines at specific wavelengths
- Moseley's law: √f ∝ (Z−1) → f ∝ (Z−1)²
- λ ∝ 1/(Z−1) (shorter wavelength for higher atomic number)
Q: H-atom electron in n=5 orbit. Find total energy, KE, PE and ionization energy.
TE = −13.6/n² = −13.6/25 = −0.544 eV; KE = +0.544 eV; PE = −1.088 eV; IE = |TE| = 0.544 eV
Ans: TE = −0.544 eV; KE = 0.544 eV; PE = −1.088 eV; IE = 0.544 eV ✓
Q: Find number of spectral lines when electron falls from n=5 to n=2.
N = (n₂−n₁)(n₂−n₁+1)/2 = (5−2)(5−2+1)/2 = 3×4/2 = 6
Ans: 6 spectral lines ✓
Q: An X-ray tube operates at 40 kV. Find minimum wavelength of X-ray produced.
λmin = hc/(eV₀) = (6.63×10⁻³⁴ × 3×10⁸)/(1.6×10⁻¹⁹ × 40×10³) = 1989×10⁻²⁶/6.4×10⁻¹⁵ ≈ 0.311 Å
Ans: λmin ≈ 0.311 Å ✓